Condensed Matter Physics - Oulu

67
Condensed Matter Physics 763628S Jani Tuorila Department of Physics University of Oulu April 17, 2012

Transcript of Condensed Matter Physics - Oulu

Page 1: Condensed Matter Physics - Oulu

Condensed Matter Physics763628S

Jani Tuorila

Department of Physics

University of Oulu

April 17 2012

General

The website of the course can be found athttpswikioulufidisplay763628SEtusivu

It includes the links to this material exercises and to theirsolutions Also the possible changes on the schedule belowcan be found on the web page

Schedule and Practicalities

All lectures and exercise sessions will be held at roomTE320

bull Lectures Monday 12-14Wednesday 12-14

bull Exercises Thursday 12-14

By solving the exercises you can earn extra points for thefinal exam The more problems you solve the higher is theraise Maximum raise is one grade point It is worthwhileto remember that biggest gain in trying to solve the exer-cise is however in the deeper and more efficient learningof the course

Literature

This course is based mainly on (selected parts)

bull M Marder Condensed Matter Physics (MM)

In addition you can read this course material

Other literature

bull E Thuneberg Solid state physics lecture notes (2012only in Finnish) (ET)

bull C Kittel Introduction to Solid State Physics old butstill usable

bull F Duan and J Guojum Introduction to CondensedMatter Physics

bull X G Wen Quantum Field Theory of Many-body Sys-tems way too complicated but has an intriguing in-troduction

bull N W Ashcroft and N D Mermin Solid StatePhysics a classic Previously used in this course

bull P Pietilainen previously used lecture notes based onthe book above only in Finnish

Contents

The course will cover the following

bull Atomic structure

ndash 2-and 3-dimensional crystals

ndash Experimental determination of crystal structure

ndash Boundaries and interfaces

ndash Complicated structures

bull Electronic structure

ndash Single-electron model

ndash Schrodinger equation and symmetry

ndash Nearly free and tightly bound electrons

ndash Electron-electron interactions

ndash Band structure

bull Mechanical properties

ndash Cohesion

ndash Phonons

bull Electronic transport phenomena

ndash Bloch electrons

ndash Transport phenomena and Fermi liquid theory

bull (Graphene)

1

1 IntroductionThe purpose of this course is to give the basic introduc-

tion on condensed matter physics The subject has a widerange and many interesting phenomena are left for othercourses or to be learned from the literature

History

The term Condensed matter includes all such largegroups of particles that condense into one phase Espe-cially the distances between particles have to be smallcompared with the interactions between them Examples

bull solids

bull amorphous materials

bull liquids

bull soft materials (foams gels biological systems)

bull white dwarfs and neutron stars (astrophysics)

bull nuclear matter (nuclear physics)

Condensed matter physics originates from the study ofsolids Previously the field was called solid state physicsuntil it was noticed that one can describe the behaviour ofliquid metals Helium and liquid crystals with same con-cepts and models This course concentrates mainly on per-fect crystals mainly due to the historical development ofthe field and the simplicity of the related theoretical mod-els

Approximately one third of the physicists in the USconsiders themselves as researchers of condensed matterphysics In the last 50 years (in 2011) there has been 22condensed-matter related Nobel prizes in physics and also5 in chemistry

bull Bardeen Cooper and Schrieffer Theory of low tem-perature superconductivity (1972)

bull Josephson Josephson effect (1973)

bull Cornell Ketterle and Wieman Bose-Einstein conden-sation in dilute gases of alkali atoms (2001)

bull Geim and Novoselov Graphene (2010)

The field has been the source of many practical applica-tions

bull Transistor (1948)

bull Magnetic recording

bull Liquid crystal displays (LCDs)

Therefore condensed matter physics is both interestingand beneficial for basic physics research and also for ap-plications

Finally a short listing on rdquohot topicsrdquo in the currentcondensed matter research

bull Mesoscopic physics and the realisation of quantumcomputation (research done in the University of Oulu)

bull Fermi liquid theory (Oulu)

bull Graphene

bull Topological insulators

Many-Body Problem

All known matter is formed by atoms The explainingof the properties of single atoms can be done with quan-tum mechanics with an astounding accuracy When oneadds more atoms to the system the number of the relevantdegrees of freedom in Schrodinger equation grows exponen-tially In principle all properties of the many atom systemcan still be found by solving the Schrodinger equation butin practise the required computational power grows veryrapidly out of reach As an example the computers in1980s could solve for the eigenvalues of the system con-sisting of 11 interacting electrons Two decades later thecomputational power had increased hundred-fold but it al-lowed to include only two extra electrons A typical many-body problem in condensed matter physics includes 1023or so electrons Therefore it is clear that it is extremelyimpractical to study such physics starting from the basicprinciples

Based on the above the condensed matter theories areso-called effective theories In principle they have to beresults of averaging the Schrodinger equation properly butin practise their form has been guessed based on symme-tries and experimental results In this way the theories ofthe condensed matter have become simple beautiful andone can use them to obtain results that are precise (aneffective theory does not have to be imprecise) and givedetailed explanations

The modest goal of condensed matter physics is to ex-plain the whole material world It overlaps with statisticalphysics material physics and liquid and solid mechanicsDue to its diversity the coherent treatment of the subjectis blurred

ldquoThe ability to reduce everything to simple fundamentallaws does not imply the ability to start from those lawsand reconstruct the universe The constructionisthypothesis breaks down when confronted by the twin diffi-culties of scale and complexity The behaviour of large andcomplex aggregates of elementary particles it turns outis not to be understood in terms of a simple extrapolationof the properties of a few particles Instead at eachlevel of complexity entirely new properties appear andthe understanding of the new behaviours require researchwhich I think is as fundamental in its nature as any otherrdquo

- P W Anderson 1972

2

2 Atomic StructureMM Chapters 1 and 2 excluding 233-236 and

262

Scanning-tunnelling-microscopic image (page 17)on NbSe2 surface in atomic resolution The dis-tance between nearest neighbours is 035 nm(httpwwwpmacaltecheduGSRcondmathtml)

Fluoride chrystal on top of quartz chrystal (image byChip Clark)

The simplest way to form a macroscopic solid is to orga-nize atoms into small basic units that repeat periodicallyThis is called crystal structure Let us recall the definitionsin the course of Solid State Physics (763333A)

21 Crystal Structure (ET)Many crystals appear in forms where flat surfaces meet

each other with constant angles Such shapes can be un-derstood based on the organization of atoms Solids areoften formed of many crystals This means that they arecomposed of many joined crystals oriented in different di-rections A single crystal can contain eg sim 1018 atomswhereas the whole macroscopic body has sim 1023 atoms

Generally speaking the structure of a solid can be ex-tremely complicated Even if it was formed by basic unitscomprising of same atoms it does not necessarily have arepeating structure Glass is an example of such a mate-rial formed by SiO2-units This kind of material is calledamorphous

Let us consider an ideal case where we ignore all possibleimperfections of lattice The description of such crystal canbe divided in two parts

1) Group of atoms that form a repeating object in the

crystal We refer to this as basis

2) Group of points in space where one has to place thebasis in order to form the crystal Such a group ispresented in form

r = n1a1 + n2a2 + n3a3 (1)

Here n1 n2 and n3 are integers Vectors a1 a2 anda3 called primitive vectors (they have to be linearlyindependent) The group of points (1) is referred toas Bravais lattice and its members as lattice points

a1a2

a3

The volume defined by the primitive vectors is calledprimitive cell The primitive cells contain the completeinformation on the whole crystal Primitive cells are notunique but they all have to have the same volume Aprimitive cell of a Bravais lattice contains exactly one lat-tice point and thus the volume of the cell is inverse of thedensity of the crystal

+ =

An example in two dimensions

a1

a2

a1

a2

The choice of the primitive vectors is not unique In thefigure one can use also the vectors aprime1 and aprime2 as primitivevectors They reproduce also all lattice points

a1

a2 a1

a2

alkeiskoppi

yksikkoumlkoppi

In certain symmetric lattices it is more practical to useorthogonal vectors instead of primitive ones (even if theydo not span the whole lattice) The volume defined by suchvectors is called unit cell The lengths of the sides of a unitcell are called lattice constants

3

22 Two-Dimensional LatticeLet us first restrict our considerations into two dimen-

sional lattices because they are easier to understand andvisualize than their three dimensional counter-parts How-ever it is worthwhile to notice that there exists genuinelytwo-dimensional lattices such as graphene that is pre-sented later in the course Also the surfaces of crystalsand interfaces between two crystals are naturally two di-mensional

Bravais Lattice

In order to achieve two-dimensional Bravais lattice weset a3 = 0 in Definition (1) One can show using grouptheory that there exist five essentially different Bravais lat-tices

bull square symmetric in reflections with respect to bothx and y -axes and with respect to 90 -rotations

bull rectangular when square lattice is squeezed it losesits rotational symmetry and becomes rectangular

bull hexagonal (trigonal) symmetric with respect two xand y -reflections and 60 -rotations

bull centered rectangular squeezed hexagonal no ro-tational symmetry By repeating the boxed structureone obtains the lattice hence the name

bull oblique arbitrary choice of primitive vectors a1 anda2 without any specific symmetries only inversionsymmetry rrarr minusr

The gray areas in the above denote the so called Wigner-Seitz cells Wigner-Seitz cells are primitive cells that areconserved in any symmetry operation that leaves the wholelattice invariant The Wigner-Seitz cell of a lattice pointis the volume that is closer to that point than any otherlattice point (cf Figures)

Example Hexagonal lattice A choice for the primi-tive vectors of the hexagonal lattice are for example

a1 = a(1 0

)a2 = a

(12 minus

radic3

2

)

where a is the lattice constant Another option is

aprime1 = a(radic

32

12

)aprime2 = a

(radic3

2 minus 12

)

Lattice and a Basis

A structure is a Bravais lattice only if it is symmetricwith respect to translations with a lattice vector (cf thedefinition in a later section) In nature the lattices areseldom Bravais lattices but lattices with a basis As anexample let us consider the honeycomb lattice which isthe ordering for the carbon atoms in graphene

Example Graphene

Geim and Kim Carbon Wonderland ScientificAmerican 90-97 April 2008

Graphene is one atom thick layer of graphite in whichthe carbon atoms are ordered in the honeycomb structureresembling a chicken wire

4

(Wikipedia)

Graphene has been used as a theoretical tool since the1950rsquos but experimentally it was rdquofoundrdquo only in 2004A Geim and K Novoselov were able to separate thin lay-ers of graphite (the material of pencils consists of stackedlayers of graphene) some of which were only one atomthick Therefore graphene is the thinnest known materialin the Universe It is also the strongest ever measured ma-terial (200 times stronger than steel) It is flexible so itis easy to mold Graphene supports current densities thatare six times of those in copper Its charge carriers behavelike massless fermions that are described with the Diracequation This allows the study of relativistic quantummechanics in graphene These and many other interest-ing properties are the reason why graphene is used as anillustrative tool in this course

As was mentioned in the above graphene takes the formof the honeycomb lattice which is a Bravais lattice with abasis The starting point is the hexagonal lattice whoseprimitive vectors are

aprime1 = a(radic

32

12

)

aprime2 = a(radic

32 minus 1

2

)

Each lattice point is then replaced with the basis definedby

v1 = a(

12radic

30)

v2 = a(minus 1

2radic

30)

In each cell the neighbours of the left- and right-handatoms are found in different directions Anyhow if oneallows π3-rotations the surroundings of any atom areidentical to any other atom in the system (exercise) Thedashed vertical line in the figure right is the so-called glideline The lattice remains invariant when it is translatedvertically by a2 and then reflected with respect to thisline Neither operation alone is enough to keep the latticeinvariant Let us return to to the properties of graphenelater

23 SymmetriesLet us then define the concept of symmetry in a more

consistent manner Some of the properties of the crys-tals observed in scattering experiments (cf the next chap-ter) are a straight consequence of the symmetries of thecrystals In order to understand these experiments it isimportant to know which symmetries are possible Alsothe behaviour of electrons in periodic crystals can only beexplained by using simplifications in the Schrodinger equa-tion permitted by the symmetries

Space Group

We are interested in such rigid operations of the crystalthat leave the lattice points invariant Examples of thoseinclude translations rotations and reflections In Bravaislattices such operations are

bull Operations defined by a (Bravais) lattice vector (trans-lations)

bull Operations that map at least one lattice point ontoitself (point operations)

bull Operations that are obtained as a sequence of trans-lations and point operations

These can be described as a mapping

y = a +Rx (2)

which first rotates (or reflects or inverses) an arbitrary vec-tor x with a matrix R and then adds a vector a to theresult In order to fulfil the definition of the symmetry op-eration this should map the whole Bravais lattice (1) ontoitself The general lattice (with a basis) has symmetry op-erations that are not of the above mentioned form They

5

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 2: Condensed Matter Physics - Oulu

General

The website of the course can be found athttpswikioulufidisplay763628SEtusivu

It includes the links to this material exercises and to theirsolutions Also the possible changes on the schedule belowcan be found on the web page

Schedule and Practicalities

All lectures and exercise sessions will be held at roomTE320

bull Lectures Monday 12-14Wednesday 12-14

bull Exercises Thursday 12-14

By solving the exercises you can earn extra points for thefinal exam The more problems you solve the higher is theraise Maximum raise is one grade point It is worthwhileto remember that biggest gain in trying to solve the exer-cise is however in the deeper and more efficient learningof the course

Literature

This course is based mainly on (selected parts)

bull M Marder Condensed Matter Physics (MM)

In addition you can read this course material

Other literature

bull E Thuneberg Solid state physics lecture notes (2012only in Finnish) (ET)

bull C Kittel Introduction to Solid State Physics old butstill usable

bull F Duan and J Guojum Introduction to CondensedMatter Physics

bull X G Wen Quantum Field Theory of Many-body Sys-tems way too complicated but has an intriguing in-troduction

bull N W Ashcroft and N D Mermin Solid StatePhysics a classic Previously used in this course

bull P Pietilainen previously used lecture notes based onthe book above only in Finnish

Contents

The course will cover the following

bull Atomic structure

ndash 2-and 3-dimensional crystals

ndash Experimental determination of crystal structure

ndash Boundaries and interfaces

ndash Complicated structures

bull Electronic structure

ndash Single-electron model

ndash Schrodinger equation and symmetry

ndash Nearly free and tightly bound electrons

ndash Electron-electron interactions

ndash Band structure

bull Mechanical properties

ndash Cohesion

ndash Phonons

bull Electronic transport phenomena

ndash Bloch electrons

ndash Transport phenomena and Fermi liquid theory

bull (Graphene)

1

1 IntroductionThe purpose of this course is to give the basic introduc-

tion on condensed matter physics The subject has a widerange and many interesting phenomena are left for othercourses or to be learned from the literature

History

The term Condensed matter includes all such largegroups of particles that condense into one phase Espe-cially the distances between particles have to be smallcompared with the interactions between them Examples

bull solids

bull amorphous materials

bull liquids

bull soft materials (foams gels biological systems)

bull white dwarfs and neutron stars (astrophysics)

bull nuclear matter (nuclear physics)

Condensed matter physics originates from the study ofsolids Previously the field was called solid state physicsuntil it was noticed that one can describe the behaviour ofliquid metals Helium and liquid crystals with same con-cepts and models This course concentrates mainly on per-fect crystals mainly due to the historical development ofthe field and the simplicity of the related theoretical mod-els

Approximately one third of the physicists in the USconsiders themselves as researchers of condensed matterphysics In the last 50 years (in 2011) there has been 22condensed-matter related Nobel prizes in physics and also5 in chemistry

bull Bardeen Cooper and Schrieffer Theory of low tem-perature superconductivity (1972)

bull Josephson Josephson effect (1973)

bull Cornell Ketterle and Wieman Bose-Einstein conden-sation in dilute gases of alkali atoms (2001)

bull Geim and Novoselov Graphene (2010)

The field has been the source of many practical applica-tions

bull Transistor (1948)

bull Magnetic recording

bull Liquid crystal displays (LCDs)

Therefore condensed matter physics is both interestingand beneficial for basic physics research and also for ap-plications

Finally a short listing on rdquohot topicsrdquo in the currentcondensed matter research

bull Mesoscopic physics and the realisation of quantumcomputation (research done in the University of Oulu)

bull Fermi liquid theory (Oulu)

bull Graphene

bull Topological insulators

Many-Body Problem

All known matter is formed by atoms The explainingof the properties of single atoms can be done with quan-tum mechanics with an astounding accuracy When oneadds more atoms to the system the number of the relevantdegrees of freedom in Schrodinger equation grows exponen-tially In principle all properties of the many atom systemcan still be found by solving the Schrodinger equation butin practise the required computational power grows veryrapidly out of reach As an example the computers in1980s could solve for the eigenvalues of the system con-sisting of 11 interacting electrons Two decades later thecomputational power had increased hundred-fold but it al-lowed to include only two extra electrons A typical many-body problem in condensed matter physics includes 1023or so electrons Therefore it is clear that it is extremelyimpractical to study such physics starting from the basicprinciples

Based on the above the condensed matter theories areso-called effective theories In principle they have to beresults of averaging the Schrodinger equation properly butin practise their form has been guessed based on symme-tries and experimental results In this way the theories ofthe condensed matter have become simple beautiful andone can use them to obtain results that are precise (aneffective theory does not have to be imprecise) and givedetailed explanations

The modest goal of condensed matter physics is to ex-plain the whole material world It overlaps with statisticalphysics material physics and liquid and solid mechanicsDue to its diversity the coherent treatment of the subjectis blurred

ldquoThe ability to reduce everything to simple fundamentallaws does not imply the ability to start from those lawsand reconstruct the universe The constructionisthypothesis breaks down when confronted by the twin diffi-culties of scale and complexity The behaviour of large andcomplex aggregates of elementary particles it turns outis not to be understood in terms of a simple extrapolationof the properties of a few particles Instead at eachlevel of complexity entirely new properties appear andthe understanding of the new behaviours require researchwhich I think is as fundamental in its nature as any otherrdquo

- P W Anderson 1972

2

2 Atomic StructureMM Chapters 1 and 2 excluding 233-236 and

262

Scanning-tunnelling-microscopic image (page 17)on NbSe2 surface in atomic resolution The dis-tance between nearest neighbours is 035 nm(httpwwwpmacaltecheduGSRcondmathtml)

Fluoride chrystal on top of quartz chrystal (image byChip Clark)

The simplest way to form a macroscopic solid is to orga-nize atoms into small basic units that repeat periodicallyThis is called crystal structure Let us recall the definitionsin the course of Solid State Physics (763333A)

21 Crystal Structure (ET)Many crystals appear in forms where flat surfaces meet

each other with constant angles Such shapes can be un-derstood based on the organization of atoms Solids areoften formed of many crystals This means that they arecomposed of many joined crystals oriented in different di-rections A single crystal can contain eg sim 1018 atomswhereas the whole macroscopic body has sim 1023 atoms

Generally speaking the structure of a solid can be ex-tremely complicated Even if it was formed by basic unitscomprising of same atoms it does not necessarily have arepeating structure Glass is an example of such a mate-rial formed by SiO2-units This kind of material is calledamorphous

Let us consider an ideal case where we ignore all possibleimperfections of lattice The description of such crystal canbe divided in two parts

1) Group of atoms that form a repeating object in the

crystal We refer to this as basis

2) Group of points in space where one has to place thebasis in order to form the crystal Such a group ispresented in form

r = n1a1 + n2a2 + n3a3 (1)

Here n1 n2 and n3 are integers Vectors a1 a2 anda3 called primitive vectors (they have to be linearlyindependent) The group of points (1) is referred toas Bravais lattice and its members as lattice points

a1a2

a3

The volume defined by the primitive vectors is calledprimitive cell The primitive cells contain the completeinformation on the whole crystal Primitive cells are notunique but they all have to have the same volume Aprimitive cell of a Bravais lattice contains exactly one lat-tice point and thus the volume of the cell is inverse of thedensity of the crystal

+ =

An example in two dimensions

a1

a2

a1

a2

The choice of the primitive vectors is not unique In thefigure one can use also the vectors aprime1 and aprime2 as primitivevectors They reproduce also all lattice points

a1

a2 a1

a2

alkeiskoppi

yksikkoumlkoppi

In certain symmetric lattices it is more practical to useorthogonal vectors instead of primitive ones (even if theydo not span the whole lattice) The volume defined by suchvectors is called unit cell The lengths of the sides of a unitcell are called lattice constants

3

22 Two-Dimensional LatticeLet us first restrict our considerations into two dimen-

sional lattices because they are easier to understand andvisualize than their three dimensional counter-parts How-ever it is worthwhile to notice that there exists genuinelytwo-dimensional lattices such as graphene that is pre-sented later in the course Also the surfaces of crystalsand interfaces between two crystals are naturally two di-mensional

Bravais Lattice

In order to achieve two-dimensional Bravais lattice weset a3 = 0 in Definition (1) One can show using grouptheory that there exist five essentially different Bravais lat-tices

bull square symmetric in reflections with respect to bothx and y -axes and with respect to 90 -rotations

bull rectangular when square lattice is squeezed it losesits rotational symmetry and becomes rectangular

bull hexagonal (trigonal) symmetric with respect two xand y -reflections and 60 -rotations

bull centered rectangular squeezed hexagonal no ro-tational symmetry By repeating the boxed structureone obtains the lattice hence the name

bull oblique arbitrary choice of primitive vectors a1 anda2 without any specific symmetries only inversionsymmetry rrarr minusr

The gray areas in the above denote the so called Wigner-Seitz cells Wigner-Seitz cells are primitive cells that areconserved in any symmetry operation that leaves the wholelattice invariant The Wigner-Seitz cell of a lattice pointis the volume that is closer to that point than any otherlattice point (cf Figures)

Example Hexagonal lattice A choice for the primi-tive vectors of the hexagonal lattice are for example

a1 = a(1 0

)a2 = a

(12 minus

radic3

2

)

where a is the lattice constant Another option is

aprime1 = a(radic

32

12

)aprime2 = a

(radic3

2 minus 12

)

Lattice and a Basis

A structure is a Bravais lattice only if it is symmetricwith respect to translations with a lattice vector (cf thedefinition in a later section) In nature the lattices areseldom Bravais lattices but lattices with a basis As anexample let us consider the honeycomb lattice which isthe ordering for the carbon atoms in graphene

Example Graphene

Geim and Kim Carbon Wonderland ScientificAmerican 90-97 April 2008

Graphene is one atom thick layer of graphite in whichthe carbon atoms are ordered in the honeycomb structureresembling a chicken wire

4

(Wikipedia)

Graphene has been used as a theoretical tool since the1950rsquos but experimentally it was rdquofoundrdquo only in 2004A Geim and K Novoselov were able to separate thin lay-ers of graphite (the material of pencils consists of stackedlayers of graphene) some of which were only one atomthick Therefore graphene is the thinnest known materialin the Universe It is also the strongest ever measured ma-terial (200 times stronger than steel) It is flexible so itis easy to mold Graphene supports current densities thatare six times of those in copper Its charge carriers behavelike massless fermions that are described with the Diracequation This allows the study of relativistic quantummechanics in graphene These and many other interest-ing properties are the reason why graphene is used as anillustrative tool in this course

As was mentioned in the above graphene takes the formof the honeycomb lattice which is a Bravais lattice with abasis The starting point is the hexagonal lattice whoseprimitive vectors are

aprime1 = a(radic

32

12

)

aprime2 = a(radic

32 minus 1

2

)

Each lattice point is then replaced with the basis definedby

v1 = a(

12radic

30)

v2 = a(minus 1

2radic

30)

In each cell the neighbours of the left- and right-handatoms are found in different directions Anyhow if oneallows π3-rotations the surroundings of any atom areidentical to any other atom in the system (exercise) Thedashed vertical line in the figure right is the so-called glideline The lattice remains invariant when it is translatedvertically by a2 and then reflected with respect to thisline Neither operation alone is enough to keep the latticeinvariant Let us return to to the properties of graphenelater

23 SymmetriesLet us then define the concept of symmetry in a more

consistent manner Some of the properties of the crys-tals observed in scattering experiments (cf the next chap-ter) are a straight consequence of the symmetries of thecrystals In order to understand these experiments it isimportant to know which symmetries are possible Alsothe behaviour of electrons in periodic crystals can only beexplained by using simplifications in the Schrodinger equa-tion permitted by the symmetries

Space Group

We are interested in such rigid operations of the crystalthat leave the lattice points invariant Examples of thoseinclude translations rotations and reflections In Bravaislattices such operations are

bull Operations defined by a (Bravais) lattice vector (trans-lations)

bull Operations that map at least one lattice point ontoitself (point operations)

bull Operations that are obtained as a sequence of trans-lations and point operations

These can be described as a mapping

y = a +Rx (2)

which first rotates (or reflects or inverses) an arbitrary vec-tor x with a matrix R and then adds a vector a to theresult In order to fulfil the definition of the symmetry op-eration this should map the whole Bravais lattice (1) ontoitself The general lattice (with a basis) has symmetry op-erations that are not of the above mentioned form They

5

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 3: Condensed Matter Physics - Oulu

1 IntroductionThe purpose of this course is to give the basic introduc-

tion on condensed matter physics The subject has a widerange and many interesting phenomena are left for othercourses or to be learned from the literature

History

The term Condensed matter includes all such largegroups of particles that condense into one phase Espe-cially the distances between particles have to be smallcompared with the interactions between them Examples

bull solids

bull amorphous materials

bull liquids

bull soft materials (foams gels biological systems)

bull white dwarfs and neutron stars (astrophysics)

bull nuclear matter (nuclear physics)

Condensed matter physics originates from the study ofsolids Previously the field was called solid state physicsuntil it was noticed that one can describe the behaviour ofliquid metals Helium and liquid crystals with same con-cepts and models This course concentrates mainly on per-fect crystals mainly due to the historical development ofthe field and the simplicity of the related theoretical mod-els

Approximately one third of the physicists in the USconsiders themselves as researchers of condensed matterphysics In the last 50 years (in 2011) there has been 22condensed-matter related Nobel prizes in physics and also5 in chemistry

bull Bardeen Cooper and Schrieffer Theory of low tem-perature superconductivity (1972)

bull Josephson Josephson effect (1973)

bull Cornell Ketterle and Wieman Bose-Einstein conden-sation in dilute gases of alkali atoms (2001)

bull Geim and Novoselov Graphene (2010)

The field has been the source of many practical applica-tions

bull Transistor (1948)

bull Magnetic recording

bull Liquid crystal displays (LCDs)

Therefore condensed matter physics is both interestingand beneficial for basic physics research and also for ap-plications

Finally a short listing on rdquohot topicsrdquo in the currentcondensed matter research

bull Mesoscopic physics and the realisation of quantumcomputation (research done in the University of Oulu)

bull Fermi liquid theory (Oulu)

bull Graphene

bull Topological insulators

Many-Body Problem

All known matter is formed by atoms The explainingof the properties of single atoms can be done with quan-tum mechanics with an astounding accuracy When oneadds more atoms to the system the number of the relevantdegrees of freedom in Schrodinger equation grows exponen-tially In principle all properties of the many atom systemcan still be found by solving the Schrodinger equation butin practise the required computational power grows veryrapidly out of reach As an example the computers in1980s could solve for the eigenvalues of the system con-sisting of 11 interacting electrons Two decades later thecomputational power had increased hundred-fold but it al-lowed to include only two extra electrons A typical many-body problem in condensed matter physics includes 1023or so electrons Therefore it is clear that it is extremelyimpractical to study such physics starting from the basicprinciples

Based on the above the condensed matter theories areso-called effective theories In principle they have to beresults of averaging the Schrodinger equation properly butin practise their form has been guessed based on symme-tries and experimental results In this way the theories ofthe condensed matter have become simple beautiful andone can use them to obtain results that are precise (aneffective theory does not have to be imprecise) and givedetailed explanations

The modest goal of condensed matter physics is to ex-plain the whole material world It overlaps with statisticalphysics material physics and liquid and solid mechanicsDue to its diversity the coherent treatment of the subjectis blurred

ldquoThe ability to reduce everything to simple fundamentallaws does not imply the ability to start from those lawsand reconstruct the universe The constructionisthypothesis breaks down when confronted by the twin diffi-culties of scale and complexity The behaviour of large andcomplex aggregates of elementary particles it turns outis not to be understood in terms of a simple extrapolationof the properties of a few particles Instead at eachlevel of complexity entirely new properties appear andthe understanding of the new behaviours require researchwhich I think is as fundamental in its nature as any otherrdquo

- P W Anderson 1972

2

2 Atomic StructureMM Chapters 1 and 2 excluding 233-236 and

262

Scanning-tunnelling-microscopic image (page 17)on NbSe2 surface in atomic resolution The dis-tance between nearest neighbours is 035 nm(httpwwwpmacaltecheduGSRcondmathtml)

Fluoride chrystal on top of quartz chrystal (image byChip Clark)

The simplest way to form a macroscopic solid is to orga-nize atoms into small basic units that repeat periodicallyThis is called crystal structure Let us recall the definitionsin the course of Solid State Physics (763333A)

21 Crystal Structure (ET)Many crystals appear in forms where flat surfaces meet

each other with constant angles Such shapes can be un-derstood based on the organization of atoms Solids areoften formed of many crystals This means that they arecomposed of many joined crystals oriented in different di-rections A single crystal can contain eg sim 1018 atomswhereas the whole macroscopic body has sim 1023 atoms

Generally speaking the structure of a solid can be ex-tremely complicated Even if it was formed by basic unitscomprising of same atoms it does not necessarily have arepeating structure Glass is an example of such a mate-rial formed by SiO2-units This kind of material is calledamorphous

Let us consider an ideal case where we ignore all possibleimperfections of lattice The description of such crystal canbe divided in two parts

1) Group of atoms that form a repeating object in the

crystal We refer to this as basis

2) Group of points in space where one has to place thebasis in order to form the crystal Such a group ispresented in form

r = n1a1 + n2a2 + n3a3 (1)

Here n1 n2 and n3 are integers Vectors a1 a2 anda3 called primitive vectors (they have to be linearlyindependent) The group of points (1) is referred toas Bravais lattice and its members as lattice points

a1a2

a3

The volume defined by the primitive vectors is calledprimitive cell The primitive cells contain the completeinformation on the whole crystal Primitive cells are notunique but they all have to have the same volume Aprimitive cell of a Bravais lattice contains exactly one lat-tice point and thus the volume of the cell is inverse of thedensity of the crystal

+ =

An example in two dimensions

a1

a2

a1

a2

The choice of the primitive vectors is not unique In thefigure one can use also the vectors aprime1 and aprime2 as primitivevectors They reproduce also all lattice points

a1

a2 a1

a2

alkeiskoppi

yksikkoumlkoppi

In certain symmetric lattices it is more practical to useorthogonal vectors instead of primitive ones (even if theydo not span the whole lattice) The volume defined by suchvectors is called unit cell The lengths of the sides of a unitcell are called lattice constants

3

22 Two-Dimensional LatticeLet us first restrict our considerations into two dimen-

sional lattices because they are easier to understand andvisualize than their three dimensional counter-parts How-ever it is worthwhile to notice that there exists genuinelytwo-dimensional lattices such as graphene that is pre-sented later in the course Also the surfaces of crystalsand interfaces between two crystals are naturally two di-mensional

Bravais Lattice

In order to achieve two-dimensional Bravais lattice weset a3 = 0 in Definition (1) One can show using grouptheory that there exist five essentially different Bravais lat-tices

bull square symmetric in reflections with respect to bothx and y -axes and with respect to 90 -rotations

bull rectangular when square lattice is squeezed it losesits rotational symmetry and becomes rectangular

bull hexagonal (trigonal) symmetric with respect two xand y -reflections and 60 -rotations

bull centered rectangular squeezed hexagonal no ro-tational symmetry By repeating the boxed structureone obtains the lattice hence the name

bull oblique arbitrary choice of primitive vectors a1 anda2 without any specific symmetries only inversionsymmetry rrarr minusr

The gray areas in the above denote the so called Wigner-Seitz cells Wigner-Seitz cells are primitive cells that areconserved in any symmetry operation that leaves the wholelattice invariant The Wigner-Seitz cell of a lattice pointis the volume that is closer to that point than any otherlattice point (cf Figures)

Example Hexagonal lattice A choice for the primi-tive vectors of the hexagonal lattice are for example

a1 = a(1 0

)a2 = a

(12 minus

radic3

2

)

where a is the lattice constant Another option is

aprime1 = a(radic

32

12

)aprime2 = a

(radic3

2 minus 12

)

Lattice and a Basis

A structure is a Bravais lattice only if it is symmetricwith respect to translations with a lattice vector (cf thedefinition in a later section) In nature the lattices areseldom Bravais lattices but lattices with a basis As anexample let us consider the honeycomb lattice which isthe ordering for the carbon atoms in graphene

Example Graphene

Geim and Kim Carbon Wonderland ScientificAmerican 90-97 April 2008

Graphene is one atom thick layer of graphite in whichthe carbon atoms are ordered in the honeycomb structureresembling a chicken wire

4

(Wikipedia)

Graphene has been used as a theoretical tool since the1950rsquos but experimentally it was rdquofoundrdquo only in 2004A Geim and K Novoselov were able to separate thin lay-ers of graphite (the material of pencils consists of stackedlayers of graphene) some of which were only one atomthick Therefore graphene is the thinnest known materialin the Universe It is also the strongest ever measured ma-terial (200 times stronger than steel) It is flexible so itis easy to mold Graphene supports current densities thatare six times of those in copper Its charge carriers behavelike massless fermions that are described with the Diracequation This allows the study of relativistic quantummechanics in graphene These and many other interest-ing properties are the reason why graphene is used as anillustrative tool in this course

As was mentioned in the above graphene takes the formof the honeycomb lattice which is a Bravais lattice with abasis The starting point is the hexagonal lattice whoseprimitive vectors are

aprime1 = a(radic

32

12

)

aprime2 = a(radic

32 minus 1

2

)

Each lattice point is then replaced with the basis definedby

v1 = a(

12radic

30)

v2 = a(minus 1

2radic

30)

In each cell the neighbours of the left- and right-handatoms are found in different directions Anyhow if oneallows π3-rotations the surroundings of any atom areidentical to any other atom in the system (exercise) Thedashed vertical line in the figure right is the so-called glideline The lattice remains invariant when it is translatedvertically by a2 and then reflected with respect to thisline Neither operation alone is enough to keep the latticeinvariant Let us return to to the properties of graphenelater

23 SymmetriesLet us then define the concept of symmetry in a more

consistent manner Some of the properties of the crys-tals observed in scattering experiments (cf the next chap-ter) are a straight consequence of the symmetries of thecrystals In order to understand these experiments it isimportant to know which symmetries are possible Alsothe behaviour of electrons in periodic crystals can only beexplained by using simplifications in the Schrodinger equa-tion permitted by the symmetries

Space Group

We are interested in such rigid operations of the crystalthat leave the lattice points invariant Examples of thoseinclude translations rotations and reflections In Bravaislattices such operations are

bull Operations defined by a (Bravais) lattice vector (trans-lations)

bull Operations that map at least one lattice point ontoitself (point operations)

bull Operations that are obtained as a sequence of trans-lations and point operations

These can be described as a mapping

y = a +Rx (2)

which first rotates (or reflects or inverses) an arbitrary vec-tor x with a matrix R and then adds a vector a to theresult In order to fulfil the definition of the symmetry op-eration this should map the whole Bravais lattice (1) ontoitself The general lattice (with a basis) has symmetry op-erations that are not of the above mentioned form They

5

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 4: Condensed Matter Physics - Oulu

2 Atomic StructureMM Chapters 1 and 2 excluding 233-236 and

262

Scanning-tunnelling-microscopic image (page 17)on NbSe2 surface in atomic resolution The dis-tance between nearest neighbours is 035 nm(httpwwwpmacaltecheduGSRcondmathtml)

Fluoride chrystal on top of quartz chrystal (image byChip Clark)

The simplest way to form a macroscopic solid is to orga-nize atoms into small basic units that repeat periodicallyThis is called crystal structure Let us recall the definitionsin the course of Solid State Physics (763333A)

21 Crystal Structure (ET)Many crystals appear in forms where flat surfaces meet

each other with constant angles Such shapes can be un-derstood based on the organization of atoms Solids areoften formed of many crystals This means that they arecomposed of many joined crystals oriented in different di-rections A single crystal can contain eg sim 1018 atomswhereas the whole macroscopic body has sim 1023 atoms

Generally speaking the structure of a solid can be ex-tremely complicated Even if it was formed by basic unitscomprising of same atoms it does not necessarily have arepeating structure Glass is an example of such a mate-rial formed by SiO2-units This kind of material is calledamorphous

Let us consider an ideal case where we ignore all possibleimperfections of lattice The description of such crystal canbe divided in two parts

1) Group of atoms that form a repeating object in the

crystal We refer to this as basis

2) Group of points in space where one has to place thebasis in order to form the crystal Such a group ispresented in form

r = n1a1 + n2a2 + n3a3 (1)

Here n1 n2 and n3 are integers Vectors a1 a2 anda3 called primitive vectors (they have to be linearlyindependent) The group of points (1) is referred toas Bravais lattice and its members as lattice points

a1a2

a3

The volume defined by the primitive vectors is calledprimitive cell The primitive cells contain the completeinformation on the whole crystal Primitive cells are notunique but they all have to have the same volume Aprimitive cell of a Bravais lattice contains exactly one lat-tice point and thus the volume of the cell is inverse of thedensity of the crystal

+ =

An example in two dimensions

a1

a2

a1

a2

The choice of the primitive vectors is not unique In thefigure one can use also the vectors aprime1 and aprime2 as primitivevectors They reproduce also all lattice points

a1

a2 a1

a2

alkeiskoppi

yksikkoumlkoppi

In certain symmetric lattices it is more practical to useorthogonal vectors instead of primitive ones (even if theydo not span the whole lattice) The volume defined by suchvectors is called unit cell The lengths of the sides of a unitcell are called lattice constants

3

22 Two-Dimensional LatticeLet us first restrict our considerations into two dimen-

sional lattices because they are easier to understand andvisualize than their three dimensional counter-parts How-ever it is worthwhile to notice that there exists genuinelytwo-dimensional lattices such as graphene that is pre-sented later in the course Also the surfaces of crystalsand interfaces between two crystals are naturally two di-mensional

Bravais Lattice

In order to achieve two-dimensional Bravais lattice weset a3 = 0 in Definition (1) One can show using grouptheory that there exist five essentially different Bravais lat-tices

bull square symmetric in reflections with respect to bothx and y -axes and with respect to 90 -rotations

bull rectangular when square lattice is squeezed it losesits rotational symmetry and becomes rectangular

bull hexagonal (trigonal) symmetric with respect two xand y -reflections and 60 -rotations

bull centered rectangular squeezed hexagonal no ro-tational symmetry By repeating the boxed structureone obtains the lattice hence the name

bull oblique arbitrary choice of primitive vectors a1 anda2 without any specific symmetries only inversionsymmetry rrarr minusr

The gray areas in the above denote the so called Wigner-Seitz cells Wigner-Seitz cells are primitive cells that areconserved in any symmetry operation that leaves the wholelattice invariant The Wigner-Seitz cell of a lattice pointis the volume that is closer to that point than any otherlattice point (cf Figures)

Example Hexagonal lattice A choice for the primi-tive vectors of the hexagonal lattice are for example

a1 = a(1 0

)a2 = a

(12 minus

radic3

2

)

where a is the lattice constant Another option is

aprime1 = a(radic

32

12

)aprime2 = a

(radic3

2 minus 12

)

Lattice and a Basis

A structure is a Bravais lattice only if it is symmetricwith respect to translations with a lattice vector (cf thedefinition in a later section) In nature the lattices areseldom Bravais lattices but lattices with a basis As anexample let us consider the honeycomb lattice which isthe ordering for the carbon atoms in graphene

Example Graphene

Geim and Kim Carbon Wonderland ScientificAmerican 90-97 April 2008

Graphene is one atom thick layer of graphite in whichthe carbon atoms are ordered in the honeycomb structureresembling a chicken wire

4

(Wikipedia)

Graphene has been used as a theoretical tool since the1950rsquos but experimentally it was rdquofoundrdquo only in 2004A Geim and K Novoselov were able to separate thin lay-ers of graphite (the material of pencils consists of stackedlayers of graphene) some of which were only one atomthick Therefore graphene is the thinnest known materialin the Universe It is also the strongest ever measured ma-terial (200 times stronger than steel) It is flexible so itis easy to mold Graphene supports current densities thatare six times of those in copper Its charge carriers behavelike massless fermions that are described with the Diracequation This allows the study of relativistic quantummechanics in graphene These and many other interest-ing properties are the reason why graphene is used as anillustrative tool in this course

As was mentioned in the above graphene takes the formof the honeycomb lattice which is a Bravais lattice with abasis The starting point is the hexagonal lattice whoseprimitive vectors are

aprime1 = a(radic

32

12

)

aprime2 = a(radic

32 minus 1

2

)

Each lattice point is then replaced with the basis definedby

v1 = a(

12radic

30)

v2 = a(minus 1

2radic

30)

In each cell the neighbours of the left- and right-handatoms are found in different directions Anyhow if oneallows π3-rotations the surroundings of any atom areidentical to any other atom in the system (exercise) Thedashed vertical line in the figure right is the so-called glideline The lattice remains invariant when it is translatedvertically by a2 and then reflected with respect to thisline Neither operation alone is enough to keep the latticeinvariant Let us return to to the properties of graphenelater

23 SymmetriesLet us then define the concept of symmetry in a more

consistent manner Some of the properties of the crys-tals observed in scattering experiments (cf the next chap-ter) are a straight consequence of the symmetries of thecrystals In order to understand these experiments it isimportant to know which symmetries are possible Alsothe behaviour of electrons in periodic crystals can only beexplained by using simplifications in the Schrodinger equa-tion permitted by the symmetries

Space Group

We are interested in such rigid operations of the crystalthat leave the lattice points invariant Examples of thoseinclude translations rotations and reflections In Bravaislattices such operations are

bull Operations defined by a (Bravais) lattice vector (trans-lations)

bull Operations that map at least one lattice point ontoitself (point operations)

bull Operations that are obtained as a sequence of trans-lations and point operations

These can be described as a mapping

y = a +Rx (2)

which first rotates (or reflects or inverses) an arbitrary vec-tor x with a matrix R and then adds a vector a to theresult In order to fulfil the definition of the symmetry op-eration this should map the whole Bravais lattice (1) ontoitself The general lattice (with a basis) has symmetry op-erations that are not of the above mentioned form They

5

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 5: Condensed Matter Physics - Oulu

22 Two-Dimensional LatticeLet us first restrict our considerations into two dimen-

sional lattices because they are easier to understand andvisualize than their three dimensional counter-parts How-ever it is worthwhile to notice that there exists genuinelytwo-dimensional lattices such as graphene that is pre-sented later in the course Also the surfaces of crystalsand interfaces between two crystals are naturally two di-mensional

Bravais Lattice

In order to achieve two-dimensional Bravais lattice weset a3 = 0 in Definition (1) One can show using grouptheory that there exist five essentially different Bravais lat-tices

bull square symmetric in reflections with respect to bothx and y -axes and with respect to 90 -rotations

bull rectangular when square lattice is squeezed it losesits rotational symmetry and becomes rectangular

bull hexagonal (trigonal) symmetric with respect two xand y -reflections and 60 -rotations

bull centered rectangular squeezed hexagonal no ro-tational symmetry By repeating the boxed structureone obtains the lattice hence the name

bull oblique arbitrary choice of primitive vectors a1 anda2 without any specific symmetries only inversionsymmetry rrarr minusr

The gray areas in the above denote the so called Wigner-Seitz cells Wigner-Seitz cells are primitive cells that areconserved in any symmetry operation that leaves the wholelattice invariant The Wigner-Seitz cell of a lattice pointis the volume that is closer to that point than any otherlattice point (cf Figures)

Example Hexagonal lattice A choice for the primi-tive vectors of the hexagonal lattice are for example

a1 = a(1 0

)a2 = a

(12 minus

radic3

2

)

where a is the lattice constant Another option is

aprime1 = a(radic

32

12

)aprime2 = a

(radic3

2 minus 12

)

Lattice and a Basis

A structure is a Bravais lattice only if it is symmetricwith respect to translations with a lattice vector (cf thedefinition in a later section) In nature the lattices areseldom Bravais lattices but lattices with a basis As anexample let us consider the honeycomb lattice which isthe ordering for the carbon atoms in graphene

Example Graphene

Geim and Kim Carbon Wonderland ScientificAmerican 90-97 April 2008

Graphene is one atom thick layer of graphite in whichthe carbon atoms are ordered in the honeycomb structureresembling a chicken wire

4

(Wikipedia)

Graphene has been used as a theoretical tool since the1950rsquos but experimentally it was rdquofoundrdquo only in 2004A Geim and K Novoselov were able to separate thin lay-ers of graphite (the material of pencils consists of stackedlayers of graphene) some of which were only one atomthick Therefore graphene is the thinnest known materialin the Universe It is also the strongest ever measured ma-terial (200 times stronger than steel) It is flexible so itis easy to mold Graphene supports current densities thatare six times of those in copper Its charge carriers behavelike massless fermions that are described with the Diracequation This allows the study of relativistic quantummechanics in graphene These and many other interest-ing properties are the reason why graphene is used as anillustrative tool in this course

As was mentioned in the above graphene takes the formof the honeycomb lattice which is a Bravais lattice with abasis The starting point is the hexagonal lattice whoseprimitive vectors are

aprime1 = a(radic

32

12

)

aprime2 = a(radic

32 minus 1

2

)

Each lattice point is then replaced with the basis definedby

v1 = a(

12radic

30)

v2 = a(minus 1

2radic

30)

In each cell the neighbours of the left- and right-handatoms are found in different directions Anyhow if oneallows π3-rotations the surroundings of any atom areidentical to any other atom in the system (exercise) Thedashed vertical line in the figure right is the so-called glideline The lattice remains invariant when it is translatedvertically by a2 and then reflected with respect to thisline Neither operation alone is enough to keep the latticeinvariant Let us return to to the properties of graphenelater

23 SymmetriesLet us then define the concept of symmetry in a more

consistent manner Some of the properties of the crys-tals observed in scattering experiments (cf the next chap-ter) are a straight consequence of the symmetries of thecrystals In order to understand these experiments it isimportant to know which symmetries are possible Alsothe behaviour of electrons in periodic crystals can only beexplained by using simplifications in the Schrodinger equa-tion permitted by the symmetries

Space Group

We are interested in such rigid operations of the crystalthat leave the lattice points invariant Examples of thoseinclude translations rotations and reflections In Bravaislattices such operations are

bull Operations defined by a (Bravais) lattice vector (trans-lations)

bull Operations that map at least one lattice point ontoitself (point operations)

bull Operations that are obtained as a sequence of trans-lations and point operations

These can be described as a mapping

y = a +Rx (2)

which first rotates (or reflects or inverses) an arbitrary vec-tor x with a matrix R and then adds a vector a to theresult In order to fulfil the definition of the symmetry op-eration this should map the whole Bravais lattice (1) ontoitself The general lattice (with a basis) has symmetry op-erations that are not of the above mentioned form They

5

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 6: Condensed Matter Physics - Oulu

(Wikipedia)

Graphene has been used as a theoretical tool since the1950rsquos but experimentally it was rdquofoundrdquo only in 2004A Geim and K Novoselov were able to separate thin lay-ers of graphite (the material of pencils consists of stackedlayers of graphene) some of which were only one atomthick Therefore graphene is the thinnest known materialin the Universe It is also the strongest ever measured ma-terial (200 times stronger than steel) It is flexible so itis easy to mold Graphene supports current densities thatare six times of those in copper Its charge carriers behavelike massless fermions that are described with the Diracequation This allows the study of relativistic quantummechanics in graphene These and many other interest-ing properties are the reason why graphene is used as anillustrative tool in this course

As was mentioned in the above graphene takes the formof the honeycomb lattice which is a Bravais lattice with abasis The starting point is the hexagonal lattice whoseprimitive vectors are

aprime1 = a(radic

32

12

)

aprime2 = a(radic

32 minus 1

2

)

Each lattice point is then replaced with the basis definedby

v1 = a(

12radic

30)

v2 = a(minus 1

2radic

30)

In each cell the neighbours of the left- and right-handatoms are found in different directions Anyhow if oneallows π3-rotations the surroundings of any atom areidentical to any other atom in the system (exercise) Thedashed vertical line in the figure right is the so-called glideline The lattice remains invariant when it is translatedvertically by a2 and then reflected with respect to thisline Neither operation alone is enough to keep the latticeinvariant Let us return to to the properties of graphenelater

23 SymmetriesLet us then define the concept of symmetry in a more

consistent manner Some of the properties of the crys-tals observed in scattering experiments (cf the next chap-ter) are a straight consequence of the symmetries of thecrystals In order to understand these experiments it isimportant to know which symmetries are possible Alsothe behaviour of electrons in periodic crystals can only beexplained by using simplifications in the Schrodinger equa-tion permitted by the symmetries

Space Group

We are interested in such rigid operations of the crystalthat leave the lattice points invariant Examples of thoseinclude translations rotations and reflections In Bravaislattices such operations are

bull Operations defined by a (Bravais) lattice vector (trans-lations)

bull Operations that map at least one lattice point ontoitself (point operations)

bull Operations that are obtained as a sequence of trans-lations and point operations

These can be described as a mapping

y = a +Rx (2)

which first rotates (or reflects or inverses) an arbitrary vec-tor x with a matrix R and then adds a vector a to theresult In order to fulfil the definition of the symmetry op-eration this should map the whole Bravais lattice (1) ontoitself The general lattice (with a basis) has symmetry op-erations that are not of the above mentioned form They

5

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 7: Condensed Matter Physics - Oulu

are known as the glide line and screw axis We will returnto them later

The goal is to find a complete set of ways to transformthe lattice in such way that the transformed lattice pointsare on top of the original ones Many of such transfor-mations can be constructed from a minimal set of simplertransformations One can use these symmetry operationsin the classification of different lattice structures and forexample to show using group theoretical arguments thatthere are only five essentially different Bravais-lattices intwo dimensions a result stated earlier

Space group (or symmetry group) G is the set of opera-tions that leave the crystal invariant (why such a set is agroup)

Translations and Point Groups

Let us consider two sub-groups of the space group Theelements in the translation group move all the points in thelattice by a vector

m1a1 +m2a2 +m3a3

and thus leave the lattice invariant according to the defi-nition of the lattice

Point group includes rotation-like operations (rotationsreflections inversions) that leave the structure invariantand in addition map one point onto itself The spacegroup is not just a product of the point and translationgroups For example the glide line (see the definition be-fore) and screw axis (translation and rotation) are bothcombinations whose parts are not elements of the spacegroup

Does the point group define the lattice No the lat-tices with the same point group belong to the same crystalsystem but they do not necessarily have the same latticestructure nor the space group The essential question iswhether the lattices can be transformed continuously toone another without breaking the symmetries along theprocess Formally this means that one must be able tomake a linear transformation S between the space groupsG and Gprime of the crystals ie

SGSminus1 = Gprime

Then there exists a set of continuous mappings from theunit matrix to matrix S

St = (1minus t)I + St

where t is between [0 1] Using this one obtains such a con-tinuous mapping from one lattice to the other that leavesthe symmetries invariant

For example there does not exist such a set of transfor-mations between the rectangular and centered rectangularlattices even though they share the same point group

When one transforms the rectangular into the centeredrectangular the reflection symmetry with respect to they-axis is destroyed

24 3-Dimensional LatticeOne has to study 3-dimensional lattices in order to de-

scribe the crystals found in Nature Based on symmetriesone can show that there exists 230 different lattices with abasis and those have 32 different point groups The com-plete listing of all of them is of course impossible to dohere Therefore we will restrict ourselves in the classifi-cation of the 3-dimensional Bravais lattices Let us firstintroduce some of the structures found in Nature

Simple cubic lattice (sc) is the simplest 3-dimensionallattice The only element that has taken this form as itsground state is polonium This is partly due to the largerdquoemptyrdquo space between the atoms the most of the ele-ments favour more efficient ways of packing

a a

a

Face-centered cubic lattice (fcc) is formed by a sim-ple cubic lattice with an additional lattice points on eachface of the cube

a1a2

a3

a

Primitive vectors defining the lattice are for example

a1 =a

2

(1 1 0

)a2 =

a

2

(1 0 1

)a3 =

a

2

(0 1 1

)

where a is the lattice constant giving the distance be-tween the corners of the cube (not with the nearest neigh-bours) Face-centered cubic lattice is often called cubicclosed-packed structure If one takes the lattice points asspheres with radius a2

radic2 one obtains the maximal pack-

ing density Fcc-lattice can be visualized as layered trigonallattices

6

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
Page 8: Condensed Matter Physics - Oulu

a

Body-centered cubic lattice (bcc) is formed by in-serting an additional lattice point into the center of prim-itive cell of a simple cubic lattice

a1

a2a3

a

An example of a choice for primitive vectors is

a1 =a

2

(1 1 minus1

)a2 =

a

2

(minus1 1 1

)a3 =

a

2

(1 minus1 1

)

Hexagonal lattice is cannot be found amongst the el-ements Its primitive vectors are

a1 =(a 0 0

)a2 =

(a2

aradic

32 0

)a3 =

a

2

(0 0 c

)

Hexagonal closed-packed lattice (hcp) is more inter-esting than the hexagonal lattice because it is the groundstate of many elements It is a lattice with a basis formedby stacking 2-dimensional trigonal lattices like in the caseof fcc-closed packing The difference to fcc is that in hcpthe lattice points of layer are placed on top of the centresof the triangles in the previous layer at a distance c2 Sothe structure is repeated in every other layer in hcp Thisshould be contrasted with the fcc-closed packing where therepetition occurs in every third layer

a

c

Hcp-lattice is formed by a hexagonal lattice with a basis

v1 =(0 0 0

)v2 =

(a2

a2radic

3c2 )

The lattice constants c and a are arbitrary but by choosingc =

radic83a one obtains the closed-packing structure

Both of the closed-packing structures are common espe-cially among the metal elements If the atoms behaved likehard spheres it would be indifferent whether the orderingwas fcc or hcp Nevertheless the elements choose alwayseither of them eg Al Ni and Cu are fcc and Mg Zn andCo are hcp In the hcp the ratio ca deviates slightly fromthe value obtained with the hard sphere approximation

In addition to the closed-packed structures the body-centered cubic lattice is common among elements eg KCr Mn Fe The simple cubic structure is rare with crys-tals (ET)

Diamond lattice is obtained by taking a copy of an fcc-lattice and by translating it with a vector (14 14 14)

The most important property of the diamond structureis that every lattice point has exactly four neighbours (com-pare with the honeycomb structure in 2-D that had threeneighbours) Therefore diamond lattice is quite sparselypacked It is common with elements that have the ten-dency of bonding with four nearest neighbours in such away that all neighbours are at same angles with respect toone another (1095) In addition to carbon also silicon(Si) takes this form

Compounds

The lattice structure of compounds has to be describedwith a lattice with a basis This is because as the namesays they are composed of at least two different elementsLet us consider as an example two most common structuresfor compounds

Salt - Sodium Chloride

The ordinary table salt ie sodium chloride (NaCl)consists of sodium and chlorine atoms ordered in an al-ternating simple cubic lattice This can be seen also as anfcc-structure (lattice constant a) that has a basis at points(0 0 0) (Na) and a2(1 0 0)

Many compounds share the same lattice structure(MM)

7

Crystal a Crystal a Crystal a Crystal a

AgBr 577 KBr 660 MnSe 549 SnTe 631AgCl 555 KCl 630 NaBr 597 SrO 516AgF 492 KF 535 NaCl 564 SrS 602BaO 552 KI 707 NaF 462 SrSe 623BaS 639 LiBr 550 NaI 647 SrTe 647BaSe 660 LiCl 513 NiO 417 TiC 432BaTe 699 LiF 402 PbS 593 TiN 424CaS 569 LiH 409 PbSe 612 TiO 424CaSe 591 LiI 600 PbTe 645 VC 418CaTe 635 MgO 421 RbBr 685 VN 413CdO 470 MgS 520 RbCl 658 ZrC 468CrN 414 MgSe 545 RbF 564 ZrN 461CsF 601 MnO 444 RbI 734FeO 431 MnS 522 SnAs 568

Lattice constants a (10minus10m) Source Wyckoff (1963-71)vol 1

Cesium Chloride

In cesium chloride (CsCl) the cesium and chlorine atomsalternate in a bcc lattice One can see this as a simplecubic lattice that has basis defined by vectors (0 0 0) anda2(1 1 1) Some other compounds share this structure(MM)

Crystal a Crystal a Crystal aAgCd 333 CsCl 412 NiAl 288AgMg 328 CuPd 299 TiCl 383AgZn 316 CuZn 295 TlI 420CsBr 429 NH4Cl 386 TlSb 384

Lattice constants a (10minus10m) Source Wyckoff (1963-71) vol 1

25 Classification of Lattices by Symme-try

Let us first consider the classification of Bravais lattices3-dimensional Bravais lattices have seven point groups thatare called crystal systems There 14 different space groupsmeaning that on the point of view of symmetry there are14 different Bravais lattices In the following the cyrstalsystems and the Bravais lattices belonging to them arelisted

bull Cubic The point group of the cube Includes the sim-ple face-centered and body-centered cubic lattices

bull Tetragonal The symmetry of the cube can be reducedby stretching two opposite sides of the cube resultingin a rectangular prism with a square base This elim-inates the 90-rotational symmetry in two directionsThe simple cube is thus transformed into a simpletetragonal lattice By stretching the fcc and bcc lat-tices one obtains the body-centered tetragonal lattice(can you figure out why)

bull Orthorombic The symmetry of the cube can befurther-on reduced by pulling the square bases of thetetragonal lattices to rectangles Thus the last 90-rotational symmetry is eliminated When the simpletetragonal lattice is pulled along the side of the basesquare one obtains the simple orthorhombic latticeWhen the pull is along the diagonal of the squareone ends up with the base-centered orthorhombic lat-tice Correspondingly by pulling the body-centeredtetragonal lattice one obtains the body-centered andface-centered orthorhombic lattices

bull Monoclinic The orthorhombic symmetry can be re-duced by tilting the rectangles perpendicular to thec-axis (cf the figure below) The simple and base-centered lattices transform into the simple monocliniclattice The face- and body-centered orthorhombiclattices are transformed into body-centered monocliniclattice

bull Triclinic The destruction of the symmetries of thecube is ready when the c-axis is tilted so that it is nolonger perpendicular with the other axes The onlyremaining point symmetry is that of inversion Thereis only one such lattice the triclinic lattice

By torturing the cube one has obtained five of the sevencrystal systems and 12 of the 14 Bravais lattices Thesixth and the 13th are obtained by distorting the cube ina different manner

bull Rhobohedral or Trigonal Let us stretch the cubealong its diagonal This results in the trigonal lat-tice regardless of which of the three cubic lattices wasstretched

The last crystal system and the last Bravais lattice arenot related to the cube in any way

bull Hexagonal Let us place hexagons as bases and per-pendicular walls in between them This is the hexag-onal point group that has one Bravais lattice thehexagonal lattice

It is not in any way trivial why we have obtained all pos-sible 3-D Bravais lattices in this way It is not necessaryhowever to justify that here At this stage it is enoughto know the existence of different classes and what belongsin them As a conclusion a table of all Bravais latticepresented above

(Source httpwwwiuetuwienacatphd

karlowatznode8html)

8

On the Symmetries of Lattices with Basis

The introduction of basis into the Bravais lattice com-plicates the classification considerably As a consequencethe number of different lattices grows to 230 and numberof point groups to 32 The complete classification of theseis not a subject on this course but we will only summarisethe basic principle As in the case of the classification ofBravais lattices one should first find out the point groupsIt can be done by starting with the seven crystal systemsand by reducing the symmetries of their Bravais latticesin a similar manner as the symmetries of the cube were re-duced in the search for the Bravais lattices This is possibledue to the basis which reduces the symmetry of the latticeThe new point groups found this way belong to the origi-nal crystal system up to the point where their symmetryis reduced so far that all of the remaining symmetry oper-ations can be found also from a less symmetrical systemThen the point group is joined into the less symmetricalsystem

The space groups are obtained in two ways Symmor-phic lattices result from placing an object correspondingto every point group of the crystal system into every Bra-vais lattice of the system When one takes into accountthat the object can be placed in several ways into a givenlattice one obtains 73 different space groups The rest ofthe space groups are nonsymmorphic They contain oper-ations that cannot be formed solely by translations of theBravais lattice and the operations of the point group Egglide line and screw axis

Macroscopic consequences of symmetries

Sometimes macroscopic phenomena reveal symmetriesthat reduce the number of possible lattice structures Letus look more closely on two such phenomenon

Pyroelectricity

Some materials (eg tourmaline) have the ability of pro-ducing instantaneous voltages while heated or cooled Thisis a consequence of the fact that pyroelectric materials havenon-zero dipole moments in a unit cell leading polarizationof the whole lattice (in the absence of electric field) In aconstant temperature the electrons neutralize this polariza-tion but when the temperature is changing a measurablepotential difference is created on the opposite sides of thecrystal

In the equilibrium the polarization is a constant andtherefore the point group of the pyroelectric lattice has toleave its direction invariant This restricts the number ofpossible point groups The only possible rotation axis isalong the polarization and the crystal cannot have reflec-tion symmetry with respect to the plane perpendicular tothe polarization

Optical activity

Some crystals (like SiO2) can rotate the plane of polar-ized light This is possible only if the unit cells are chiral

meaning that the cell is not identical with its mirror image(in translations and rotations)

26 Binding Forces (ET)Before examining the experimental studies of the lattice

structure it is worthwhile to recall the forces the bind thelattice together

At short distances the force between two atoms is alwaysrepulsive This is mostly due to the Pauli exclusion prin-ciple preventing more than one electron to be in the samequantum state Also the Coulomb repulsion between elec-trons is essential At larger distances the forces are oftenattractive

r00

E(r)

A sketch of the potential energy between two atoms asa function of their separation r

Covalent bond Attractive force results because pairsof atoms share part of their electrons As a consequencethe electrons can occupy larger volume in space and thustheir average kinetic energy is lowered (In the ground stateand according to the uncertainty principle the momentump sim ~d where d is the region where the electron can befound and the kinetic energy Ekin = p22m On the otherhand the Pauli principle may prevent the lowering of theenergy)

Metallic bond A large group of atoms share part oftheir electrons so that they are allowed to move through-out the crystal The justification otherwise the same as incovalent binding

Ionic bond In some compounds eg NaCl a sodiumatom donates almost entirely its out-most electron to achlorine atom In such a case the attractive force is dueto the Coulomb potential The potential energy betweentwo ions (charges n1e and n2e) is

E12 =1

4πε0

n1n2e2

|r1 minus r2| (3)

In a NaCl-crystal one Na+-ion has six Clminus-ions as itsnearest neighbours The energy of one bond between anearest neighbour is

Ep1 = minus 1

4πε0

e2

R (4)

where R is the distance between nearest neighbours Thenext-nearest neighbours are 12 Na+-ions with a distanced =radic

2R

9

6 kpld = R

12 kpld = 2 R

Na+Cl-

tutkitaantaumlmaumln

naapureita

8 kpld = 3 R

The interaction energy of one Na+ ion with other ions isobtained by adding together the interaction energies withneighbours at all distances As a result one obtains

Ep = minus e2

4πε0R

(6minus 12radic

2+

8radic3minus

)= minus e2α

4πε0R (5)

Here α is the sum inside the brackets which is called theMadelung constant Its value depends on the lattice andin this case is α = 17627

In order to proceed on has to come up with form for therepulsive force For simplicity let us assume

Eprepulsive =βe2

4πε0Rn (6)

The total potential energy is thus

Ep(R) = minus e2

4πε0

Rminus β

Rn

) (7)

By finding its minimum we result in β = αRnminus1n and inenergy

Ep = minus αe2

4πε0R

(1minus 1

n

) (8)

The binding energy U of the whole lattice is the negativeof this multiplied with the number N of the NaCl-pairs

U =Nαe2

4πε0R

(1minus 1

n

) (9)

By choosing n = 94 this in correspondence with the mea-surements Notice that the contribution of the repulsivepart to the binding energy is small sim 10

Hydrogen bond Hydrogen has only one electronWhen hydrogen combines with for example oxygen themain part of the wave function of the electron is centerednear the oxygen leaving a positive charge to the hydrogenWith this charge it can attract some third atom The re-sulting bond between molecules is called hydrogen bondThis is an important bond for example in ice

Van der Waals interaction This gives a weak attrac-tion also between neutral atoms Idea Due to the circularmotion of the electrons the neutral atoms behave like vi-brating electric dipoles Instantaneous dipole moment inone atom creates an electric field that polarizes the otheratom resulting in an interatomic dipole-dipole force Theinteraction goes with distance as 1r6 and is essential be-tween eg atoms of noble gases

In the table below the melting temperatures of somesolids are listed (at normal pressure) leading to estimatesof the strengths between interatomic forces The lines di-vide the materials in terms of the bond types presentedabove

material melting temperature (K) lattice structureSi 1683 diamondC (4300) diamond

GaAs 1511 zincblendeSiO2 1670

Al2O3 2044Hg 2343Na 371 bccAl 933 fccCu 1356 fccFe 1808 bccW 3683 bcc

CsCl 918NaCl 1075H2O 273He -Ne 245 fccAr 839 fccH2 14O2 547

In addition to diamond structure carbon as also anotherform graphite that consists of stacked layers of grapheneIn graphene each carbon atom forms a covalent bond be-tween three nearest neighbours (honeycomb structure) re-sulting in layers The interlayer forces are of van der Waals-type and therefore very weak allowing the layers to moveeasily with respect to each other Therefore the rdquoleadrdquo inpencils (Swedish chemist Carl Scheele showed in 1779 thatgraphite is made of carbon instead of lead) crumbles easilywhen writing

A model of graphite where the spheres represent the car-bon atoms (Wikipedia)

Covalent bonds are formed often into a specific direc-tion Covalent crystals are hard but brittle in other wordsthey crack when they are hit hard Metallic bonds arenot essentially dependent on direction Thus the metallicatoms can slide past one another still retaining the bondTherefore the metals can be mold by forging

27 Experimental Determination of Crys-tal Structure

MM Chapters 31-32 33 main points 34-344except equations

10

In 1912 German physicist Max von Laue predicted thatthe then recently discovered (1895) X-rays could scatterfrom crystals like ordinary light from the diffraction grat-ing1 At that time it was not known that crystals haveperiodic structures nor that the X-rays have a wave char-acter With a little bit of hindsight the prediction wasreasonable because the average distances between atomsin solids are of the order of an Angstrom (A=10minus10 m)and the range of wave lengths of X-rays settles in between01-100 A

The idea was objected at first The strongest counter-argument was that the inevitable wiggling of the atoms dueto heat would blur the required regularities in the latticeand thus destroy the possible diffraction maxima Nowa-days it is of course known that such high temperatureswould melt the crystal The later experimental results havealso shown that the random motion of the atoms due toheat is only much less than argued by the opponents Asan example let us model the bonds in then NaCl-latticewith a spring The measured Youngrsquos modulus indicatesthat the spring constant would have to be of the orderk = 10 Nm According to the equipartition theoremthis would result in average thermal motion of the atoms〈x〉 =

radic2kBTk asymp 2 middot 10minus11 m which is much less than

the average distance between atoms in NaCl-crystal (cfprevious table)

Scattering Theory of Crystals

The scattering experiment is conducted by directing aplane wave towards a sample of condensed matter Whenthe wave reaches the sample there is an interaction be-tween them The outgoing (scattered) radiation is mea-sured far away from the sample Let us consider here asimple scattering experiment and assume that the scatter-ing is elastic meaning that the energy is not transferredbetween the wave and the sample In other words the fre-quencies of the incoming and outgoing waves are the sameThis picture is valid whether the incoming radiation con-sists of photons or for example electrons or neutrons

The wave ψ scattered from an atom located at origintakes the form (cf Quantum Mechanics II)

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r

r

] (10)

In fact this result holds whether the scattered waves aredescribed by quantum mechanics or by classical electrody-namics

One assumes in the above that the incoming radiation isa plane wave with a wave vector k0 defining the directionof propagation The scattering is measured at distances rmuch greater than the range of the interaction between theatom and the wave and at angle 2θ measured from the k0-axis The form factor f contains the detailed informationon the interaction between the scattering potential and the

1von Laue received the Nobel prize from the discovery of X-raydiffraction in 1914 right after his prediction

scattered wave Note that it depends only on the directionof the r-vector

The form factor gives the differential cross section of thescattering

Iatom equivdσ

dΩatom= |f(r)|2 (11)

The intensity of the scattered wave at a solid angle dΩ atdistance r from the sample is dΩtimes Iatomr

2

(Source MM) Scattering from a square lattice (25atoms) When the radiation k0 comes in from the rightdirection the waves scattered from different atoms inter-fere constructively By measuring the resulting wave k oneobserves an intensity maximum

There are naturally many atoms in a lattice and it isthus necessary to study scattering from multiple scatterersLet us assume in the following that we know the form factorf The angular dependence of the scattering is due to twofactors

bull Every scatterer emits radiation into different direc-tions with different intensities

bull The waves coming from different scatterers interfereand thus the resultant wave contains information onthe correlations between the scatterers

Let us assume in the following that the origin is placedat a fixed lattice point First we have to find out howEquation (10) is changed when the scatterer is located at adistance R from the origin This deviation causes a phasedifference in the scattered wave (compared with a wave

11

scattered at origin) In addition the distance travelled bythe scattered wave is |rminusR| Thus we obtain

ψ asymp Aeminusiωt[eik0middotr + eik0middotRf(r)

eik0|rminusR|

|rminusR|

] (12)

We have assumed in the above that the point of obser-vation r is so far that the changes in the scattering anglecan be neglected At such distances (r R) we can ap-proximate (up to first order in rR)

k0|rminusR| asymp k0r minus k0r

rmiddotR (13)

Let us then define

k = k0r

r

q = k0 minus k

The wave vector k points into the direction of the measure-ment device r has the magnitude of the incoming radiation(elastic scattering) The quantity q describes the differencebetween the momenta of the incoming and outgoing raysThus we obtain

ψ asymp Aeminusiωt[eik0middotr + f(r)

eik0r+iqmiddotR

r

] (14)

The second term in the denominator can be neglected inEquation (13) However one has to include into the ex-ponential function all terms that are large compared with2π

The magnitude of the change in momentum is

q = 2k0 sin θ (15)

where 2θ is the angle between the incoming and outgoingwaves The angle θ is called the Braggrsquos angle Assumingspecular reflection (Huygensrsquo principle valid for X-rays)the Bragg angle is the same as the angle between the in-coming ray and the lattice planes

Finally let us consider the whole lattice and assumeagain that the origin is located somewhere in the middle ofthe lattice and that we measure far away from the sampleBy considering that the scatterers are sparse the observedradiation can be taken to be sum of the waves producedby individual scatterers Furthermore let us assume thatthe effects due to multiple scattering events and inelasticprocesses can be ignored We obtain

ψ asymp Aeminusiωt[eik0middotr +

suml

fl(r)eik0r+iqmiddotRl

r

] (16)

where the summation runs through the whole lattice

When we study the situation outside the incoming ray(in the region θ 6= 0) we can neglect the first term Theintensity is proportional to |ψ|2 like in the one atom case

When we divide with the intensity of the incoming ray |A|2we obtain

I =sumllprime

flflowastlprimeeiqmiddot(RlminusRlprime ) (17)

We have utilized here the property of complex numbers|suml Cl|2 =

sumllprime ClC

lowastlprime

Lattice Sums

Let us first study scattering from a Bravais lattice Wecan thus assume that the scatterers are identical and thatthe intensity (or the scattering cross section)

I = Iatom

∣∣∣∣∣suml

eiqmiddotRl

∣∣∣∣∣2

(18)

where Iatom is the scattering cross section of one atomdefined in Equation (11)

In the following we try to find those values q of thechange in momentum that lead to intensity maxima Thisis clearly occurs if we can choose q so that exp(iq middotRl) = 1at all lattice points In all other cases the phases of thecomplex numbers exp(iq middot Rl) reduce the absolute valueof the sum (destructive interference) Generally speakingone ends up with similar summations whenever studyingthe interaction of waves with a periodic structure (like con-duction electrons in a lattice discussed later)

We first restrict the summation in the intensity (18) inone dimension and afterwards generalize the procedureinto three dimensions

One-Dimensional Sum

The lattice points are located at la where l is an integerand a is the distance between the points We obtain

Σq =

Nminus1suml=0

eilaq

where N is the number of lattice points By employing theproperties of the geometric series we get

|Σq|2 =sin2Naq2

sin2 aq2 (19)

When the number of the lattice points is large (as isthe case in crystals generally) the plot of Equation (19)consists of sharp and identical peaks with a (nearly) zerovalue of the scattering intensity in between

12

Plot of Equation (19) Normalization with N is introducedso that the effect of the number of the lattice points onthe sharpness of the peaks is more clearly seen

The peaks can be found at those values of the momentumq that give the zeroes of the denominator

q = 2πla (20)

These are exactly those values that give real values for theexponential function (= 1) As the number of lattice pointsgrows it is natural to think Σq as a sum of delta functions

Σq =

infinsumlprime=minusinfin

cδ(q minus 2πlprime

a

)

where

c =

int πa

minusπadqΣq =

2πN

L

In the above L is length of the lattice in one dimensionThus

Σq =2πN

L

infinsumlprime=minusinfin

δ(q minus 2πlprime

a

) (21)

It is worthwhile to notice that essentially this Fourier trans-forms the sum in the intensity from the position space intothe momentum space

Reciprocal Lattice

Then we make a generalization into three dimensions InEquation (18) we observe sharp peaks whenever we chooseq for every Bravais vectors R as

q middotR = 2πl (22)

where l is an integer whose value depends on vector RThus the sum in Equation (18) is coherent and producesa Braggrsquos peak The set of all wave vectors K satisfyingEquation (22) is called the reciprocal lattice

Reciprocal lattice gives those wave vectors that result incoherent scattering from the Bravais lattice The magni-tude of the scattering is determined analogously with theone dimensional case from the equationsum

R

eiRmiddotq = N(2π)3

VsumK

δ(qminusK) (23)

where V = L3 is the volume of the lattice

Why do the wave vectors obeying (22) form a latticeHere we present a direct proof that also gives an algorithmfor the construction of the reciprocal lattice Let us showthat the vectors

b1 = 2πa2 times a3

a1 middot a2 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1(24)

b3 = 2πa1 times a2

a3 middot a1 times a2

are the primitive vectors of the reciprocal lattice Becausethe vectors of the Bravais lattice are of form

R = n1a1 + n2a2 + n3a3

we obtain

bi middotR = 2πni

where i = 1 2 3

Thus the reciprocal lattice contains at least the vectorsbi In addition the vectors bi are linearly independent If

eiK1middotR = 1 = eiK2middotR

then

ei(K1+K2)middotR = 1

Therefore the vectors K can be written as

K = l1b1 + l2b2 + l3b3 (25)

where l1 l2 and l3 are integers This proves in fact thatthe wave vectors K form a Bravais lattice

The requirement

K middotR = 0

defines a Braggrsquos plane in the position space Now theplanes K middotR = 2πn are parallel where n gives the distancefrom the origin and K is the normal of the plane Thiskind of sets of parallel planes are referred to as familiesof lattice planes The lattice can be divided into Braggrsquosplanes in infinitely many ways

Example By applying definition (24) one can show thatreciprocal lattice of a simple cubic lattice (lattice constanta) is also a simple cubic lattice with a lattice constant 2πaCorrespondingly the reciprocal lattice of an fcc lattice is abcc lattice (lattice constant 4πa) and that of a bcc latticeis an fcc lattice (4πa)

Millerrsquos Indices

The reciprocal lattice allows the classification of all pos-sible families of lattice planes For each family there existperpendicular vectors in the reciprocal lattice shortest ofwhich has the length 2πd (d is separation between theplanes) Inversely For each reciprocal lattice vector thereexists a family of perpendicular lattice planes separatedwith a distance d from one another (2πd is the length of

13

the shortest reciprocal vector parallel to K) (proof exer-cise)

Based on the above it is natural to describe a given lat-tice plane with the corresponding reciprocal lattice vectorThe conventional notation employs Millerrsquos indices in thedescription of reciprocal lattice vectors lattice planes andlattice points Usually Millerrsquos indices are used in latticeswith a cubic or hexagonal symmetry As an example letus study a cubic crystal with perpendicular coordinate vec-tors x y and z pointing along the sides of a typical unitcell (=cube) Millerrsquos indices are defined in the followingway

bull [ijk] is the direction of the lattice

ix+ jy + kz

where i j and k are integers

bull (ijk) is the lattice plane perpendicular to vector [ijk]It can be also interpreted as the reciprocal lattice vec-tor perpendicular to plane (ijk)

bull ijk is the set of planes perpendicular to vector [ijk]and equivalent in terms of the lattice symmetries

bull 〈ijk〉 is the set of directions [ijk] that are equivalentin terms of the lattice symmetries

In this representation the negative numbers are denotedwith a bar minusi rarr i The original cubic lattice is calleddirect lattice

The Miller indices of a plane have a geometrical propertythat is often given as an alternative definition Let us con-sider lattice plane (ijk) It is perpendicular to reciprocallattice vector

K = ib1 + jb2 + kb3

Thus the lattice plane lies in a continuous plane K middotr = Awhere A is a constant This intersects the coordinate axesof the direct lattice at points x1a1 x2a2 and x3a3 Thecoordinates xi are determined by the condition Kmiddot(xiai) =A We obtain

x1 =A

2πi x2 =

A

2πj x3 =

A

2πk

We see that the points of intersection of the lattice planeand the coordinate axes are inversely proportional to thevalues of Millerrsquos indices

Example Let us study a lattice plane that goes throughpoints 3a1 1a2 and 2a3 Then we take the inverses of thecoefficients 1

3 1 and 12 These have to be integers so

we multiply by 6 So Millerrsquos indices of the plane are(263) The normal to this plane is denoted with the samenumbers but in square brackets [263]

a1

3a1

2a3

(263)a3

a2

yksikkouml-koppi

If the lattice plane does not intersect with an axis it cor-responds to a situation where the intersection is at infinityThe inverse of that is interpreted as 1

infin = 0

One can show that the distance d between adjacent par-allel lattice planes is obtained from Millerrsquos indices (hkl)by

d =aradic

h2 + k2 + l2 (26)

where a is the lattice constant of the cubic lattice

(100) (110) (111)

In the figure there are three common lattice planesNote that due to the cubic symmetry the planes (010)and (001) are identical with the plane (100) Also theplanes (011) (101) and (110) are identical

Scattering from a Lattice with Basis

The condition of strong scattering from a Bravais lat-tice was q = K This is changed slightly when a basis isintroduced to the lattice Every lattice vector is then ofform

R = ul + vlprime

where ul is a Bravais lattice vector and vlprime is a basis vectorAgain we are interested in the sumsum

R

eiqmiddotR =

(suml

eiqmiddotul

)(sumlprime

eiqmiddotvlprime

)

determining the intensity

I prop |sumR

eiqmiddotR|2 =

(sumjjprime

eiqmiddot(ujminusujprime )

)(sumllprime

eiqmiddot(vlminusvlprime )

)

(27)The intensity appears symmetric with respect to the latticeand basis vectors The difference arises in the summationswhich for the lattice have a lot of terms (of the order 1023)whereas for the basis only few Previously it was shownthat the first term in the intensity is non-zero only if qbelongs to the reciprocal lattice The amplitude of thescattering is now modulated by the function

Fq =

∣∣∣∣∣suml

eiqmiddotvl

∣∣∣∣∣2

(28)

14

caused by the introduction of the basis This modulationcan even cause an extinction of a scattering peak

Example Diamond Lattice The diamond lattice isformed by an fcc lattice with a basis

v1 = (0 0 0) v2 =a

4(1 1 1)

It was mentioned previously that the reciprocal lattice ofan fcc lattice is a bcc lattice with a lattice constant 4πaThus the reciprocal lattice vectors are of form

K = l14π

2a(1 1 minus 1) + l2

2a(minus1 1 1) + l3

2a(1 minus 1 1)

Therefore

v1 middotK = 0

and

v2 middotK =π

2(l1 + l2 + l3)

The modulation factor is

FK =∣∣∣1 + eiπ(l1+l2+l3)2

∣∣∣2 (29)

=

4 l1 + l2 + l3 = 4 8 12 2 l1 + l2 + l3 is odd0 l1 + l2 + l3 = 2 6 10

28 Experimental MethodsNext we will present the experimental methods to test

the theory of the crystal structure presented above Letus first recall the obtained results We assumed that weare studying a sample with radiation whose wave vector isk0 The wave is scattered from the sample into the direc-tion k where the magnitudes of the vectors are the same(elastic scattering) and the vector q = k0 minus k has to be inthe reciprocal lattice The reciprocal lattice is completelydetermined by the lattice structure and the possible basisonly modulates the magnitudes of the observed scatteringpeaks (not their positions)

Problem It turns out that monochromatic radiationdoes not produce scattering peaks The measurement de-vice collects data only from one direction k This forcesus to restrict ourselves into a two-dimensional subspace ofthe scattering vectors This can be visualized with Ewaldsphere of the reciprocal lattice that reveals all possiblevalues of q for the give values of the incoming wave vector

Two-dimensional cross-cut of Ewald sphere Especiallywe see that in order to fulfil the scattering condition (22)the surface of the sphere has to go through at least tworeciprocal lattice points In the case above strong scatter-ing is not observed The problem is thus that the pointsin the reciprocal lattice form a discrete set in the three di-mensional k-space Therefore it is extremely unlikely thatany two dimensional surface (eg Ewald sphere) would gothrough them

Ewald sphere gives also an estimate for the necessarywave length of radiation In order to resolve the atomicstructure the wave vector k has to be larger than the lat-tice constant of the reciprocal lattice Again we obtain theestimate that the wave length has to be of the order of anAngstrom ie it has to be composed of X-rays

One can also use Ewald sphere to develop solutions forthe problem with monochromatic radiation The most con-ventional ones of them are Laue method rotating crystalmethod and powder method

Laue Method

Let us use continuous spectrum

Rotating Crystal Method

Let us use monochromatic radi-ation but also rotate the cyrstal

15

Powder Method (Debye-Scherrer Method)

Similar to the rotating crystal method We usemonochromatic radiation but instead of rotating a sam-ple consisting of many crystals (powder) The powder isfine-grained but nevertheless macroscopic so that theyare able to scatter radiation Due to the random orienta-tion of the grains we observe the same net effect as whenrotating a single crystal

29 Radiation Sources of a Scattering Ex-periment

In addition to the X-ray photons we have studied so farthe microscopic structure of matter is usually studied withelectrons and neutrons

X-Rays

The interactions between X-rays and condensed matterare complex The charged particles vibrate with the fre-quency of the radiation and emit spherical waves Becausethe nuclei of the atoms are much heavier only their elec-trons participate to the X-ray scattering The intensity ofthe scattering depends on the number density of the elec-trons that has its maximum in the vicinity of the nucleus

Production of X-rays

The traditional way to produce X-rays is by collidingelectrons with a metal (eg copper in the study of structureof matter wolfram in medical science) Monochromaticphotons are obtained when the energy of an electron islarge enough to remove an electron from the inner shellsof an atom The continuous spectrum is produced whenan electron is decelerated by the strong electric field of thenucleus (braking radiation bremsstrahlung) This way ofproducing X-ray photons is very inefficient 99 of theenergy of the electron is turned into heat when it hits themetal

In a synchrotron X-rays are produced by forcing elec-trons accelerate continuously in large rings with electro-magnetic field

Neutrons

The neutrons interact only with nuclei The interac-tion depends on the spin of the nucleus allowing the studyof magnetic materials The lattice structure of matter isstudied with so called thermal neutrons (thermal energyET asymp 3

2kBT with T = 293 K) whose de Broglie wave

length λ = hp asymp 1 A It is expensive to produce neutronshowers with large enough density

Electrons

The electrons interact with matter stronger than photonsand neutrons Thus the energies of the electrons have tobe high (100 keV) and the sample has to be thin (100 A)in order to avoid multiple scattering events

X-rays Neutrons ElectronsCharge 0 0 -e

Mass 0 167 middot 10minus27 kg 911 middot 10minus31 kgEnergy 12 keV 002 eV 60 keV

Wave length 1 A 2 A 005 AAttenuation length 100 microm 5 cm 1 microm

Form factor f 10minus3 A 10minus4 A 10 A

Typical properties of different sources of radiation in scat-tering experiments (Source MM Eberhart Structuraland Chemical Analysis of Materials (1991))

210 Surfaces and InterfacesMM Chapter 4 not 422-423

Only a small part of the atoms of macroscopic bodiesare lying on the surface Nevertheless the study of sur-faces is important since they are mostly responsible forthe strength of the material and the resistance to chemi-cal attacks For example the fabrication of circuit boardsrequires good control on the surfaces so that the conduc-tion of electrons on the board can be steered in the desiredmanner

The simplest deviations from the crystal structure oc-cur when the lattice ends either to another crystal or tovacuum This instances are called the grain boundary andthe surface In order to describe the grain boundary oneneeds ten variables three for the relative location betweenthe crystals six for their interfaces and one for the angle inbetween them The description of a crystal terminating tovacuum needs only two variables that determine the planealong which the crystal ends

In the case of a grain boundary it is interesting to knowhow well the two surfaces adhere especially if one is form-ing a structure with alternating crystal lattices Coherentinterface has all its atoms perfectly aligned The grow-ing of such structure is called epitaxial In a more generalcase the atoms in the interfaces are aligned in a largerscale This kind of interface is commensurate

Experimental Determination and Creation ofSurfaces

Low-Energy Electron Diffraction

Low-energy electron diffraction (LEED) was used in thedemonstration of the wave nature of electrons (Davissonand Germer 1927)

16

In the experiment the electrons are shot with a gun to-wards the sample The energy of the electrons is small (lessthan 1 keV) and thus they penetrate only into at most fewatomic planes deep Part of the electrons is scattered backwhich are then filtered except those whose energies havechanged only little in the scattering These electrons havebeen scattered either from the first or the second plane

The scattering is thus from a two-dimensional latticeThe condition of strong scattering is the familiar

eiqmiddotR = 1

where R is now a lattice vector of the surface and l is aninteger that depends on the choice of the vector R Eventhough the scattering surface is two dimensional the scat-tered wave can propagate into any direction in the threedimensional space Thus the strong scattering conditionis fulfilled with wave vectors q of form

q = (KxKy qz) (30)

where Kx and Ky are the components of a reciprocal lat-tice vector K On the other hand the component qz isa continuous variable because the z-component of the twodimensional vector R is zero

In order to observe strong scattering Ewald sphere hasto go through some of the rods defined by condition (30)This occurs always independent on the choice of the in-coming wave vector or the orientation of the sample (com-pare with Laue rotating crystal and powder methods)

Reflection High-Energy Electron Diffraction

RHEED

Electrons (with energy 100 keV) are reflected from thesurface and they are studied at a small angle The lengthsof the wave vectors (sim 200 Aminus1) are large compared withthe lattice constant of the reciprocal lattice and thus thescattering patterns are streaky The sample has to be ro-tated in order to observe strong signals in desired direc-tions

Molecular Beam Epitaxy MBE

Molecular Beam Epitaxy enables the formation of a solidmaterial by one atomic layer at a time (the word epitaxyhas its origin in Greek epi = above taxis=orderly) TheTechnique allows the selection and change of each layerbased on the needs

A sample that is flat on the atomic level is placed ina vacuum chamber The sample is layered with elementsthat are vaporized in Knudsen cells (three in this example)When the shutter is open the vapour is allowed to leavethe cell The formation of the structure is continuouslymonitored using the RHEED technique

Scanning Tunnelling Microscope

The (Scanning Tunnelling Microscope) is a thin metal-lic needle that can be moved in the vicinity of the studiedsurface In the best case the tip of the needle is formedby only one atom The needle is brought at a distance lessthan a nanometer from the conducting surface under studyBy changing the electric potential difference between theneedle and the surface one can change the tunnelling prob-ability of electrons from the needle to the sample Thenthe created current can be measured and used to map thesurface

17

The tunnelling of an electron between the needle and thesample can be modelled with a potential wall whose heightU(x) depends on the work required to free the electron fromthe needle In the above figure s denotes the distancebetween the tip of the needle and studied surface Thepotential wall problem has been solved in the course ofquantum mechanics (QM II) and the solution gave thewave function outside the wall to be

ψ(x) prop exp[ i~

int x

dxprimeradic

2m(E minus U(xprime))]

Inside the wall the amplitude of the wave function dropswith a factor

exp[minus sradic

2mφ~2]

The current is dependent on the square of the wave func-tion and thus depend exponentially on the distance be-tween the needle and the surface By recording the currentand simultaneously moving the needle along the surfaceone obtains a current mapping of the surface One canreach atomic resolution with this method (figure on page3)

neulankaumlrki

tutkittavapinta

The essential part of the functioning of the device is howto move the needle without vibrations in atomic scale

Pietzoelectric crystal can change its shape when placedin an electric field With three pietzoelectric crystals onecan steer the needle in all directions

Atomic Force Microscope

Atomic force microscope is a close relative to the scan-ning tunnelling microscope A thin tip is pressed slightlyon the studied surface The bend in the lever is recordedas the tip is moved along the surface Atomic force micro-scope can be used also in the study of insulators

Example Graphene

(Source Novoselov et al PNAS 102 10451 (2005))Atomic force microscopic image of graphite Note the colorscale partly the graphite is only one atomic layer thick iegraphene (in graphite the distance between graphene layersis 335 A)

(Source Li et al PRL 102 176804 (2009)) STM imageof graphene

Graphene can be made in addition to previously men-tioned tape-method by growing epitaxially (eg on ametal) A promising choice for a substrate is SiC whichcouples weakly with graphene For many years (after itsdiscovery in 2004) graphene has been among the most ex-pensive materials in the world The production methodsof large sheets of graphene are still (in 2012) under devel-opment in many research groups around the world

211 Complex StructuresMM Chapters 51-53 541 (not Correlation

functions for liquid) 55 partly 56 names 58main idea

The crystal model presented above is an idealization andrarely met in Nature as such The solids are seldom in athermal equilibrium and equilibrium structures are notalways periodic In the following we will study shortlyother forms of condensed matter such as alloys liquidsglasses liquid crystals and quasi crystals

Alloys

The development of metallic alloys has gone hand in

18

hand with that of the society For example in the BronzeAge (in Europe 3200-600 BC) it was learned that by mixingtin and copper with an approximate ratio 14 one obtainsan alloy (bronze) that is stronger and has a lower meltingpoint than either of its constituents The industrial revolu-tion of the last centuries has been closely related with thedevelopment of steel ie the adding of carbon into iron ina controlled manner

Equilibrium Structures

One can always mix another element into a pure crystalThis is a consequence of the fact that the thermodynamicalfree energy of a mixture has its minimum with a finiteimpurity concentration This can be seen by studying theentropy related in adding of impurity atoms Let us assumethat there are N points in the lattice and that we addM N impurity atoms The adding can be done in(

N

M

)=

N

M (N minusM)asymp NM

M

different ways The macroscopic state of the mixture hasthe entropy

S = kB ln(NMM ) asymp minuskBN(c ln cminus c)

where c = MN is the impurity concentration Each impu-rity atom contributes an additional energy ε to the crystalleading to the free energy of the mixture

F = E minus TS = N [cε+ kBT (c ln cminus c)] (31)

In equilibrium the free energy is in its minimum Thisoccurs at concentration

c sim eminusεkBT (32)

So we see that at finite temperatures the solubility is non-zero and that it decreases exponentially when T rarr 0 Inmost materials there are sim 1 of impurities

The finite solubility produces problems in semiconduc-tors since in circuit boards the electrically active impuri-ties disturb the operation already at concentrations 10minus12Zone refining can be used to reduce the impurity concen-tration One end of the impure crystal is heated simulta-neously moving towards the colder end After the processthe impurity concentration is larger the other end of thecrystal which is removed and the process is repeated

Phase Diagrams

Phase diagram describes the equilibrium at a given con-centration and temperature Let us consider here espe-cially system consisting of two components Mixtures withtwo substances can be divided roughly into two groupsThe first group contains the mixtures where only a smallamount of one substance is mixed to the other (small con-centration c) In these mixtures the impurities can eitherreplace a lattice atom or fill the empty space in betweenthe lattice points

Generally the mixing can occur in all ratios Intermetal-lic compound is formed when two metals form a crystalstructure at some given concentration In a superlatticeatoms of two different elements find the equilibrium in thevicinity of one another This results in alternating layersof atoms especially near to some specific concentrationsOn the other hand it is possible that equilibrium requiresthe separation of the components into separate crystalsinstead of a homogeneous mixture This is called phaseseparation

Superlattices

The alloys can be formed by melting two (or more) ele-ments mixing them and finally cooling the mixture Whenthe cooling process is fast one often results in a similar ran-dom structure as in high temperatures Thus one does notobserve any changes in the number of resonances in egX-ray spectroscopy This kind of cooling process is calledquenching It is used for example in the hardening of steelAt high temperatures the lattice structure of iron is fcc(at low temperatures it is bcc) When the iron is heatedand mixed with carbon the carbon atoms fill the centersof the fcc lattice If the cooling is fast the iron atoms donot have the time to replace the carbon atoms resultingin hard steel

Slow cooling ie annealing results in new spectralpeaks The atoms form alternating crystal structures su-perlattices With many combinations of metals one obtainssuperlattices mostly with mixing ratios 11 and 31

Phase Separation

Let us assume that we are using two substances whosefree energy is of the below form when they are mixed ho-mogeneously

(Source MM) The free energy F(c) of a homogeneousmixture of two materials as a function of the relative con-centration c One can deduce from the figure that whenthe concentration is between ca and cb the mixture tendsto phase separate in order to minimize the free energy Thiscan be seen in the following way If the atoms divide be-tween two concentrations ca lt c and cb gt c (not necessarilythe same as in the figure) the free energy of the mixtureis

Fps = fF(ca) + (1minus f)F(cb)

where f is the fraction of the mixture that has the con-centration ca Correspondingly the fraction 1minus f has theconcentration cb The fraction f is not arbitrary because

19

the total concentration has to be c Thus

c = fca + (1minus f)cb rArr f =cminus cbca minus cb

Therefore the free energy of the phase separated mixtureis

Fps =cminus cbca minus cb

F(ca) +ca minus cca minus cb

F(cb) (33)

Phase separation can be visualized geometrically in thefollowing way First choose two points from the curve F(c)and connect the with a straight line This line describesthe phase separation of the chosen concentrations In thefigure the concentrations ca and cb have been chosen sothat the phase separation obtains the minimum value forthe free energy We see that F(c) has to be convex in orderto observe a phase separation

A typical phase diagram consists mostly on regions witha phase separation

(Source MM) The phase separation of the mixture ofcopper and silver In the region Ag silver forms an fcc lat-tice and the copper atoms replace silver atoms at randomlattice points Correspondingly in the region Cu silver re-places copper atoms in an fcc lattice Both of them arehomogeneous solid alloys Also the region denoted withrdquoLiquidrdquo is homogeneous Everywhere else a phase separa-tion between the metals occurs The solid lines denote theconcentrations ca and cb (cf the previous figure) as a func-tion of temperature For example in the region rdquoAg+Lrdquoa solid mixture with a high silver concentration co-existswith a liquid with a higher copper concentration than thesolid mixture The eutectic point denotes the lowest tem-perature where the mixture can be found as a homogeneousliquid

(Source MM) The formation of a phase diagram

Dynamics of Phase Separation

The heating of a solid mixture always results in a ho-mogeneous liquid When the liquid is cooled the mixtureremains homogeneous for a while even though the phaseseparated state had a lower value of the free energy Letus then consider how the phase separation comes about insuch circumstances as the time passes

The dynamics of the phase separation can be solved fromthe diffusion equation It can be derived by first consider-ing the concentration current

j = minusDnablac (34)

The solution of this equation gives the atom current jwhose direction is determined by the gradient of the con-centration Due to the negative sign the current goes fromlarge to small concentration The current is created by therandom thermal motion of the atoms Thus the diffusionconstant D changes rapidly as a function of temperatureSimilarly as in the case of mass and charge currents we candefine a continuity equation for the flow of concentrationBased on the analogy

partc

partt= Dnabla2c (35)

This is the diffusion equation of the concentration Theequation looks innocent but with a proper set of boundaryconditions it can create complexity like in the figure below

20

(Source MM) Dendrite formed in the solidification pro-cess of stainless steel

Let us look as an example a spherical drop of iron carbidewith an iron concentration ca We assume that it grows ina mixture of iron and carbon whose iron concentrationcinfin gt ca The carbon atoms of the mixture flow towardsthe droplet because it minimizes the free energy In thesimplest solution one uses the quasi-static approximation

partc

parttasymp 0

Thus the concentration can be solved from the Laplaceequation

nabla2c = 0

We are searching for a spherically symmetric solution andthus we can write the Laplace equation as

1

r2

part

partr

(r2 partc

partr

)= 0

This has a solution

c(r) = A+B

r

At the boundary of the droplet (r = R) the concentrationis

c(R) = ca

and far away from the droplet (r rarrinfin)

limrrarrinfin

c(r) = cinfin

These boundary conditions determine the coefficients Aand B leading to the unambiguous solution of the Laplaceequation

c(r) = cinfin +R

r(ca minus cinfin)

The gradient of the concentration gives the current density

j = minusDnablac = DR(cinfin minus ca)nabla1

r= minusDR(cinfin minus ca)

r

r2

For the total current into the droplet we have to multi-ply the current density with the surface area 4πR2 of thedroplet This gives the rate of change of the concentrationinside the droplet

partc

partt= 4πR2(minusjr) = 4πDR(cinfin minus ca)

On the other hand the chain rule of derivation gives therate of change of the volume V = 4πR33 of the droplet

dV

dt=partV

partc

partc

partt

By denoting v equiv partVpartc we obtain

R =vDR

(cinfin minus ca) rArr R propradic

2vD(cinfin minus ca)t

We see that small nodules grow the fastest when measuredin R

Simulations

When the boundary conditions of the diffusion equationare allowed to change the non-linear nature of the equa-tion often leads to non-analytic solutions Sometimes thecalculation of the phase separations turns out to be difficulteven with deterministic numerical methods Then insteadof differential equations one has to rely on descriptions onatomic level Here we introduce two methods that are incommon use Monte Carlo and molecular dynamics

Monte Carlo

Monte Carlo method was created by von NeumannUlam and Metropolis in 1940s when they were workingon the Manhattan Project The name originates from thecasinos in Monte Carlo where Ulamrsquos uncle often went togamble his money The basic principle of Monte Carlo re-lies on the randomness characteristic to gambling Theassumption in the background of the method is that theatoms in a solid obey in equilibrium at temperature T theBoltzmann distribution exp(minusβE) where E is position de-pendent energy of an atom and β = 1kBT If the energydifference between two states is δE then the relative occu-pation probability is exp(minusβδE)

The Monte Carlo method presented shortly

1) Assume that we have N atoms Let us choose theirpositions R1 RN randomly and calculate their en-ergy E(R1 RN ) = E1

2) Choose one atom randomly and denote it with indexl

3) Create a random displacement vector eg by creatingthree random numbers pi isin [0 1] and by forming avector

Θ = 2a(p1 minus1

2 p2 minus

1

2 p3 minus

1

2)

21

In the above a sets the length scale Often one usesthe typical interatomic distance but its value cannotaffect the result

4) Calculate the energy difference

δE = E(R1 Rl + Θ RN )minus E1

The calculation of the difference is much simpler thanthe calculation of the energy in the position configu-ration alone

5) If δE lt 0 replace Rl rarr Rl + Θ Go to 2)

6) If δE gt 0 accept the displacement with a probabilityexp(minusβδE) Pick a random number p isin [0 1] If pis smaller than the Boltzmann factor accept the dis-placement (Rl rarr Rl + Θ and go to 2) If p is greaterreject the displacement and go to 2)

In low temperatures almost every displacement that areaccepted lower the systems energy In very high temper-atures almost every displacement is accepted After suffi-cient repetition the procedure should generate the equilib-rium energy and the particle positions that are compatiblewith the Boltzmann distribution exp(minusβE)

Molecular Dynamics

Molecular dynamics studies the motion of the singleatoms and molecules forming the solid In the most gen-eral case the trajectories are solved numerically from theNewton equations of motion This results in solution ofthe thermal equilibrium in terms of random forces insteadof random jumps (cf Monte Carlo) The treatment givesthe positions and momenta of the particles and thus pro-duces more realistic picture of the dynamics of the systemapproaching thermal equilibrium

Let us assume that at a given time we know the positionsof the particles and that we calculate the total energy ofthe system E The force Fl exerted on particle l is obtainedas the gradient of the energy

Fl = minus partEpartRl

According to Newtonrsquos second law this force moves theparticle

mld2Rl

dt2= Fl

In order to solve these equations numerically (l goesthrough values 1 N where N is the number of parti-cles) one has to discretize them It is worthwhile to choosethe length of the time step dt to be shorter than any of thescales of which the forces Fl move the particles consider-ably When one knows the position Rn

l of the particle lafter n steps the position after n+1 steps can be obtainedby calculating

Rn+1l = 2Rn

l minusRnminus1l +

Fnlml

dt2 (36)

As was mentioned in the beginning the initial state de-termines the energy E that is conserved in the processThe temperature can be deduced only at the end of thecalculation eg from the root mean square value of thevelocity

The effect of the temperature can be included by addingterms

Rl =Flmlminus bRl + ξ(t)

into the equation of motion The first term describes thedissipation by the damping constant b that depends on themicroscopic properties of the system The second term il-lustrates the random fluctuations due to thermal motionThese additional terms cause the particles to approach thethermal equilibrium at the given temperature T The ther-mal fluctuations and the dissipation are closely connectedand this relationship is described with the fluctuation-dissipation theorem

〈ξα(0)ξβ(t)〉 =2bkBTδαβδ(t)

ml

The angle brackets denote the averaging over time or al-ternatively over different statistical realizations (ergodichypothesis) Without going into any deeper details weobtain new equations motion by replacing in Equation (36)

Fnl rarr Fnl minus bmlRnl minusRnminus1

l

dt+ Θ

radic6bmlkBTdt

where Θ is a vector whose components are determined byrandom numbers pi picked from the interval [0 1]

Θ = 2(p1 minus

1

2 p2 minus

1

2 p3 minus

1

2

)

Liquids

Every element can be found in the liquid phase Thepassing from eg the solid into liquid phase is called thephase transformation Generally the phase transforma-tions are described with the order parameter It is definedin such way that it non-zero in one phase and zero in allthe other phases For example the appearance of Braggrsquospeaks in the scattering experiments of solids can be thoughtas the order parameter of the solid phase

Let us define the order parameter of the solid phase morerigorously Consider a crystal consisting of one elementGenerally it can be described with a two particle (or vanHove) correlation function

n2(r1 r2 t) =

langsuml 6=lprime

δ(r1 minusRl(0))δ(r2 minusRlprime(t))

rang

where the angle brackets mean averaging over temperatureand vectors Rl denote the positions of the atoms If anatom is at r1 at time t1 the correlation function gives theprobability of finding another particle at r2 at time t1 + t

22

Then we define the static structure factor

S(q) equiv I

NIatom=

1

N

sumllprime

langeiqmiddot(RlminusRlprime )

rang (37)

where the latter equality results from Equation (18) Theexperiments take usually much longer than the time scalesdescribing the movements of the atoms so they measureautomatically thermal averages We obtain

S(q) =1

N

sumllprime

intdr1dr2e

iqmiddot(r1minusr2)langδ(r1 minusRl)δ(r2 minusRlprime)

rang= 1 +

1

N

intdr1dr2n(r1 r2 0)eiqmiddot(r1minusr2)

= 1 +VNn2(q) (38)

where

n2(q) =1

V

intdrdrprimen2(r + rprime r 0)eiqmiddotr

prime(39)

and V is the volume of the system

We see that the sharp peaks in the scattering experimentcorrespond to strong peaks in the Fourier spectrum of thecorrelation function n2(r1 r2 0) When one wants to definethe order parameter OK that separates the solid and liquidphases it suffices to choose any reciprocal lattice vectorK 6= 0 and set

OK = limNrarrinfin

VN2

n2(K) (40)

In the solid phase the positions Rl of the atoms are lo-cated at the lattice points and because K belongs to thereciprocal lattice the Fourier transformation (39) of thecorrelation function is of the order N(N minus 1)V Thusthe order parameter OK asymp 1 The thermal fluctuations ofthe atoms can be large as long as the correlation functionpreserves the lattice symmetry

The particle locations Rl are random in liquids and theintegral vanishes This is in fact the definition of a solidIe there is sharp transition in the long-range order It isworthwhile to notice that locally the surroundings of theatoms change perpetually due to thermal fluctuations

Glasses

Glasses typically lack the long range order which sepa-rates them from solids On the other hand they show ashort-range order similar to liquids The glasses are differ-ent to liquids in that the atoms locked in their positionslike someone had taken a photograph of the liquid Theglassy phase is obtained by a fast cooling of a liquid Inthis way the atoms in the liquid do not have the time toorganize into a lattice but remain disordered This resultsin eg rapid raise in viscosity Not a single known mate-rial has glass as its ground state On the other hand it isbelieved that all materials can form glass if they are cooledfast enough For example the cooling rate for a typical

window glass (SiO2) is 10 Ks whereas for nickel it is 107

Ks

Liquid Crystals

Liquid crystal is a phase of matter that is found in cer-tain rod-like molecules The mechanical properties of liq-uid crystals are similar to liquids and the locations of therods are random but especially the orientation of the rodsdisplays long-range order

Nematics

Nematic liquid crystal has a random distribution for thecenters of its molecular rods The orientation has never-theless long-range order The order parameter is usuallydefined in terms of quadrupole moment

O =lang

3 cos2 θ minus 1rang

where θ is the deviation from the optic axis pointing in thedirection of the average direction of the molecular axesThe average is calculated over space and time One cannotuse the dipole moment because its average vanishes whenone assumes that the molecules point rsquouprsquo and rsquodownrsquo ran-domly The defined order parameter is practical because itobtains the value O = 1 when the molecules point exactlyto the same direction (solid) Typical liquid crystal hasO sim 03 08 and liquid O = 0

Cholesterics

Cholesteric liquid crystal is made of molecules that arechiral causing a slow rotation in the direction n of themolecules in the liquid crystal

nx = 0

ny = cos(q0x)

nz = sin(q0x)

Here the wave length of the rotation λ = 2πq0 is muchlarger than the size of the molecules In addition it canvary rapidly as a function of the temperature

Smectics

Smectic liquid crystals form the largest class of liquidcrystals In addition to the orientation they show long-range order also in one direction They form layers that canbe further on divided into three classes (ABC) in termsof the direction between their mutual orientation and thevector n

Quasicrystals

Crystals can have only 2- 3- 4- and 6-fold rotationalsymmetries (cf Exercise 1) Nevertheless some materialshave scattering peaks due to other like 5-fold rotationalsymmetries These peaks cannot be due to periodic latticeThe explanation lies in the quasiperiodic organization ofsuch materials For example the two-dimensional planecan be completely covered with two tiles (Penrose tiles)

23

resulting in aperiodic lattice but whose every finite arearepeats infinitely many times

(Source MM) Penrose tiles The plane can be filled bymatching arrow heads with same color

3 Electronic StructureMM Chapter 6

A great part of condensed matter physics can be encap-sulated into a Hamiltonian operator that can be written inone line (notice that we use the cgs-units)

H =suml

P 2

2Ml+

1

2

suml 6=lprime

qlqlprime

|Rl minus Rlprime | (41)

Here the summation runs through all electrons and nucleiin the matter Ml is the mass and ql the charge of lth

particle The first term describes the kinetic energy ofthe particles and the second one the Coulomb interactionsbetween them It is important to notice that R and Pare quantum mechanical operators not classical variablesEven though the Hamiltonian operator looks simple itsSchrodinger equation can be solved even numerically withmodern computers for only 10 to 20 particles Normallythe macroscopic matter has on the order of 1023 particlesand thus the problem has to be simplified in order to besolved in finite time

Let us start with a study of electron states of the matterConsider first the states in a single atom The positivelycharge nucleus of the atom forms a Coulomb potential forthe electrons described in the figure below Electrons canoccupy only discrete set of energies denoted with a b andc

a

bc energia

(Source ET)

When two atoms are brought together the potential bar-rier between them is lowered Even though the tunnellingof an electron to the adjacent atom is possible it does nothappen in state a because the barrier is too high This is asignificant result the lower energy states of atoms remainintact when they form bonds

In state b the tunnelling of the electrons is noticeableThese states participate to bond formation The electronsin states c can move freely in the molecule

In two atom compound every atomic energy state splits

24

in two The energy difference between the split states growsas the function of the strength of the tunnelling betweenthe atomic states In four atom chain every atomic state issplit in four In a solid many atoms are brought togetherN sim 1023 Instead of discrete energies one obtains energybands formed by the allowed values of energy In betweenthem there exist forbidden zones which are called the en-ergy gaps In some cases the bands overlap and the gapvanishes

The electrons obey the Pauli exclusion principle therecan be only one electron in a quantum state Anotherway to formulate this is to say that the wave function ofthe many fermion system has to be antisymmetric withrespect exchange of any two fermions (for bosons it hasto be symmetric) In the ground state of an atom theelectrons fill the energy states starting from the one withlowest energy This property can be used to eg explainthe periodic table of elements

Similarly in solids the electrons fill the energy bandsstarting from the one with lowest energy The band that isonly partially filled is called conduction band The electricconductivity of metals is based on the existence of sucha band because filled bands do not conduct as we willsee later If all bands are completely filled or empty thematerial is an insulator

The electrons on the conduction band are called con-duction electrons These electrons can move quite freelythrough the metal In the following we try to study theirproperties more closely

31 Free Fermi GasLet us first consider the simplest model the free Fermi

gas The Pauli exclusion principle is the only restrictionlaid upon electrons Despite the very raw assumptions themodel works in the description of the conduction electronof some metals (simple ones like alkali metals)

The nuclei are large compared with the electrons andtherefore we assume that they can be taken as static par-ticles in the time scales set by the electronic motion Staticnuclei form the potential in which the electrons move Inthe free Fermi gas model this potential is assumed tobe constant independent on the positions of the electrons(U(rl) equiv U0 vector rl denote the position of the electronl) and thus sets the zero of the energy In addition weassume that the electrons do not interact with each otherThus the Schrodinger equation of the Hamiltonian (41) isreduced in

minus ~2

2m

Nsuml=1

nabla2lΨ(r1 rN ) = EΨ(r1 rN ) (42)

Because the electrons do not interact it suffices to solvethe single electron Schrodinger equation

minus ~2nabla2

2mψl(r) = Elψl(r) (43)

The number of conduction electrons is N and the totalwave function of this many electron system is a product

of single electron wave functions Correspondingly theeigenenergy of the many electron system is a sum of singleelectron energies

In order to solve for the differential equation (43) onemust set the boundary conditions Natural choice wouldbe to require that the wave function vanishes at the bound-aries of the object This is not however practical for thecalculations It turns out that if the object is large enoughits bulk properties do not depend on what is happeningat the boundaries (this is not a property just for a freeFermi gas) Thus we can choose the boundary in a waythat is the most convenient for analytic calculations Typ-ically one assumes that the electron is restricted to movein a cube whose volume V = L3 The insignificance of theboundary can be emphasized by choosing periodic bound-ary conditions

Ψ(x1 + L y1 z1 zN ) = Ψ(x1 y1 z1 zN )

Ψ(x1 y1 + L z1 zN ) = Ψ(x1 y1 z1 zN )(44)

which assume that an electron leaving the cube on one sidereturns simultaneously at the opposite side

With these assumptions the eigenfunctions of a free elec-tron (43) are plane waves

ψk =1radicVeikmiddotr

where the prefactor normalizes the function The periodicboundaries make a restriction on the possible values of thewave vector k

k =2π

L(lx ly lz) (45)

where li are integers By inserting to Schrodinger equationone obtains the eigenenergies

E0k =

~2k2

2m

The vectors k form a cubic lattice (reciprocal lattice)with a lattice constant 2πL Thus the volume of theWigner-Seitz cell is (2πL)3 This result holds also forelectrons in a periodic lattice and also for lattice vibrations(phonons) Therefore the density of states presented inthe following has also broader physical significance

Ground State of Non-Interacting Electrons

As was mentioned the wave functions of many non-interacting electrons are products of single electron wavefunctions The Pauli principle prevents any two electronsto occupy same quantum state Due to spin the state ψk

can have two electrons This way one can form the groundstate for N electron system First we set two electrons intothe lowest energy state k = 0 Then we place two elec-trons to each of the states having k = 2πL Because theenergies E0

k grow with k the adding of electrons is done byfilling the empty states with the lowest energy

25

Let us then define the occupation number fk of the statek It is 1 when the corresponding single electron statebelongs to the ground state Otherwise it is 0 When thereare a lot of electrons fk = 1 in the ground state for all k ltkF and fk = 0 otherwise Fermi wave vector kF definesa sphere into the k-space with a radius kF The groundstate of the free Fermi gas is obtained by occupying allstates inside the Fermi sphere The surface of the spherethe Fermi surface turns out to be a cornerstone of themodern theory of metals as we will later see

Density of States

The calculation of thermodynamic quantities requiressums of the form sum

k

Fk

where F is a function defined by the wave vectors k (45) Ifthe number of electrons N is large then kF 2πL andthe sums can be transformed into integrals of a continuousfunction Fk

The integrals are defined by dividing the space intoWigner-Seitz cells by summing the values of the functioninside the cell and by multiplying with the volume of thecell int

dkFk =sumk

(2π

L

)3

Fk (46)

We obtain sumk

Fk =V

(2π)3

intdkFk (47)

where V = L3 Despite this result is derived for a cubicobject one can show that it holds for arbitrary (large)volumes

Especially the delta function δkq should be interpretedas

(2π)3

Vδ(kminus q)

in order that the both sides of Equation (47) result in 1

Density of states can be defined in many different waysFor example in the wave vector space it is defined accord-ing to Equation (47)sum

Fk = V

intdkDkFk

where the density of states

Dk = 21

(2π)3

and the prefactor is a consequence of including the spin(σ)

The most important one is the energy density of statesD(E) which is practical when one deals with sums thatdepend on the wave vector k only via the energy Eksum

F (Ek) = V

intdED(E)F (E) (48)

The energy density of states is obtained by using theproperties of the delta functionsum

F (Ek) = V

intdkDkF (Ek)

= V

intdEintdkDkδ(E minus Ek)F (E)

rArr D(E) =2

(2π)3

intdkδ(E minus Ek) (49)

Results of Free Electrons

The dispersion relation for single electrons in the freeFermi gas was

E0k =

~2k2

2m

Because the energy does not depend on the direction ofthe wave vector by making a transformation into sphericalcoordinates in Equation (49) results in

D(E) =2

(2π)3

intdkδ(E minus E0

k)

= 4π2

(2π)3

int infin0

dkk2δ(E minus E0k) (50)

The change of variables k rarr E gives for the free Fermi gas

D(E) =m

~3π2

radic2mE (51)

The number of electrons inside the Fermi surface can becalculated by using the occupation number

N =sumkσ

fk

=2V

(2π)3

intdkfk (52)

Because fk = 0 when k gt kF we can use the Heavisidestep function in its place

θ(k) =

1 k ge 00 k lt 0

We obtain

N =2V

(2π)3

intdkθ(kF minus k) (53)

=V k3

F

3π2 (54)

26

So the Fermi wave number depends on the electron densityn = NV as

kF = (3π2n)13 (55)

One often defines the free electron sphere

3r3s =

V

N

where VN is the volume per an electron

The energy of the highest occupied state is called theFermi energy

EF =~2k2

F

2m (56)

The Fermi surface is thus formed by the wave vectors kwith k = kF and whose energy is EF

Fermi velocity is defined as

vF =~kFm

Density of States at Fermi Surface

Almost all electronic transport phenomena like heatconduction and response to electric field are dependenton the density of states at the Fermi surface D(EF ) Thestates deep inside the surface are all occupied and there-fore cannot react on disturbances by changing their statesThe states above the surface are unoccupied at low tem-peratures and cannot thus explain the phenomena due toexternal fields This means that the density of states at theFermi surface is the relevant quantity and for free Fermigas it is

D(EF ) =3n

2EF

1- and 2- Dimensional Formulae

When the number of dimensions is d one can define thedensity of states as

Dk = 2( 1

)d

In two dimensions the energy density of states is

D(E) =m

π~2

and in one dimension

D(E) =

radic2m

π2~2E

General Ground State

Let us assume for a moment that the potential due tonuclei is not a constant but some position dependent (egperiodic) function U(r) Then the Schrodinger equationof the system is

Nsuml=1

(minus ~2

2mnabla2l + U(rl)

)Ψ(r1 rN ) = EΨ(r1 rN )

(57)

When the electrons do not interact with each other itis again sufficient to solve the single electron Schrodingerequation (

minus ~2nabla2

2m+ U(r)

)ψl(r) = Elψl(r) (58)

By ordering the single electron energies as

E0 le E1 le E2

the ground state ofN electron system can still be formed byfilling the lowest energies E0 EN The largest occupiedenergy is still called the Fermi energy

Temperature Dependence of Equilibrium

The ground state of the free Fermi gas presented aboveis the equilibrium state only when the temperature T = 0Due to thermal motion the electrons move faster whichleads to the occupation of the states outside the Fermisurface One of the basic results of the statistic physics isthe Fermi-Dirac distribution

f(E) =1

eβ(Eminusmicro) + 1 (59)

It gives the occupation probability of the state with energyE at temperature T In the above β = 1kBT and micro is thechemical potential which describes the change in energy inthe system required when one adds one particle and keepsthe volume and entropy unchanged

(Source MM) Fermi-Dirac probabilities at differenttemperatures The gas of non-interacting electrons is atclassical limit when the occupation probability obeys theBoltzmann distribution

f(E) = CeminusβE

This occurs when

f(E) 1 rArr eβ(Eminusmicro) 1

Due to spin every energy state can be occupied by twoelectrons at maximum When this occurs the state is de-generate At the classical limit the occupation of everyenergy state is far from double-fold and thus the elec-trons are referred to as non-degenerate According to theabove condition this occurs when kBT micro When thetemperature drops (or the chemical potential micro ie thedensity grows) the energy states start degenerate starting

27

from the lowest Also the quantum mechanical phenom-ena begin to reveal themselves

When the temperature T rarr 0

f(E)rarr θ(microminus E)

In other words the energy states smaller than the chemicalpotential are degenerate and the states with larger ener-gies are completely unoccupied We see that at the zerotemperature the equilibrium state is the ground state ofthe free Fermi gas and that micro = EF

It is impossible to define generally what do the rdquohighrdquoand rdquolowrdquo temperatures mean Instead they have to bedetermined separately in the system at hand For examplefor aluminium one can assume that the three electrons of itsoutmost electron shell (conduction electrons) form a Fermigas in the solid phase This gives the electron density ntogether with the density of aluminium ρ = 27middot103 kgm3Thus we obtain an estimate for the Fermi temperature ofaluminium

TF =EFkBasymp 135000 K

which is much larger than its melting temperature Fermitemperature gives a ball park estimate for the energyneeded to excite the Fermi gas from ground state In met-als the Fermi temperatures are around 10000 K or largerThis means that at room temperature the conduction elec-trons in metals are at very low temperature and thus forma highly degenerate Fermi gas Therefore only a smallfraction of the electrons located near the Fermi surface isactive This is the most important single fact of metalsand it is not changed when the theory is expanded

Sommerfeld expansion

Before the quantum age in the beginning of 20th centurythere was a problem in the theory of electrons in metalThomson estimated (1907) that every electron proton andneutron increases the specific heat of the metal with thefactor 3kBT according to the equipartition theorem Themeasured values for specific heats were only half of thisvalue The correction due to quantum mechanics and es-pecially the Pauli principle is presented qualitatively inthe above Only the electrons near the Fermi surface con-tribute to the specific heat

The specific heat is a thermodynamic property of matterIn metals the melting temperatures are much smaller thanthe Fermi temperature which suggests a low temperatureexpansion for thermodynamic quantities This expansionwas derived first by Sommerfeld (1928) and it is based onthe idea that at low temperatures the energies of the ther-modynamically active electrons deviate at most by kBTfrom the Fermi energy

Formal Derivation

Assume that H(E) is an arbitrary (thermodynamic)function In the Fermi gas its expectation value is

〈H〉 =

int infinminusinfin

dEH(E)f(E)

where f(E) is the Fermi-Dirac distribution When one as-sumes that the integrand vanishes at the infinity we obtainby partial integration

〈H〉 =

int infinminusinfin

dE

[int Eminusinfin

dE primeH(E prime)

][partfpartmicro

]

where minuspartfpartE = partfpartmicro Having this function in the in-tegral is an essential part of the expansion It contains theidea that only the electrons inside the distance kBT fromthe Fermi surface are active

(Source MM) We see that the function partfpartmicro is non-zero only in the range kBT Thus it is sufficient to considerthe integral in square brackets only in the vicinity of thepoint E = micro Expansion into Taylor series givesint Eminusinfin

dE primeH(E prime) asympint micro

minusinfindE primeH(E prime) (60)

+ H(micro)(E minus micro) +1

2H prime(micro)(E minus micro)2 + middot middot middot

By inserting this into the expectation value 〈H〉 we seethat the terms with odd powers of E minus micro vanish due tosymmetry since partfpartmicro an even power of the same functionWe obtain

〈H〉 =

int micro

minusinfindEH(E) (61)

+π2

6[kBT ]2H prime(micro) +

7π4

360[kBT ]4H primeprimeprime(micro) + middot middot middot

The expectation value can be written also in an algebraicform but in practice one seldom needs terms that are be-yond T 2 The above relation is called the Sommerfeld ex-pansion

Specific Heat at Low Temperatures

Let us apply the Sommerfeld expansion in the determi-nation of the specific heat due to electrons at low temper-atures The specific heat cV describes the change in theaverage electron energy density EV as a function of tem-perature assuming that the number of electrons N and thevolume V are kept fixed

cV =1

V

partEpartT

∣∣∣∣∣NV

28

According to the previous definition (48) of the energy den-sity of states the average energy density is

EV

=

intdE primef(E prime)E primeD(E prime)

=

int micro

minusinfindE primeE primeD(E prime) +

π2

6(kBT )2

d(microD(micro)

)dmicro

(62)

The latter equality is obtained from the two first terms ofthe Sommerfeld expansion for the function H(E) = ED(E)

By making a Taylor expansion at micro = EF we obtain

EV

=

int EFminusinfin

dE primeE primeD(E prime) + EFD(EF )(microminus EF

)+

π2

6(kBT )2

(D(EF ) + EFDprime(EF )

) (63)

An estimate for the temperature dependence of micro minus EF isobtained by calculating the number density of electronsfrom the Sommerfeld expansion

N

V=

intdEf(E)D(E) (64)

=

int EFminusinfin

dED(E) +D(EF )(microminus EF

)+π2

6(kBT )2Dprime(EF )

where the last equality is again due to Taylor expandingWe assumed that the number of electrons and the volumeare independent on temperature and thus

microminus EF = minusπ2

6(kBT )2D

prime(EF )

D(EF )

By inserting this into the energy density formula and thenderivating we obtain the specific heat

cV =π2

3D(EF )k2

BT (65)

The specific heat of metals has two major componentsIn the room temperature the largest contribution comesfrom the vibrations of the nuclei around some equilibrium(we will return to these later) At low temperatures theselattice vibrations behave as T 3 Because the conductionelectrons of metals form a free Fermi gas we obtain for thespecific heat

cV =π2

2

(kBEF

)nkBT =

π2

2nkB

T

TF (66)

where n is the density of conduction electrons in metalAt the temperatures around 1 K one observes a lineardependence on temperature in the measured specific heatsof metals This can be interpreted to be caused by theelectrons near the Fermi surface

Because the Fermi energy is inversely proportional toelectron mass the linear specific heat can be written as

cV =mkF3~2

k2BT

If there are deviations in the measured specific heats oneinterpret them by saying that the matter is formed by effec-tive particles whose masses differ from those of electronsThis can be done by replacing the electron mass with theeffective mass mlowast in the specific heat formula When thebehaviour as a function of temperature is otherwise simi-lar this kind of change of parameters allows the use of thesimple theory What is left to explain is the change in themass For example for iron mlowastmel sim 10 and for someheavy fermion compounds (UBe13 UPt3) mlowastmel sim 1000

32 Schrodinger Equation and SymmetryMM Chapters 7-722

The Sommerfeld theory for metals includes a fundamen-tal problem How can the electrons travel through the lat-tice without interacting with the nuclei On the otherhand the measured values for resistances in different ma-terials show that the mean free paths of electrons are largerthan the interatomic distances in lattices F Bloch solvedthese problems in his PhD thesis in 1928 where he showedthat in a periodic potential the wave functions of electronsdeviate from the free electron plane waves only by peri-odic modulation factor In addition the electrons do notscatter from the lattice itself but from its impurities andthermal vibrations

Bloch Theorem

We will still assume that the electrons in a solid do notinteract with each other but that they experience the pe-riodic potential due to nuclei

U(r + R) = U(r) (67)

The above relation holds for all Bravais lattice vectorsRAs before it is sufficient to consider a single electron whoseHamiltonian operator is

H =P 2

2m+ U(R) (68)

and the corresponding Schrodinger equation is(minus ~2nabla2

2m+ U(r)

)ψ(r) = Eψ(r) (69)

Again the wave function of the many electron system isa product of single electron wave functions One shouldnotice that even though the Hamiltonian operator H isperiodic it does not necessarily result in periodic eigen-functions ψ(r)

In order to describe a quantum mechanical system com-pletely one has to find all independent operators thatcommute with the Hamiltonian operator One can as-sign a quantum number for each such an operator andtogether these number define the quantum state of the sys-tem Thus we will consider the translation operator

TR = eminusiP middotR~

29

where P is the momentum operator and R is the Bravaislattice vector (cf Advanced Course in Quantum Mechan-ics) The translation operator shifts the argument of anarbitrary function f(r) with a Bravais lattice vector R

TRf(r) = f(r + R)

Due to the periodicity of the potential all operators TRcommute with the Hamiltonian operator H Thus theyhave common eigenstates |ψ〉 One can show that thereare no other essentially different operators that commutewith the Hamiltonian Therefore we need two quantumnumbers n for the energy and k for the translations

Consider the translations We obtain

T daggerR|ψ〉 = CR|ψ〉

When this is projected into position space (inner productwith the bra-vector 〈r|) we obtain

ψ(r + R) = CRψ(r) (70)

On the other hand if we calculate the projection intothe momentum space we get

eikmiddotR〈k|ψ〉 = CR〈k|ψ〉

rArr joko CR = eikmiddotR tai 〈k|ψ〉 = 0

So we see that the eigenstate |ψ〉 overlaps only with oneeigenstate of the momentum (|k〉) The vector k is calledthe Bloch wave vector and the corresponding momentum~k the crystal momentum The Bloch wave vector is usedto index the eigenstates ψk

For a fixed value of a Bloch wave vector k one has manypossible energy eigenvalues Those are denoted in the fol-lowing with the band index n Thus the periodicity hasallowed the classification of the eigenstates

H|ψnk〉 = Enk|ψnk〉 (71)

T daggerR|ψnk〉 = eikmiddotR|ψnk〉 (72)

The latter equation is called the Bloch theorem Let usstudy that and return later to the determining of the energyquantum number

The Bloch theorem is commonly represented in two al-ternative forms According to Equation (70) one obtains

ψnk(r + R) = eikmiddotRψnk(r) (73)

On the other hand if we define

unk(r) = eminusikmiddotrψnk(r) (74)

we see that u is periodic

u(r + R) = u(r)

andψnk(r) = eikmiddotrunk(r) (75)

Thus we see that the periodic lattice causes modulationin the amplitude of the plane wave solutions of the freeelectrons The period of this modulation is that of thelattice

Allowed Bloch Wave Vectors

The finite size of the crystal restricts the possible valuesof the Bloch wave vector k In the case of a cubic crystal(volume V = L3) we can use the previous values for thewave vectors of the free electrons (45) The crystals areseldom cubes and it is more convenient to look at thingsin a primitive cell of the Bravais lattice

Let us first find the Bloch wave vectors with the help ofthe reciprocal lattice vectors bi

k = x1b1 + x2b2 + x3b3

The coefficients xi are at this point arbitrary Let us thenassume that the lattice is formed by M = M1M2M3 prim-itive cells Again if we assume that the properties of thebulk do not depend on the boundary conditions we cangeneralize the periodic boundary conditions leading to

ψ(r +Miai) = ψ(r) (76)

where i = 1 2 3 According to the Bloch theorem we havefor all i = 1 2 3

ψnk(r +Miai) = eiMikmiddotaiψnk(r)

Because bl middot alprime = 2πδllprime then

e2πiMixi = 1

andxi =

mi

Mi

where mi are integers Thus the general form of the al-lowed Bloch wave vectors is

k =3suml=1

ml

Mlbl (77)

We can make the restriction 0 le ml lt Ml because weare denoting the eigenvalues (72) of the operator TR Bydoing this the wave vectors differing by a reciprocal latticevector have the same eigenvalue (exp(iKmiddotR) = 1) and theycan be thought to be physically identical In addition weobtain that the eigenstates and eigenvalues of the periodicHamiltonian operator (68) are periodic in the reciprocallattice

ψnk+K(r) = ψnk(r) (78)

Enk+K = Enk (79)

We have thus obtained that in a primitive cell in thereciprocal space we have M1M2M3 different states whichis also the number of the lattice points in the whole crystalIn other words we have obtained a very practical and gen-eral result The number of the physically different Bloch

30

wave vectors is the same as the number of the lattice pointsin the crystal

Brillouin Zone

An arbitrary primitive cell in the reciprocal space is notunique and does not necessarily reflect the symmetry ofthe whole crystal Both of these properties are obtained bychoosing the Wigner-Seitz cell of the origin as the primitivecell It is called the (first) Brillouin zone

Crystal Momentum

The crystal momentum ~k is not the momentum of anelectron moving in the periodic lattice This is a conse-quence of the fact that the eigenstates ψnk are not eigen-states of the momentum operator p = minusi~nabla

minus i~nablaψnk = minusi~nabla(eikmiddotrunk(r)

)= ~kminus i~eikmiddotrnablaunk(r) (80)

In the following we will see some similarities with themomentum p especially when we study the response ofthe Bloch electrons to external electric field For now theBloch wave vector k has to be thought merely as the quan-tum number describing the translational symmetry of theperiodic potential

Eigenvalues of Energy

We showed in the above that the eigenvalues of the trans-lation operator TR are exp(ikmiddotR) For each quantum num-ber k of the translation operator there are several eigenval-ues of energy Insert the Bloch wave function (75) into theSchrodinger equation which leads to an eigenvalue equa-tion for the periodic function unk(r + R) = unk(r)

Hkunk =~2

2m

(minus inabla+ k

)2

unk + Uunk = Enkunk (81)

Because u is periodic we can restrict the eigenvalue equa-tion into a primitive cell of the crystal In a finite volumewe obtain an infinite and discrete set of eigenvalues thatwe have already denoted with the index n

It is worthwhile to notice that although the Bloch wavevectors k obtain only discrete values (77) the eigenenergiesEnk = En(k) are continuous functions of the variable kThis is because k appears in the Schrodinger equation (81)only as a parameter

Consider a fixed energy quantum number n Becausethe energies are continuous and periodic in the reciprocallattice (En(k + K) = En(k)) they form a bound set whichis called the energy band The energy bands Enk determinewhether the material is a metal semiconductor or an insu-lator Their slopes give the velocities of the electrons thatcan be used in the explaining of the transport phenomenaIn addition one can calculate the minimum energies of thecrystals and even the magnetic properties from the shapesof the energy bands We will return to some of these laterin the course

Calculation of Eigenstates

It turns out that due to the periodicity it is enough tosolve the Schrodinger equation in only one primitive cellwith boundary conditions

ψnk(r + R) = eikmiddotRψnk(r)

n middot nablaψnk(r + R) = eikmiddotRn middot nablaψnk(r) (82)

This saves the computational resources by a factor 1023

Uniqueness of Translation States

Because ψnk+K = ψnk the wave functions can be in-dexed in several ways

1) Reduced zone scheme Restrict to the first Brillouinzone Then for each Bloch wave vector k exists aninfinite number of energies denoted with the index n

2) Extended zone scheme The wave vector k has valuesfrom the entire reciprocal space We abandon the in-dex n and denote ψnk rarr ψk+Kn

3) Repeated zone scheme Allow the whole reciprocalspace and keep the index n

In the first two cases we obtain a complete and linearlyindependent set of wave functions In the third optioneach eigenstate is repeated infinitely many times

(Source MM) The classification of the free electron k-states in one dimensional space The energies are of theform

E0nk =

~2(k + nK)2

2m

where K is a primitive vector of the reciprocal lattice

Density of States

As in the case of free electrons one sometimes need tocalculate sums over wave vectors k We will need the vol-ume per a wave vector in a Brillouin zone so that we cantransform the summations into integrals We obtain

b1 middot (b2 times b3)

M1M2M3= =

(2π)3

V

31

Thus the sums over the Brillouin zone can be transformedsumkσ

Fk = V

intdkDkFk

where the density of states Dk = 2(2π)3 similar to freeelectrons One should notice that even though the den-sities of states are the same one calculates the sums overthe vectors (77) in the Brillouin zone of the periodic lat-tice but for free electrons they are counted over the wholereciprocal space

Correspondingly one obtains the energy density of states

Dn(E) =2

(2π)3

intdkδ(E minus Enk)

Energy Bands and Electron Velocity

Later in the course we will show that the electron in theenergy band Enk has a non-zero average velocity

vnk =1

~nablakEnk (83)

This is a very interesting result It shows that an electronin a periodic potential has time-independent energy statesin which the electron moves forever with the same averagevelocity This occurs regardless of but rather due to theinteractions between the electron and the lattice This isin correspondence with the definition of group velocity v =partωpartk that is generally defined for the solutions of thewave equation

Bloch Theorem in Fourier Space

Let us assume that the function U(r) is periodic in theBravais lattice ie

U(r + R) = U(r)

for all vectors R in the lattice A periodic function canalways be represented as a Fourier series Due to the peri-odicity each Fourier component has to fulfil

eikmiddot(r+R) = eikmiddotr

Thus the only non-zero components are obtained whenk = K is taken from the reciprocal lattice

Consider the same property in a little more formal man-ner We count the Fourier transform of the function U(r)int

dreminusiqmiddotrU(r) =sumR

intunitcell

dreminusiqmiddotRU(r + R)eminusiqmiddotr

= ΩsumR

eminusiqmiddotRUq (84)

where Ω is the volume of the unit cell and

Uq equiv1

Ω

intunitcell

dreminusiqmiddotrU(r) (85)

By generalizing the previous results for the one dimensionalsumΣq we obtain the relationsum

R

eminusiqmiddotR = NsumK

δqK

Thus we can writeintdreminusiqmiddotrU(r) = V

sumK

δqKUK (86)

where V = NΩ The inverse Fourier transform gives

U(r) =sumK

eiKmiddotrUK (87)

where the sum is calculated over the whole reciprocal lat-tice not just over the first Brillouin zone

Earlier we stated that the eigenstate of the Hamiltonianoperator (68) is not necessarily periodic Nevertheless wecan write by employing the periodicity of the function u

ψnk(r) = eikmiddotrunk(r)

=sumK

(unk)Kei(k+K)middotr (88)

Thus we see that the Bloch state is a linear superposi-tion of the eigenstates of the momentum with eigenvalues~(k + K) In other words the periodic potential mixes themomentum states that differ by a reciprocal lattice vec-tor ~K This was seen already before when we studiedscattering from a periodic crystal

Let us study this more formally The wave function canbe presented in the V as linear combination of plane wavessatisfying the periodic boundary condition (76)

ψ(r) =1

V

sumq

ψ(q)eiqmiddotr

Then the Schrodinger equation for the Hamiltonian oper-ator (68) can be written in the form

0 =(H minus E

) 1

V

sumqprime

ψ(qprime)eiqprimemiddotr

=1

V

sumqprime

[E0qprime minus E + U(r)

]ψ(qprime)eiq

primemiddotr

=1

V

sumqprime

[(E0

qprime minus E)eiqprimemiddotr +

sumK

ei(K+qprime)middotrUK

]ψ(qprime)(89)

Next we will multiply the both sides of the equationwith the factor exp(minusiq middotr) and integrate over the positionspace

0 =sumqprime

intdr

V

[(E0

qprime minus E)ei(qprimeminusq)middotr +

sumK

ei(K+qprimeminusq)middotrUK

]ψ(qprime)

=sumqprime

[(E0

qprime minus E)δqprimeq +sumK

δK+qprimeminusq0UK

]ψ(qprime)

rArr 0 = (E0q minus E)ψ(q) +

sumK

UKψ(qminusK) (90)

32

We have used the propertyintdreiqmiddotr = V δq0

When we consider that k belongs to the first Bril-louin zone we have obtained an equation that couples theFourier components of the wave function that differ by areciprocal lattice vector Because the Brillouin zone con-tains N vectors k the original Schrodinger equation hasbeen divided in N independent problems The solution ofeach is formed by a superposition of such plane waves thatcontain the wave vector k and wave vectors that differfrom k by a reciprocal lattice vector

We can write the Hamiltonian using the Dirac notation

H =sumqprime

E0qprime |qprime〉〈qprime|+

sumqprimeK

UK|qprime〉〈qprime minusK| (91)

When we calculate 〈q|H minus E|ψ〉 we obtain Equation (90)

Representing the Schrodinger equation in the Fourierspace turns out to be one of the most beneficial tools ofcondensed matter physics An equivalent point of viewwith the Fourier space is to think the wave functions ofelectrons in a periodic lattice as a superposition of planewaves

33 Nearly Free and Tightly Bound Elec-trons

MM Chapter 8

In order to understand the behaviour of electrons in aperiodic potential we will have to solve the Schrodingerequation (90) In the most general case the problem is nu-merical but we can separate two limiting cases that can besolved analytically Later we will study the tight bindingapproximation in which the electrons stay tightly in thevicinity of one nucleus However let us first consider theother limit where the potential experienced by the elec-trons can be taken as a small perturbation

Nearly Free Electrons

When we study the metals in the groups I II III andIV of the periodic table we observe that their conductionelectrons behave as they were moving in a nearly constantpotential These elements are often called the rdquonearly freeelectronrdquo metals because they can be described as a freeelectron gas perturbed with a weak periodic potential

Let us start by studying an electron moving in a weak pe-riodic potential When we studied scattering from a crystallattice we noticed that strong scattering occurs when

k0 minus k = K

In the above k0 is the wave vector of the incoming ray andk that of the scattered one The vector K belongs to thereciprocal lattice If we consider elastic scattering we havethat k0 = k and thus

E0k = E0

k+K (92)

Such points are called the degeneracy points

A degeneracy point of a one-dimensional electron in therepeated zone scheme

Above discussion gives us the reason to expect thatthe interaction between the electron and the lattice is thestrongest at the degeneracy points Let us consider thisformally by assuming that the potential experienced bythe electron can be taken as a weak perturbation

UK = ∆wK (93)

where ∆ is a small dimensionless parameter Then wesolve the Schrodinger (90)

(E0q minus E)ψ(q) +

sumK

UKψ(qminusK) = 0

by assuming that the solutions can be presented as poly-nomials of the parameter ∆

ψ(q) = ψ(0)(q) + ψ(1)(q)∆ +

E = E(0) + E(1)∆ + (94)

The Schrodinger equation should be then fulfilled for eachpower of the parameter ∆

Zeroth Order

We insert Equations (94) into the Schrodinger equation(90) and study the terms that are independent on ∆

rArr ψ(0)(q)[E0q minus E(0)

]= 0

In the extended zone scheme the wave vector can havevalues from the entire reciprocal lattice In addition foreach value of the wave vector k we have one eigenvalue ofenergy Remembering that

T daggerRψ(0)k (q) = eikmiddotRψ

(0)k (q)

and

ψ(0)k (r) =

1

V

sumq

ψ(0)k (q)eiqmiddotr

we must have

ψ(0)k (q) = δkq rArr ψ

(0)k (r) = eikmiddotr rArr E(0)

k = E0k

33

Correspondingly in the reduced zone scheme the wavevector k is restricted into the first Brillouin zone and thusthe wave function ψ needs the index n for the energy Weobtain

ψ(0)nk (q) = δk+Knq rArr ψ

(0)nk (r) = ei(k+Kn)middotr rArr E(0)

nk = E0k+Kn

In the above the reciprocal lattice vector Kn is chose sothat qminusKn is in the first Brillouin zone

First Order

Consider the terms linear in ∆ in the perturbation theory[E0qminusE

(0)k

(1)k (q) +

sumK

wKψ(0)k (qminusK)minusE(1)

k ψ(0)k (q) = 0

When q = k according to the zeroth order calculation

E(1)k = w0

By using this we obtain the first order correction for thewave function

ψ(1)k (q) =

sumK6=0

wKδkqminusKE0k minusE0

k+K

rArr ψk(q) asymp δqk +sumK 6=0

UKδkqminusKE0k minus E0

k+K

(95)

Equation (95) is extremely useful in many calculationsin which the potential can be considered as a weak pertur-bation The region where the perturbation theory does notwork is also very interesting As one can see from Equa-tion (95) the perturbative expansion does not convergewhen

E0k = E0

k+K

The perturbation theory thus shows that the interac-tion between an electron and the periodic potential is thestrongest at the degeneracy points as was noted alreadywhen we studied scattering Generally speaking the stateswith the same energy mix strongly when they are per-turbed and thus one has to choose their superposition asthe initial state and use the degenerate perturbation theory

Degenerate Perturbation Theory

Study two eigenstates ψ1 and ψ2 of the unperturbedHamiltonian H0 (= P 22m in our case) that have (nearly)same energies E1 sim E2 (following considerations can be gen-eralized also for larger number of degenerate states) Thesestates span a subspace

ψ = c1ψ1 + c2ψ2

whose vectors are degenerate eigenstates of the Hamilto-nian H0 when E1 = E2 Degenerate perturbation theorydiagonalizes the Hamiltonian operator H = H0 + U(R) inthis subspace

Write the Schrodinger equation into the form

H|ψ〉 = E|ψ〉 rArr (H minus E)|ψ〉

By calculating the projections to the states |ψ1〉 and |ψ2〉we obtain sum

j

〈ψi|(H minus E)|ψj〉cj = 0

This is a linear group of equations that has a solution whenthe determinant of the coefficient matrix

Heffij = 〈ψj |(H minus E)|ψj〉 (96)

is zero A text-book example from this kind of a situationis the splitting of the (degenerate) spectral line of an atomin external electric or magnetic field (StarkZeeman effect)We will now study the disappearance of the degeneracy ofan electron due to a weak periodic potential

In this case the degenerate states are |ψ1〉 = |k〉 and|ψ2〉 = |k+K〉 According to Equation (91) we obtain thematrix representation(

〈k|H minus E|k〉 〈k|H minus E|k + K〉〈k + K|H minus E|k〉 〈k + K|H minus E|k + K〉

)=

(E0k + U0 minus E UminusK

UK E0k+K + U0 minus E

) (97)

We have used the relation UminusK = UlowastK that follows fromEquation (85) and the fact that the potential U(r) is realWhen we set the determinant zero we obtain as a resultof straightforward algebra

E = U0 +E0k + E0

k+K

2plusmn

radic(E0

k minus E0k+K)2

4+ |UK|2 (98)

At the degeneracy point (E0k+K = E0

k) we obtain

E = E0k + U0 plusmn |UK|

Thus we see that the degeneracy is lifted due to the weakperiodic potential and there exists an energy gap betweenthe energy bands

Eg = 2|UK| (99)

One-Dimensional Case

Assume that the electrons are in one-dimensional latticewhose lattice constant is a Thus the reciprocal latticevectors are of form K = n2πa Equation (92) is fulfilledat points k = nπa

34

In the extended zone scheme we see that the periodic po-tential makes the energy levels discontinuous and that themagnitude of those can be calculated in the weak potentialcase from Equation (99) When the electron is acceleratedeg with a (weak) electric field it moves along the con-tinuous parts of these energy bands The energy gaps Egprevent its access to other bands We will return to thislater

The reduced zone scheme is obtained by moving the en-ergy levels in the extended zone scheme with reciprocallattice vectors into the first Brillouin zone

van Hove Singularities

From the previous discussion we see that the energybands Enk are continuous in the k-space and thus theyhave extrema at the Brillouin zone boundaries (minimaand maxima) and inside the zone (saddle points) Ac-cordingly one can see divergences in the energy densityof states called the van Hove singularities For examplein one dimension the density of states can be written as

D(E) =

intdk

2

2πδ(E minus Ek)

=1

π

int infin0

2dEk|dEkdk|

δ(E minus Ek)

=2

π

1

|dEkdk|∣∣∣Ek=E

(100)

where we have used the relation Ek = Eminusk According tothe previous section the density of states diverges at theBrillouin zone boundary

Correspondingly one can show (cf MM) that the den-

sity of states of d-dimensional system is

D(E) =2

(2π)d

intdkδ(E minus Ek)

=2

(2π)d

intdΣ

|nablakEk| (101)

integrated over the energy surface E = Ek

We will return to the van Hove singularities when westudy the thermal vibrations of the lattice

Brillouin Zones

When the potential is extremely weak the energies of theelectron are almost everywhere as there were no potentialat all Therefore it is natural to use the extended zonewhere the states are classified only with the wave vectork Anyhow the energy bands are discontinuous at thedegeneracy points of the free electron Let us study in thefollowing the consequences of this result

At the degeneracy point

E0k = E0

k+K rArr k2 = k2 + 2k middotK +K2

rArr k middot K = minusK2

We see that the points fulfilling the above equation forma plane that is exactly at the midpoint of the origin andthe reciprocal lattice point K perpendicular to the vectorK Such planes are called the Bragg planes In the scat-tering experiment the strong peaks are formed when theincoming wave vector lies in a Bragg plane

By crossing a Bragg plane we are closer to the point Kthan the origin When we go through all the reciprocallattice points and draw them the corresponding planeswe restrict a volume around the origin whose points arecloser to the origin than to any other point in the reciprocallattice The Bragg planes enclose the Wigner-Seitz cell ofthe origin ie the first Brillouin zone

The second Brillouin zone is obtained similarly as theset of points (volume) to whom the origin is the secondclosest reciprocal lattice point Generally the nth Brillouinzone is the set of points to whom the origin is the nth

closest reciprocal lattice point Another way to say thesame is that the nth Brillouin zone consists of the pointsthat can be reached from the origin by crossing exactlynminus1 Bragg planes One can show that each Brillouin zoneis a primitive cell and that they all have the same volume

35

Six first Brillouin zones of the two-dimensional squarelattice The Bragg planes are lines that divide the planeinto areas the Brillouin zones

Effect of the Periodic Potential on Fermi Surface

As has been already mentioned only those electrons thatare near the Fermi surface contribute to the transport phe-nomena Therefore it is important to understand how theFermi surface is altered by the periodic potential In theextended zone scheme the Fermi surface stays almost thesame but in the reduced zone scheme it changes drasti-cally

Let us consider as an example the two-dimensionalsquare lattice (lattice constant a) and assume that eachlattice point has two conduction electrons We showedpreviously that the first Brillouin zone contains the samenumber of wave vectors k as there are points in the latticeBecause each k- state can hold two electrons (Pauli princi-ple + spin) the volume filled by electrons in the reciprocallattice has to be equal to that of the Brillouin zone Thevolume of a Brillouin zone in a square lattice is 4π2a2 Ifthe electrons are assumed to be free their Fermi surface isa sphere with the volume

πk2F =

4π2

a2rArr kF =

2radicπ

π

aasymp 1128

π

a

Thus we see that the Fermi surface is slightly outside theBrillouin zone

A weak periodic potential alters the Fermi surfaceslightly The energy bands are continuous in the reducedzone scheme and one can show that also the Fermi sur-face should be continuous and differentiable in the reducedzone Thus we have alter the Fermi surface in the vicinityof Bragg planes In the case of weak potential we can showthat the constant energy surfaces (and thus the Fermi sur-face) are perpendicular to a Bragg plane The basic idea isthen to modify the Fermi surface in the vicinity of Braggplanes Then the obtained surface is translated with re-ciprocal lattice vectors into the first Brillouin zone

The Fermi surface of a square lattice with weak periodicpotential in the reduced zone scheme Generally the Fermisurfaces of the electrons reaching multiple Brillouin zonesare very complicated in three dimensions in the reducedzone scheme even in the case of free electrons (see exampleson three-dimensional Fermi surfaces in MM)

Tightly Bound Electrons

So far we have assumed that the electrons experiencethe nuclei as a small perturbation causing just slight devi-ations into the states of the free electrons It is thus nat-ural to study the rdquooppositerdquo picture where the electronsare localized in the vicinity of an atom In this approxi-mation the atoms are nearly isolated from each other andtheir mutual interaction is taken to be small perturbationWe will show in the following that the two above men-tioned limits complete each other even though they are inan apparent conflict

Let us first define the Wannier functions as a superpo-sition of the Bloch functions

wn(R r) =1radicN

sumk

eminusikmiddotRψnk(r) (102)

where N is the number of lattice points and thus the num-ber of points in the first Brillouin zone The summationruns through the first Brillouin zone These functions canbe defined for all solids but especially for insulators theyare localized in the vicinity of the lattice points R

The Wannier functions form an orthonormal setintdrwn(R r)wlowastm(Rprime r)

=

intdrsumk

sumkprime

1

NeminusikmiddotR+ikprimemiddotRprimeψnk(r)ψlowastmkprime(r)

=1

N

sumkkprime

eminusikmiddotR+ikprimemiddotRprimeδmnδkkprime

= δRRprimeδnm (103)

where the Bloch wave functions ψnk are assumed to beorthonormal

With a straightforward calculation one can show that the

36

Bloch states can be obtained from the Wannier functions

ψnk(r) =1radicN

sumR

eikmiddotRwn(R r) (104)

In quantum mechanics the global phase has no physicalsignificance Thus we can add an arbitrary phase into theBloch functions leading to Wannier functions of form

wn(R r) =1radicN

sumk

eminusikmiddotR+iφ(k)ψnk(r)

where φ(k) is an arbitrary real phase factor Even thoughthe phase factors do not influence on the Bloch states theyalter significantly the Wannier functions This is becausethe phase differences can interfere constructively and de-structively which can be seen especially in the interferenceexperiments Thus the meaning is to optimize the choicesof phases so that the Wannier functions are as centred aspossible around R

Tight Binding Model

Let us assume that we have Wannier functions the de-crease exponentially when we leave the lattice point RThen it is practical to write the Hamiltonian operator inthe basis defined by the Wannier functions Consider theband n and denote the Wannier function wn(R r) withthe Hilbert space ket vector |R〉 We obtain the represen-tation for the Hamiltonian

H =sumRRprime

|R〉〈R|H|Rprime〉〈Rprime| (105)

The matrix elements

HRRprime = 〈R|H|Rprime〉 (106)

=

intdrwlowastn(R r)

[minus ~2nabla2

2m+ U(r)

]wn(Rprime r)

tell how strongly the electrons at different lattice pointsinteract We assume that the Wannier functions have verysmall values when move over the nearest lattice points Inother words we consider only interactions between nearestneighbours Then the matrix elements can be written as

HRRprime =1

N

sumk

Enkeikmiddot(RminusRprime) (107)

Chemists call these matrix elements the binding energiesWe see that they depend only on the differences betweenthe lattice vectors

Often symmetry implies that for nearest neighboursHRRprime = t = a constant In addition when R = Rprime wecan denote HRRprime = U = another constant Thus we ob-tain the tight binding Hamiltonian

HTB =sumRδ

|R〉t〈R + δ|+sumR

|R〉U〈R| (108)

In the above δ are vectors that point from R towards itsnearest neighbours The first term describes the hopping

of the electrons between adjacent lattice points The termU describes the energy that is required to bring an electroninto a lattice point

The eigenproblem of the tight binding Hamiltonian (108)can be solved analytically We define the vector

|k〉 =1radicN

sumR

eikmiddotR|R〉 (109)

where k belongs to the first Brillouin zone Because this isa discrete Fourier transform we obtain the Wannier func-tions with the inverse transform

|R〉 =1radicN

sumk

eminusikmiddotR|k〉 (110)

By inserting this into the tight binding Hamiltonian weobtain

HTB =sumk

Ek|k〉〈k| (111)

whereEk = U + t

sumδ

eikmiddotδ (112)

If there are w nearest neighbours the maximum value ofthe energy Ek is U+ |t|w and the minimum Uminus|t|w Thuswe obtain the width of the energy band

W = 2|t|w (113)

One can show that the localization of the Wannier func-tion is inversely proportional to the size of the energy gapTherefore this kind of treatment suits especially for insu-lators

Example 1-D lattice

In one dimension δ = plusmna and thus

Ek = U + t(eika + eminusika) = U + 2t cos(ka) (114)

This way we obtain the energy levels in the repeated zonescheme

k

E

πaminusπa 0 2πaminus2πa

Example Graphene

The tight binding model presented above works for Bra-vais lattices Let us then consider how the introduction ofbasis changes things We noted before that graphene isordered in hexagonal lattice form

a1 = a(radic

32

12

)37

a2 = a(radic

32 minus 1

2

)

with a basis

vA = a(

12radic

30)

vB = a(minus 1

2radic

30)

In addition to the honeycomb structure graphene canthought to be formed of two trigonal Bravais lattices lo-cated at points R + vA and R + vB

In graphene the carbon atoms form double bonds be-tween each three nearest neighbours Because carbon hasfour valence electrons one electron per a carbon atom isleft outside the double bonds For each point in the Bra-vais lattice R there exists two atoms at points R + vAand R + vB Assume also in this case that the electronsare localized at the lattice points R + vi where they canbe denoted with a state vector |R + vi〉 The Bloch wavefunctions can be written in form

|ψk〉 =sumR

[A|R + vA〉+B|R + vB〉

]eikmiddotR (115)

The meaning is to find such values for coefficients A andB that ψk fulfils the Schrodinger equation Therefore theHamiltonian operator is written in the basis of localizedstates as

H =sumijRRprime

|Rprime + vi〉〈Rprime + vi|H|R + vj〉〈R + vj | (116)

where i j = AB Here we assume that the localized statesform a complete and orthonormal basis in the Hilbertspace similar to the case of Wannier functions

Operate with the Hamiltonian operator to the wave func-tion (115)

H|ψk〉 =sumRRprimei

|Rprime + vi〉〈Rprime + vi|H

[A|R + vA〉+B|R + vB〉

]eikmiddotR

Next we count the projections to the vectors |vA〉 and|vB〉 We obtain

〈vA|H|ψk〉 =sumR

[AγAA(R) +BγAB(R)

]eikmiddotR(117)

〈vB |H|ψk〉 =sumR

[AγBA(R) +BγBB(R)

]eikmiddotR(118)

where

γii(R) = 〈vi|H|R + vi〉 i = AB (119)

γij(R) = 〈vi|H|R + vj〉 i 6= j (120)

In the tight binding approximation we can set

γii(0) = U (121)

γAB(0) = γAB(minusa1) = γAB(minusa2) = t (122)

γBA(0) = γBA(a1) = γBA(a2) = t (123)

The couplings vanish for all other vectors R We see thatin the tight binding approximation the electrons can jumpone nucleus to another only by changing the trigonal lat-tice

Based on the above we can write the Schrodinger equa-tion into the matrix form U t

(1 + eikmiddota1 + eikmiddota2

)t(

1 + eminusikmiddota1 + eminusikmiddota2

)U

(AB

) U t

[1 + 2ei

radic3akx2 cos

(aky2

)]t[1 + 2eminusi

radic3akx2 cos

(aky2

)]U

(AB

)

= E(AB

)(124)

The solutions of the eigenvalue equation give the energiesin the tight binding model of graphene

E = U plusmn

radic1 + 4 cos2

(aky2

)+ 4 cos

(aky2

)cos(radic3akx

2

)

(125)

We see that in the (kx ky)-plane there are points wherethe energy gap is zero It turns out that the Fermi level ofgraphene sets exactly to these Dirac points In the vicinityof the Dirac points the dispersion relation of the energy islinear and thus the charge carriers in graphene are mass-less Dirac fermions

34 Interactions of ElectronsMM Chapter 9 (not 94)

Thus far we have considered the challenging problem ofcondensed matter physics (Hamiltonian (41)) in the sin-gle electron approximation Because the nuclei are heavy

38

compared with the electrons we have ignored their motionand considered them as static classical potentials (Born-Oppenheimer approximation) In addition the interactionsbetween electrons have been neglected or at most includedinto the periodic potential experienced by the electrons asan averaged quantity

Next we will relax the single electron approximation andconsider the condensed matter Hamiltonian (41) in a moredetailed manner In the Born-Oppenheimer approximationthe Schrodinger equation can be written as

HΨ = minusNsuml=1

[~2

2mnabla2lΨ + Ze2

sumR

1

|rl minusR|Ψ

]

+1

2

sumiltj

e2

|ri minus rj |Ψ

= EΨ (126)

The nuclei are static at the points R of the Bravais latticeThe number of the interacting electrons is N (in condensedmatter N sim 1023) and they are described with the wavefunction

Ψ = Ψ(r1σ1 rNσN )

where ri and σi are the place and the spin of the electroni respectively

The potential is formed by two terms first of which de-scribes the electrostatic attraction of the electron l

Uion(r) = minusZe2sumR

1

|rminusR|

that is due to the nuclei The other part of potential givesthe interaction between the electrons It is the reason whythe Schrodinger equation of a condensed matter system isgenerally difficult to solve and can be done only when N 20 In the following we form an approximative theory thatincludes the electron-electron interactions and yet gives(at least numerical) results in a reasonable time

Hartree Equations

The meaning is to approximate the electric field of thesurrounding electrons experienced by a single electron Thesimplest (non-trivial) approach is to assume that the otherelectrons form a uniform negative distribution of chargewith the charge density ρ In such field the potential en-ergy of an electron is

UC(r) =

intdrprime

e2n(r)

|rminus rprime| (127)

where the number density of electrons is

n(r) =ρ(r)

e=suml

|ψl(r)|2

The above summation runs through the occupied singleelectron states

By plugging this in the Schrodinger equation of the con-densed matter we obtain the Hartree equation

minus ~2

2mnabla2ψl + [Uion(r) + UC(r)]ψl = Eψl (128)

that has to be solved by iteration In practise one guessesthe form of the potential UC and solves the Hartree equa-tion Then one recalculates the UC and solves again theHartree equation Hartree In an ideal case the methodis continued until the following rounds of iteration do notsignificantly alter the potential UC

Later we will see that the averaging of the surroundingelectrons is too harsh method and as a consequence thewave function does not obey the Pauli principle

Variational Principle

In order to improve the Hartree equation one shouldfind a formal way to derive it systematically It can bedone with the variational principle First we show thatthe solutions Ψ of the Schrodinger equation are extrema ofthe functional

F(Ψ) = 〈Ψ|H|Ψ〉 (129)

We use the method of Lagrangersquos multipliers with a con-straint that the eigenstates of the Hamiltonian operatorare normalized

〈Ψ|Ψ〉 = 1

The value λ of Lagrangersquos multiplier is determined from

δ(F minus λ〈Ψ|Ψ〉) = 〈δΨ|(H minus λ)|Ψ〉+ 〈Ψ|(H minus λ)|δΨ〉= 〈δΨ|(E minus λ)|Ψ〉+ 〈Ψ|(E minus λ)|δΨ〉= 0 (130)

which occurs when λ = E We have used the hermiticityof the Hamiltonian operator Hdagger = H and the Schrodingerequation H|Ψ〉 = E|Ψ〉

As an exercise one can show that by choosing

Ψ =

Nprodl=1

ψl(rl) (131)

the variational principle results in the Hartree equationHere the wave functions ψi are orthonormal single elec-tron eigenstates The wave function above reveals why theHartree equation does not work The true many electronwave function must vanish when two electrons occupy thesame place In other words the electrons have to obey thePauli principle Clearly the Hartree wave function (131)does not have this property Previously we have notedthat in metals the electron occupy states whose energiesare of the order 10000 K even in ground state Thus wehave to use the Fermi-Dirac distribution (not the classi-cal Boltzmann distribution) and the quantum mechanicalproperties of the electrons become relevant Thus we haveto modify the Hartree wave function

Hartree-Fock Equations

39

Fock and Slater showed in 1930 that the Pauli principleholds when we work in the space spanned by the antisym-metric wave functions By antisymmetry we mean that themany-electron wave function has to change its sign whentwo of its arguments are interchanged

Ψ(r1σ1 riσi rjσj rNσN )

= minusΨ(r1σ1 rjσj riσi rNσN ) (132)

The antisymmetricity of the wave function if the funda-mental statement of the Pauli exclusion principle The pre-vious notion that the single electron cannot be degeneratefollows instantly when one assigns ψi = ψj in the function(132) This statement can be used only in the indepen-dent electron approximation For example we see that theHartree wave function (131) obeys the non-degeneracy con-dition but is not antisymmetric Thus it cannot be thetrue many electron wave function

The simplest choice for the antisymmetric wave functionis

Ψ(r1σ1 rNσN )

=1radicN

sums

(minus1)sψs1(r1σ1) middot middot middotψsN (rNσN ) (133)

=

∣∣∣∣∣∣∣∣∣∣∣

ψ1(r1σ1) ψ1(r2σ2) ψ1(rNσN )

ψN (r1σ1) ψN (r2σ2) ψN (rNσN )

∣∣∣∣∣∣∣∣∣∣∣

where the summation runs through all permutations of thenumbers 1 N The latter form of the equation is calledthe Slater determinant We see that the wave function(133) is not anymore a simple product of single electronwave functions but a sum of such products Thus theelectrons can no longer be dealt with independently Forexample by changing the index r1 the positions of allelectrons change In other words the Pauli principle causescorrelations between electrons Therefore the spin has tobe taken into account in every single electron wave functionwith a new index σ that can obtain values plusmn1 If theHamiltonian operator does not depend explicitly on spinthe variables can be separated (as was done in the singleelectron case when the Hamiltonian did not couple theelectrons)

ψl(riσi) = φl(ri)χl(σi) (134)

where the spin function

χl(σi) =

δ1σi spin ylosδminus1σi spin alas

Hartree-Fock equations result from the expectation value

〈Ψ|H|Ψ〉

in the state (133) and by the application of the varia-tional principle Next we will go through the laboriousbut straightforward derivation

Expectation Value of Kinetic Energy

Let us first calculate the expectation value of the kineticenergy

〈T 〉 = 〈Ψ|T |Ψ〉

=sum

σ1σN

intdNr

1

N

sumssprime

(minus1)s+sprime

[prodj

ψlowastsj (rjσj)

]

timessuml

minus~2nabla2l

2m

[prodi

ψsprimei(riσi)

] (135)

We see that because nablal operates on electron i = l thenthe integration over the remaining terms (i 6= l) gives dueto the orthogonality of the wave functions ψ that s = sprime

and thus

〈T 〉 =suml

sumσl

intdrl

1

N

sums

ψlowastsl(rlσl)minus~2nabla2

l

2mψsl(rlσl)

(136)

The summation over the index s goes over all permuta-tions of the numbers 1 N In our case the summationdepends only on the value of the index sl Thus it is prac-tical to divide the sum in twosum

s

rarrsumj

sumssl=j

The latter sum produces only the factor (N minus 1) Becausewe integrate over the variables rl and σl we can make thechanges of variables

rl rarr r σl rarr σ

We obtain

〈T 〉 =suml

sumσ

intdr

1

N

sumlprime

ψlowastlprime(rσ)minus~2nabla2

2mψlprime(rσ) (137)

The summation over the index l produces a factor N Bymaking the replacement lprime rarr l we end up with

〈T 〉 = =sumσ

intdrsuml

ψlowastl (rσ)minus~2nabla2

2mψl(rσ)

=

Nsuml=1

intdrφlowastl (r)

[minus~2nabla2

2m

]φl(r) (138)

where the latter equality is obtained by using Equation(134) and by summing over the spin

Expectation Value of Potential Energy

The expectation value of the (periodic) potential Uiondue to nuclei is obtained similarly

〈Uion〉 =

Nsuml=1

intdrφlowastl (r)Uion(r)φl(r) (139)

What is left is the calculation of the expectation value ofthe Coulomb interactions between the electrons Uee For

40

that we will use a short-hand notation rlσl rarr l We resultin

〈Uee〉

=sum

σ1σN

intdNr

sumssprime

1

N

sumlltlprime

e2(minus1)s+sprime

|rl minus rlprime |

timesprodj

ψlowastsj (j)prodi

ψsprimei(i) (140)

The summations over indices l and lprime can be transferredoutside the integration In addition by separating the elec-trons l and lprime from the productprodji

ψlowastsj (j)ψsprimei(i) = ψlowastsl(l)ψlowastslprime

(lprime)ψsprimel(l)ψsprimelprime (lprime)prod

ji 6=llprimeψlowastsj (j)ψsprimei(i)

we can integrate over other electrons leaving only two per-mutations sprime for each permutation s We obtain

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sums

1

N

e2

|rl minus rlprime |(141)

times ψlowastsl(l)ψlowastslprime

(lprime)[ψsl(l)ψslprime (l

prime)minus ψslprime (l)ψsl(lprime)]

Again the summation s goes through all permutations ofthe numbers 1 N The sums depend only on the valuesof the indices sl and slprime so it is practical to writesum

s

rarrsumi

sumj

sumssl=islprime=j

where latter sum produces the factor (N minus 2) Thus

〈Uee〉

=sumlltlprime

sumσlσlprime

intdrldrlprime

sumij

1

N(N minus 1)

e2

|rl minus rlprime |(142)

times ψlowasti (l)ψlowastj (lprime)[ψi(l)ψj(l

prime)minus ψj(l)ψi(lprime)]

We integrate over the positions and spins and thus wecan replace

rl rarr r1 σl rarr σ1

rlprime rarr r2 σlprime rarr σ2

In addition the summation over indices l and lprime gives thefactor N(N minus 1)2 We end up with the result

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

sumij

e2

|r1 minus r2|(143)

times ψlowasti (1)ψlowastj (2)[ψi(1)ψj(2)minus ψj(1)ψi(2)

]

The terms with i = j result in zero so we obtain

〈Uee〉

=1

2

sumσ1σ2

intdr1dr2

e2

|r1 minus r2|(144)

timessumiltj

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

=1

2

intdr1dr2

e2

|r1 minus r2|(145)

timessumiltj

[|φi(r1)|2|φj(r2)|2

minusφlowasti (r1)φlowastj (r2)φj(r1)φi(r2)δσiσj

]

We see that the interactions between electrons produce twoterms into the expectation value of the Hamiltonian oper-ator The first is called the Coulomb integral and it isexactly the same as the averaged potential (127) used inthe Hartree equations The other term is the so-called ex-change integral and can be interpreted as causing the in-terchange between the positions of electrons 1 and 2 Thenegative sign is a consequence of the antisymmetric wavefunction

Altogether the expectation value of the Hamiltonian op-erator in state Ψ is

〈Ψ|H|Ψ〉

=sumi

sumσ1

intdr1

[ψlowasti (1)

minus~2nabla2

2mψi(1) + Uion(r1)|ψi(1)|2

]+

intdr1dr2

e2

|r1 minus r2|(146)

timessum

iltjσ1σ2

[|ψi(1)|2|ψj(2)|2 minus ψlowasti (1)ψlowastj (2)ψj(1)ψi(2)

]

According to the variational principle we variate this ex-pectation value with respect to all orthonormal single elec-tron wave functions ψlowasti keeping ψi fixed at the same timeThe constriction is now

Eisumσ1

intdr1ψ

lowasti (1)ψi(1)

As a result of the variation we obtain the Hartree-Fockequations [

minus ~2nabla2

2m+ Uion(r) + UC(r)

]φi(r)

minusNsumj=1

δσiσjφj(r)

intdrprime

e2φlowastj (rprime)φi(r

prime)

|rminus rprime|

= Eiφi(r) (147)

Numerical Solution

Hartree-Fock equations are a set of complicated coupledand non-linear differential equations Thus their solutionis inevitably numerical When there is an even number of

41

electrons the wave functions can sometimes be divided intwo groups spin-up and spin-down functions The spatialparts are the same for the spin-up and the correspondingspin-down electrons In these cases it is enough to calcu-late the wave functions of one spin which halves the com-putational time This is called the restricted Hartree-FockIn the unrestricted Hartree-Fock this assumption cannotbe made

The basic idea is to represent the wave functions φi inthe basis of such functions that are easy to integrate Suchfunctions are found usually by rdquoguessingrdquo Based on expe-rience the wave functions in the vicinity of the nucleus lare of the form eminusλi|rminusRl| The integrals in the Hartree-Fock- equations are hard to solve In order to reduce thecomputational time one often tries to fit the wave functioninto a Gaussian

φi =

Ksumk=1

Blkγk

where

γl =sumlprime

Allprimeeminusalprime (rminusRlprime )

2

and the coefficients Allprime and alprime are chosen so that theshape of the wave function φi is as correct as possible

The functions γ1 γK are not orthonormal Becauseone has to be able to generate arbitrary functions the num-ber of these functions has to be K N In addition tothe wave functions also the functions 1|r1 minus r2| must berepresented in this basis When such representations areinserted into the Hartree-Fock equations we obtain a non-linear matrix equation for the coefficients Blk whose sizeis KtimesK We see that the exchange and Coulomb integralsproduce the hardest work because their size is K4 Thegoal is to choose the smallest possible set of functions γkNevertheless it is easy to see why the quantum chemistsuse the largest part of the computational time of the worldssupercomputers

The equation is solved by iteration The main principlepresented shortly

1 Guess N wave functions

2 Present the HF-equations in the basis of the functionsγi

3 Solve the so formed K timesK-matrix equation

4 We obtain K new wave functions Choose N withthe lowest energy and return to 2 Repeat until thesolution converges

(Source MM) Comparison between the experimentsand the Hartree-Fock calculation The studied moleculescontain 10 electrons and can be interpreted as simple testsof the theory The effects that cannot be calculated in theHartree-Fock approximation are called the correlation Es-pecially we see that in the determining the ionization po-tentials and dipole moments the correlation is more than10 percent Hartree-Fock does not seem to be accurateenough for precise molecular calculations and must be con-sidered only directional Chemists have developed moreaccurate approximations to explain the correlation Thoseare not dealt in this course

Hartree-Fock for Jellium

Jellium is a quantum mechanical model for the interact-ing electrons in a solid where the positively charged nucleiare consider as a uniformly distributed background (or asa rigid jelly with uniform charge density)

Uion(r) = minusNV

intdrprime

e2

|rminus rprime|= constant (148)

where N is the number of electrons and V the volume of thesystem This enables the focus on the phenomena due tothe quantum nature and the interactions of the electronsin a solid without the explicit inclusion of the periodiclattice

It turns out that the solutions of the Hartree-Fock equa-tions (147) for jellium are plane waves

φi(r) =eikmiddotrradicV (149)

This is seen by direct substitution We assume the periodicboundaries leading to the kinetic energy

~2k2i

2mφi

The Coulomb integral cancels the ionic potential Weare left with the calculation of the exchange integral

Iexc = minusNsumj=1

δσiσjφj(r)

intdr2

e2φlowastj (r2)φi(r2)

|rminus r2|

= minuse2Nsumj=1

δσiσjeikj middotrradicV

intdr2

V

ei(kiminuskj)middotr2

|rminus r2|

= minuse2φi(r)

Nsumj=1

δσiσj

intdrprime

V

ei(kiminuskj)middotrprime

rprime (150)

42

where in the last stage we make a change of variables rprime =r2 minus r Because the Fourier transform of the function 1ris 4πk2 we obtain

Iexc = minuse2φi(r)1

V

Nsumj=1

δσiσj4π

|ki minus kj |2 (151)

We assume then that the states are filled up to the Fermiwave number kF and we change the sum into an integralThe density of states in the case of the periodic boundarycondition is the familiar 2(2π)3 but the delta functionhalves this value Thus the exchange integral gives

Iexc = minuse2φi(r)

int|k|ltkF

dk

(2π)3

k2i + k2 minus 2k middot ki

= minus2e2kFπ

F

(k

kF

)φi(r) (152)

where the latter equality is left as an Exercise and

F (x) =1

4x

[(1minus x2) ln

∣∣∣∣∣1 + x

1minus x

∣∣∣∣∣+ 2x

](153)

is the Lindhard dielectric function

We see that in the case of jellium the plane waves arethe solutions of the Hartree-Fock equations and that theenergy of the state i is

Ei =~2k2

i

2mminus 2e2kF

πF

(k

kF

) (154)

Apparently the slope of the energy diverges at the Fermisurface This is a consequence of the long range of theCoulomb interaction between two electrons which is in-versely proportional to their distance Hartree-Fock ap-proximation does not take into account the screening dueto other electrons which created when the electrons be-tween the two change their positions hiding the distantpair from each other

The total energy of jellium is

E =suml

[~2k2

l

2mminus e2kF

πF

(klkF

)](155)

= N

[3

5EF minus

3

4

e2kFπ

] (156)

where the correction to the free electron energy is only halfof the value give by the Hartree-Fock calculation (justifi-cation in the Exercises) The latter equality follows bychanging the sum into an integral

Density Functional Theory

Instead of solving the many particle wave function fromthe Schrodinger equation (eg by using Hartree-Fock) wecan describe the system of interacting electrons exactlywith the electron density

n(r) = 〈Ψ|Nsuml=1

δ(rminusRl)|Ψ〉 (157)

= N

intdr2 drN |Ψ(r r2 rN )|2 (158)

This results in the density functional theory (DFT) whichis the most efficient and developable approach in the de-scription of the electronic structure of matter The rootsof the density functional theory lie in the Thomas-Fermimodel (cf next section) but its theoretical foundation waslaid by P Hohenberg and W Kohn2 in the article publishedin 1964 (Phys Rev 136 B864) In the article Hohenbergand Kohn proved two basic results of the density functionaltheory

1st H-K Theorem

The electron density of the ground state determines theexternal potential (up to a constant)

This is a remarkable result if one recalls that two differ-ent many-body problems can differ by potential and num-ber of particles According to the first H-K theorem bothof them can be deduced from the electron density whichtherefore solves the entire many-body problem We provethe claim by assuming that it is not true and showing thatthis results into a contradiction Thus we assume thatthere exists two external potentials U1 and U2 which resultin the same density n Let then H1 and H2 be the Hamil-tonian operators corresponding to the potentials and Ψ1

and Ψ2 their respective ground states For simplicity weassume that the ground states are non-degenerate (the re-sult can be generalized for degenerate ground states butthat is not done here) Therefore we can write that

E1 = 〈Ψ1|H1|Ψ1〉lt 〈Ψ2|H1|Ψ2〉= 〈Ψ2|H2|Ψ2〉+ 〈Ψ2|H1 minus H2|Ψ2〉

= E2 +

intdrn(r)

[U1(r)minus U2(r)

] (159)

Correspondingly we can show starting from the groundstate of the Hamiltonian H2 that

E2 lt E1 +

intdrn(r)

[U2(r)minus U1(r)

] (160)

By adding the obtained inequalities together we obtain

E1 + E2 lt E1 + E2 (161)

which is clearly not true Thus the assumption we madein the beginning is false and we have shown that the elec-tron density determines the external potential (up to a con-stant) The electrons density contains in principle the same

2Kohn received the Nobel in chemistry in 1998 for the developmentof density functional theory

43

information as the wave function of entire electron sys-tem Especially the density function describing the groupof electrons depends only on three variables (x y z) in-stead of the previous 3N

Based on the above result it is easy to understand thatall energies describing the many electron system can berepresented as functionals of the electron density

E [n] = T [n] + Uion[n] + Uee[n] (162)

2nd H-K Theorem

The density n of the ground state minimizes the func-tional E [n] with a constraintint

drn(r) = N

If we can assume that for each density n we can assign awave function Ψ the result is apparently true

The energy functional can be written in the form

E [n] =

intdrn(r)Uion(r) + FHK [n] (163)

whereFHK [n] = T [n] + Uee[n] (164)

We see that FHK does not depend on the ionic potentialUion and thus defines a universal functional for all sys-tems with N particles By finding the shape of this func-tional one finds a solution to all many particle problemswith all potentials Uion So far this has not been doneand it is likely that we never will Instead along the yearsmany approximations have been developed of which wewill present two

Thomas-Fermi Theory

The simplest approximation resulting in an explicit formof the functionals F [n] and E [n] is called the Thomas-Fermitheory The idea is to find the energy of the electrons asa function of the density in a potential that is uniformlydistributed (jellium) Then we use this functional of thedensity in the presence of the external potential

We will use the results obtained from the Hartree-Fockequations for jellium and write them as functions of thedensity By using Equation (55) we obtain the result forthe kinetic energy functional

T [n] =sumkσ

~2k2

2m=

3

5NEF = V

~2

2m

3

5(3π2)23n53 (165)

The Coulomb integral due to the electron-electron interac-tions is already in the functional form

UC [n] =1

2

intdr1dr2

e2n(r1)n(r2)

|r1 minus r2| (166)

In the case of jellium this cancelled the ionic potentialbut now it has to be taken into account because the ionicpotential is not assumed as constant in the last stage

In addition the exchange integral caused an extra terminto the energy of jellium

Uexc[n] = minusN 3

4

e2kFπ

= minusV 3

4

(3

π

)13

e2n43 (167)

In the Thomas-Fermi theory we assume that in a systemwhose charge density changes slowly the above terms canbe evaluated locally and integrated over the volume Thekinetic energy is then

T [n] =

intdr

~2

2m

3

5(3π2)23n53(r) (168)

Correspondingly the energy due to the exchange integralis

Uexc[n] = minusintdr

3

4

(3

π

)13

e2n43(r) (169)

The energy functional can then be written as

E [n] =~2

2m

3

5(3π2)23

intdrn53(r) +

intdrn(r)Uion(r)

+e2

2

intdr1dr2

n(r1)n(r2)

|r1 minus r2|

minus 3

4

(3

π

)13

e2

intdrn43(r) (170)

When the last term due to the exchange integral is ignoredwe obtain the Thomas-Fermi energy functional The func-tional including the exchange term takes into account alsothe Pauli principle which results in the Thomas-Fermi-Dirac theory

Conventional Thomas-Fermi-Dirac equation is obtainedwhen we variate the functional (170) with respect to theelectron density n The constriction isint

drn(r) = N

and Lagrangersquos multiplier for the density is the chemicalpotential

micro =δEδn(r)

(171)

The variational principle results in the Thomas-Fermi-Dirac equation

~2

2m(3π2)23n23(r) + Uion(r) +

intdr2

e2n(r2)

|rminus r2|

minus

(3

π

)13

e2n13(r) = micro (172)

When the last term on the left-hand side is neglected weobtain the Thomas-Fermi equation

The Thomas-Fermi equation is simple to solve but notvery precise For example for an atom whose atomic num-ber is Z it gives the energy minus15375Z73 Ry For small

44

atoms this is two times too large but the error decreases asthe atomic number increases The results of the Thomas-Fermi-Dirac equation are even worse The Thomas-Fermitheory assumes that the charge distribution is uniform lo-cally so it cannot predict the shell structure of the atomsFor molecules both theories give the opposite result to thereality by moving the nuclei of the molecule further apartone reduces the total energy

Despite of its inaccuracy the Thomas-Fermi theory isoften used as a starting point of a many particle problembecause it is easy to solve and then one can deduce quicklyqualitative characteristics of matter

Kohn-Sham Equations

The Thomas-Fermi theory has two major flaws It ne-glects the fact apparent in the Schrodinger equation thatthe kinetic energy of an electron is large in the regionswhere its gradient is large In addition the correlation isnot included Due to this shortcomings the results of theThomas-Fermi equations are too inaccurate On the otherhand the Hartree-Fock equations are more accurate butthe finding of their solution is too slow W Kohn and LJ Sham presented in 1965 a crossing of the two which isnowadays used regularly in large numerical many particlecalculations

As we have seen previously the energy functional hasthree terms

E [n] = T [n] + Uion[n] + Uee[n]

The potential due to the ionic lattice is trivial

Uion[n] =

intdrn(r)Uion(r)

Kohn and Sham suggested that the interacting systemof many electrons in a real potential is replace by non-interacting electrons moving in an effective Kohn-Shamsingle electron potential The idea is that the potentialof the non-interacting system is chosen so that the elec-tron density is the same as that of the interacting systemunder study With these assumptions the wave functionΨNI of the non-interacting system can be written as theSlater determinant (133) of the single electron wave func-tions resulting in the electron density

n(r) =

Nsuml=1

|ψl(r)|2 (173)

The functional of the kinetic energy for the non-interactingelectrons is of the form

TNI [n] = minus ~2

2m

suml

intdrψlowastl (r)nabla2ψl(r) (174)

Additionally we can assume that the classical Coulombintegral produces the largest component UC [n] of theelectron-electron interactions Thus the energy functionalcan be reformed as

E [n] = TNI [n] + Uion[n] + UC [n] + Uexc[n] (175)

where we have defined the exchange-correlation potential

Uexc[n] = (T [n]minus TNI [n]) + (Uee[n]minus UC [n]) (176)

The potential Uexc contains the mistakes originating in theapproximation where we use the kinetic energy of the non-interacting electrons and deal with the interactions be-tween the electrons classically

The energy functional has to be variated in terms of thesingle electron wave function ψlowastl because we do not knowhow to express the kinetic energy in terms of the densityn The constriction is again 〈ψl|ψl〉 = 1 and we obtain

minus ~2nabla2

2mψl(r) +

[Uion(r)

+

intdrprime

e2n(rprime)

|rminus rprime|+partUexc(n)

partn

]ψl(r)

= Elψl(r) (177)

We have obtained the Kohn-Sham equations which arethe same as the previous Hartree-Fock equations Thecomplicated non-local exchange-potential in the Hartree-Fock has been replaced with a local and effective exchange-correlation potential which simplifies the calculations Inaddition we include effects that are left out of the rangeof the Hartree-Fock calculations (eg screening) Oneshould note that the correspondence between the manyparticle problem and the non-interacting system is exactonly when the functional Uexc is known Unfortunatelythe functional remains a mystery and in practise one hasto approximate it somehow Such calculations are calledthe ab initio methods

When the functional Uexc is replaced with the exchange-correlation potential of the uniformly distributed electrongas one obtains the local density approximation

35 Band CalculationsMM Chapters 101 and 103

We end this Chapter by returning to the study of theband structure of electrons The band structure calcula-tions are done always in the single electron picture wherethe effects of the other electrons are taken into account witheffective single particle potentials ie pseudopotentials (cflast section) In this chapter we studied simple methods toobtain the band structure These can be used as a startingpoint for the more precise numerical calculations For ex-ample the nearly free electron model works for some metalsthat have small interatomic distances with respect to therange of the electron wave functions Correspondingly thetight binding model describes well the matter whose atomsare far apart from each other Generally the calculationsare numerical and one has to use sophisticated approxima-tions whose development has taken decades and continuesstill In this course we do not study in detail the methodsused in the band structure calculations Nevertheless theunderstanding of the band structure is important becauseit explains several properties of solids One example is the

45

division to metals insulators and semiconductors in termsof electric conductivity

(Source MM) In insulators the lowest energy bands arecompletely occupied When an electric field is turned onnothing occurs because the Pauli principle prevents the de-generacy of the single electron states The only possibilityto obtain a net current is that the electrons tunnel from thecompletely filled valence band into the empty conductionband over the energy gap Eg An estimate for the probabil-ity P of the tunnelling is obtained from the Landau-Zenerformula

P prop eminuskF EgeE (178)

where E is the strength of the electric field This formulais justified in the chapter dealing with the electronic trans-port phenomena

Correspondingly the metals have one band that is onlypartially filled causing their good electric conductivityNamely the electrons near the Fermi surface can make atransition into nearby k-states causing a shift of the elec-tron distribution into the direction of the external fieldproducing a finite net momentum and thus a net currentBecause the number of electron states in a Brillouin zoneis twice the number of primitive cells in a Bravais lattice(the factor two comes from spin) the zone can be com-pletely occupied only if there are an odd number electronsper primitive cell in the lattice Thus all materials withan odd number of electrons in a primitive cell are met-als However it is worthwhile to recall that the atoms inthe columns 7A and 5A of the periodic table have an oddnumber of valence electrons but they are nevertheless in-sulators This due to the fact that they do not form aBravais lattice and therefore they can have an even num-ber of atoms and thus electrons in their primitive cells

In addition to metals and insulators one can form twomore classes of solids based on the filling of the energybands and the resulting conduction properties Semicon-ductors are insulators whose energy gaps Eg are smallcompared with the room temperature kBT This leadsto a considerable occupation of the conduction band Nat-urally the conductivity of the semiconductors decreaseswith temperature

Semimetals are metals with very few conduction elec-trons because their Fermi surface just barely crosses the

Brillouin zone boundary

46

4 Mechanical PropertiesMM Chapters 111 Introduction 13-1332

134-1341 135 main idea

We have studied the electron structure of solids by as-suming that the locations of the nuclei are known and static(Born-Oppenheimer approximation) The energy neededto break this structure into a gas of widely separated atomsis called the cohesive energy The cohesive energy itself isnot a significant quantity because it is not easy to de-termine in experiments and it bears no relation to thepractical strength of matter (eg to resistance of flow andfracture)

The cohesive energy tells however the lowest energystate of the matter in other words its lattice structureBased on that the crystals can be divided into five classesaccording to the interatomic bonds These are the molec-ular ionic covalent metallic and hydrogen bonds Wehad discussion about those already in the chapter on theatomic structure

41 Lattice VibrationsBorn-Oppenheimer approximation is not able to explain

all equilibrium properties of the matter (such as the spe-cific heat thermal expansion and melting) the transportphenomena (sound propagation superconductivity) norits interaction with radiation (inelastic scattering the am-plitudes of the X-ray scattering and the background radi-ation) In order to explain these one has to relax on theassumption that the nuclei sit still at the points R of theBravais lattice

We will derive the theory of the lattice vibrations byusing two weaker assumptions

bull The nuclei are located on average at the Bravais latticepoints R For each nucleus one can assign a latticepoint around which the nucleus vibrates

bull The deviations from the equilibrium R are small com-pared with the internuclear distances

Let us then denote the location of the nucleus l with thevector

rl = Rl + ul (179)

The purpose is to determine the minimum of the energy ofa solid

E = E(u1 uN ) (180)

as a function of arbitrary deviations ul of the nuclei Ac-cording to our second assumption we can express the cor-rections to the cohesive energy Ec due to the motions ofthe nuclei by using the Taylor expansion

E = Ec +sumαl

partEpartulα

ulα +sumαβllprime

ulαΦllprime

αβulprime

β + middot middot middot (181)

where α β = 1 2 3 and

Φllprime

αβ =part2E

partulαpartulprimeβ

(182)

is a symmetric 3 times 3-matrix Because the lowest energystate has its minimum as a function of the variables ul thelinear term has to vanish The first non-zero correction isthus quadratic If we neglect the higher order correctionswe obtain the harmonic approximation

Periodic Boundary Conditions

Similarly as in the case of moving electrons we requirethat the nuclei obey the periodic boundary conditions Inother words the assumption is that every physical quantity(including the nuclear deviations u) obtains the same valuein every nth primitive cell With this assumption everypoint in the crystal is equal at the equilibrium and thecrystal has no boundaries or surfaces Thus the matrixΦllprime

αβ can depend only on the distance Rl minusRlprime

Thus the classical equations of motion of the nucleus lbecomes

M ul = minusnablalE = minussumlprime

Φllprimeulprime (183)

where M is the mass of the nucleus One should note thatthe matrix Φll

primehas to be symmetric in such translations

of the crystal that move each nucleus with a same vectorTherefore the energy of the crystal cannot change andsum

lprime

Φllprime

= 0 (184)

Clearly Equation (183) describes the coupled vibrations ofthe nuclei around their equilibria

Normal Modes

The classical equations of motion are solved with thetrial function

ul = εeikmiddotRlminusiωt (185)

where ε is the unit vector that determines the polarisationof the vibrations This can be seen by inserting the trialinto Equation (183)

Mω2ε =sumlprime

Φllprimeeikmiddot(R

lprimeminusRl)ε

= Φ(k)ε (186)

whereΦ(k) =

sumlprime

eikmiddot(RlminusRlprime )Φll

prime (187)

The Fourier series Φ(k) does not depend on index lbecause all physical quantities depend only on the differ-ence Rlprime minus Rl In addition Φ(k) is symmetric and real3 times 3-matrix that has three orthogonal eigenvectors foreach wave vector k Here those are denoted with

εkν ν = 1 2 3

and the corresponding eigenvalues with

Φν(k) = Mω2kν (188)

As usual for plane waves the wave vector k deter-mines the direction of propagation of the vibrations In

47

an isotropic crystals one can choose one polarization vec-tor to point into the direction of the given wave vector k(ie ε1 k) when the other two are perpendicular to thedirection of propagation (ε2 ε3 perp k) The first vector iscalled the longitudinal vibrational mode and the latter twothe transverse modes

In anisotropic matter this kind of choice cannot alwaysbe made but especially in symmetric crystals one polar-ization vector points almost into the direction of the wavevector This works especially when the vibration propa-gates along the symmetry axis Thus one can still use theterms longitudinal and transverse even though they workexactly with specific directions of the wave vector k

Because we used the periodic boundary condition thematrix Φ(k) has to be conserved in the reciprocal latticetranslations

Φ(k + K) = Φ(k) (189)

where K is an arbitrary vector in the reciprocal lattice Sowe can consider only such wave vectors k that lie in thefirst Brillouin zone The lattice vibrations are waves likethe electrons moving in the periodic potential caused bythe nuclei located at the lattice points Rl Particularlywe can use the same wave vectors k in the classification ofthe wave functions of both the lattice vibrations and theelectrons

The difference with electrons is created in that the firstBrillouin zone contains all possible vibrational modes ofthe Bravais lattice when the number of the energy bandsof electrons has no upper limit

Example One-Dimensional Lattice

Consider a one-dimensional lattice as an example andassume that only the adjacent atoms interact with eachother This can be produced formally by defining thespring constant

K = minusΦll+1

In this approximation Φll is determined by the transla-tional symmetry (184) of the whole crystal The otherterms in the matrix Φll

primeare zero Thus the classical equa-

tion of motion becomes

Mul = K(ul+1 minus 2ul + ulminus1) (190)

By inserting the plane wave solution

ul = eiklaminusiωt

one obtains the dispersion relation for the angular fre-quency ie its dependence on the wave vector

Mω2 = 2K(1minus cos ka) = 4K sin2(ka2)

rArr ω = 2

radicKM

∣∣∣ sin(ka2)∣∣∣ (191)

This simple example contains general characteristicsthat can be seen also in more complicated cases Whenthe wave vector k is small (k πa) the dispersion rela-tion is linear

ω = c|k|

where the speed of sound

c =partω

partkasymp a

radicKM k rarr 0 (192)

We discussed earlier that the energy of lattice has to be con-served when all nuclei are translated with the same vectorThis can be seen as the disappearance of the frequency atthe limit k = 0 because the large wave length vibrationslook like uniform translations in short length scales

As in discrete media in general we start to see deviationsfrom the linear dispersion when the wave length is of theorder of the interatomic distance In addition the (group)velocity has to vanish at the Brillouin zone boundary dueto condition (189)

Vibrations of Lattice with Basis

The treatment above holds only for Bravais latticesWhen the number of atoms in a primitive cell is increasedthe degrees of freedom in the lattice increases also Thesame happens to the number of the normal modes If oneassumes that there are M atoms in a primitive cell thenthere are 3M vibrational modes The low energy modesare linear with small wave vectors k and they are calledthe acoustic modes because they can be driven with soundwaves In acoustic vibrations the atoms in a primitive cellmove in same directions ie they are in the same phase ofvibration

In addition to acoustic modes there exists modes inwhich the atoms in a primitive cell move into opposite di-rections These are referred to as the optical modes be-cause they couple strongly with electromagnetic (infrared)radiation

The changes due to the basis can be formally includedinto the theory by adding the index n into the equation ofmotion in order to classify the nuclei at points determinedby the basis

Mnuln = minussumlprimenprime

Φlnlprimenprimeul

primenprime (193)

48

As in the case of the Bravais lattice the solution of theclassical equation of motion can be written in the form

uln = εneikmiddotRlnminusiωt (194)

We obtain

rArrMnω2εn =

sumnprime

Φnnprime(k)εn

prime (195)

Example One-Dimensional Lattice with Basis

Consider as an example the familiar one-dimensional lat-tice with a lattice constant a and with two alternatingatoms with masses M1 and M2

When we assume that each atom interacts only withits nearest neighbours we obtain similar to the one-dimensional Bravais lattice the equations of motion

M1ul1 = K

(ul2 minus 2ul1 + ulminus1

2

)M2u

l2 = K

(ul+1

1 minus 2ul2 + ul1

) (196)

By inserting the trial function (194) into the equations ofmotion we obtain

rArr minusω2M1ε1eikla = K

(ε2 minus 2ε1 + ε2e

minusika)eikla

minusω2M2ε2eikla = K

(ε1e

ika minus 2ε2 + ε1

)eikla(197)

This can be written in matrix form as

rArr(ω2M1 minus 2K K(1 + eminusika)K(1 + eika) ω2M2 minus 2K

)(ε1ε2

)= 0 (198)

The solutions of the matrix equation can be found againfrom the zeroes of the determinant of the coefficient matrixwhich are

rArr ω =radicK

radicM1 +M2 plusmn

radicM2

1 +M22 + 2M1M2 cos(ka)

M1M2

(199)

With large wave lengths (k πa) we obtain

ω(k) =

radicK

2(M1 +M2)ka

ε1 = 1 ε2 = 1 + ika2 (200)

ω(k) =

radic2K(M1 +M2)

M1M2

ε1 = M2 ε2 = minusM1(1 + ika2)

Clearly at the limit k rarr 0 in the acoustic mode the atomsvibrate in phase (ε1ε2 = 1) and the optical modes in theopposite phase (ε1ε2 = minusM2M1)

Example Graphene

Graphene has two carbon atoms in each primitive cellThus one can expect that it has six vibrational modes

(Source L A Falkovsky Journal of Experimentaland Theoretical Physics 105(2) 397 (2007)) The acous-tic branch of graphene consists of a longitudinal (LA) andtransverse (TA) modes and of a mode that is perpendic-ular to the graphene plane (ZA) Correspondingly the op-tical mode contains the longitudinal (LO) and transverse(TA) modes and the mode perpendicular to the latticeplane (ZO) The extrema of the energy dispersion relationhave been denoted in the figure with symbols Γ (k = 0)M (saddle point) and K (Dirac point)

49

42 Quantum Mechanical Treatment ofLattice Vibrations

So far we dealt with the lattice vibrations classicallyIn the harmonic approximation the normal modes can beseen as a group of independent harmonic oscillators Thevibrational frequencies ωkν are the same in the classicaland the quantum mechanical treatments The differencebetween the two is created in the vibrational amplitudewhich can have arbitrary values in according to the clas-sical mechanics Quantum mechanics allows only discretevalues of the amplitude which is seen as the quantizationof the energy (the energy of the oscillator depends on theamplitude as E prop |A|2)

~ωkν

(n+

1

2

) (201)

The excited states of a vibrational mode are calledphonons3 analogous to the excited states (photons) of theelectromagnetic field

Similar to photons an arbitrary number of phonons canoccupy a given k-state Thus the excitations of the latticevibrations can described as particles that obey the Bose-Einstein statistics It turns out that some of the propertiesof matter such as specific heat at low temperatures de-viate from the classically predicted values Therefore onehas to develop formally the theory of the lattice vibrations

Harmonic Oscillator

Simple harmonic oscillator is described by the Hamilto-nian operator

H =P 2

2M+

1

2Mω2R2 (202)

where R is the operator (hermitian Rdagger = R) that describesthe deviation from the equilibrium and P the (hermitian)operator of the corresponding canonical momentum Theyobey the commutation relations

[R R] = 0 = [P P ]

[R P ] = i~

In the quantum mechanical treatment of the harmonicoscillator one often defines the creation and annihilationoperators

adagger =

radicMω

2~Rminus i

radic1

2~MωP

a =

radicMω

2~R+ i

radic1

2~MωP (203)

Accordingly to their name the creationannihilation op-erator addsremoves one quantum when it operates on anenergy eigenstate The original operators for position andmomentum can be written in form

R =

radic~

2Mω(adagger + a) (204)

P = i

radic~Mω

2(adagger minus a) (205)

3Greek φωνη (phone)

Especially the Hamiltonian operator of the harmonic os-cillator simplifies to

H = ~ω(adaggera+

1

2

)= ~ω

(n+

1

2

) (206)

The number operator n = adaggera describes the number ofquanta in the oscillator

Phonons

We return to the lattice vibrations in a solid As previ-ously we assume that the crystal is formed by N atomsIn the second order the Hamiltonian operator describingthe motion of the atoms can be written as

H =

Nsuml=1

P 2l

2M+

1

2

sumllprime

ulΦllprimeulprime+ (207)

The operator ul describes the deviation of the nucleus lfrom the equilibrium Rl One can think that each atom inthe lattice behaves like a simple harmonic oscillator andthus the vibration of the whole lattice appears as a col-lective excitation of these oscillators This can be madeexplicit by defining the annihilation and creation operatorof the lattice vibrations analogous to Equations (203)

akν =1radicN

Nsuml=1

eminusikmiddotRl

εlowastkν

[radicMωkν

2~ul + i

radic1

2~MωkνP l]

adaggerkν =1radicN

Nsuml=1

eikmiddotRl

εkν

[radicMωkν

2~ul minus i

radic1

2~MωkνP l]

We see that the operator adaggerkν is a sum of terms that excitelocally the atom l It creates a collective excited state thatis called the phonon

Commutation relation [ul P l] = i~ leads in

[akν adaggerkν ] = 1 (208)

In addition the Hamiltonian operator (207) can be writtenin the simple form (the intermediate steps are left for theinterested reader cf MM)

H =sumkν

~ωkν

(adaggerkν akν +

1

2

) (209)

If the lattice has a basis one has to add to the above sum anindex describing the branches On often uses a shortenednotation

H =sumi

~ωi(ni +

1

2

) (210)

where the summation runs through all wave vectors k po-larizations ν and branches

Specific Heat of Phonons

We apply the quantum theory of the lattice vibrationsto the calculation of the specific heat of the solids We cal-culated previously that the electronic contribution to the

50

specific heat (cf Equation (65)) depends linearly on thetemperature T In the beginning of the 20th century theproblem was that one had only the prediction of the classi-cal statistical physics at disposal stating that the specificheat should be kB2 for each degree of freedom Becauseeach atom in the crystal has three degrees of freedom forboth the kinetic energy and the potential energy the spe-cific heat should be CV = 3NkB ie a constant indepen-dent on temperature This Dulong-Petit law was clearly incontradiction with the experimental results The measuredvalues of the specific heats at low temperatures were de-pendent on temperature and smaller than those predictedby the classical physics

Einstein presented the first quantum mechanical modelfor the lattice vibrations He assumed that the crystal hasN harmonic oscillators with the same natural frequency ω0In addition the occupation probability of each oscillatorobeys the distribution

n =1

eβ~ω0 minus 1

which was at that time an empirical result With these as-sumptions the average energy of the lattice at temperatureT is

E =3N~ω0

eβ~ω0 minus 1 (211)

The specific heat tells how much the average energychanges when temperature changes

CV =partEpartT

∣∣∣∣∣V

=3N(~ω0)2eβ~ω0(kBT

2)

(eβ~ω0 minus 1)2 (212)

The specific heat derived by Einstein was a major im-provement to the Dulong-Petit law However at low tem-peratures it is also in contradiction with the experimentsgiving too small values for the specific heat

In reality the lattice vibrates with different frequenciesand we can write in the spirit of the Hamiltonian operator(210) the average energy in the form

E =sumi

~ωi(ni +

1

2

) (213)

where the occupation number of the mode i is obtainedfrom the Bose-Einstein distribution

ni =1

eβ~ωi minus 1 (214)

One should note that unlike electrons the phonons arebosons and thus do not obey the Pauli principle There-fore their average occupation number can be any non-negative integer Accordingly the denominator in theBose-Einstein distribution has a negative sign in front ofone

The specific heat becomes

CV =sumi

~ωipartnipartT

(215)

Similarly as one could determine the thermal properties ofthe electron gas from the electron density of states one canobtain information on the thermal properties of the latticefrom the density of states of the phonons The density ofstates is defined as

D(ω) =1

V

sumi

δ(ω minus ωi) =1

V

sumkν

δ(ω minus ωkν)

=

intdk

(2π)3

sumν

δ(ω minus ωkν) (216)

For example one can us the density of states and write thespecific heat (215) in form

CV = V

int infin0

dωD(ω)part

partT

~ωeβ~ω minus 1

(217)

It is illustrative to consider the specific heat in two limitingcases

High Temperature Specific Heat

When the temperature is large (kBT ~ωmax) we ob-tain the classical Dulong-Petit law

CV = 3NkB

which was a consequence of the equipartition theorem ofstatistical physics In the point of view of the classicalphysics it is impossible to understand results deviatingfrom this law

Low Temperature Specific Heat

At low temperatures the quantum properties of mattercan create large deviations to the classical specific heatWhen ~ωi kBT the occupation of the mode i is verysmall and its contribution to the specific heat is minorThus it is enough to consider the modes with frequenciesof the order

νi kBT

hasymp 02 THz

when one assumes T = 10 K The speed of sound in solidsis c = 1000ms or more Therefore we obtain an estimatefor the lower bound of the wave length of the vibration

λmin = cν ge 50 A

This is considerably larger than the interatomic distancein a lattice (which is of the order of an Angstrom) Thuswe can use the long wave length dispersion relation

ωkν = cν(k)k (218)

where we assume that the speed of sound can depend onthe direction of propagation

We obtain the density of states

D(ω) =

intdk

(2π)3

sumν

δ(ω minus cν(k)k)

=3ω2

2π2c3 (219)

51

where the part that is independent on the frequency isdenoted with (integration is over the directions)

1

c3=

1

3

sumν

intdΣ

1

cν(k) (220)

Thus the specific heat can be written in form

CV = Vpart

partT

3(kBT )4

2π2(c~)3

int infin0

dxx3

ex minus 1= V

2π2k4B

5~3c3T 3 (221)

where x = β~ω

There is large region between the absolute zero and theroom temperature where low and high temperature ap-proximations do not hold At these temperatures one hasto use the general form (217) of the specific heat Insteadof a rigorous calculation it is quite common to use inter-polation between the two limits

Einstein model was presented already in Equation (211)and it can be reproduced by approximating the density ofstates with the function

D(ω) =3N

Vδ(ω minus ω0) (222)

Debye model approximates the density of states with thefunction

D(ω) =3ω2

2π2c3θ(ωD minus ω) (223)

where the density of states obtains the low temperaturevalue but is then cut in order to obtain a right number ofmodes The cut is done at the Debye frequency ωD whichis chosen so that the number of modes is the same as thatof the vibrating degrees of freedom

3N = V

int infin0

dωD(ω) rArr n =ω3D

6π2c3 (224)

where n is the number density of the nuclei

With the help of the Debye frequency one can define theDebye temperature ΘD

kBΘD = ~ωD (225)

where all phonons are thermally active The specific heatbecomes in this approximation

CV = 9NkB

(T

ΘD

)3 int ΘDT

0

dxx4ex

(ex minus 1)2(226)

Debye temperatures are generally half of the meltingtemperature meaning that at the classical limit the har-monic approximation becomes inadequate

Specific Heat Lattice vs Electrons

Because the lattice part of the specific heat decreases asT 3 in metals it should at some temperature go below linearcomponent due to electrons However this occurs at verylow temperatures because of the difference in the orderof magnitude between the Fermi and Debye temperaturesThe specific heat due to electrons goes down like TTF (cfEquation (66)) and the Fermi temperatures TF ge 10000K The phonon part behaves as (TΘD)3 where ΘD asymp100 500 K Thus the temperature where the specificheats of the electrons and phonons are roughly equal is

T = ΘD

radicΘDTF asymp 10 K

It should also be remembered that insulators do nothave the electronic component in specific heat at lowtemperatures because their valence bands are completelyfilled Thus the electrons cannot absorb heat and the spe-cific heat as only the cubic component CV prop T 3 due to thelattice

43 Inelastic Neutron Scattering fromPhonons

The shape of the dispersion relations ων(k) of the nor-mal modes can be resolved by studying energy exchangebetween external particles and phonons The most infor-mation can be acquired by using neutrons because theyare charge neutral and therefore interact with the elec-trons in the matter only via the weak coupling betweenmagnetic moments The principles presented below holdfor large parts also for photon scattering

The idea is to study the absorbedemitted energy of theneutron when it interacts with the lattice (inelastic scat-tering) Due to the weak electron coupling this is causedby the emissionabsorption of lattice phonons By study-ing the directions and energies of the scattered neutronswe obtain direct information on the phonon spectrum

Consider a neutron with a momentum ~k and an energy~2k22mn Assume that after the scattering the neutronmomentum is ~kprime When it propagates through the latticethe neutron can emitabsorb energy only in quanta ~ωqν where ωqν is one of the normal mode frequencies of thelattice Thus due to the conservation of energy we obtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωqν (227)

In other words the neutron travelling through the lat-tice can create(+)annihilate(-) a phonon which is ab-sorbedemitted by the lattice

For each symmetry in the Hamiltonian operator thereexists a corresponding conservation law (Noether theorem)For example the empty space is symmetric in translations

52

This property has an alternative manifestation in the con-servation of momentum The Bravais lattice is symmetriconly in translations by a lattice vector Thus one can ex-pect that there exists a conservation law similar to that ofmomentum but a little more constricted Such was seenin the case of elastic scattering where the condition forstrong scattering was

kminus kprime = K

This is the conservation law of the crystal momentumIn elastic scattering the crystal momentum is conservedprovided that it is translated into the first Brillouin zonewith the appropriate reciprocal lattice vector K We willassume that the conservation of crystal momentum holdsalso in inelastic scattering ie

k = kprime plusmn q + K rArr kminus kprime = plusmnq + K (228)

Here ~q is the crystal momentum of a phonon which isnot however related to any true momentum in the latticeIt is just ~ times the wave vector of the phonon

By combining conservations laws (227) and (228) weobtain

~2k2

2mn=

~2(kprime)2

2mnplusmn ~ωkminuskprimeν (229)

By measuring k and kprime one can find out the dispersionrelations ~ων(q)

The previous treatment was based on the conservationlaws If one wishes to include also the thermal and quan-tum fluctuations into the theory one should study the ma-trix elements of the interaction potential Using those onecan derive the probabilities for emission and absorption byusing Fermirsquos golden rule (cf textbooks of quantum me-chanics) This kind of treatment is not done in this course

44 Mossbauer EffectDue to the lattice one can observe phenomena that are

not possible for free atomic nuclei One example is theMossbauer effect It turns out that the photons cannotcreate excited states for free nuclei The reason is thatwhen the nucleus excites part of the energy ck of the photonis transformed into the recoil energy of the nucleus

~ck = ∆E +~2k2

2m

due to the conservation of momentum In the above ∆Eis the energy difference of two energy levels in the nucleusIf the life time of the excited state is τ the uncertaintyprinciple of energy and time gives an estimate to the linewidth W

W asymp ~τ

In nuclear transitions

~2k2

2mgt W

and the free nuclei cannot absorb photons

Mossbauer observed in 1958 that the crystal lattice canabsorb the momentum of the photon leaving (almost) allof the energy into the excitation of the nucleus This aconsequence of the large mass of the crystal

Erecoil =~2k2

2Mcrystalasymp 0

With the spectroscopy based on the Mossbauer effect onecan study the effects of the surroundings on the atomicnuclei

45 Anharmonic CorrectionsIn our model of lattice vibrations we assume that the

deviations from the equilibrium positions of the nuclei aresmall and that we can use the harmonic approximationto describe the phenomena due to vibrations It turns outthat the including of the anharmonic terms is essential inthe explanations of several phenomena We list couple ofthose

bull Thermal expansion cannot be explained with a theorythat has only second order corrections to the cohesiveenergy This is because the size of the harmonic crystalin equilibrium does not depend on temperature

bull Heat conductivity requires also the anharmonic termsbecause that of a completely harmonic (insulating)crystal is infinite

bull The specific heat of a crystal does not stop at theDulong-Petit limit as the temperature grows but ex-ceeds it This is an anharmonic effect

In this course we do not study in more detail the anhar-monic lattice vibrations

53

5 Electronic Transport PhenomenaMM Chapters 16-1621 1623 1631 (Rough

calculation) 164 (no detailed calculations) 17-1722 174-1746 1748 1751

One of the most remarkable achievements of the con-densed matter theory is the explanation of the electric con-ductivity of matter All dynamical phenomena in mattercan be explained with the interaction between the electronsand the lattice potential The central idea is that the en-ergy bands Enk determine the response of the matter tothe external electromagnetic field The first derivatives ofthe energy bands give the velocities of electrons and thesecond derivatives describe the effective masses A largepart of the modern electronics is laid on this foundation

Drude Model

In 1900 soon after the discovery of electron4 P Drudepresented the first theory of electric and heat conductivityof metals It assumes that when the atoms condense intosolid their valence electrons form a gas that is responsiblefor the conduction of electricity and heat In the followingthese are called the conduction electrons or just electronswhen difference with the passive electrons near the nucleidoes not have to be highlighted

The Drude model has been borrowed from the kinetictheory of gases in which the electrons are dealt with asidentical solid spheres that move along straight trajecto-ries until they hit each other the nuclei or the boundariesof the matter These collisions are described with the re-laxation time τ which tells the average time between colli-sions The largest factor that influences the relaxation timecomes from the collisions between electrons and nuclei Inbetween the collisions the electrons propagate freely ac-cording to Newtonrsquos laws disregarding the surroundingelectrons and nuclei This is called the independent andfree electron approximation Later we will see that that inorder to obtain the true picture of the conduction in met-als one has to treat especially the periodic potential dueto nuclei more precisely

In the presence of external electromagnetic field (E B)the Drude electrons obey the equation of motion

mv = minuseEminus ev

ctimesBminusmv

τ (230)

When the external field vanishes we see that the move-ment of the electrons ceases This can be seen by assumingan initial velocity v0 for the electrons and by solving theequation of motion The result

v(t) = v0eminustτ (231)

shows that the relaxation time describes the decay of anyfluctuation

Assume then that electrons are in a static electric fieldE Thus an electron with initial velocity v0 propagates as

v(t) = minusτem

E +[v0 +

τe

mE]eminustτ

4J J Thomson in 1897

When the electric field has been present for a long time(compared with the relaxation time) we obtain

v = minusτem

E

Based on the above we obtain the current density cre-ated by the presence of the electric field

j = minusnev =nτe2

mE = σE (232)

where n is the density of the conduction electrons Theabove defined electric conductivity σ is the constant ofproportionality between the electric current and field

The electric conductivity is presented of in terms of itsreciprocal the resistivity ρ = σminus1 By using the typi-cal values for the resistivity and the density of conductionelectrons we obtain an estimate for the relaxation time

τ =m

ne2ρasymp 10minus14 s (233)

Heat Conductivity

The Drude model gets more content when it is appliedto another transport phenomenon which can be used todetermine the unknown τ Consider the heat conductiv-ity κ which gives the energy current jE as a function oftemperature gradient

jE = minusκnablaT (234)

Let us look at the electrons coming to the point x Elec-trons travelling along positive x-axis with the velocity vxhave travelled the average distance of vxτ after the previousscattering event Their energy is E(x minus vxτ) whence theenergy of the electrons travelling into the direction of thenegative x-axis is E(x+ vxτ) The density of electrons ar-riving from both directions is approximately n2 becausehalf of the electrons are travelling along the positive andhalf along the negative x-axis We obtain an estimate forthe energy flux

jE =n

2vx[E(xminus vxτ)minus E(x+ vxτ)

]asymp minusnv2

xτpartEpartx

= minusnv2xτpartEpartT

partT

partx= minus2n

m

1

2mv2

xcV τpartT

partx

= minusτnm

3k2BT

2

partT

partx (235)

We have used in the above the classical estimates for thespecific heat cV = partEpartT = 3kB2 and for the kineticenergy mv2

x2 = kBT2

We obtain

rArr κ

σT=

3

2

(kBe

)2

= 124 middot 10minus20 J

cmK2 (236)

Clearly we see that the above discussion is not plausible(electrons have the same average velocity vx but different

54

amount of kinetic energy) The obtained result is howevermore or less correct It says that the heat conductivity di-vided by the electric conductivity and temperature is aconstant for metals (Wiedemann-Franz law) Experimen-tally measured value is around 23 middot 10minus20 Jcm K2

The improvement on the Drude model requires a littlebit of work and we do it in small steps

51 Dynamics of Bloch ElectronsThe first step in improving the Drude model is to relax

the free electron assumption Instead we take into consid-eration the periodic potential due to the lattice In manycases it is enough to handle the electrons as classical par-ticles with a slightly unfamiliar equations of motion Wewill first list these laws and then try to justify each of them

Semiclassical Electron Dynamics

bull Band index n is a constant of motion

bull Position of an electron in a crystal obeys the equationof motion

r =1

~partEkpartk

(237)

bull Wave vector of an electron obeys the equation of mo-tion

~k = minuseEminus e

crtimesB (238)

We have used the notation partpartk = nablak =(partpartkx partpartky partpartkz) Because the energies and wavefunctions are periodic in the k-space the wave vectorsk and k + K are physically equivalent

Bloch Oscillations

In spite of the classical nature of the equations of motion(237) and (238) they contain many quantum mechanicaleffects This is a consequence of the fact that Enk is peri-odic function of the wave vector k and that the electronstates obey the Fermi-Dirac distribution instead of theclassical one As an example let us study the semiclassicaldynamics of electrons in the tight binding model which inone dimension gave the energy bands (114)

Ek = minus2t cos ak (239)

In a constant electric field E we obtain

~k = minuseE (240)

rArr k = minuseEt~ (241)

rArr r = minus2ta

~sin(aeEt

~

)(242)

rArr r =2t

eEcos(aeEt

~

) (243)

We see that the position of the electron oscillates in timeThese are called the Bloch oscillations Even though the

wave vector k grows without a limit the average locationof the electron is a constant Clearly this is not a com-mon phenomenon in metals because otherwise the elec-trons would oscillate and metals would be insulators Itturns out that the impurities destroy the Bloch oscilla-tions and therefore they are not seen except in some spe-cial materials

Justification of Semiclassical Equations

The rigorous justification of the semiclassical equationsof motion is extremely hard In the following the goalis to make the theory plausible and to give the guidelinesfor the detailed derivation The interested reader can referto the books of Marder and Ashcroft amp Mermin and thereferences therein

Let us first compare the semiclassical equations with theDrude model In the Drude model the scattering of theelectrons was mainly from the nuclei In the semiclassicalmodel the nuclei have been taken into account explicitlyin the energy bands Ek As a consequence the electron inthe band n has the velocity given by Equation (237) andin the absence of the electromagnetic field it maintains itforever This is a completely different result to the one ob-tained from the Drude model (231) according to which thevelocity should decay in the relaxation time τ on averageThis can be taken as a manifestation of the wave nature ofthe electrons In the periodic lattice the electron wave canpropagate without decay due to the coherent constructiveinterference of the scattered waves The finite conductiv-ity of metals can be interpreted to be caused by the lat-tice imperfections such as impurities missing nuclei andphonons

Analogy with Free Electron Equations

The first justification is made via an analogy Considerfirst free electrons They have the familiar dispersion rela-tion

E =~2k2

2m

The average energy of a free electron in the state definedby the wave vector k can thus be written in the form

v =~k

m=

1

~partEpartk

(244)

We see that the semiclassical equation of motion of theelectron (237) is a straightforward generalization of that ofthe free electron

Also the equation determining the evolution of the wavevector has exactly the same form for both the free andBloch electrons

~k = minuseEminus e

crtimesB

It is important to remember that in the free electron case~k is the momentum of the electron whereas for Blochelectrons it is just the crystal momentum (cf Equation(80)) Particularly the rate of change (238) of the crys-tal momentum does not depend on the forces due to the

55

lattice so it cannot describe the total momentum of anelectron

Average Velocity of Electron

The average velocity (237) of an electron in the energyband n is obtained by a perturbative calculation Assumethat we have solved Equation (81)

H0kunk(r) =

~2

2m

(minusinabla+k

)2

unk(r)+Uunk(r) = Enkunk(r)

with some value of the wave vector k Let us then increasethe wave vector to the value k + δk meaning that theequation to solve is

~2

2m

(minus inabla+ k + δk

)2

u(r) + Uu(r)

=~2

2m

[(minus inabla+ k

)2

+ δk middot(δkminus 2inabla+ 2k

)]u(r)

+ Uu(r)

= (H0k + H1

k)u(r) = Eu(r) (245)

where

H1k =

~2

2mδk middot

(δkminus 2inabla+ 2k

)(246)

is considered as a small perturbation for the HamiltonianH0

k

According to the perturbation theory the energy eigen-values change like

Enk+δk = Enk + E(1)nk + E(2)

nk + (247)

The first order correction is of the form

E(1)nk =

~2

m

langunk

∣∣δk middot (minus inabla+ k)∣∣unkrang (248)

=~2

m

langψnk

∣∣eikmiddotrδk middot (minus inabla+ k)eminusikmiddotr

∣∣ψnkrang=

~m

langψnk

∣∣δk middot P ∣∣ψnkrang (249)

where the momentum operator P = minusi~nabla We have usedthe definition of the Bloch wave function

unk = eminusikmiddotrψnk

and the result(minus inabla+ k

)eminusikmiddotr

∣∣ψnkrang = minusieminusikmiddotrnabla∣∣ψnkrang

By comparing the first order correction with the Taylorexpansion of the energy Enk+δk we obtain

partEnkpart~k

=1

m

langψnk

∣∣P ∣∣ψnkrang (250)

= 〈v〉 = vnk (251)

where the velocity operator v = P m Thus we haveshown that in the state ψnk the average velocity of an

electron is obtained from the first derivative of the energyband

Effective Mass

By calculating the second order correction we arrive atthe result

1

~2

part2Enkpartkαpartkβ

= (252)

1

mδαβ +

1

m2

sumnprime 6=n

〈ψnk|Pα|ψnprimek〉〈ψnprimek|Pβ |ψnk〉+ cc

Enk minus Enkprime

where cc denotes the addition of the complex conjugateof the previous term The sum is calculated only over theindex nprime

We get an interpretation for the second order correctionby studying the acceleration of the electron

d

dt〈vα〉 =

sumβ

part〈vα〉partkβ

partkβpartt

(253)

where vα is the α-component of the velocity operator (inthe Cartesian coordinates α = x y or z) This accelerationcan be caused by eg the electric or magnetic field and asa consequence the electrons move along the energy bandIn order the perturbative treatment to hold the field haveto be weak and thus k changes slowly enough

By combining the components of the velocity into a sin-gle matrix equation we have that

d

dt〈v〉 = ~Mminus1k (254)

where we have defined the effective mass tensor

(Mminus1)αβ =1

~2

part2Enkpartkαpartkβ

(255)

Generally the energy bands Enk are not isotropic whichmeans that the acceleration is not perpendicular with thewave vector k An interesting result is also that the eigen-values of the mass tensor can be negative For example inthe one-dimensional tight binding model

Ek = minus2t cos ak

and the second derivative at k = πa is negative

According to Equation (238) the wave vector k growsin the opposite direction to the electric field When theeffective mass is negative the electrons accelerate into theopposite direction to the wave vector In other words theelectrons accelerate along the electric field (not to the op-posite direction as usual) Therefore it is more practicalto think that instead of negative mass the electrons havepositive charge and call them holes

Equation of Wave Vector

56

The detailed derivation of the equation of motion of thewave vector (238) is an extremely difficult task How-ever we justify it by considering an electron in a time-independent electric field E = minusnablaφ (φ is the scalar poten-tial) Then we can assume that the energy of the electronis

En(k(t))minus eφ(r(t)) (256)

This energy changes with the rate

partEnpartkmiddot kminus enablaφ middot r = vnk middot (~kminus enablaφ) (257)

Because the electric field is a constant we can assume thatthe total energy of the electron is conserved when it movesinside the field Thus the above rate should vanish Thisoccurs at least when

~k = minuseEminus e

cvnk timesB (258)

This is the equation of motion (238) The latter term canalways be added because it is perpendicular with the ve-locity vnk What is left is to prove that this is in fact theonly possible extra term and that the equation works alsoin time-dependent fields These are left for the interestedreader

Adiabatic Theorem and Zener Tunnelling

So far we have presented justifications for Equations(237) and (238) In addition to those the semiclassicaltheory of electrons assumes that the band index n is a con-stant of motion This holds only approximatively becausedue to a strong electric field the electrons can tunnel be-tween bands This is called the Zener tunnelling

Let us consider this possibility by assuming that the mat-ter is in a constant electric field E Based on the courseof electromagnetism we know that the electric field can bewritten in terms of the scalar and vector potentials

E = minus1

c

partA

parttminusnablaφ

One can show that the potentials can be chosen so thatthe scalar potential vanishes This is a consequence of thegauge invariance of the theory of electromagnetism (cf thecourse of Classical Field Theory) When the electric fieldis a constant this means that the vector potential is of theform

A = minuscEt

From now on we will consider only the one-dimensionalcase for simplicity

According to the course of Analytic Mechanics a par-ticle in the vector potential A can be described by theHamiltonian operator

H =1

2m

(P +

e

cA)2

+ U(R) =1

2m

(P minus eEt

)2

+ U(R)

(259)The present choice of the gauge leads to an explicit timedependence of the Hamiltonian

Assume that the energy delivered into the system by theelectric field in the unit time is small compared with allother energies describing the system Then the adiabatictheorem holds

System whose Hamiltonian operator changes slowly intime stays in the instantaneous eigenstate of the timeindependent Hamiltonian

The adiabatic theory implies immediately that the bandindex is a constant of motion in small electric fields

What is left to estimate is what we mean by small electricfield When the Hamiltonian depends on time one gener-ally has to solve the time-dependent Schrodinger equation

H(t)|Ψ(t)〉 = i~part

partt|Ψ(t)〉 (260)

In the case of the periodic potential we can restrict to thevicinity of the degeneracy point where the energy gap Eghas its smallest value One can show that in this regionthe tunnelling probability from the adiabatic state Enk isthe largest and that the electron makes transitions almostentirely to the state nearest in energy

In the nearly free electron approximation the instancewhere we can truncate to just two free electron states wasdescribed with the matrix (97)

H(t) =

(E0k(t) Eg2Eg2 E0

kminusK(t)

) (261)

where E0k(t) are the energy bands of the free (uncoupled)

electrons with the time-dependence of the wave vector ofthe form

k = minuseEt~

The Hamiltonian operator has been presented in the freeelectron basis |k〉 |k minus K〉 The solution of the time-dependent Schrodinger equation in this basis is of the form

|Ψ(t)〉 = Ck(t)|k〉+ CkminusK(t)|k minusK〉 (262)

C Zener showed in 19325 that if the electron is initiallyin the state |k〉 it has the finite asymptotic probabilityPZ = |Ck(infin)|2 to stay in the same free electron stateeven though it passes through the degeneracy point

5In fact essentially the same result was obtained independentlyalso in the same year by Landau Majorana and Stuckelberg

57

In other words the electron tunnels from the adiabaticstate to another with the probability PZ The derivationof the exact form of this probability requires so much workthat it is not done in this course Essentially it depends onthe shape of the energy bands of the uncoupled electronsand on the ratio between the energy gap and the electricfield strength Marder obtains the result

PZ = exp(minus 4E32

g

3eE

radic2mlowast

~2

) (263)

where mlowast is the reduced mass of the electron in the lattice(cf the mass tensor) For metals the typical value forthe energy gap is 1 eV and the maximum electric fieldsare of the order E sim 104 Vcm The reduced masses formetals are of the order of that of an electron leading tothe estimate

PZ asymp eminus103

asymp 0

For semiconductors the Zener tunnelling is a generalproperty because the electric fields can reach E sim 106

Vcm reduced masses can be one tenth of the mass of anelectron and the energy gaps below 1 eV

Restrictions of Semiclassical Equations

We list the assumptions that reduce the essentially quan-tum mechanical problem of the electron motion into a semi-classical one

bull Electromagnetic field (E and B) changes slowly bothin time and space Especially if the field is periodicwith a frequency ω and the energy splitting of twobands is equal to ~ω the field can move an electronfrom one band to the other by emitting a photon (cfthe excitation of an atom)

bull Magnitude E of the electric field has to be small inorder to adiabatic approximation to hold In otherwords the probability of Zener tunnelling has to besmall ie

eE

kF Eg

radicEgEF

(264)

bull Magnitude B of the magnetic field also has to be smallSimilar to the electric field we obtain the condition

~ωc EgradicEgEF

(265)

where ωc = eBmc is the cyclotron frequency

As the final statement the semiclassical equations can bederived neatly in the Hamilton formalism if one assumesthat the Hamilton function is of the form

H = En(p + eAc)minus eφ(r) (266)

where p is the crystal momentum of the electron Thisdeviates from the classical Hamiltonian by the fact thatthe functions En are obtained from a quantum mechanical

calculation and that they include the nuclear part of thepotential As an exercise one can show that Equations(237) and (238) follow from Hamiltonrsquos equations of motion

r =partHpartp

p = minuspartHpartr

(267)

52 Boltzmann Equation and Fermi Liq-uid Theory

The semiclassical equations describe the movement of asingle electron in the potential due to the periodic latticeand the electromagnetic field The purpose is to develop atheory of conduction so it is essential to form such equa-tions that describe the movement of a large group of elec-trons We will present two phenomenological models thatenable the description of macroscopic experiments with afew well-chosen parameters without the solution of theSchrodinger equation

Boltzmann Equation

The Boltzmann equation is a semiclassical model of con-ductivity It describes any group of particles whose con-stituents obey the semiclassical Hamiltonrsquos equation of mo-tion (267) and whose interaction with the electromagneticfield is described with Hamiltonrsquos function (266) Thegroup of particles is considered as ideal non-interactingelectron Fermi gas taking into account the Fermi-Diracstatistics and the influences due to the temperature Theexternal fields enable the flow in the gas which is never-theless considered to be near the thermal equilibrium

Continuity Equation

We start from the continuity equation Assume thatg(x) is the density of the particles at point x and thatthe particles move with the velocity v(x) We study theparticle current through the surface A perpendicular tothe direction of propagation x The difference between theparticles flowing in and out of the volume Adx in a timeunit dt is

dN = Adxdg = Av(x)g(x)dtminusAv(x+ dx)g(x+ dx)dt(268)

We obtain that the particle density at the point x changeswith the rate

partg

partt= minus part

partx

(v(x)g(x t)

) (269)

In higher dimensions the continuity equation generalizesinto

partg

partt= minus

suml

part

partxl

(vl(r)g(r t)

)= minus part

partrmiddot(rg(r t)

) (270)

where r = (x1 x2 x3)

The continuity equation holds for any group of parti-cles that can be described with the average velocity v that

58

depends on the coordinate r We will consider particu-larly the occupation probability grk(t) of electrons thathas values from the interval [0 1] It gives the probabilityof occupation of the state defined by the wave vector kat point r at time t Generally the rk-space is called thephase space In other words the number of electrons in thevolume dr and in the reciprocal space volume dk is

grk(t)drDkdk =2

(2π)3dkdrgrk(t) (271)

By using the electron density g one can calculate the ex-pectation value of an arbitrary function Grk by calculating

G =

intdkdrDkgrkGrk (272)

Derivation of Boltzmann Equation

The electron density can change also in the k-space inaddition to the position Thus the continuity equation(270) is generalized into the form

partg

partt= minus part

partrmiddot(rg)minus part

partkmiddot(kg)

= minusr middot partpartrg minus k middot part

partkg (273)

where the latter equality is a consequence of the semiclas-sical equations of motion

Boltzmann added to the equation a collision term

dg

dt

∣∣∣coll (274)

that describes the sudden changes in momentum causedby eg the impurities or thermal fluctuations This waythe equation of motion of the non-equilibrium distributiong of the electrons is

partg

partt= minusr middot part

partrg(r t)minus k middot part

partkg(r t) +

dg

dt

∣∣∣coll (275)

which is called the Boltzmann equation

Relaxation Time Approximation

The collision term (274) relaxes the electrons towardsthe thermal equilibrium Based on the previous considera-tions we know that in the equilibrium the electrons obey(locally) the Fermi-Dirac distribution

frk =1

eβr(Ekminusmicror) + 1 (276)

The simplest collision term that contains the above men-tioned properties is called the relaxation time approxima-tion

dg

dt

∣∣∣coll

= minus1

τ

[grk minus frk

] (277)

The relaxation time τ describes the decay of the electronmotion towards the equilibrium (cf the Drude model)Generally τ can depend on the electron energy and the di-rection of propagation For simplicity we assume in the fol-lowing that τ depends on the wave vector k only through

the energy Ek Thus the relaxation time τE can be consid-ered as a function of only energy

Because the distribution g is a function of time t positionr and wave vector k its total time derivative can be writtenin the form

dg

dt=

partg

partt+ r middot partg

partr+ k middot partg

partk

= minusg minus fτE

(278)

where one has used Equations (275) and (277)

The solution can be written as

grk(t) =

int t

minusinfin

dtprime

τEf(tprime)eminus(tminustprime)τE (279)

which can be seen by inserting it back to Equation (278)We have used the shortened notation

f(tprime) = f(r(tprime)k(tprime)

)

where r(tprime) and k(tprime) are such solutions of the semiclassicalequations of motion that at time tprime = t go through thepoints r and k

The above can be interpreted in the following way Con-sider an arbitrary time tprime and assume that the electronsobey the equilibrium distribution f(tprime) The quantitydtprimeτE can be interpreted as the probability of electronscattering in the time interval dtprime around time tprime Onecan show (cf Ashcroft amp Mermin) that in the relaxationtime approximation

dtprime

τEf(tprime)

is the number of such electrons that when propagatedwithout scattering in the time interval [tprime t] reach the phasespace point (rk) at t The probability for the propagationwithout scattering is exp(minus(tminus tprime)τE) Thus the occupa-tion number of the electrons at the phase space point (rk)at time t can be interpreted as a sum of all such electronhistories that end up to the given point in the phase spaceat the given time

By using partial integration one obtains

grk(t)

=

int t

minusinfindtprimef(tprime)

d

dtprimeeminus(tminustprime)τE

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE df(tprime)

dtprime(280)

= frk minusint t

minusinfindtprimeeminus(tminustprime)τE

(rtprime middot

part

partr+ ktprime middot

part

partk

)f(tprime)

where the latter term describes the deviations from thelocal equilibrium

The partial derivatives of the equilibrium distributioncan be written in the form

partf

partr=

partf

partE

[minusnablamicrominus (E minus micro)

nablaTT

](281)

partf

partk=

partf

partE~v (282)

59

By using the semiclassical equations of motion we arriveat

grk(t) = frk minusint t

minusinfindtprimeeminus(tminustprime)τE (283)

timesvk middot(eE +nablamicro+

Ek minus microTnablaT)partf(tprime)

dmicro

When micro T and E change slowly compared with the relax-ation time τE we obtain

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

(284)

When one studies the responses of solids to the electricmagnetic and thermal fields it is often enough to use theBoltzmann equation in form (284)

Semiclassical Theory of Conduction

The Boltzmann equation (284) gives the electron dis-tribution grk in the band n Consider the conduction ofelectricity and heat with a fixed value of the band indexIf more than one band crosses the Fermi surface the effectson the conductivity due to several bands have to be takeninto account by adding them together

Electric Current

We will consider first a solid in an uniform electric fieldThe electric current density j is defined as the expectationvalue of the current function minusevk in k-space

j = minuseintdkDkvkgrk (285)

where the integration is over the first Brillouin zone Inthe Drude model the conductivity was defined as

σ =partj

partE

Analogously we define the general conductivity tensor

σαβ equivpartjαpartEβ

= e2

intdkDkτEvαvβ

partf

partmicro (286)

where we have used the Boltzmann equation (284) Thusthe current density created by the electric field is obtainedfrom the equation

j = σE (287)

In order to understand the conductivity tensor we canchoose from the two approaches

1) Assume that the relaxation time τE is a constantThus one can write

σαβ = minuse2τ

intdkDkvα

partf

part~kβ (288)

Because v and f are periodic in k-space we obtainwith partial integration and by using definition (255)

of the mass tensor that

σαβ = e2τ

intdkDkfk

partvαpart~kβ

= e2τ

intdkDkfk(Mminus1)αβ (289)

When the lattice has a cubic symmetry the conductiv-ity tensor is diagonal and we obtain a result analogousto that from the Drude model

σ =ne2τ

mlowast (290)

where1

mlowast=

1

3n

intdkDkfkTr(Mminus1) (291)

2) On the other hand we know that for metals

partf

partmicroasymp δ(E minus EF )

(the derivative deviates essentially from zero inside thedistance kBT from the Fermi surface cf Sommerfeldexpansion) Thus by using result (101)

σαβ = e2

intdΣ

4π3~vτEvαvβ (292)

where the integration is done over the Fermi surfaceand ~v = |partEkpartk| Thus we see that only the elec-trons at the Fermi surface contribute to the conductiv-ity of metals (the above approximation for the deriva-tive of the Fermi function cannot be made in semicon-ductors)

Filled Bands Do Not Conduct

Let us consider more closely the energy bands whosehighest energy is lower than the Fermi energy When thetemperature is much smaller than the Fermi temperaturewe can set

f = 1 rArr partf

partmicro= 0

Thusσαβ = 0 (293)

Therefore filled bands do not conduct current The rea-son is that in order to carry current the velocities of theelectrons would have to change due to the electric field Asa consequence the index k should grow but because theband is completely occupied there are no empty k-statesfor the electron to move The small probability of the Zenertunnelling and the Pauli principle prevent the changes inthe states of electrons in filled bands

Effective Mass and Holes

Assume that the Fermi energy settles somewhere insidethe energy band under consideration Then we can write

σαβ = e2τ

intoccupiedlevels

dkDk(Mminus1)αβ (294)

= minuse2τ

intunoccupiedlevels

dkDk(Mminus1)αβ (295)

60

This is a consequence of the fact that we can write f =1minus (1minusf) and the integral over one vanishes contributingnothing to the conductivity

When the Fermi energy is near to the minimum of theband Ek the dispersion relation of the band can be ap-proximated quadratically

Ek = E0 +~2k2

2mlowastn (296)

This leads to a diagonal mass tensor with elements

(Mminus1)αβ =1

mlowastnδαβ

Because integral (294) over the occupied states gives theelectron density n we obtain the conductivity

σ =ne2τ

mlowastn(297)

Correspondingly if the Fermi energy is near to the max-imum of the band we can write

Ek = E0 minus~2k2

2mlowastp (298)

The conductivity is

σ =pe2τ

mlowastp (299)

where p is the density of the empty states The energystates that are unoccupied behave as particles with a pos-itive charge and are called the holes Because the conduc-tivity depends on the square of the charge it cannot revealthe sign of the charge carriers In magnetic field the elec-trons and holes orbit in opposite directions allowing oneto find out the sign of the charge carriers in the so-calledHall measurement (we will return to this later)

Electric and Heat Current

Boltzmann equation (284)

grk = frk minus τEvk middot(eE +nablamicro+

Ek minus microTnablaT)partfdmicro

can be interpreted so that the deviations from the equi-librium distribution frk are caused by the electrochemicalrdquoforcerdquo and that due to the thermal gradient

G = E +nablamicroe

and

minusnablaTT

As usual these forces move the electrons and create a cur-rent flow Assume that the forces are weak which leads toa linear response of the electrons An example of such aninstance is seen in Equation (287) of the electric conduc-tivity One should note that eg the thermal gradient canalso create an electric current which means that generally

the electric and heat current densities j and jQ respec-tively are obtained from the pair of equations

j = L11G + L12

(minus nablaT

T

)

jQ = L21G + L22

(minus nablaT

T

) (300)

By defining the electric current density as in Equation(285) and the heat current density as

j =

intdkDk(Ek minus micro)vkgrk (301)

we obtain (by using Boltzmann equation (284)) the matri-ces Lij into the form

L11 = L(0) L21 = L12 = minus1

eL(1) L22 =

1

e2L(2) (302)

In the above(L(ν)

)αβ

= e2

intdkDkτE

partf

partmicrovαvβ(Ek minus micro)ν (303)

We define the electric conductivity tensor

σαβ(E) = τEe2

intdkDkvαvβδ(E minus Ek) (304)

and obtain(L(ν)

)αβ

=

intdE partfpartmicro

(E minus micro)νσαβ(E) (305)

When the temperature is much lower than the Fermi tem-perature we obtain for metals

partf

partmicroasymp δ(E minus EF )

Thus (L(0)

)αβ

= σαβ(EF ) (306)(L(1)

)αβ

=π2

3(kBT )2σprimeαβ(EF ) (307)(

L(2))αβ

=π2

3(kBT )2σαβ(EF ) (308)

where in the second equation we have calculated the linearterm of the Taylor expansion of the tensor σ at the Fermienergy EF leading to the appearance of the first derivative

σprime =partσ

partE

∣∣∣E=EF

(309)

Equations (300) and (306-309) are the basic results of thetheory of the thermoelectric effects of electrons They re-main the same even though more than one energy band ispartially filled if one assumes that the conductivity tensorσαβ is obtained by summing over all such bands

61

Wiedemann-Franz law

The theory presented above allows the study of severalphysical situations with electric and thermal forces presentHowever let us first study the situation where there is noelectric current

j = L11G + L12

(minus nablaT

T

)= 0

rArr G =(L11)minus1

L12nablaTT (310)

We see that one needs a weak field G in order to cancel theelectric current caused by the thermal gradient In a finitesample this is created by the charge that accumulates atits boundaries

We obtain the heat current

jQ =[L21(L11)minus1

L12 minus L22]nablaTT

= minusκnablaT (311)

where the heat conductivity tensor

κ =L22

T+O

(kBT

EF

)2

(312)

Because L21 and L12 behave as (kBTEF )2 they can beneglected

We have obtained the Wiedemann-Franz law of theBloch electrons

κ =π2k2

BT

3e2σ (313)

where the constant of proportionality

L0 =π2k2

B

3e2= 272 middot 10minus20JcmK2 (314)

is called the Lorenz number

The possible deviations from the Wiedemann-Franz lawthat are seen in the experiments are due to the inadequacyof the relaxation time approximation (and not eg of thequantum phenomena that the semiclassical model cannotdescribe)

Thermoelectric Effect

Thermoelectric effect is

bull Electric potential difference caused by a thermal gra-dient (Seebeck effect)

bull Heat produced by a potential difference at the inter-face of two metals (Peltier effect)

bull Dependence of heat dissipation (heatingcooling) of aconductor with a thermal current on the direction ofthe electric current (Thomson effect)

Seebeck Effect

Consider a conducting metallic rod with a temperaturegradient According to Equations (300) this results in heat

flow from the hot to the cold end Because the electronsare responsible on the heat conduction the flow containsalso charge Thus a potential difference is created in be-tween the ends and it should be a measurable quantity(cf the justification of the Wiedemann-Franz law) A di-rect measurement turns out to be difficult because thetemperature gradient causes an additional electrochemicalvoltage in the voltmeter It is therefore more practical touse a circuit with two different metals

Because there is no current in an ideal voltmeter weobtain

G = αnablaT (315)

rArr α = (L11)minus1 L12

T= minusπ

2k2BT

3eσminus1σprime (316)

We see that the thermal gradient causes the electrochem-ical field G This is called the Seebeck effect and the con-stant of proportionality (tensor) α the Seebeck coefficient

Peltier Effect

Peltier Effect means the heat current caused by the elec-tric current

jQ = Πj (317)

rArr Π = L12(L11)minus1 = Tα (318)

where Π is the Peltier coefficient

Because different metals have different Peltier coeffi-cients they carry different heat currents At the junctionsof the metals there has to be heat transfer between the sys-

62

tem and its surroundings The Peltier effect can be usedin heating and cooling eg in travel coolers

Semiclassical Motion in Magnetic Field

We consider then the effect of the magnetic field on theconductivity The goal is to find out the nature of thecharge carriers in metals Assume first that the electricfield E = 0 and the magnetic field B = Bz is a constantThus according to the semiclassical equations of motionthe wave vector of the electron behaves as

~k = minusec

partEpart~k

timesB (319)

Thus

k perp z rArr kz = vakio

and

k perp partEpartkrArr E = vakio

Based on the above the electron orbits in the k-space canbe either closed or open

In the position space the movement of the electrons isobtained by integrating the semiclassical equation

~k = minusecrtimesB

So we have obtained the equation

~(k(t)minus k(0)

)= minuse

c

(rperp(t)minus rperp(0)

)timesB (320)

where only the component of the position vector that isperpendicular to the magnetic field is relevant By takingthe cross product with the magnetic field on both sides weobtain the position of the electron as a function of time

r(t) = r(0) + vztz minusc~eB2

Btimes(k(t)minus k(0)

) (321)

We see that the electron moves with a constant speed alongthe magnetic field and in addition it has a componentperpendicular to the field that depends whether the k-space orbit is closed or open

de Haas-van Alphen Effect

It turns out that if the electron orbits are closed in k-space (then also in the position space the projections of theelectron trajectories on the plane perpendicular to the fieldare closed) the magnetization of the matter oscillates as afunction of the magnetic field B This is called the de Haas-van Alphen effect and is an indication of the inadequacyof the semiclassical equations because not one classicalproperty of a system can depend on the magnetic field atthermal equilibrium (Bohr-van Leeuwen theorem)

The electron orbits that are perpendicular to the mag-netic field are quantized The discovery of these energylevels is hard in the general case where the electrons ex-perience also the periodic potential due to the nuclei in

addition to the magnetic field At the limit of large quan-tum numbers one can use however the Bohr-Sommerfeldquantization condition

∮dQ middotP = 2π~l (322)

where l is an integer Q the canonical coordinate of theelectron and P the corresponding momentum The samecondition results eg to Bohrrsquos atomic model

In a periodic lattice the canonical (crystal) momentumcorresponding to the electron position is

p = ~kminus eA

c

where A is the vector potential that determines the mag-netic field Bohr-Sommerfeld condition gives

2πl =

∮dr middot

(kminus eA

~c

)(323)

=

int τ

0

dtr middot

(kminus eA

~c

)(324)

=e

2~c

int τ

0

dtB middot rtimes r (325)

=eB

~cAr =

~ceBAk (326)

We have used the symmetric gauge

A =1

2Btimes r

In addition τ is the period of the orbit and Ar is the areaswept by the electron during one period One can showthat there exists a simple relation between the area Arand the area Ak swept in the k-space which is

Ar =( ~ceB

)2

Ak

When the magnetic field B increases the area Ak has togrow also in order to keep the quantum number l fixed Onecan show that the magnetization has its maxima at suchvalues of the magnetic field 1B where the quantizationcondition is fulfilled with some integer l and the area Akis the extremal area of the Fermi surface By changing thedirection of the magnetic field one can use the de Haas-vanAlphen effect to measure the Fermi surface

Hall Effect

In 1879 Hall discovered the effect that can be used in thedetermination of the sign of the charge carriers in metalsWe study a strip of metal in the electromagnetic field (E perpB) defined by the figure below

63

We assume that the current j = jxx in the metal isperpendicular to the magnetic field B = Bz Due to themagnetic field the velocity of the electrons obtains a y-component If there is no current through the boundariesof the strip (as is the case in a voltage measurement) thenthe charge accumulates on the boundaries and creates theelectric field Eyy that cancels the current created by themagnetic field

According to the semiclassical equations we obtain

~k = minuseEminus e

cv timesB (327)

rArr Btimes ~k + eBtimesE = minusecBtimes v timesB

= minusecB2vperp (328)

rArr vperp = minus ~ceB2

Btimes k minus c

B2BtimesE (329)

where vperp is component of the velocity that is perpendicularto the magnetic field

The solution of the Boltzmann equation in the relaxationtime approximation was of form (283)

g minus f = minusint t

minusinfindtprimeeminus(tminustprime)τE

(vk middot eE

partf

dmicro

)tprime

By inserting the velocity obtained above we have

g minus f =~cB2

int t

minusinfindtprimeeminus(tminustprime)τE ktprime middotEtimesB

partf

dmicro(330)

=~cB2

(kminus 〈k〉

)middot[EtimesB

]partfpartmicro

(331)

The latter equality is obtained by partial integration

int t

minusinfindtprimeeminus(tminustprime)τE ktprime = kminus 〈k〉 (332)

where

〈k〉 =1

τE

int t

minusinfindtprimeeminus(tminustprime)τEktprime (333)

Again it is possible that the electron orbits in the k-space are either closed or open In the above figure thesehave been sketched in the reduced zone scheme In thefollowing we will consider closed orbits and assume inaddition that the magnetic field is so large that

ωcτ 1 (334)

where ωc is the cyclotron frequency In other words theelectron completes several orbits before scattering allowingus to estimate

〈k〉 asymp 0 (335)

Thus the current density is

j = minuseintdkDkvkgrk (336)

= minuse~cB2

intdkpartEkpart~k

partf

partmicrok middot (EtimesB) (337)

=e~cB2

intdkDk

partf

part~kk middot (EtimesB) (338)

=ec

B2

intdkDk

part

partk

(fk middot (EtimesB)

)minusnecB2

(EtimesB) (339)

where the electron density

n =

intdkDkf (340)

Then we assume that all occupied states are on closedorbits Therefore the states at the boundary of the Bril-louin zone are empty (f = 0) and the first term in thecurrent density vanishes We obtain

j = minusnecB2

(EtimesB) (341)

On the other hand if the empty states are on the closedorbits we have 1 minus f = 0 at the Brillouin zone boundaryand obtain (by replacing f rarr 1minus (1minus f))

j =pec

B2(EtimesB) (342)

where

p =

intdkDk(1minus f) (343)

is the density of holes

64

Current density (341) gives

jx = minusnecBEy

Thus by measuring the voltage perpendicular to the cur-rent jx we obtain the Hall coefficient of the matter

R =EyBjx

=

minus 1nec electrons1pec holes

(344)

In other words the sign of the Hall coefficient tells whetherthe charge carriers are electrons or holes

In the presence of open orbits the expectation value 〈k〉cannot be neglected anymore Then the calculation of themagnetoresistance ie the influence of the magnetic fieldon conductivity is far more complicated

Fermi Liquid Theory

The semiclassical theory of conductivity presented abovedeals with the electrons in the single particle picture ne-glecting altogether the strong Coulomb interactions be-tween the electrons The Fermi liquid theory presentedby Landau in 1956 gives an explanation why this kind ofmodel works The Fermi liquid theory was initially devel-oped to explain the liquid 3He but it has been since thenapplied in the description of the electron-electron interac-tions in metals

The basic idea is to study the simple excited states of thestrongly interacting electron system instead of its groundstate One can show that these excited states behave likeparticles and hence they are called the quasiparticlesMore complex excited states can be formed by combiningseveral quasiparticles

Let us consider a thought experiment to visualize howthe quasiparticles are formed Think of the free Fermi gas(eg a jar of 3He or the electrons in metal) where the in-teractions between the particles are not rdquoturned onrdquo As-sume that one particle is excited slightly above the Fermisurface into a state defined by the wave vector k and en-ergy E1 All excited states of the free Fermi gas can becreated in this way by exciting single particles above theFermi surface Such excited states are in principle eternal(they have infinite lifetime) because the electrons do notinteract and cannot thus be relaxed to the ground state

It follows from the the adiabatic theorem (cf Zenertunnelling) that if we can turn on the interactions slowlyenough the excited state described above evolves into aneigenstate of the Hamiltonian of the interacting systemThese excited states of the interacting gas of particles arecalled the quasiparticles In other words there exists aone-to-one correspondence between the free Fermi gas andthe interacting system At low excitation energies and tem-peratures the quantum numbers of the quasiparticles arethe same as those of the free Fermi gas but due to the in-teractions their dynamic properties such as the dispersionrelation of energy and effective mass have changed

When the interactions are on the excited states do notlive forever However one can show that the lifetimes of the

quasiparticles near to the Fermi surface approach infinityas (E1minusEF )minus2 Similarly one can deduce that even stronginteractions between particles are transformed into weakinteractions between quasiparticles

The quasiparticle reasoning of Landau works only attemperatures kBT EF In the description of electronsin metals this does not present a problem because theirFermi temperatures are of the order of 10000 K The mostimportant consequence of the Fermi liquid theory for thiscourse is that we can represent the conductivity in thesingle electron picture as long as we think the electrons asquasielectrons whose energies and masses have been alteredby the interactions between the electrons

65

6 ConclusionIn this course we have studied condensed matter in

terms of structural electronic and mechanical propertiesand transport phenomena due to the flow of electrons Theresearch field of condensed matter physics is extremelybroad and thus the topics handled in the course coveronly its fraction The purpose has been especially to ac-quire the knowledge needed to study the uncovered mate-rial Such include eg the magnetic optical and supercon-ducting properties of matter and the description of liquidmetals and helium These are left upon the interest of thereader and other courses

The main focus has been almost entirely on the definitionof the basic concepts and on the understanding of the fol-lowing phenomena It is remarkable how the behaviour ofthe complex group of many particles can often be explainedby starting with a small set of simple observations Theperiodicity of the crystals and the Pauli exclusion prin-ciple are the basic assumptions that ultimately explainnearly all phenomena presented in the course togetherwith suitably chosen approximations (Bohr-Oppenheimerharmonic semiclassical ) Finally it is worthwhile toremember that an explanation is not a scientific theoryunless one can organize an experiment that can prove itwrong Thus even though the simplicity of an explana-tion is often a tempting criterion to evaluate its validitythe success of a theory is measured by comparing it withexperiments We have had only few of such comparisonsin this course In the end I strongly urge the reader ofthese notes read about the experimental tests presentedin the books by Marder and Ashcroft amp Mermin and thereferences therein and elsewhere

66

  • Introduction
    • Atomic Structure
      • Crystal Structure (ET)
      • Two-Dimensional Lattice
      • Symmetries
      • 3-Dimensional Lattice
      • Classification of Lattices by Symmetry
      • Binding Forces (ET)
      • Experimental Determination of Crystal Structure
      • Experimental Methods
      • Radiation Sources of a Scattering Experiment
      • Surfaces and Interfaces
      • Complex Structures
        • Electronic Structure
          • Free Fermi Gas
          • Schroumldinger Equation and Symmetry
          • Nearly Free and Tightly Bound Electrons
          • Interactions of Electrons
          • Band Calculations
            • Mechanical Properties
              • Lattice Vibrations
              • Quantum Mechanical Treatment of Lattice Vibrations
              • Inelastic Neutron Scattering from Phonons
              • Moumlssbauer Effect
                • Anharmonic Corrections
                  • Electronic Transport Phenomena
                    • Dynamics of Bloch Electrons
                    • Boltzmann Equation and Fermi Liquid Theory
                      • Conclusion
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