Introduction to Condensed Matter Physics

Introduction to Condensed Matter Physics

Antonio H. Castro NetoDepartment of Physics

Boston University

February 28, 2006

Contents

1 Atoms and Molecules 71.1 Interactions in Isolated Atoms . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 71.2 Atomic Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 11

1.2.1 Thermal Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 151.3 Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 16

1.3.1 The H+2 molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.3.2 The H2 molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.3.3 Ionic interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 23

1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 23

2 Crystals 272.1 Discrete Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 28

2.1.1 The reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 312.1.2 The one-dimensional chain . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 33

2.2 Elastic scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 342.2.1 Experimental constraints . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 39

2.3 Defects in solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 392.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 41

2.4.1 Appendix: Dirac and Kronecker delta functions . . . . . .. . . . . . . . . . . . . . 44

3 Elasticity Theory 453.1 Elastic properties of crystals . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 463.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 53

4 Atoms in motion 554.1 The one-dimensional chain . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 554.2 Phonons in higher dimensions . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 574.3 Phonons at finite temperature . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 634.4 Inelastic scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 674.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 71

4.5.1 Appendix: Thermal averages for bosons . . . . . . . . . . . . .. . . . . . . . . . . 72

5 Electrons in Solids 755.1 The one-dimensional chain . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 755.2 Dimensions higher than one . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 875.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 94

3

4 CONTENTS

6 The Free Electron Gas 956.1 Elementary excitations . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 986.2 Free electron gas in a magnetic field . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 101

6.2.1 Zeeman effect and Pauli susceptibility . . . . . . . . . . . .. . . . . . . . . . . . . 1026.2.2 Landau levels and de Hass-van Alphen effect . . . . . . . . .. . . . . . . . . . . . 1056.2.3 Flux quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 110

6.3 Free electrons at finite temperature . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 1126.3.1 Specific heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 1166.3.2 Correction to the Pauli susceptibility . . . . . . . . . . . .. . . . . . . . . . . . . . 1186.3.3 Landau diamagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 119

6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 121

7 Disordered electronic systems 1237.1 Weak scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1237.2 Strong scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 1277.3 Scaling Theory of Localization . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 1307.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 133

8 Semiconductors 1358.1 Intrinsic Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 1378.2 Extrinsic Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 1388.3 The p-n junction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1418.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 144

9 Electron-phonon interactions 1459.1 Boson operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1459.2 Fermion operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 147

9.2.1 Free electrons in a box . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 1509.2.2 Free electrons in a lattice . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 151

9.3 Electron-phonon interaction . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 1529.3.1 The optical case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 1549.3.2 The polaron problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 1569.3.3 The acoustic case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 1579.3.4 Solitons in one dimension . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1589.3.5 Retarded Interactions . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 1609.3.6 Cooper pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 163

9.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 165

10 Magnetism 16710.1 Crystal fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 16710.2 The Heitler-London theory . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 17110.3 Coulomb interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 17210.4 Magnetic anisotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 17510.5 Localized Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 176

10.5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 17610.6 Magnetic interactions in metallic alloys . . . . . . . . . . .. . . . . . . . . . . . . . . . . 17710.7 The Double Exchange problem . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 182

10.7.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 184

CONTENTS 5

10.8 The Anderson Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 18510.9 The Kondo problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 191

10.9.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 20010.10Itinerant Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 201

10.10.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 205

11 Magnetic phase transitions 20711.1 Spontaneous symmetry breaking . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 207

11.1.1 Critical exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 21011.2 Mean field approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 21311.3 Mean field and beyond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 217

11.3.1 Interactions with infinite range . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 21811.3.2 Local interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 219

11.4 Continuous symmetries and the Goldstone theorem . . . . .. . . . . . . . . . . . . . . . . 22311.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 225

12 Magnetic excitations 22712.1 The one-dimensional Ising model . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 22712.2 The Heisenberg-Ising Hamiltonian . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 232

12.2.1 The ferromagnetic case . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 23212.2.2 The antiferromagnetic case . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 235

12.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 239

13 The Interacting Electron Gas 24113.1 The Thomas-Fermi approach . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 24213.2 The Random Phase Approximation: RPA . . . . . . . . . . . . . . . .. . . . . . . . . . . 24413.3 Response of the one-dimensional electron gas . . . . . . . .. . . . . . . . . . . . . . . . . 24913.4 The Kohn anomaly and the Peierls distortion . . . . . . . . . .. . . . . . . . . . . . . . . . 25013.5 Electron-electron interactions . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 25313.6 Landau’s Theory of the Fermi Liquid: An Introduction . .. . . . . . . . . . . . . . . . . . 25413.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 260

6 CONTENTS

Chapter 1

Atoms and Molecules

1.1 Interactions in Isolated Atoms

In order to understand the properties of solids one has to be able to describe their constituents and how theyinteract with each other. Solids are made out of arrays of atoms that are composed of electrons, protons andneutrons. Most of the properties of solids depend on the behavior of electrons and protons. These are entitiesthat have opposite charge, a large mass difference (the massof the proton,mp, is approximately1000 timesthe electron mass,me), and different spatial position in the atom. The nucleus occupies the center of theatom and its volume (of order of10−45 m3) is much smaller than the surrounding electronic cloud thatisextended over a much larger volume (10−30 m3). Since the characteristic distances in solids are of orderofa few angstroms (1 Å= 10−10m) it is the electron that plays a major role in the propertiesof solids.

The basic physics of atoms can be understood starting from the Hydrogen atom. The Hamiltonian thatdescribes the Hydrogen atom is given by

H =p2P

2mp+

p2e

2me− e2

|rP − re|(1.1)

wherepP (pe) is the momentum,rP (re) the position of the proton (electron) ande is the electric charge.In quantum mechanics, momentum and position are operators that act on a Hilbert space of functions.Moreover, momentum and position are conjugated so that theyobey commutation rules, namely,

[xi, pj ] = ihδi,j (1.2)

where we have introduced the components of the vector asr = (x1, x2, x3) andp = (p1, p2, p3) (δi,j = 1if i = j and0). Moreover, the operators for the electron and proton commute among themselves since theyare particles with different quantum numbers. The state of the system can be represented in terms of thepositions of electron and proton by|rP , re, σe, σP 〉 whereσ is the spin degree of freedom of each one ofthese particles. Protons and electrons have spin1/2 and are called fermions. In this caseσ can only havetwo possible projections on a fixed axis, that is, up (↑) or down (↓). Observe that although|rP , re, σe, σP 〉is a legitimate state of the problem and span the whole Hilbert space, it does not represent an eigenstate ofthe Hamiltonian. The reason is that the momentum operator that appears in the kinetic term of (1.1) doesnot commute with the position operator and therefore it induces transitions between states with differentpositions (that is, the electron and proton move around).

As usual in any problem in quantum mechanics one has to find thebasis that properly describes thesystem of interest. The obvious thing to do, as in classical mechanics, is to transform the Hamiltonian (1.1)

7

8 CHAPTER 1. ATOMS AND MOLECULES

to the center of mass and relative coordinate system:

R =mprP +mere

mp +me

r = rP − re . (1.3)

In terms of these new coordinates the Hamiltonian reads:

H =P2

2M+

p2

2µ− e2

r(1.4)

whereP = pP + pe is the total momentum operator,p = (mppe −mepP )/M is the relative momentumoperator,M = me+mp is the total mass andµ = 1/me+1/mp is the reduced mass of the system. Observethat sincemp ≈ 1000me we can rewriteM ≈ mp andµ ≈ me with good accuracy. Hamiltonian (1.4)already represents a major simplification in regards to (1.1). First of all we observe that the center of massmotion decouples from the relative motion sinceR does not appear explicitly in (1.4). Notice that almostall kinetic energy of the center of mass is carried by the proton that is much heavier. It implies that we candiagonalize the problem in the basis of total momentum operator P:

P|K〉 = hK|K〉 (1.5)

that is equivalent say that the electron-proton system moves freely in space. The non-trivial part of theproblem is the solution of the relative motion. Because the potential is central the problem has a symmetryof rotations in space.

Rotations in space are generated by the angular momentum operatorL = r × p that can be rewrittenin a convenient form:Li =

∑

j,k ǫi,j,kxjpk whereǫi,j,k is the Levi-Civita tensor that is completely anti-symmetric (that is,ǫi,j,k = 0 if i = j or i = k or j = k and the other components are defined such thatǫ1,2,3 = +1 as well as all other cyclic permutations,ǫ2,3,1 = ǫ3,1,2 = +1 and all the non-cyclic permutationsare−1: ǫ1,3,2 = −1). The angular momentum operators obey the so-called Lie algebra

[Li, Lj ] = ihǫi,j,kLk (1.6)

that you can easily show from definition ofLi and (1.2). The fact that the angular momentum operatorsdo not commute among themselves implies that one cannot classify the states in terms of these operatorsindependently. Instead we use one of them, say,L3 (Lz) and its amplitudeL2 =

∑

i L2i . It is indeed trivial

to show that the Hamiltonian (1.4) commutes with these operators,[H,L2] = [H,L3] = 0 and therefore thestates of the system can be classified according to the eigenvalues of these operators, namely,

L2|l,m〉 = h2l(l + 1)|l,m〉L3|l,m〉 = hm|l,m〉 (1.7)

wherem = −l,−l + 1, l..., l − 1, l in unit steps. The Schrödinger equation for the problem determinesthe principal quantum numbern. Thus, we can define an eigenstate of the Hamiltonian has the form|K, l,m, n, σe, σP 〉 so that,

H|K, n, l,m, σe, σP 〉 =

(

h2K2

2M− 13.6(eV )

n2

)

|K, n, l,m, σe, σP 〉 (1.8)

is the full solution of the problem. Observe that the solution of this problem does not involvel = 0, 1, 2..., n−1 (these are usually calleds, p, d, f, ... states) ,m = −l, ...,+l or the spin degrees of freedom. Thus the

1.1. INTERACTIONS IN ISOLATED ATOMS 9

states of the system are degenerate in this quantum numbers (there is more than one state of the systemwith the same energy). Moreover, it is obvious that the ground state corresponds to occupy an electron statewith n = 1 andK = 0 corresponding to a static Hydrogen atom. The eigenstates ofthe problem can berepresented in real space by projecting|Kn, l,m, σe, σP 〉, that is,

ψK,n,l,m(R, r, θ, φ) = 〈R, r, θ, φ|K, l,m, n〉∝ eiK·RYl,m(θ, φ)Rn,l(r) (1.9)

whereYl,m(θ, φ) is a spherical harmonic andRn,l(r) is the so-called radial wavefunction that extends overa distance of order of the Bohr radius,a0 = h2/(µe2). The shape of some of these functions is depicted inFig.1.1.

ψ

r

n=1,l=0

n=2,l=1

n=2,l=0

−

+

−−

+

+

l=0 l=1 l=2

Figure 1.1: Radial and angular dependence of some of the wavefunctions of the H atom.

Things are relatively simple in the H atom because the protononly acts as an external potential. In atomswith more than one electron the situation is not so simple because electrons interact among themselves viathe Coulomb interaction. Consider, for instance, the case of the He atom that has 2 electrons and 2 protons.The nucleus of the atom has mass2mp and the full Hamiltonian of the problem reads

H =p2P

4mp+

p21

2me+

p22

2me− 2e2

|rP − r1|− 2e2

|rP − r2|+

e2

|r1 − r2|(1.10)

wherer1,2 is the position of the two electrons. This is a quite complicated problem. The wavefunctionof the system is a function of the coordinates,Ψ(rP , r1, r2). What we need, however, is not the completesolution of the problem but an approximate solution that provides qualitative understanding and a reasonablequantitative description. Since the mass of the proton is somuch larger than the mass of the electron oneexpects the kinetic energy of the protons to be small compared to the other terms in the Hamiltonian, thatis, we expect the electrons to move distances much larger than the protons in the same time interval. Duringthis time interval the protons appear static for the fast electron. If the protons are static then we are back toour one-body quantum mechanics problem where the electronsmove under a potential field created by thetwo protons. This potential is parameterized by the distance between the protons (this implies that we can


solve the Schrödinger equation for each configuration of theprotons). If this is the case the wavefunction ofthe electrons can be written asψrp(r1, r2), and depends on the proton coordinates as a parameter. Therefore,it is intuitive to look for a variational solution of the quantum mechanical problem with the form:

Ψ(rp, r1, r2) = ψrp(r1, r2)φ(rp) (1.11)

whereφ(rP ) is the nucleus wavefunction. Eq.(1.11) is known as the Born-Oppenheimer approximation. Inorder for this wavefunction to make any sense one imposes that ψrp(r1, r2) is an eigenstate of

HS(rP ) =p2

1

2me+

p22

2me− 2e2

|rP − r1|− 2e2

|rP − r2|+

e2

|r1 − r2|(1.12)

with energyEα(rP ) whereα labels the possible quantum numbers of (1.12). Once the eigenenergies of(1.12) are known the nucleus wavefunctionφ(rp) is an eigenstate of

HN =p2P

4mp+ Eα(rP ) . (1.13)

This separation of energy scales between nucleus and electrons is natural and it will be discussed in moredetail later.

There is no exact analytical solution for the quantum mechanical problem of three bodies (contrary toits classical counterpart). We can understand the physics of the many-electron atom qualitatively based onwhat we know about the H atom and Pauli’s exclusion principlethat states that there are no two electronswith the same quantum numbers. As you probably know this is a consequence of the Fermi-Dirac statisticsobeyed by electrons. Let us go back to He atom in the Born-Oppenheimer approximation (1.12) and setrP = 0 for simplicity, that is,

HS =p2

1

2me+

p22

2me− 2e2

|r1|− 2e2

|r2|+

e2

|r1 − r2|. (1.14)

that describes two electrons moving in the Coulomb potential of a charge+2e and interacting with eachother. Observe that because the electrons are practically in the same volume, the direct interaction betweenthe electrons and the nucleus is approximately2 times larger than the electron-electron interaction and hasto be considered first.

At short distances the electron experiences the full potential of the nucleus with charge+2e while atlarger distances it “feels” a smaller effective charge because of the other electron, that is, it sees a Coulombpotential with charge2e − e = +e. This effect only occurs because the electrons are free to move andadapt to the changes in the Coulomb potential. In a static system (that is, a problem whereme → ∞)this is not so and the electron actually experiences the fullcharge of the nucleus. The process in whichthe kinetic energy of the electrons leads to a smaller effective nuclear charge is called screening. What weare proposing is that, beyond the Hamiltonian (1.10), thereis a simpler Hamiltonian or effective theory thatdescribes the problem. For the moment being this effective theory is hidden due to our current ignoranceand lack of sophistication. Nevertheless, due to the symmetry of the problem, the form of the potential doesnot change substantially with distance since the charge distribution is spherically symmetric. Thereforethe wavefunction of the problem looks like a H-like state with a slightly different energy than the H atom.Thus, the He atom is obtained by filling up the1s state (n = 1 andl = 0) of the H atom with electrons ofopposite spin (Pauli’s principle). Observe, moreover, that because the first shell is filled, the He atom is notvery reactive since, as seen from far away, it looks like a neutral object to a foreign electron. Indeed, theionization energy of the He atom is24.6 eV instead of the13.6 eV of the H atom (1 eV ≈ 104 K).

1.2. ATOMIC MAGNETISM 11

Consider now a somewhat more complicated situation: the Li atom that has charge+3e in the nucleus.Again using our effective theory we conclude that the wavefunctions are H-like and that in the ground statewe can occupy the first allowed1s state with two electrons with opposite spins. That is, we form first a He+

atom. Where does the second electron go? Naturally it shouldgo to an = 2 state that can be in thes orp state (l = 0 or l = 1, respectively). In the H atom these two states are degenerate but is this true for theLi atom? The reason for the non-degeneracy of the Li atom is related with the charge of the nucleus andthe shape of the wavefunctions in thes andp states. In Fig.1.2 we show the result of the combination ofthe radial and angular part of the wavefunction as shown in Fig.1.1. Observe that in thes state the electronis closer to the nucleus that has charge+3e, while in thep orbital the electronic charge is distant fromnucleus. Thus, thes state has lower electron-proton Coulomb energy than thep state and is occupied first.Again, it is the preponderance of the Coulomb interaction (that in the case of Li is 3 times larger than theelectron-electron interaction) that determines the ground state properties. Observe that since the first1s shellif filled this state is H-like and the ionization energy for this electron is5.4 eV showing that Li is chemicallyreactive.

Most of the atoms in the periodic table can be understood by simple arguments like these ones. Theunderstanding of the atoms and how one describes their ground state is fundamental for the understandingof solids. The formation of a solid depends on how the protonsinteract with the electrons and how theelectrons interact among themselves.

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB


BBBBBBBBBBBBBBBBBBBBBBBBBBBBBB


BBBBBBBBBB

BBBBBBBBBBBBBBB

s p d

Figure 1.2: The shaded areas are proportional to the probability of finding the electron in each orbital.

1.2 Atomic Magnetism

While the order and classification of energy levels in atoms is determined by the gross value of the Coulombinteraction between the electrons and the nucleus (which isof order of10 eV) the magnetic behavior ofisolated atoms depends on a delicate balance of energy scales. When we talk about magnetism what wereally mean is the magnetic response of the atoms to an applied magnetic field. In the presence of a magneticfield the Hamiltonian of the problem reads

H = H0 +HI

H0 =N∑

i=1

1

2m

(

pi +e

cA(ri)

)2+ µBg0

N∑

i=1

B · Si/h (1.15)

whereri andpi are the coordinate and momentum of theith electron in the atom,A(r) is the vector potential(B = ∇ × A), the last term is the Zeeman energy of the electron spin in a fieldB, Si is the spin of theith


electron,µB = eh/(2mc) is the Bohr magneton, andg0 ≈ 2 is the g-factor. In (1.15) we have

HI = −N∑

i=1

Ze2

|ri|+

N∑

i,j=1,i6=j

e2

|ri − rj |, (1.16)

describes the Coulomb energy for the interaction between the electron and the nucleus (Z is the nuclearcharge) and between the electrons themselves. We assume that the magnetic field is applied in thez directionand choose an electromagnetic gauge such that

A =B

2(xy − yx) (1.17)

that is called the symmetric gauge (notice thatB = Bz). Substituting (1.17) into (1.15) we get

H0 =∑

i

p2i

2m+ µBB

∑

i

(Lz,i + g0Sz,i)/h+e2B2

8mc2

∑

i

(

x2i + y2

i

)

(1.18)

We can show that for practical purposes the terms in the Hamiltonian where the magnetic field appearsexplicitly are small when compared to the terms that are fieldindependent. In this case the magnetic fieldcan be well treated in perturbation theory. Thus, assuming the magnetic field to be small we find that, up tosecond order in perturbation theory, the change in the ground state energy due to the magnetic field is givenby:

δE0 = µBB〈0|∑

i

(Lz,i + g0Sz,i)/h|0〉 + (µBB)2∑

m6=0

|〈m|∑

i(Lz,i + g0Sz,i)/h|0〉|2E0 − Em

+e2B2

8mc2〈0|∑

i

(x2i + y2

i )|0〉 . (1.19)

We can immediately estimate the size of these terms. The firstorder correction isµBB〈n|∑

i(Lz,i +g0Sz,i)/h|n〉 ≈ µBB ≈ hωc where

ωc =eB

mc(1.20)

is the cyclotron frequency. This term is of order of10−4 eV (≈ 1 K) for a field of 1 Tesla (T). In the thirdterm we havee2B2/(8mc2)〈n|∑i r

2i |n〉 ≈ (e2B2a2

0)(8mc2) ≈ (hωc)

2/(e2/a0) that is of order of10−9

eV (≈ 10−5 K!) for a field of 1 T. Indeed these terms are small when compared to the characteristic atomicenergies that are of order of104 K.

In order to calculate the perturbative shift in energy givenin (1.19) one needs to know the nature of theground state of the atom in the absence of the field. In the absence of interactions among the electrons thisis given by the energy levels of the H atom. The system is degenerate because the problem has rotationalsymmetry and for each statel there are2l + 1 degenerate states corresponding to the possible projectionsof l, that is, to the quantum numbersm = −l, ..., l. In a system withN electrons we can distribute theelectrons among2(2l + 1) states (the factor of 2 comes from the two possible spin orientations) withoutchanging the energy of the state. The total number of possible combinations of quantum numbers is[2(2l+1)]!/(N ![2(2l + 1) −N ]!).

Firstly, let us consider the problem of an atom with filled shell, that is, withN = 2l + 1. This is thecase of noble gases such as He and Ar. It is clear that in the ground state we haveLz|0〉 = Sz|0〉 = 0 andthe only term that matters is the last one in (1.19). Using that fact that〈x2

i 〉 = 〈y2i 〉 = 〈z2

i 〉 = 〈r2i 〉/3 for the


case of a spherically symmetric potential we have

δE =Ne2B2

12mc2a2 (1.21)

where

a2 =1

N

∑

i

〈r2i 〉 . (1.22)

The magnetization per unit of volume in the system forNa atoms is given by

M = −Na

V

∂δE

∂B= −Ne

2a2Bρ

6mc2(1.23)

that has a negative sign implying that the response is diamagnetic, that is, in responding to an applied fieldthe atoms want to reduce its value (ρ = Na/V is the atomic density). The magnetic susceptibility is givenby

χ =∂M

∂B= −Ne

2a2ρ

6mc2. (1.24)

The situation described above is simple because we are dealing with a system with a closed shell. Mostatoms have only partially filled shells. In this case the matrix elements in (1.19) do not vanish. The inter-action among the electrons becomes important and the degeneracy of the orbitals is lifted by the Coulombinteraction. We observe that because the Coulomb interaction is spherically symmetric the total angularmomentum,L, and the total spin,S, are constants of motion (that is, they commute with the Hamiltonian)and the states can still be classified in terms of the eigenstates of these two operators, that is, we can simplifythe problem by working with the basis:|L,S,Lz , Sz〉. Moreover, the system is degenerate since for a fixedvalue ofL andS we can have(2L + 1)(2S + 1) states for different values ofLz andSz corresponding todifferent projections ofL andS. This is called a multiplet. Consider, for instance, an atomwith a configu-ration like4f2. Thef orbital can comport14 electrons. Since there are two electron they can be organizedinto the orbitals in14!/(2!12!) = 91 different ways! In this case we haveS = 0, 1 andL = 0, 1, 2, 3, 4, 5and for each value ofS andL there are(2S + 1)(2L + 1) degenerate states.

The relevant question is: what state|S,L〉 has the lowest energy? In order to answer this questionone needs a quantitative calculation that involves the coupling of all the electrons via the Coulomb term.Observe, however, that the exclusion principle does not allow two electrons with the same spin to be in thesame place in space. Thus, electrons with the same spin effectively repel each other (we will give a naturalexplanation for this phenomenon later when we consider the Coulomb interaction in more detail). Naturallythis effect lowers the Coulomb energy. Thus, electrons in anatom tend to have all their spins aligned. Thisis called Hund’s first rule: in an atom the electrons maximizethe total spinS.

The first rule solves the problem of the spin but not of the orbital angular momentum. Again the nameof the game here is: minimize the Coulomb energy between electrons. On the one hand, in states with smallL (like an s-state) the electrons spend much of their time close to the nucleus and therefore pay the energeticprice of the repulsion among themselves. On the other hand, in states with largeL the electrons are apartfrom the nucleus and feel a weaker Coulomb repulsion (they explore a larger volume of space). This givesrise to Hund’s second rule: For a maximum value ofS the energy is minimized by the largest value ofL.

The first and second Hund’s rules specify the values ofL andS for which the energy is minimum butstill for fixed L andS we have(2L + 1)(2S + 1) degenerate states. Is this degeneracy real in an atom?The answer is no! The reason being that the orbital motion is coupled to the spin of the electron by the


so-called spin-orbit effect. Consider one electron movingwith velocity v around a nucleus. From the pointof view of the electron (that is, looking at the problem at theframe co-moving with the electron) the nucleusmoves around with velocity−v. Since the nucleus is charged, we can imagine the nucleus motion as a littlecurrent of charge circulating around the electron. This current generates a magnetic field at the positionof the electron that is proportional tor × v ∝ L. Thus, the Zeeman energy created by this field is alsoproportional toL · S, that is, the spin-orbit coupling. One usually writes the Hamiltonian associated withthis coupling as

HSO = λL · S (1.25)

whereλ depends on details of the atomic problem and can be obtained from the atomic spectra. Indeed, ifone adds (1.25) to our original Hamiltonian (1.15) we see that L andS do not commute with the Hamiltonianseparately. However, the total angular momentum

J = L + S (1.26)

commutes with the Hamiltonian. It means that we can still classify the eigenstates of the problem in terms ofJ, that is, the states can be labeled as:|J, Jz , L, S〉. This implies that the degeneracy of the(2L+1)(2S+1)states is lifted and the states split into(2J + 1) degenerate states corresponding to different values ofJz.Indeed we can rewrite (1.25) in terms of this operator as

HSO =λ

2

(

J2 − L2 − S2)

(1.27)

since the Hamiltonian commutes withJ, L2 andS2 . Observe that due to (1.26) the allowed values ofJ gofrom |L−S| toL+S in unit steps and the allowed valued ofJz go from−J to J (that gives the degeneracyof 2J + 1). The energy associated with each one of these states is

E(J, Jz , S, L) =λ

2[J(J + 1) − L(L+ 1) − S(S + 1)] . (1.28)

It is experimentally observed thatλ > 0 for shells that are less than half filled (N ≤ 2l + 1) andλ < 0for shells that are more than half filled (N ≥ 2l + 1). If λ > 0 it is clear from (1.28) that the energy isminimized for a givenS andL for a configuration with smallestJ , that is,J = |L−S|. Otherwise, ifλ < 0the energy is minimized for the largest value ofJ , that is,J = L + S. This is the so-called Hund’s thirdrule.

Hund’s rules can guide us to understand the response of an atom to a magnetic field. Consider a systemwhereJ = 0. In this case we can use the Wigner-Eckart theorem and show that the first term in the energyin (1.19) actually vanishes. We are left with the two other terms. The third term is just the diamagneticresponse we studied in the case of atoms with filled shell. Thesecond term is negative (remember thatsecond order perturbation theory always lowers the energy of the ground state) and therefore it gives rise toa susceptibility with positive sign. This is the so-called Van Vleck paramagnetic response.

If J 6= 0 then the first term in (1.19) does not vanish and it is the largest contribution to the energy shift.We can rewrite this term in a more appropriate form

HZ = µBB · (L + g0S) = µBB · (J + (g0 − 1)S) (1.29)

that allows us to interpretJ + (g0 − 1)S as the effective magnetic moment of the atom,

M = −µB(J + (g0 − 1)S) (1.30)


J

S

L

Θ

Figure 1.3: Angular momentum geometry

that is not a constant of motion sinceS is not conserved. It turns out, however, thatJ is a constant ofmotion and we can think ofJ as being fixed and thatL andS precess aroundJ as in Fig.1.3. Thus themagnetic moment is given by the component ofL + g0S parallel toJ. This is the only component ofthe magnetic moment that contributes in first order perturbation theory since the components ofS in thedirection transverse toJ introduce transition between different values ofJz and give zero average when wecalculate with an unperturbed state|J, Jz , L, S〉. The parallel component ofS can be calculated from theangle betweenJ andS,

SJ =(J · S)

J2J

=(J2 − L2 + S2)

2J2J (1.31)

which, in a state|J, Jz, S, L〉 has an expectation value:

〈SJ〉 =[J(J + 1) − L(L+ 1) + S(S + 1)]

2J(J + 1)J (1.32)

and therefore the effective magnetic moment in the direction of J is given in (1.30)

M = −gµBJ (1.33)

where

g(J,L, S) =(g0 + 1)

2+

(g0 − 1)

2

S(S + 1) − L(L+ 1)

J(J + 1)(1.34)

is the Landé g-factor. The energy associated with (1.29) is

δE(J, Jz , S, L) = gJzµBB (1.35)

whereJz = −J, ..., J . This final expression gives the splitting between the energy levels of a multi-electronatom in the lowest order approximation.

1.2.1 Thermal Effects

One interesting application of (1.35) is the calculation ofthe magnetic response of a set of isolated identicalatoms in a magnetic field. Observe that the partition function for the problem defined in (1.35) is given by


(β = 1/(kBT ))

Z =

J∑

Jz=−J

e−βE(J,Jz,S,L) =

J∑

Jz=−J

e−βg(J,L,S)JzµBB

=sinh [βgµBB(J + 1/2)]

sinh [βgµBB/2](1.36)

where in the last line we used the sum of a geometric series. The magnetization is obtained like in (1.23)and it is given by

M(B) = ρgµBJBJ(βgJµBB) (1.37)

where

BJ(x) =(2J + 1)

2Jcoth

(

(2J + 1)x

2J

)

− 1

2Jcoth

( x

2J

)

(1.38)

is the so-called Brillouin function. Notice that forβgµBB ≫ 1 (that is, low temperatures) the magnetizationsaturates atM → ρgµBJ as expected since all the moments are aligned. At high temperatures, that is,βgµBB ≪ 1 we find

M ≈ ρJ(J + 1)(gµB)2B

3kBT(1.39)

and gives a magnetic susceptibility

χ(T ) = ρJ(J + 1)(gµB)2

3kBT(1.40)

that is known as the Curie susceptibility and it is importantbecause it allows an experimentalist to extract theeffective moment of the atom,J . Notice thatg, µB andkB are universal quantities andρ can be measuredindependently. Thus, by plottingχ−1(T ) as a function ofT one should get a straight line where the slope isessentiallyJ(J + 1).

1.3 Molecules

1.3.1 The H+2 molecule

We have seen that the physics of many-electron atom can be quite complex, but qualitatively explainable interms of basic quantum mechanics. Let us consider a problem one notch higher in complexity, namely, theformation a molecule. Many of the concepts that lead to the understanding of the formation of moleculesare useful in order to understand the stability of matter.

To get a qualitative understanding of the problem let us consider the problem of two protons and oneelectron, that is, the H+2 molecule. The Hamiltonian of this problem can be written as

H =p2e

2me+

p21 + p2

2

2mp− e2

(

1

|re − R1|+

1

|re − R2|− 1

|R1 − R2|

)

(1.41)

wherere is the position of the electron andR1 andR2 are the positions of the two protons. The firstthree terms are the kinetic energies of each one of these particles and the last three their interaction energy.

1.3. MOLECULES 17

This is a quite complicated problem. The wavefunction of thesystem is a function of all the coordinates,Ψ(re,R1,R2).

In order to simplify the problem we make use of the Born-Oppenheimer approximation and assumethat the protons are static during the time of motion of the electron. Suppose that initially the protons areinfinitely apart. For simplicity assumeR1 = 0 andR2 = R. In this case the problem has two solutionsthat are degenerate with each other, that is, either the electron is bound to proton1 or to proton2 withthe same energyE0. The wavefunction of the electron,ψ0, is well localized in each proton (it is a H-likewavefunction).

Let us consider a simplification of the problem that disregards all the other states of the problem exceptthe states where the electron is localized in one of the two protons as in Fig.1.4. Let〈1| (〈2|) be the state ofthe electron bound to proton1(2). If R→ ∞ one has

H0|1〉 = E0|1〉H0|2〉 = E0|2〉 (1.42)

where〈1|2〉 = 0 and〈1|1〉 = 〈2|2〉 = 1.

−

+ +

+ +

−

|1>

|2>

R

R

Figure 1.4: Two states retained in the two-level system approximation of H+2 .

Equation (1.42) can be rewritten as

H0 = E0 (|1〉〈1| + |2〉〈2|) . (1.43)

As the protons get closer to each other there is a finite probability that the electron hops from proton1 toproton2 and vice-versa. This is the so-called quantum tunneling andis depicted in Fig.1.5. The tunnelingdepends on the amount of overlap between the wavefunction ofthe electron in the two different protons.There is an energy scale associated with the tunneling that we are going to callt. t is a function ofR,vanishing whenR → ∞, and becomes large whenR → 0. Thus, in order to incorporate tunneling into theproblem one has to add a perturbation that mixes the two states. This perturbation we callHT and it has tobe such that

HTψ1 ∝ ψ2 (1.44)

and of courseH2T ∝ I since if we hop the electron twice it has to return to the same atom. It is obvious that

the tunneling Hamiltonian must have the form:

HT = −t (|1〉〈2| + |2〉〈1|) . (1.45)


Any eigenstate ofH = H0 +HT has to be a linear combination of states〈1| and〈2|. This problem can bestudied by rewriting the Hamiltonian in matrix form. The matrix elements are:

〈1|H|1〉 = 〈2|H|2〉 = E0

〈1|H|2〉 = 〈2|H|1〉 = −t (1.46)

that can be rewritten in matrix form:

[H] =

[

E0 −t−t E0

]

. (1.47)

Observe that in terms of the matrix formulation the states are represented by vectors

[Ψ1] =

(

10

)

(1.48)

[Ψ2] =

(

01

)

. (1.49)

Our goal is to solve the Schrödinger equation

H|ψ〉 = E|ψ〉 . (1.50)

From basic quantum mechanics the most general solution of this problem has to be a linear combinationof base states, that is,|ψ〉 = a|1〉 + b|2〉. Normalization of the wavefunction (or probability) insures that〈ψ|ψ〉 = 1 = |a|2 + |b|2. In the language of (1.47) we have

[Ψ] =

(

ab

)

(1.51)

and the Schrödinger equation becomes a simple eigenvalue problem

[H][Ψ] = E[Ψ]

([H] − E[I])[Ψ] = 0 (1.52)

where

[I] =

[

1 00 1

]

(1.53)

is the identity matrix. It is a simple exercise to show that eigenstates of the problem are:

H|A〉 = (E0 + t)|A〉H|B〉 = (E0 − t)|B〉 . (1.54)

1.3. MOLECULES 19

where

|A〉 =1√2

(|1〉 − |2〉) =1√2

(

1−1

)

|B〉 =1√2

(|1〉 + |2〉) =1√2

(

11

)

. (1.55)

These states are called anti-bonding and bonding, respectively. Observe that the bonding state is the groundstate of the problem. In the representation of position, these two states can be easily written as

〈R, r|B〉 = ψB,R(r) = ψ0(r) + ψ0(r − R)

〈R, r|A〉 = ψA,R(r) = ψ0(r) − ψ0(r − R) (1.56)

whereψ0 is the ground state wavefunction of the H atom (ψn=0,l=0,m=0(r)). Observe that the anti-bondingstate has a node in the middle position between the protons (ψA,R(r = R/2) = 0) while the bondingstate is always finite. Thus in the bonding state the amount of“charge" between the protons is larger (theprobability of finding the electron in the middle of the protons is higher) than in the anti-bonding state. Thatis the reason its energy is lowest: the electrons act as a “glue” between protons. In the anti-bonding case theelectronic charge is mostly centered around the two protonsthat are therefore "shielded" from each other.Thus, the anti-bonding state is unstable. A plot of square ofthe wavefunction for each one of these states isshown on Fig. 1.6.

+ ++

V(r)

R

r

−

Figure 1.5: Potential energy for H+2 molecule showing the overlap of the electron wavefunction in the twodifferent protons.

Another way to understand this problem is to realize that at finiteR we can expand the potential term in(1.41) in powers ofr/R and it is clear that the first term is of orderr/R2. Since this term is small, we canperform perturbation theory. The first order term cancels due to symmetry〈ψ0|r|ψ0〉 = 0. The second orderterm has the formδE =

∑

n 6=0 |〈ψ|V |ψn〉|2/(E0 − En). This term is always negative ifE0 is the groundstate. From the previous argument it is clear that|t| = −δE > 0. Thus, in this way, we relate our originalproblem with the two-level system calculation. On the one hand, the energy of the ground state (in this casea bonding state) has to decrease as we decrease the distance between atoms. On the other hand, the firstexcited state (the anti-bonding state) increases in energyas we decrease the distance between the protons(sincet → 0 asR → ∞). When the distance between atoms goes to zero the energy hasto diverge sincethe Coulomb energy term1/R diverges. Thus we conclude that there must be a minimum of thegroundstate energy at some distanceR0 (see Fig.1.7). In a first approximation the energy close to the minimum isparabolic withR and is quantized in units ofhω0 ≈ E0. Moreover, since this corresponds to the potential


where the electron is trapped we can write

Ee(R) ≈ E0 +kmeE

20

2h2 (R − R0)2 (1.57)

wherek is a constant of order unit. Thus, by direct substitution in (1.13) we see that the protons undergoharmonic motion around the equilibrium position with a frequencyω0 ≈

√

me/mpE0/h.

(a)

(b)

Figure 1.6: Contour plot for the square of the wavefunction for: (a) Bonding state; (b) Anti-bonding state.

At this point it is interesting to check the validity of the Born-Oppenheimer approximation. By directsubstitution of (1.11) and (1.13) into the Schrödinger equation we find that the have neglected terms ofthe form−h2∇1ψ · ∇1φ/mp. Since variations inR1 produces variations inr we can write this term asapproximately|pe||pp|/(mp). Thus, in order for our approximation to be valid we have to requirep2

p/mp ≫|pe||pp|/mp or |pp| ≫ |pe|. The momentum of the electron in a bound state of the H-atom isapproximately

pe ≈√meE0 while for the protons undergoing harmonic motion we havepp ≈

√

mp

√

me/mpE0. Thusthe Born-Oppenheimer approximation is reasonable when

m1/4e ≪ m1/4

p (1.58)

that is a good approximation in most cases ((me/mp)1/4 ≈ 0.16). Thus we have shown that within these

approximations the protons undergo harmonic motion and within a period of oscillation of the protons theelectron can be found in a bonding state. This is an example ofa covalent bond. The term covalent impliesthat the electron is equally shared by the protons.

1.3.2 The H2 molecule

Let us consider the H2 molecule which is a neutral molecule that is found commonly in nature. Each Hatom has one electron that is tightly bound to each proton. Wecan still use the same staring point of the H+

2

molecule and use perturbation theory by imagining that the atoms are far apart from each other. This is theso-called atomic limit. In this case, since the atoms are neutral, the electric monopole terms of H2 vanish andone has to consider the next term in the multipole expansion of the Coulomb potential. The next contributioncomes from a dipole term that behaves like1/R3. Thus, if we perform perturbation theory on this term, thefirst order correction vanishes and the second order perturbation energy correction for the ground state is

1.3. MOLECULES 21

E

R

R0

Bonding state

Anti−bonding state

Figure 1.7: Energy of the H+2 system as a function of the distance between the protonsR.

negative. Thus, second order perturbation theory producesan effective interaction that behaves like

HV dW = −α1α2

R6(1.59)

whereα1,2 are the polarizabilities of the atoms. This is a short rangedinteraction if compared with the bareCoulomb interaction and furthermore, it is attractive. This is known as the Van der Waals interaction.

Our reasoning regarding the ground state of the problem is similar to the one in the problem of the H+2

system. Within the Born-Oppenheimer approximation we consider the wavefunction of the problem withtwo electrons and two protons separated by a fixed distanceR. We have various possibilities among thevarious arrangements of the electrons around the protons. We can have one electron around each protonor we can have the two electrons around one proton. In this last situation the Coulomb repulsion betweenthe electrons is large. Therefore, the ground state has to besuch that one has on average one electron perproton. This is the so-called Heitler-London approximation. Thus, we have two states that are degeneratedwhen the protons are infinitely apart, that is, with the electrons in their original position or with the electronsexchanged between the protons. These two states are depicted on Fig.1.8.

−

+ +

+ +

−

|1>

|2>

R

R

−

−

Figure 1.8: Two states retained in the two-level system approximation of H2.

The situation described here is completely equivalent to the H+2 system and as one imagines the protons

approaching each other they can exchange their electrons. Let r1,2 be the position of each electron. The


most general wavefunction for this problem has the form

Ψ(r1, r2) = αψ1(r1)ψ2(r2) + βψ1(r2)ψ2(r1) (1.60)

whereα, β are coefficients andψi(rj) is the state of electronj on atomi. If the distance between the atomsis large any linear combination of the type (1.60) is a solution of the problem. As the distance betweenH atoms is reduced, the electrons from each atom can tunnel from one proton to another. Using the two-level system approach we find that the states that matter are the bonding and anti-bonding states that can bewritten as,

ΨA(r1, r2) = NA (ψ1(r1)ψ2(r2) − ψ1(r2)ψ2(r1)) ,

ΨB(r1, r2) = NB (ψ1(r1)ψ2(r2) + ψ1(r2)ψ2(r1)) , (1.61)

whereNA,B are normalization constants. Thus, in principle, we have found the solution of the problemwhich is entirely analogous with the H+2 molecule.

Nevertheless, there is a possible inconsistency with our solution. The Pauli principle requires thatwhen we exchange the electrons the wavefunction has to change sign (indeed, one must have,Ψ(r1, r2) =−Ψ(r2, r1)) and therefore the wavefunction of two fermions has to vanish at equal position (Ψ(r1, r1) = 0).Thus, stateΨB (that we concluded to be the ground state) has the wrong statistics. What is wrong with thispicture? What is wrong is that we forgot the electron spin. Wecan still have a wavefunction that is symmet-ric with respect to the coordinates if we have a spin part thatis anti-symmetric. Thus, from the four statesavailable for the two electron problem, the only spin statesthat matter are:

χs(s1, s2) = |s1 =↑, s2 =↓〉 − |s1 =↓, s2 =↑〉 ,χt(s1, s2) = |s1 =↑, s2 =↓〉 + |s1 =↓, s2 =↑〉 , (1.62)

that are singlet and triplet combinations of the spins, respectively. Thus, accordingly to the Pauli principlethe states allowed are

Ψs(r1, s1; r2, s2) = ΨB(r1, r2)χs(s1, s2)

Ψt(r1, s1; r2, s2) = ΨA(r1, r2)χt(s1, s2) . (1.63)

In this particular case the singlet state must be the ground state of the problem.

It is clear that the Pauli principle has strong consequenceshere. The tunneling of electrons betweenatoms favors the bonding state but if the electrons have the same spin projection this state not allowed. Thusatoms with the same spin projection repel each other. As we bring two H atoms together we are tryingto form a He atom. If the spins were the same the electronic configuration should be1s12s1 by the Pauliprinciple. If we had the atoms with electrons with opposite spin then we would get1s2 that is lower inenergy. For the anti-bonding state we can have the electronswith the same spin but in this case the atom isnot stable (as shown in Fig.1.7). Thus, the ground state of the H2 atom has to be a singlet state.

This discussion implies that the Pauli principle acts as an interaction between the electrons, that is, thereis a quantum mechanical repulsion between the atoms with thesame spin. There is no classical analogueto this effect. Moreover, at shorter distances the Coulomb repulsion between the electronic clouds becomeslarge and this effect also increases the repulsion. It is hard to calculate the combined effect of all interactionsfrom first principles and one usually uses the a phenomenological approach and introduce a repulsive termof the form1/R12 that is short ranged and is called Lennard-Jones potential so that the atom has a minimumat someR0 as before. Thus the energy of the molecule again looks very much like in Fig.1.7.

1.4. PROBLEMS 23

1.3.3 Ionic interactions

Another important interaction between atoms is the so-called ionic bond. This interaction occurs when theatoms involved have strong tendency to form a filled electronic shell (that is, there is gain of energy bybinding strongly one electron to the nucleus instead of sharing it with the other atom). Standard examplesare the combinations of the elements of the column 1A and 7A ofthe periodic table such as, NaCl. Inorder to understand this type of bonding, observe that the charge distribution of Na is1s22s22p63s1 whileCl is 1s22s22p63s23p5. Thus, if Na donates one electron entirely to Cl it has the electronic configurationof Ne while Cl has the configuration of Ar that are noble gases.Thus, in the most stable configuration themolecule of NaCl has ions Na+Cl−. In this case there is almost no screening of the electronic clouds andthe atoms actually feel the bare Coulomb interaction between them. At short distances, as we saw before,there is a strong repulsion between the atoms that can be accounted by the Lennard-Jones potential. Thus,for a molecule of NaCl the energy of the system looks again like in Fig.1.7 with some minimum atR0. Ifwe bring another molecule of NaCl close to the first one Coulomb attraction (repulsion) between the atomsbind the molecules together.

1.4 Problems

1. Verify the following identities related with the Levi-Civita tensor:

(1) ǫi,j,kǫl,m,n = δi,lδj,mδk,n + δi,mδj,nδk,l + δi,nδj,lδk,m − δi,nδj,mδk,l − δi,mδj,lδk,n − δi,lδj,nδk,m;

(2)∑

k ǫi,j,kǫk,l,m = δi,lδj,m − δi,mδj,l;

Use the definition ofLi and (1.2) and the properties of the Levi-Civita tensor to prove (1.6).

2. The ionization energy of the oxygen is13.6 eV and it is lower than the energy of its neighbors on theperiodic table (the ionization energy of N is14.5 eV and of F is17.4 eV). Explain qualitatively whythis is so in terms of the interaction between the electrons.Hint: Start the problem by thinking whathappens to Boron (B) and go up in the atomic number.

3. Argon (A) is a noble gas and has filled shell with a configuration 3 p6. The next atom in the periodictable is Potassium (K) that has the configuration of lowest energy with 4 s1 instead of 3 d1! Providean argument that explains this observation.

4. Show that equation (1.18) is correct.

5. Show that the total angular momentum defined in (1.26) commutes with the Hamiltonian in the pres-ence of spin-orbit coupling.

6. Use Hund’s rules to find the configuration of lowest energy for an atom in which the last incompleteshell has a configuration: 1) d8; 2) f9. What is the value of thetotal magnetic moment in each one ofthese configurations.

7. What is the condition forJ = 0 in terms ofN andl?

8. Use the Wigner-Eckart theorem to show that forJ = 0 the first term in (1.19) vanishes.

9. Use the algebra of angular momentum (Clebsh-Gordan coefficients) and prove (1.33) and (1.34).

10. Prove equation (1.35).




13. Two localized spins1/2 interact via an exchange mechanism that is described by a Hamiltonian

H = JS1 · S2 . (1.64)

i) What are all the possible configurations of the two spins in the basis ofS1, S1Z , S2, S2Z?ii ) What are the energies and their respective eigenstates of the system whenJ > 0?iii ) What are the energies and their respective eigenstates of the system whenJ < 0?iv) Suppose a magnetic field,B, is applied to the system so that we have to add the Zeeman energy tothe Hamiltonian in (1.64):

HB = −µBB (S1Z + S2Z) (1.65)

whereµB > 0 is the effective Bohr magneton. Make a plot of the energy of the states you found onitem 2) as a function of magnetic field. What is the state with lowest energy whenB → ∞? What isits physical meaning?

14. Consider the Schrödinger equation (1.50). Assume that|Ψ〉 = a|1〉 + b|2〉 and calculatea andb bydirect substitution.

15. Solve the eigenvalue problem of equation (1.52) in matrix form and show that the solution can bewritten in terms of (1.54) and (1.55).

16. Consider a one dimensional molecule described by the Hamiltonian

H = − h2

2M

∂2

∂R2− h2

2m

∂2

∂r2+mω2

2

(

|r| − R

2

)2

(1.66)

whereR > 0 is the relative coordinate of the nuclei with mass M,m is the mass of the electronandr is the coordinate of the electron relative to the center of mass. Using the Born-Oppenheimerapproximation, that is, assuming

Ψ(r,R) = ψR(r)φ(R) (1.67)

find, by direct substitution of (1.67) into (1.66) what termsare neglected. Assuming that the energyhas a minimum close toR0 show that the nuclei oscillate with a frequencyω

√

m/M . From this resultshow that we can use the Born-Oppenheimer approximation when (m/M)1/4 ≪ 1.

17. Show that the dominant interaction between two H atoms has a dipole form at large values of theseparation between them (call it R).

18. Now consider a molecule made out of three atoms as shown ofFig.1.9.

1.4. PROBLEMS 25

R

R

Rtt

t

Figure 1.9: Three atom molecule as a triangle of sideR and hopping energyt.

In this case the states of the electrons localized in each oneof the atoms can be written as:

ψ1 =

100

ψ2 =

010

ψ3 =

001

.

(i) If tunneling between the atoms is allowed show that the Hamiltonian is written as

H = −t

0 1 11 0 11 1 0

.

(ii ) Diagonalize the Hamiltonian and show that the eigenvalue problem gives the following energies−2t andt with eigenvectors,

ψG =1√3

111

ψE1 =1√2

−101

ψE2 =1√2

−110

(1.68)

respectively. Observe thatψE1 andψE2 are not orthogonal to each other and one has to orthogonalizethem. Use the Gram-Schmidt method and find an orthogonal basis.

Note: Observe that the degeneracy of the problem was lifted by the tunneling. However, two of the


states are still degenerate. This is because the problem hasan extra symmetry that is due to rotationof π/3. A similar thing happens in the Benzene molecule that has 6 C atoms.

Chapter 2

Crystals

We have seen that by sharing or exchanging electrons stable molecules of atoms can be formed. Dependingon external conditions such as temperature or pressure, as atoms or molecules get closer to each other, asolid can be formed. Solids are highly symmetric structuresthat can be found in many different shapes. Theshapes depend on the type of orbitals that participate in thebinding between atoms.

As an illustration, let us consider the example of the famoushigh temperature superconductors (HTC)that layered materials formed by planes of CuO2. The atoms in these planes are arranged in squares as inFig.2.1. An isolated O atom has an electronic configuration1s22s22p4 while Cu has1s22s22p63s23p63d104s1.Thus, by “stealing” two electrons the O atom can close its p-shell and acquire the same configuration of Ne.We say that O has valence−2. Therefore, in the CuO2 planes we have Cu+2 and O−2 configuration. TheO atom has a close p shell and the Cu is in a3d9 configuration. It means that there is a place for a singleelectron in the d shell of Cu (there is one unpaired electron). It is very reasonable to imagine that the bondbetween O and Cu is done by ahybridization(or mixing) of the p orbital of the O with the d orbital of Cu.Since these orbitals have an anisotropic structure and are oriented 90 degrees in respect to each other we ex-pect a square lattice such the one in Fig.2.1(a). The orbitals overlap like in Fig.2.1(b). It is interesting to notethat this simple orbital structure is probably responsiblefor the remarkable properties of these materials.

A striking property of a crystalline solid is its symmetry. Imagine yourself walking over a lattice ofatoms which show a periodic structure such as the one in Fig.2.1(a). You immediately note that as you movefrom atom to atom you see exactly the same structure. Moreover, as you look at the system from somespecific angles it looks the same. Thus, one expects based on this observation that in perfect crystals thephysical properties are the same at each point of the lattice, that is, the physical properties of the system areinvariant under symmetry operations. These symmetry operations can be mathematically defined and helpus to predict many different properties of crystals.

In terms of quantum mechanics symmetries can be expressed bythe fact that there are quantum mechan-ical operatorsO that generate them. For instance, the operator that generates translations by an amountR

is

OT = eiP·R/h (2.1)

whereP is the momentum operator. It is very simple to show that this is true. Suppose we apply thisoperator to a wavefunctionΨ(r) and suppose theR = δr is an infinitesimal quantity. Then,

OTΨ(r) ≈ Ψ(r) + δr · ∇Ψ(r) ≈ Ψ(r + δr) (2.2)

where we have expanded the exponential and used thatP = −ih∇.For systems with translation symmetry described by a HamiltonianH, the operatorOT must be a con-

27

28 CHAPTER 2. CRYSTALS

stant, that is, it must commute with the Hamiltonian:[H,OT ] = 0. In this way, we know from the funda-mentals of quantum mechanics that the wavefunctions of the Hamiltonian can be classified according to theeigenstates of the operatorOT . This operator can be diagonalized straightforwardly since its eigenstates arethe momentum eigenstates,

OT |k〉 = eik·R/h |k〉 (2.3)

Thus, even when we do not know how to calculate exactly the eigenstates of the Hamiltonian (and we don’tin most cases), we know that the momentum is a good quantum number and therefore the wavefunctionscan be labeled by the momentum. This is a major advantage since we can make predictions based on simplecalculations. Translation is a simple example of a continous symmetry. There are many other symmetriesthat can be expressed in terms of operators as well.

(a)

− Cu

− O

(b)

p−orbital

d−orbital

Figure 2.1: (a) Spatial structure of a CuO2 plane; (b) Atomic orbitals involved in the binding.

2.1 Discrete Symmetries

The two most important symmetry operations in crystals are:1) Translations by multiples of the lattice spacing;2) Point operations: rotations and reflections.In order to define a translation we have to define first two mathematical concepts: lattice and basis.

Lattice is a periodic array of points which can be described by a translation vector

T = n1a1 + n2a2 + n3a3 (2.4)

wherea1,2,3 are three independent vectors andn1,2,3 three arbitrary integers (in two dimensions we have,of course, only two of them). A periodic array of points is called a lattice if givenR on the array thenR′ = R + T is also on the array. Thus by choosing different sets of integers we can generate the wholelattice. For each lattice point we can assign different atoms. In this case we have a basis. LetRα be thecoordinate of these atoms with respect to a lattice point. Here,α = 1, 2, ...,Nb, whereNb is the number of

2.1. DISCRETE SYMMETRIES 29

atoms in a basis. A crystal structure is combination of a lattice plus a basis, that is, any distance betweentwo atoms on a lattice can be written asT + Rα.

Observe that there is no unique way to define a lattice but it iscommon to define the primitive quantitiesas the most economic way to describe the crystal. We call the primitive translation vectors as the smallesta1,2,3 that still allow the definition of a lattice. An example of a two-dimensional lattice is shown on Fig.2.2.

a 1

a 2

Figure 2.2: Example of a two-dimensional lattice with a basis.

We also define what is called as the unit cell as a certain volume that fill out the space when translated byall possibleT. It is clear that this definition is not unique. We can define a primitive unit cell that is the onewith the smallest volume or the Wigner-Seitz unit cell whichis obtained by linking the nearest neighboringatoms together and then cutting these lines in the middle by planes (see Fig.2.3). Observe that ifai are theprimitive vectors then the volume of the primitive unit cellis simply

V0 = |a1 · (a2 × a3)| . (2.5)

(a)

(b)

Figure 2.3: Conventional unit cell; (b) Wigner-Seitz cell.

Together with the translation symmetry the point symmetries define what is known as the Bravais lat-tices. To each symmetry we have an operator which changes thecoordinates of the system around a pointon the lattice with an axis through it. The principal axis is the axis with the highest symmetry (that is, theone with the largest number of symmetry operations). The symmetry operations are: (i) Identity (R → R);(ii ) Inversion (R → −R); (iii ) Rotations,Cn, by an angle of2π/n wheren is an integer; (iv) Reflectionby a plane; (v) Improper rotations which are combinations of a rotation and a reflection through a planeperpendicular to a principal axis. These operations can be easily identified by inspection.

In two dimensions there are only five types of Bravais lattices that can be obtained from the operationsabove. They are shown in Fig.2.4. It is easy to show that in twodimensions it is not possible to have aBravais lattice with 5 fold symmetry (that is, it is not possible to fill out the plane with pentagons) and thereis no Bravais lattice with rotations higher than six-fold symmetry. Thus, we are only left with the Bravaislattices of Fig.2.4: (1) Oblique (which is symmetric only underC2); (2) Rectangular (which is symmetric


underC2 and has two reflection planes); (3) Rectangular face centered; (4) Square (which is symmetricunderC4 and three reflection planes); (5) Hexagonal (which is symmetric underC6, C3, and six reflectionplanes).

Oblique Rectangular

RectangularFace centered

Square

Hexagonal

Figure 2.4: Possible Bravais lattices in two-dimensions.

Observe that the hexagonal lattice is the most symmetric of all Bravais lattices in two dimensions (thatis, is invariant under the largest number of symmetry operations). Moreover, the hexagonal lattice is specialbecause it is close packed, that is, it is a lattice that has the densest packing of hard spheres. Indeed, inthe hexagonal lattice we can densely fill the lattice by placing spheres of radiusa/2 (wherea is the latticespacing) in the center of a triangular lattice as in Fig.2.5.

Figure 2.5: Close packing structure for the hexagonal lattice.

In three dimensions there are fourteen Bravais lattices in seven different types of structures. One of themost important is the cubic structure that has three Bravaislattices: Simple Cubic (SC); Body CenteredCubic (BCC); Face Centered Cubic (FCC). These Bravais lattices can be again classified by the symmetryoperations described before and are shown in Fig.2.6.


Triclinic−P Monoclinic−P Monoclinic−B

Orthorhombic−P −C −I

Orthorhombic−F

Hexagonal−P

Tetragonal−P −I

Trigonal−R Cubic−P

Cubic−I Cu bic−F

Figure 2.6: Bravais lattices in three dimensions.

2.1.1 The reciprocal lattice

As we said before, symmetries are very useful in helping to describe the various physical properties ofmaterials. In a perfect crystal the properties of the systemshould not vary through the lattice. One of themost important physical properties is the density of particles,ρ(r). Given a set ofN atoms at positionsRi

with i = 1, ..., N the density is simply

ρ(r) =N∑

i=1

δ(r − Ri) (2.6)

(observe that if we integrate the above expression in all space we obtainN which is the total number ofatoms). Observe that from our previous discussion we can rewrite for a crystal

Ri = T + Rα . (2.7)

Therefore, for a crystal, one can rewrite the density as

ρ(r) =∑

T

∑

α

δ(r − T − Rα) . (2.8)

Observe therefore that if we haveNs unit cells withNb atoms in a basis we must haveN = NsNb.The translation symmetry of the crystal requires that

ρ(r) = ρ(r + T) (2.9)

for all T as one can easily see if we use (2.8) and notice that the sum of two lattice vectors is another latticevector. Observe that this property has strong consequencesif we write the Fourier series of the density:

ρ(r) =∑

q

ρ(q)eiq·r (2.10)


then from (2.9) we must haveq = G where

G ·T = 2πn (2.11)

wheren is an integer. It is clear from the definition ofT in (2.4) that we can always defineG by

G = m1b1 +m2b2 +m3b3 (2.12)

wherem1,2,3 are three arbitrary integer andb1,2,3 three independent vectors such that

bi · aj = 2πδij (2.13)

δii = 1 andδij = 0 if i 6= j is a Kronecker delta. It is simple to show that

b1 = 2πa2 × a3

a1 · (a2 × a3)(2.14)

and the other vectors are just cyclic permutations of the above relation. By the same token one has

a1 = 2πb2 × b3

b1 · (b2 × b3). (2.15)

Observe that these new vectors span a new lattice. This lattice is called reciprocal lattice.

Everything we said before in regards to Bravais lattices is also valid for reciprocal lattices. In particular,the Wigner-Seitz cell of the reciprocal lattice is called Brillouin zone. Moreover, observe that (2.11) definesplanes in the real space such that each plane is perpendicular to G (see Fig.2.7). In order to see that this istrue remember that the density is written as:

ρ(r) =∑

G

ρGeiG·r , (2.16)

thus, for eachG the density is given by a plane wave of the formcos(G · r) in the direction ofG. Eachmaximum of the wave corresponds to a plane of atoms (since thedensity is maximum at these planes) wherethe empty space between the planes corresponds to a minimum in the wave. Since the wavelength of thiswave is2π/|G| and the distance between planes perpendicular toG is d we must have

d =2π

|G| . (2.17)

Another way to see this relation if through (2.11). Considertwo points in neighboring planes, sayRn andRn−1, such that|Rn − Rn−1| = d is the distance between planes. Then from (2.11) one has

|G| =2π

d. (2.18)

ThusG labels an infinite number of parallel planes in real space. The classification of these planes inparticularly useful in crystallography. Since we saw thatG = hb1 + kb2 + lb3 we can label planes by theset of number[hkl] which are called Miller indices. For instance, for a set of planes in thex direction wehave(100) and for a set of planes in thex + y direction we would have(110), etc. Observe that the planes(200) and(−100) = (100), for instance, are parallel to the planes(100). In this way it is very simple tothink at the planes in real space as labeled by reciprocal lattice vectors.

We have seen that the density can be written in terms of a Fourier series of reciprocal lattice vectors


G

n+1

n

n−1

d

Figure 2.7: Series of planes and its associated reciprocal lattice vector.

G, as in (2.16), which guarantees the periodic properties required by symmetry. The Fourier componentsρG can be calculated by the inverse transform. This is done by multiplying (2.16) bye−iK·r whereK is areciprocal lattice vector and integrating in the volume of the unit cell,V0. Observe therefore that we are leftwith the integral

∫

V0

ddrei(G−K)·r = V0δG,K . (2.19)

In order to prove that this is indeed true we first remember that the sum of two reciprocal lattice vectorsis another reciprocal lattice vector that can be written in the form (2.12). The result is trivial ifG = K.Moreover, observe that the integrand has the periodicity ofthe lattice and its integral over a unit cell cannotdepend on the choice of cell. In particular, it cannot changeif we translate the unit cell by an arbitrary vectorR. Therefore, from this result we have

ρG =1

V0

∫

V0

ddre−iG·rρ(r) . (2.20)

2.1.2 The one-dimensional chain

Consider a circular chain ofN atoms as depicted on Fig.2.8. The atomic density along the chain (which weparameterize byx) is simply

ρ(x) =

N∑

j=0

δ(x − ja) (2.21)

wherea is the lattice spacing. Eq. (2.21) has a simple Fourier series

ρ(x) =∑

k

eikxnk (2.22)

and because the system has discrete translational symmetryby a, that is,n(x + a) = n(x) (in (2.21) theshift bya is just equivalent to a renumbering of the lattice sites!), we must have

eika = 1

kj =2πj

a(2.23)

Thus, the values allowed by symmetry form a lattice with lattice spacing2π/awhich is the reciprocal lattice.The reciprocal lattice vectors are simplyG = 2πn

a . Also it is very simple to show that (2.19) and (2.20) are


correct. Let us rewrite (2.22) in terms of the allowed valuesof k

ρ(x) =∑

j

e2πijxa ρj (2.24)

where the sum runs over all integers (positive and negative). Multiply both sides of the above equation bye−2πimx

a and integrate from0 to a∫ a

0dxe−2πimx

a ρ(x) =∑

j

ρj

∫ a

0dxe2πi(j−m)x

a . (2.25)

Now observe that forj 6= m one has

∫ a

0dxe2πi(j−m)x

a = ae2πi(j−m) − 1

2πi(j −m)= 0 (2.26)

becausee2πi(j−m) = 1. If j = m the integral obviously givesa. Thus, we conclude that,∫ a

0dxe2πi(j−m)x

a = aδm,j (2.27)

which is the one-dimensional version of (2.19). Going back to (2.25) one finds

ρj =1

a

∫ a

0dxe−2πij x

a ρ(x) (2.28)

which is the one-dimensional version of (2.20).

12

3N−1

N

x

Figure 2.8: Model for a one-dimensional crystal.

2.2 Elastic scattering

In order to illustrate the importance of the reciprocal lattice let us consider the problem of scattering of acrystal lattice by light with wavelengthλ. Imagine we send light into a crystal with a wave-vectork such that

2.2. ELASTIC SCATTERING 35

k = 2π/λ. The atoms in the crystal absorb the light and re-emit it in spherical waves as in Fig.2.9. Thesewaves interfere with each other and an observer located atR observes a scattered wave with wave-vectork′. Here we consider only theelasticscattering by the crystal so that no energy is lost in the scattering ofthe light with the crystal, that is,

k′ = k = 2π/λ . (2.29)

Consider an atom located at a positionr in the crystal. The amplitude of the electric field of incident lightat that position is

E(r) = E0eik·r (2.30)

whereE0 is a constant vector. The atom atr re-emits the light so that the contribution of this atom to theobserved light atR is

E(R) ∝ E(r)eik′·r′ (2.31)

wherer′ = R − r (see Fig.2.9). Thus,

E(R) ∝ E0eir·(k−k′)eik

′·R . (2.32)

The contribution from the whole crystal is obtained by integrating (2.32) over the entire volume:

E(R) ∝ E0eik′·R

∫

crystaldV ρ(r) eir·δk (2.33)

whereδk = k − k′ andρ(r) is the density of atoms. Observe that the first factor in (2.33) is just a phasefactor which is not important since it does not include the superposition of all the fields created by all atoms.As we have seen, the crystal has a periodic structure that needs to be taken into account. Since the crystal iscomposed byN unit cells we do not have to integrate over all crystal, we just have to integrate over one celland sum over all the other cells. This means that we can rewrite r = T + r′ wherer′ describes the positionof each atom in the unit cell. Thus, we can rewrite (whole) as

E(R) ∝ E0

∑

T

∫

CelldV ′ρ(T + r′)ei(T+r′)·δk

∝ E0

∑

T

eiT·δk

∫

CelldV ′ρ(r′)eir

′·δk (2.34)

where we have used the translational symmetry of the problem, that is,ρ(T + r′) = ρ(r′).

Let us now focus on the first term of (2.34). For an arbitraryδk the sums of the exponential is onlydifferent from zero if

T · δk = 2πn (2.35)

wheren is an integer. This condition means that in order to have a constructive interference we have torequire that

δk = G , (2.36)

is a reciprocal lattice vector. Thus, (2.36) tells us is thatthere is only scattering ifk′ = k − G, that


kk’

r R

ρ

Origin

Figure 2.9: Geometry for scattering of light by a crystal.

this, the incoming and the out-going plane waves can only differ by a reciprocal lattice vector. This hasprofound consequences. In the elastic scattering the energy of the incident and scattered beam is the sameand therefore|k′| = |k|. Thus, as a consequence we have,

(k′)2 = (k)2 + (G)2 − 2k ·G2|k| cos φ = |G| (2.37)

whereφ is the angle betweenk andG. But we know from (2.18) that associated with each vectorG wehave a set of planes such that|G| = 2π/d and given a plane wave we also have|k| = 2π/λ whereλ isthe wavelength of the beam. If instead ofφ we use the angle between the incident beam and the plane (seeFig.2.10) the above equation can be written as

2 d sin θ = λ (2.38)

which is the Bragg law for elastic scattering.

Another way to compute the same result is based on the use of basic quantum mechanics by calculat-ingthe differential cross-section for scattering of a plane wave with wave-vectork into another wave-vectork′. This is given in the Born approximation by

d2σ

dΩ≈ 2π

h|〈k|V |k′〉|2 (2.39)

whereV is the scattering potential. In a condensed systems (gas, liquid or solid) this potential is the sum ofthe atomic potential of each atom. Thus we can write

V (r) =

N∑

i=1

Vi(r − Ri) (2.40)

whereRi is the position of each atom. Using the fact that this potential is periodic we can write

V (r) =∑

G

VGeiG·r (2.41)

2.2. ELASTIC SCATTERING 37

where the Fourier components can be obtained as in (2.20). Thus the matrix element can be written as

〈k|V |k′〉 =1

V

∫

ddre−iq·rV (r) =1

V

∑

G

VG

∫

ddre−i(q−G)·r

=(2π)d

V

∑

G

VGδ(q − G) (2.42)

whereq = k−k′. What this last equation tells us is that there is only scattering if k′ = k−G, that this, theincoming and the out-going plane waves can only differ by a reciprocal lattice vector in complete agreementwith our previous discussion.

G

n+1

n

n−1

d

k

θ

k’

Figure 2.10: Bragg reflection through a set of planes.

We can now rewrite the differential cross-section as

d2σ

dΩ∝ |〈k|V |k′〉|2 =

1

(2π)d

∑

G

|VG|2δq,G , (2.43)

where we used the results of Appendix (2.5.1) for the substitution of the Dirac delta function for the Kro-necker delta. This equation is quite interesting because ittells us that the intensity of the reflection atq = G

is proportional to|VG|2. Thus, the Bragg condition is not sufficient in order to see elastic scattering at somevectorG but also it is required that the potential has a finite Fouriercomponent at this wavevector.

We can get even more insight into (2.43) if we rewrite the Fourier components of the potential in termsof the potential created by isolated atoms. In this case, within the unit cell, we write

V (r) =∑

α

Uαδ(r − Rα) (2.44)

whereUα is the strength of the potential for a particular atomα. Using (2.20) we find

VG =1

V0

∑

α

Uαe−iG·Rα (2.45)

which depends only on the atoms on a unit cell. In this case (2.43) becomes

d2σ

dΩ∝ 1

V 20

∑

G

∑

α,γ

U∗αUγe

iG·(Rα−Rγ)

=1

V0

∑

α,γ

U∗αUγF (Rα −Rγ) (2.46)


where

F (Rα) =1

V0

∑

G

e−iG·Rα . (2.47)

This result implies that the scattering depends on the localstructure of the atoms on the unit cell.

These equations simply considerably if we assume that the atoms are the same. In this case one canwrite

V (r) = U0

∑

α

δ(r − Rα) (2.48)

and therefore

VG = U0ρG (2.49)

where

ρG =1

V0

∑

α

e−iG·Rα (2.50)

is the Fourier transform of the density. It is straightforward to conclude from the above equations that

d2σ

dΩ∝ |U0|2S(q) (2.51)

where

S(q) = (2π)d∑

G

|ρG|2δq,G (2.52)

is called the static structure factor. Observe that a scattering at some vectorq = G is only possible ifρG 6= 0. This factor is the same that appears in equation (2.34). Thus, even if the scattering is allowed thepositions of the atoms on the unit cell determines if there isany intensity for that particular scattering.

The scattering in a disordered system such as a glass (or liquid very viscous fluid) can also be immedi-ately obtained from these equations. Since the system is disordered the unit cell is the volume of the systemitself. Moreover, the sum in (2.50) is a sum of random phases that leads to destructive interference if thephases vary wildly. Thus, from (2.50),

ρG =1

V

∑

α

e−iG·Rα = nδG,0 (2.53)

wheren = N/V is the average density of the system. By direct substitutionof this expression in (2.52) wefind

S(q) = (2π)dn2δq,0 (2.54)

shows that in this case the system only hasforward scattering.

2.3. DEFECTS IN SOLIDS 39

2.2.1 Experimental constraints

So far we have discussed scattering but we did not specify what type of scattering we are talking about, thatis, what kind of probes we can actually use. Observe that the only condition for scattering is given by theBragg law, (2.38), that requires that the wavelength of the probe must be such that,λ < 2d. Sinced is oforder of1 we need probes with very short wavelengths. The first probe that comes to mind is light. Lightinteract with the charged particles in the system via what iscalled minimal coupling, that is, we replace themomentump of the charged particle byp− eA/c whereA is the vector potential. The energy of light canbe written as

E = hω = hc|k| =hc

λ. (2.55)

Forλ ≈ 1 one needsE ≈ 104 eV which is the X-ray part of the spectrum.Another possibility is scattering by electron waves. In this case the scattering process is due to the

electron-electron interaction in the system. The energy ofan electron is given by

E =h2k2

2me=

h2

2meλ2. (2.56)

Forλ ≈ 1 one needsE ≈ 100 eV.However, the most precise measurement of crystal lattices is done by neutron scattering. Neutrons have

no electric charge and therefore are insensitive to charge degrees of freedom in the solid. They interact withthe magnetic moments in the solid (nuclear and/or electronic). The energy has the same form as above forelectrons but since its mass is a thousand times larger the relevant energies are a thousand times smaller, thatis,E ≈ 0.1 eV.

Although we can have different probes it is easy to see that each one of them measures the system atdifferent scales of energy. Thus, the choice of probe depends strongly on what kind of energy scale onewants to probe. In condensed matter physics one is usually interested in energies of the order of a few meVwhich is the energy scale of the neutrons. Neutron scattering is also particularly important because it probesdirectly the magnetic excitations. Electron scattering isa complex probe because the scattered electron tendsto interact strongly with the other electrons in the system (this is called multiple scattering) and our simpleBorn approximation formula is not valid. We really need a probe that interacts weakly with the system ofinterest. X-rays have an energy that is usually orders of magnitude larger than the energy scales of interest.It turns out, however, that they are excellent in order to measure static properties such as,S(q), via elasticscattering.

2.3 Defects in solids

So far we have discussed only perfect crystalline structures where the atoms occupy sites of a periodiclattice. It turns out, however, that real crystals are not perfect and contain a series of different defects. Themost common types of defects are vacancies and interstitials that are called point defects since they involvethe subtraction of isolated atoms. There are also line defects which involve the entire displacement of planesor lines of atoms and for this reason are called dislocations. These defects are common in most materialsand are responsible for many effects observed in crystals.

At finite temperatures all crystals contain a certain numberof vacancies or interstitials. This is due topurely entropic effects. The simplest type of defect is depicted on Fig.2.11 and it is called a Schottky defect.This kind of defect is essentially due to an atom that is missing from its original position. The missing atomin this case creates a "hole" in the crystal and therefore a local defect.


Figure 2.11: Vacancy in a crystal.

In order to understand the entropic nature of such a defect consider a perfect lattice made out ofNatoms andM vacancies which are randomly distributed (we are considering that there is no clustering ofthese vacancies and that the energy of the vacancies do not depend whether there are other vacancies in itsimmediate neighborhood). In this case it is clear that the system has many equivalent configurations. Thenumber of these configurations is simply(N + M)!/(N !M !). Thus, the total entropy at zero temperatureis:

S = kB ln

(

(N +M)!

N !M !

)

. (2.57)

Here we are going to consider the case whereN,M ≫ 1 butM/N is finite. Thus we can use the Stirlingapproximation

ln(N !) ≈ N ln(N) −N . (2.58)

Now consider the free energy of the system which is given by

F (M) = U(M) − TS(M) (2.59)

whereU is the internal energy of the system withM vacancies. In the case under consideration the energyof createM vacancies is justM times the energy to create a single vacancy since we are disregardingvacancy-vacancy interactions. In this case we have

U(M) = ǫ0M (2.60)

whereǫ0 is the energy required to create a single vacancy. For a fixed value ofN the equilibrium is attainedwhen the free energy is a minimum with respect to variations of M . Thus, we have to minimize

F (M) ≈ ǫ0M − kBT ln

[

(N +M)N+M

NNMM

]

(2.61)

with respect toM keepingN +M = Ns the number of sites constant. We write

F (M) ≈ ǫ0M − kBT ln

[

NNss

(Ns −M)Ns−MMM

]

, (2.62)

2.4. PROBLEMS 41

and the derivative becomes

∂F

∂M= ǫ0 − kBT ln

(

Ns −M

M

)

, (2.63)

and thus

ǫ0 ≈ kBT ln

(

Ns −M

M

)

, (2.64)

or

M(T ) =Ns

eǫ0

kBT + 1, (2.65)

which shows that there is always a finite number of vacancies in the crystal. Observe that whenkBT ≫ ǫ0we haveM(T ) ≈ Ns/2. However, in normal materialsǫ0 is of order of 1 eV (10, 000K) and therefore oneusually has the opposite limit, that is,kBT ≪ ǫ0, where

M(T )/Ns ≈ e−

ǫ0kBT (2.66)

that is, there is an exponentially small number of vacanciesin the crystal.

Figure 2.12: Interstitials in a crystal.

Another type of defect is the interstitial which is similar to a vacancy but the atom, instead of leaving thebulk of the crystal, moves to an intermediary position between other atoms. This is shown on Fig.2.12 and itis called a Frenkel defect. This type of defects are more common in ionic crystal where a positively chargedion can move in between negatively charged ions in a crystalline matrix. In this case charge neutralityrequires that an equal number of negative and positive defects are generated.

It is also possible to replace a negative ion by an electron localized at the defect. In this case the electronis localized in a quantum well. Int his case the absorption oflight in the crystal changes due to the existenceof vacancies since the electrons in a quantum well absorb light at different frequencies than in a perfectcrystal. It implies that the crystal changes its color. For this reason this kind of defects are called colorcenters. An example is depicted on Fig.2.13.

2.4 Problems

1. Prove that forn = 5 it is not possible to define a smallest vectora1 that generates the lattice. Theproof of the non-existence of Bravais lattices forn ≥ 7 is analogous.


+ − + + +

+ + + +

+ + + +

+ + + +

+ + +

+ + + +

+ + + +

+ + + +

− − −

−−−−

− − −

−−−

− − − −

−−−

− − − −

−−−−

+

−

−

−

Figure 2.13: Two types of color centers in an ionic crystal.

2. Prove that the Wigner-Seitz unit cell of a planar hexagonal lattice of lattice spacing2R has an area of2√

3R2. Show that the fractional area occupied by the spheres relative to this unit cell area is0.907.

3. Prove that the vectorsbi can be written as in (2.14).

4. Prove that (2.15) is indeed correct.

5. Prove that the volume of the Brillouin zone,VB = (2π)3/V0 whereV0 is the unit cell volume.

6. Show that by translating the unit cell to another cellV ′0 in (2.19) the integral is zero ifG 6= K.

7. We have seen that a reciprocal lattice vector labels a series of planes(hkl). Show that a equivalentway to label the planes in real space consists of two steps: 1)find the intercept of the plane with theaxes in terms of the vectorsa1,a2,a3; 2)take the reciprocal of these numbers and then reduce to threeintegers having the same ratio, usually the smallest three integers. The result, that is,(hkl) is theindex of the plane.

8. The primitive vectors for a bcc lattice are

a1 =a

2(−x + y + z)

a2 =a

2(x− y + z)

a3 =a

2(x + y − z) . (2.67)

Find the reciprocal lattice vectors. What lattice does it form? Make a drawing of the two lattices.

9. The primitive vectors for a fcc lattice are

a1 =a

2(y + z)

a2 =a

2(x + z)

a3 =a

2(x + y) . (2.68)

Find the reciprocal lattice vectors. What lattice does it form? Make a drawing of the two lattices.

2.4. PROBLEMS 43

10. In this problem we are going to study the effect of the second factor in (2.34), the static form factor.We write it as:

SG =

∫

CelldV n(r)eir·G (2.69)

sinceδk = G is the Bragg condition. When there areNb atoms in the unit cell we can write

n(r) =

Nb∑

j=1

nj(r − rj) (2.70)

wherenj now depends on the particular atom. Show that

SG =

Nb∑

j=1

fjeirj ·G (2.71)

where

fj =

∫

CelldV nj(r)e

ir·G (2.72)

is calledatomic form factorwhich depends only on the kind of atom is participating on thelatticeformation. Consider now thatrj can be written in terms of the primitive vectors as

rj = xja1 + yja2 + zja3 (2.73)

wherexj , yj, zj areany real number. Using thatG =∑3

i=1mibi show that

SG =

Nb∑

j=1

fje2πi(m1xj+m2yj+m3zj) . (2.74)

11. CalculateSG in previous problem for a fcc lattice with one atom per unit cell assuming that the fcclattice can be thought of a simple cubic lattice with a basis given by the vectors(0, 0, 0), (0, 1/2, 1/2),(1/2, 0, 1/2), (1/2, 1/2, 0). (ii ) Assume thatfj = f and show that ifm1,m2,m3 are all even or allodd we haveSG = 4f . (iii ) Show that if one of themj is even and the other two are odd thenSG = 0. (iv) Show that if one of themj is odd and the other two are even thenSG = 0. (v) Comparethe scattered reflections that you would get from a fcc lattice without a basis and the ones you got fromassuming that the fcc lattice can be seen as a simple cubic lattice with a basis. Do you get identicalresults?

12. Consider a two dimensional crystal in a square lattice with two different types of atoms with differentcross-sections (a two dimensional version of NaCl). What would be a result of a neutron scatteringexperiment in such a system? What are the allowed values of momentum for scattering?

13. Consider an ionic crystal with positive and negative charges. Assuming that the energy required tocreate an interstitial with positive (negative) charge requires an energyǫp (ǫn) and that the system hascharge neutrality show that the number of positive and negative interstitials is

np = nn =√

NpNne−β

ǫp+ǫn2


whereNp(Nn) is the number of sites with positive (negative) charge.

2.4.1 Appendix: Dirac and Kronecker delta functions

There is a little caveat about equation (2.42). Observe thatq is defined in the continuum (that is, it canvary continuously from0 to∞) while the reciprocal lattice vectorG is defined over a discrete set. Thus, inorder for the summation in (2.42) to make sense we should havea Kronecker delta instead of a Dirac deltafunction. The solution for this little problem is given by the quantization of the scattering states in a box.If we require the plane waves outside the material to have periodic boundary conditions in a box of sizeL,then we must have

ki =2πniL

(2.75)

wherei = x, y, z. Thus a Dirac delta can be written as

δ(k) = δ(kx)δ(ky)δ(kz) = δ(2πnx/L)δ(2πny/L)δ(2πnz/L)

=V

(2π)dδk,0 (2.76)

whereδk,0 is a Kronecker delta andV = L3 is the volume of the quantization box. In this way one canreplace sums by integrals,

∑

q

→ V

(2π)d

∫

ddq . (2.77)

Another way to rewrite this expression is due to the result ofProblem 5: ifN is the number of primitivecells andVB is their volume we have immediately,

∑

q

→ N

VB

∫

ddq . (2.78)

Chapter 3

Elasticity Theory

Point defects are not the only type of defects that occur in crystals. There are extended defects that areresponsible for very important mechanical properties of crystalline systems. Line or plane defects in crystalsare possible under application of pressure or stress. When adeformation due to stress is reversible we callit a elastic deformation, otherwise, when it is not reversible, we call it a plastic deformation. In order tounderstand the difference between plastic and elastic deformations consider the simple toy model proposedby Frenkel for the shear of a perfect crystal as shown in Fig.3.1. Suppose a stress,σ, is applied to a planeof atoms which is displaced by another plane of atoms by an amount x. Under Hook’s law the stress,σ, islinearly related to the displacement by the shear modulusG

σ(x) ≈ Gx

d(3.1)

whered is the distance between planes. Of course this equation is valid for small displacements. Supposewe apply a strong shear stress so that all the atoms move one lattice spacinga. In this case we are back tothe original situation with the atoms in their equilibrium position in zero stress. This implies that the stressmust be a periodic function of the lattice spacing. For simplicity let us consider a harmonic variation withposition:

σ(x) = Ga

2πdsin

(

2πx

a

)

(3.2)

that turns into (3.1) in the limit ofx ≪ a/(2π). Observe that (3.2) shows that the maximum stress occursfor x = a/4 where

σc = σ(a/4) = Ga

2πd. (3.3)

If the stress applied is larger thanσc the upper plane in Fig.3.1 has to move freely and never returnto theoriginal unstressed situation. This would imply plastic flow of the crystal. Sincea ≈ d ≈ 1 the criticalstress is approximatelyG/6. It turns out that this prediction is completely at odds withthe experiments. Forinstance for a single crystal of Al we find that experimentally σc ≈ G/60, 000! Thus, the conclusion is thatplastic deformations occur at a much smaller stress than thetheory predict and they must be generated notby elastic deformations of the crystal but by defects. Thesedefects are called dislocations.

Plastic deformations in crystals occur when planes of atomsslide over each other due the presence ofdislocations. The simplest types of dislocations are edge dislocations and screw dislocations. In a edgedislocation there is a mismatch of planes as if an extra planes of atoms have been inserted in the crystalwhile in a screw dislocation there is a mismatch between planes of atoms as shown in Fig.3.2.

45

46 CHAPTER 3. ELASTICITY THEORY

d

x

a

σ

Figure 3.1: Simple model for shear stress in a perfect crystal.

(a)

(b)

LowerPlane

UpperPlane

Figure 3.2: (a) Edge dislocation; (b) Screw dislocation.

3.1 Elastic properties of crystals

As we have discussed in the last section the position of thei-th atom in a crystals is given byRi = T+Rα

whereT is a vector in the direct lattice andRα is a vector in the unit cell. Moreover, in Chapter 1 wehave seen that the equilibrium position of atoms is given by the various electronic interactions. Usually onefinds that the competition between a long range attractive force and a short range repulsive force leads to amean distance between atoms that we call lattice spacing. Inthis section we are going to consider the elasticenergy required to deform the atoms from their original equilibrium positions. Since the perfect crystal isthe lowest energy state for the system, local deformations cost energy that is called elastic energy. Let usassume that some part of the crystal is displaced by a quantity u(r) that indicates a displacement of|u| in thedirection ofu at positionr. Obviously the free energy of the system due to this displacement is a functionof u(r). The form of the free energy can, many times, be obtained fromsymmetry considerations.

The first obvious symmetry is the translations of all atoms ofthe crystal by a fix valueu(r) = u0. It isclear that the free energy of the system cannot depend on the value ofu0 and therefore it can only dependon derivatives ofu(r) which we denote by

∂iuj =∂uj∂xi

(3.4)

3.1. ELASTIC PROPERTIES OF CRYSTALS 47

wherei, j = 1, 2, 3, ..., d corresponding tox, y, z.... Observe that∂iuj can be seen as a matrix, or moreformally, a tensor withd × d components. From this argument we conclude that the free energy has to bewritten asF = F [∂iuj].

Another symmetry of the problem is inversion symmetry, thatis the free energy has to be invariant underu → −u. It implies that linear terms are not allowed and thereforeF = F [(∂iuj)

2].Now let us consider another important symmetry which is the symmetry of rigid rotation of the lattice

by an angleδθ. Consider, for instance, a plane of atoms labeled by a reciprocal lattice vectorG which isrotated by this angle. Along this plane all the atoms are displaced byu(r) = |r|δθ as shown in Fig.3.3.Another way to write this displacement is to consider the vector δ~θ oriented anti-clockwise in the rotationdirection. It is clear that

u(r) = δ~θ × r . (3.5)

This relation can be easily inverted with the help of differential calculus to

δ~θ =1

4∇× u(r) (3.6)

which expresses the angle of rotation in terms of the displacement. Since the whole crystal is rigidly rotatedthe free energy cannot depend ofδ~θ, that is, the system is invariant under rigid rotations. Observe that in(3.6) the displacements appear in combinations of the form∂iuj − ∂jui which is the anti-symmetric part ofthe tensor∂iuj. Thus, due to the rotation symmetry the free energy can only depend on the symmetric partof the tensor∂iuj, that is,

uij =1

2(∂iuj + ∂jui) (3.7)

which is called thestrain tensor.In summary, due to rigid translations, inversions and rotations the free energy due to elastic deformations

of a crystal can only be a function ofuijukl.

r

u(r)

δθ

Figure 3.3: Rigid rotation of an atomic plane.

It is very simple to understand the physical meaning of each component of the strain tensor. Considerfor instance a compression or dilation of the system along the directionx. If Lx is the size of the crystal inthis direction then

δx =δLxLx

(3.8)

gives the relative compression in that direction (obviously for compressionδx < 0 and for dilationδx > 0).


In this case the volume of the whole crystal is changed byδV . By the same token the volume of the unitcell, V0 = |a1 · (a2 × a3)| is changed byδV0 becauseai changes under compression. It turns out, however,that sinceV = NV0 whereN is the total number of atoms we must have

δV

V=δV0

V0. (3.9)

This relation is valid because we are considering the crystal in the absence of vacancies and interstitials. Inthe presence of defects (3.9) has to be modified. Moreover, since the reciprocal lattice vectors are tied to thedirect lattice vectors byai · bj = 2πδij it is clear that a compression in real space will lead to a dilation inreciprocal space and vice-versa.

Consider the set ofN planes separated by a distanced in the x direction before the application ofpressure. In this case we haveLx = Nd. After pressure is applied the distance between planes becomesd′

so thatL′x = Nd′. Therefore

L′x − LxLx

=d′ − d

d= δx . (3.10)

We label the displacement of thenth plane of the crystal byun. Let us assume that the first plane is fixedbefore and after pressure is applied so thatu0 = 0. It is simple to see in Fig. 3.4 that second plane isdisplaced byu1 = d′ − d = δxd, the second plane is displaced byu2 = 2d′ − 2d = 2δxd and so on. Ingeneral

un = nδxd . (3.11)

If we label the position of each plane byx = nd we see that this last expression can be rewritten as

u(x) = δxx

∂xu(x) = δx . (3.12)

d

d’

u u u u12 3 4

Figure 3.4: Compression of a series of planes.

In a more general way we can consider a set of planes labeled bythe reciprocal lattice vectorG. Weknow that|G| = 2π/d whered is the distance between planes. After compression or dilation we must have


d′ = (1 + δx)d and therefore the reciprocal lattice vector changes for an infinitesimal value ofδx by

G′ = (1 − δx)G

G′ − G = −δxG . (3.13)

Moreover, in this case, the displacement of the atoms due to the compression or dilation can be written as

G · u(r) = (G − G′) · r= δxG · x , (3.14)

and therefore

u = δxx (3.15)

leading to:

uxx = δx , (3.16)

and all other derivatives vanish. Thus, it is clear thatuii measures the lattice compression or dilation in thedirection of the vectorxi.

The total volume of the system isV = LxLyLz and therefore we must haveδV = δLxLyLz +LxδLyLz + LxLyδLz. Thus, it is simple to show that

δV

V=∑

i

uii (3.17)

that is, the relative change in the volume of the solid is given by the trace of the strain tensor. Equation(3.17) has to be interpreted carefully. Observe thatuii(r) is a local function of the position and of courseδV is a global change in the system. Thus (3.17) is really only valid in a perfect crystal without defectslike vacancies where changes in volume can be non-uniform. In general the diagonal elements of the straintensor are related to the local changes in the unit cell volume such that,

δV0(r)

V0(r)=∑

i

uii(r) (3.18)

which is a local relation. Of course, for ordered systems equation (3.9) is valid and (3.17) is identical to(3.18).

We can always decompose the distortion of a solid in terms of volume changes and shear (that neveraffects the unit cell volume). We have seen that volume changes have to do with the diagonal elements of thestrain tensor. The off-diagonal components of the strain tensor represent the shear distortion of the systemas you can easily show. Since shear does not change the volumeof the crystal it is represented by a tracelesstensor

sij = uij − δij

∑

k ukkd

. (3.19)

Notice that indeed∑

i sii = 0 since∑

i δii = d.

From now on we are going to consider only small distortions ofthe crystal and from the symmetry


arguments the free energy has to be written as

F =1

2

∫

dr∑

i,j,k,l

Cijkluijukl (3.20)

whereCijkl are the so-called elastic constants of the crystal which in principle can haved4 components.Now observe thatuij = uji and therefore we must have the following symmetries:

Cijkl = Cjikl = Cijlk = Cjilk = Cklij . (3.21)

In addition the free energy has extra symmetries that dependon the point symmetries of the crystal itself. Ahighly symmetric crystal has less independent elastic constants than less symmetric crystals. For instance, athree dimensional cubic crystal has3 independent elastic constants while a triclinic crystal has 21 indepen-dent elastic constants. Moreover, it is very simple to estimate the order of magnitude of the elastic constantsif we remember thatuij is a dimensionless quantity and thereforeCijkl has dimensions of energy dividedby lengthd, that is, energy density. The energy here is just the biding energy per atom of the solid which isof order of a few electron volts while the length is of order ofthe lattice spacing, that is, a few angstroms.

Things simplify considerably in isotropic solids where compression and shear stress are independent ofthe direction they are applied. It must be clear that in this case there are only two elastic constants: oneassociated with compressions and dilations (the so-calledbulk modulus,B) and another associated withshear distortions (the so-called shear modulus,G). Since we have seen that compressions have to do withuii and shear are related tosij the free energy has to be written as

F =1

2

∫

dr

B

(

∑

i

uii

)2

+ 2G∑

ij

s2ij

. (3.22)

In order to see that this expression has indeed the form of (3.20) we use (3.19) and rewrite (3.22) as

F =1

2

∫

dr

(

B − 2G

d

)

(

∑

i

uii

)2

+ 2G∑

i,j

u2ij

(3.23)

which has the form of (3.20) with

Cijkl =

(

B − 2G

d

)

δijδkl +G (δilδjk + δikδjl) (3.24)

that has the symmetry properties as required.

Of course the expression for the free energy as given by (3.20) is useful if we can relate the internaldistortions (the strain) with the external agents such as pressure or strain. In order to do that one has tocalculate the work done on the system by an external force. Consider an external forceF applied to anelement of volumeδV . If the interaction between the atoms is short ranged this force is transmitted by theneighboring volumes through the surfaceδS that surrounds the volumeδV . The total force in the volumein theith direction is

Fi =

∫

δVdrfi (3.25)

wheref is the force per unit of volume of the undistorted solid. Since the force applied is a local function


of the position it can be expressed as a gradient with respectto the undistorted crystal and we write

fi =∑

j

∂jσij (3.26)

whereσij is thestress tensor. Substitution of (3.26) into (3.25) leads to

Fi =

∫

δS

∑

j

dSjσij (3.27)

where we used Gauss theorem. Observe that the stress tensorσij gives the force per unit of area in thedirection i exerted by the surrounding medium on a volume element acrossits surface oriented on thedirectionj. Consider a solid surrounded by an isotropic fluid at pressure P . This solid will experience anstress given by the hydrostatic pressure which is−P . In this case the stress tensor is simply

σij = −Pδij . (3.28)

Suppose uniaxial pressureT is applied along thex axis of a crystal. In this case it is obvious that

σxx = T (3.29)

and all other components are zero. In the more generic case consider the work done by a force densityf

which displaces the volume elements of a crystal byu(r). This work is simply

δW = −PV =

∫

drf · u(r)

=

∫

dr∑

i,j

∂jσijui(r)

= −∫

dr∑

i,j

σijuij (3.30)

where we have integrated by parts and neglected the surface terms. The change in the free energy due to thechanges in the strain field is the negative of the work done by the internal forces, thus,

δF = −δW =

∫

dr∑

i,j

σijuij (3.31)

Observe that this leads to the important relation:

σij =∂F

∂uij(3.32)

which relates the strain tensor with the stress tensor.

For the case of the isotropic crystal, accordingly to (3.22), we have:

σij = B∑

l

ullδij + 2Gsij (3.33)


that we can invert to write the strain in terms of the stress. Firstly we observe that fori 6= j we have

σij = 2Gsij = 2Guij (3.34)

where we have used (3.19). Secondly, fori = j we have

σii = 2Guii +

(

B − 2G

d

)

∑

l

ull (3.35)

and therefore∑

i σii = dB∑

i uii. Thus, we conclude that

uij = δij

∑

l σlld2B

+1

2G

(

σij − δij

∑

l σlld

)

. (3.36)

This last equation is very useful. Consider a solid subject to a hydrostatic pressureP . Then, from (3.36)one has

∑

i

uii =1

dB

∑

i

σii = −PB. (3.37)

From (3.17) one finds immediately

δV

V= −P

B1

B= − 1

V

δV

P(3.38)

which is the usual thermodynamic definition of the bulk modulus as expected.

Suppose that uniaxial stressT is applied in thez direction. Again we have

uzz =T

d

(

1

dB+d− 1

2G

)

uxx = uyy = −Td

(

1

2G− 1

dB

)

. (3.39)

Observe that the change in the size of the system alongz always follows the applied force while in thetransverse directions it will depend on the rationdB/(2G).

The Young modulus,Y , of the system is defined as

Y =T

uzz

Y =2d2GB

2G+ d(d− 1)B(3.40)

and the Poisson’s rationν is defined as

ν = −uxxuzz

=dB − 2G

d(d− 1)B + 2G. (3.41)

3.2. PROBLEMS 53

3.2 Problems

1. Write down the elastic free energy for a three dimensionalcrystal with cubic symmetry. Show thatthe elastic tensorCijkl has only 3 independent components, namely,C1111, C1122 andC1212 (1, 2, 3refers tox, y, z, respectively). Calculate: 1)the bulk modulus;

2)the Poisson ratio for stresses along one of the symmetry axes in terms of the elastic constants.

Chapter 4

Atoms in motion

In the previous section we studied the problem of static deformation of a crystal. As we have seen previouslya crystalline structure each time we take an atom from its position we have to pay an energetic price which forelastic deformations. For small deformations the energy isa simple quadratic function of the displacement.It is known from basic quantum mechanics that even at zero temperature motion does not to cease to existentirely due to quantum fluctuations. The basic example of this effect is the harmonic oscillator problemwhich is described by a potentialV (x) = mω2x2/2 wherem is the oscillator mass andω its oscillationfrequency. In quantum mechanics the ground state of this problem has finite energyhω/2 and therefore evenin the ground state the oscillator is not static but fluctuates with amplitudeA ≈ h/(mω) (sinceV (A) ≈hω/2). In classical mechanics the lowest energy state have zero energy and thereforeA = 0 (indeed, whenh → 0 we recover the classical case). Therefore, in order to understand the behavior of solids at very lowtemperatures one has to take into consideration the kineticenergy of the oscillations in the solid.

4.1 The one-dimensional chain

As an example let us consider the simplest case of atoms with massM attached to each other by springswith strengthκ as shown in Fig.4.1. Classically the lowest state of the problem has all atoms at rest at somecharacteristic distancea between them. This characteristic distance can be traced back to the equilibriumposition of atoms in a molecule as we discussed in Chapter 1. Let us now displace thenth atom by a smallvalueun. Because the atoms are tied to each other by electrostatic forces the displacement of one of themwill cause the displacement of the others. This energy, as discussed in the last chapter, can be written as

U =∑

n

κ

2(un − un+1)

2 (4.1)

since we are assuming that only the nearest neighbor atoms are coupled. IfU was everything we would haveit is clear that the configuration of lowest energy hasun = 0 for all n. But in quantum mechanics we haveto include the kinetic energy of the atoms which is given by

K =∑

n

p2n

2m(4.2)

wherepn is the momentum of thenth atom. The quantization condition for this problem is that the dis-placement and the momentum are canonically conjugated and therefore have well defined commutation

55

56 CHAPTER 4. ATOMS IN MOTION

relations:

[un, pm] = ihδn,m . (4.3)

As we all know from basic quantum mechanics the fact that two operators do not commute imply theHeisenberg uncertainty principle, that is,δpnδun > h, which implies that even we are certain that theoscillator is at certain positionun from equilibrium we lose completely the information about its momentum.This is exactly what causes the harmonic oscillator to have aground state with finite amplitude.

There are many ways to study the HamiltonianH = K + U . Here we are going to study the problemvia the equations of motion by using the Heisenberg representation for the problem. In this representationthe operators evolve in time accordingly to:

ih∂un∂t

= [un,H]

ih∂pn∂t

= [pn,H] (4.4)

which can be easily calculated by using (4.3) and the Hamiltonian’:

∂un∂t

=pnM

∂pn∂t

= −κ (un+1 − 2un + un−1) (4.5)

which can be rewritten by substituting the first equation into the second one:

∂2un∂t2

= − κ

M(un+1 − 2un + un−1) (4.6)

which is the equation for the time evolution of the operatorun. This equation is simply a simple secondorder linear differential equation and can be solved by assuming un(t) has a simple harmonic form, that is,un(t) = une

iωt which leads to

(ω2 + 2κ/M)un − κ/M (un+1 + un−1) = 0 . (4.7)

Notice that the above equation relates the displacement atnth atom with the displacement atn+1 andn−1atoms. The solution to this problem is given again by a simpleharmonic solution:

un = ueikna (4.8)

whereu is a constant. Direct substitution of (4.8) into (4.7) requires that

ω(k) = 2

√

κ

M|sin(ka/2)| . (4.9)

Which shows that there is a one to one correspondence betweenthe frequency of oscillation andk the wave-number of the oscillation. Observe that the periodicity of the chain requires thatuN+1 = u1 which, by (4.8)requires that

eikNa = 1

k(m) =2πm

Na(4.10)

4.2. PHONONS IN HIGHER DIMENSIONS 57

wherem is an integer. Therefore that the wave-numbers are quantized in units of2π/(Na) and these give theallowed quantum states of the problem. Notice, however, that the total number of states in the problem hasto be conserved. In the absence of interactions between the atoms there areN allowed states in the problemcorresponding to the rotation of the chain byN×2π/N . We have to end up with the same number of states inmomentum space, as well. IfN is even the allowed values ofm arem = 0,±1,±2, ...,±(N/2 − 1),N/2and forN odd,m = 0,±1,±2, ...,±(N − 1)/2. . Thus, we see thatk defined above varies between−(π/(2a)(1 − 1/N) ≤ k ≤ π/(2a). In the limit of a macroscopic number of atoms,N → ∞, thedistance between the allowed states shrinks to zero and the wave-numbers form a continuum in the interval−π/(2a) < k ≤ π/(2a). This is exactly the Brillouin zone for a one dimensional system, as expected fromthe periodicity of the problem.

Finally, we should point out that in the limit of very long wavelengths, that is, whenk ≪ 1/a we canexpand (4.9) as

ω(k) ≈ a

√

κ

M|k| (4.11)

and we see that the frequency of oscillation is linear with the wavenumber. This relationship is identicalto the dispersion of photons:ω(k) = c|k| wherec is the light velocity. Like photons the oscillations weare discussing propagate through the solid which a characteristic velocity cs = a

√

κM which is called the

sound velocity. Indeed, from basic quantum mechanics we know that the group velocity of a wavev(k) isgiven by hv(k) = dEk/dk. Remember that in quantum mechanics there is no distinctionbetween waves(or oscillations) and particles. Indeed we could say that wehave discovered a new particle which has beennamed acoustic phonon which propagates with the sound velocity. At longer wave-lengthsk ≈ π/(2a)the phonon does not propagate sinceω(k) ≈ 2

√

κM does not depend onk and therefore its group velocity

vanishes. Observe that phonons are an effect of the interaction between atoms (whenκ → 0 we findω(k) = 0) and they do not exist outside of the many-body system. This is the major difference betweencondensed matter physics and high energy physics where particles exist in the vacuum.

M

M M1 2

uu un n+1n−1

(a)

(b)

a

Figure 4.1: One-dimensional lattice: (a) one atom per basis; (b) two atoms per basis.

4.2 Phonons in higher dimensions

We can now generalize the discussion of the one dimensional problem to higher dimensions quite easily.We label each atom in a Bravais lattice by a vectorR + Rα. If we allow the atoms to move by an amountuα(R) then the instantaneous position of the atom is

rα(R) = R + Rα + uα(R) . (4.12)


Since the atoms undergo harmonic oscillation the potentialenergy associated with the displacement has thegeneric form

U =1

2

∑

R,R′

∑

α,β

∑

µ,ν

uα,µ(R)Kµ,να,β(R −R′)uβ,ν(R

′) (4.13)

whereKµ,να,β with µ, ν = x, y, z andα, β = 1, ...,Nb is a matrixdNb × dNb which tells us how the energy

of the system change when displace an atomα in some unit cell at positionR in the directionµ, relative toanother atomβ in the unit cell located atR′ in the directionν. Observe that this matrix can only dependon the relative position of the atoms. This is an important property as we are going to see. In the onedimensional problem discussed previously where we have just one atom in the unit cell and just one directionto move them this matrix has a very simple form which you can directly check, namely,

K(n −m) = κ(2δn−m,0 − δn−m,1 − δn−m,−1 . (4.14)

The kinetic energy of the atoms is simply

K =∑

R,α

P2α(R)

2Mα(4.15)

whereMα is the mass of each atom in the basis andPα the momentum of each atom. The momentum andthe displacement are canonically conjugated, that is,

[

uµα(R), P νβ (R′)]

= ihδµ,νδα,βδR,R′ . (4.16)

The commutation relation leads to a quantum problem we want to solve.

Notice that the Hamiltonian of the systemH = K + U is rather complex and will change dependingon the type of lattice we are working on. In order to simplify the problem one has to look for genericproperties which are common to all Bravais lattices. There are two of these properties: (1) the harmonicinteraction between atoms only depends on the relative distance between the atoms; (2) all Bravais latticeshave inversion symmetry.

BecauseK only depends on the relative distance between the atoms it isconvenient to Fourier transformthe displacement operators:

uα(R) =1√N

∑

k

eik·Ruα(k) (4.17)

wherek is the wave-vector of the problem. If, like in the one dimensional case we studied previously, weimpose periodic boundary conditions in all directions, that is uα(x+Nxax, y, z) = uα(x, y +Nyay, z) =uα(x, y, z+Nzaz) = uα(x, y, z), whereNµ is the number of atoms in each direction (observe that the totalnumber of atoms isN = NxNyNz) andaµ is the lattice spacing in each direction, we will have quantizedvalues for the components ofk:

kµ =2πnµNµaµ

. (4.18)

where−Nµ/2 < nµ ≤ Nµ/2 whenNµ → ∞.


In this case the potential term in (4.13) becomes

U =1

2

∑

α,β,µ,ν

∑

k

uµα(k)Kµ,να,β(k)uνβ(−k) (4.19)

where

Kµ,να,β(k) =

∑

R

Kµ,να,β(R)e−ik·R . (4.20)

The kinetic term transforms accordingly

K =∑

k,α

Pα(k) · Pα(−k)

2Mα. (4.21)

Notice that (4.19) is diagonal in momentum space. Thus part of our problem is solved and we just haveto diagonalize the problem in the discrete indicesµ = x, y, z andα = 1, ...,Nb. We would like to find atransformation of coordinates such that (4.19) is diagonal. Therefore there must be a unitary transformationU (U−1 = UT ) betweenuµα and a new set of coordinatesqµα, that is,

uµα(k) =∑

ν,β

Uµ,να,β qνβ(k) (4.22)

such that∑

µ′,ν′,α′,β′

[

U−1]µ,µ′

α,α′ Kµ′,ν′

α′,β′Uν′,ν

β′,β = Mαω2α,µδα,βδµ,ν . (4.23)

This problem involvesdNb×dNb matrices and therefore we will finddNb values ofωα,µ. Instead of workingwith two indices,α andµ it is common to use a short notation and introduce a single index s = 1, ..., dNb

for the eigenvaluesωs. To each eigenvalue one has a corresponding eigenvector. Each eigenvector is relatedto a different polarization vectores(k). From (4.22) we see that each polarization vector corresponds to arow (or column) of the matrixU . Thus (4.22) can be rewritten as

uα(k) =∑

s

es,α(k)qs(k) (4.24)

where the polarization vectors are orthogonal. The orthogonality condition is simply given by the conditionthatUUT = I, which can be written as

dNb∑

s=1

eµs,α(k)eνs,β(−k) = δα,βδµ,ν (4.25)

or conversely,

Nb∑

α=1

es,α(k) · es′,α(−k) = δs,s′ . (4.26)


It is clear that after the diagonalization that the Hamiltonian of the system is written

H =∑

s

∑

k

(

ps(k)ps(−k)

2M+

1

2Mω2

s(k)qs(k)qs(−k)

)

(4.27)

whereps is the momentum canonically conjugated toqs. Thus we have reduced the problem to decoupledharmonic oscillators as one would expect from the beginningsince we used the fact that the forces amongthe atoms are harmonic. Therefore, the excitation spectrumof the system is given by

Es(k) = hωs(k)

(

ns(k) +1

2

)

(4.28)

to each mode of this spectrum we associate a elementary excitation that we call the phonon. In condensedmatter physics we are interested in obtaining these elementary excitations. The problem of vibrating atomsis particularly simple since it involves only harmonic forces.

Let us now go back to the problem of obtaining the frequenciesthrough eq. (4.23). Multiplying thisequation by the left byU one obtains

[

K −Mω2I]

U = 0 (4.29)

which only has a non-trivial solution (U 6= 0) if

det[

K(k) −Mω2s(k)I

]

= 0 (4.30)

which is the equation that defines the eigenmodes and it involves the diagonalization of adNb× dNb matrixand therefore, for each value ofs, the eigenvectors have dimensiondNb. Each one of these eigenvectorsrepresent the polarization of the phonon waves in the solid.In order to proceed we need more informationaboutK(k). Since we are not dealing with a specific lattice we have to usethe most general properties ofthese systems, that is, we have to use their symmetries.

The first important symmetry which is true for any crystal is the inversion symmetry, that is,

K(R) = K(−R) . (4.31)

This property implies, from definition (4.20), that

K(k) =∑

R

K(R) cos (k ·R) (4.32)

which is an even function of the momentum, that is,K(k) = K(−k). Another important property of thesystem is that if we translate all the atoms by the same fixed but arbitrary amount, sayuα(R) = R0, thenthe energy of the system has to be the same. This is called Galilean invariance. Thus, from (4.13), one finds

∑

R,R′,α,β

R0 · Kα,β(R − R′) ·R0 = 0 (4.33)

which, for arbitraryR0 implies

∑

R,α,β

Kα,β(R) = 0 (4.34)


which in terms of the Fourier transform (4.20) implies

∑

α,β

Kα,β(k = 0) = 0 . (4.35)

This condition imposes even stronger constraints on the form of K.

Let us consider, as an illustration, the case of a basis with just one type of atom (Nb = 1). It implies thatwe will haved modes in the system only. In this case (4.35) becomesK(k = 0) = 0. Thus, using (4.32)one can write

K(k) =∑

R

K(R) [cos (k · R) − 1]

= −2∑

R

K(R) sin2 (k ·R/2) . (4.36)

Therefore, for|k| ≪ 1/d, we can rewrite

K(k) ≈ −k2

2

∑

R

(nk · R)2 K(R) (4.37)

wherenk is the unit vector on the direction ofk. The matrix

C(nk) = − 1

2M

∑

R

(nk ·R)2 K(R) (4.38)

has no information on the amplitude ofk but only its direction. Then, by direct substitution in (4.30), weobtain

ωs(k) = cs(nk)k (4.39)

where

det[

C(nk) − c2s(nk)I]

= 0 (4.40)

determine the phonon velocitycs(nk) in the direction ofk. Phonons which have a linear dispersion withmomentum, such the ones described by (4.39) are called acoustical phonons. Our approximation in (4.37)requires that the sum in (4.38) to converge. We can estimate this sum by replacing it by an integral overRand we see thatC ∝

∫

dRRd−1R2K(R) only converges ifK behaves at least like1/Rd+2 at large distanceswhich is usually the case in solids.

For atoms with more than one atom in the basis the matrixK does not necessarily vanish atk = 0although the sum of the components of the matrix must vanish as explicit in (4.35). However, the form(4.32) is still valid. It implies that whenk = 0 we are going to have solutions such that the frequencyωs(k = 0) 6= 0. These modes can be obtained directly from (4.30) by settingk = 0,

det[

K(k = 0) −Mω2O,sI

]

= 0 . (4.41)

These modes which are dispersionless at very smallk are called optical phonons. We have concluded thatin a lattice with a basis we must haved branches of acoustical phonons andd(Nb − 1) branches of opticalphonons. The typical dispersion for the phonon modes is shown on Fig.4.2. We have to plot the dispersiononly in the unit cell of the reciprocal lattice, that is, on the Brillouin zone, because of the periodicity of the


system. Moreover, from the fact that the crystal has inversion symmetry (4.32) we know thatK(k) = K(−k)and therefore we only need to know half of the Brillouin zone.

k

ω s(k)

G/2

Acoustical

Optical

0

Figure 4.2: Typical dispersions for a three-dimensional with two atoms in the basis.

We have shown that for atoms interacting through harmonic forces the problem reduces to a set ofdecoupled harmonic operators as given in the Hamiltonian (4.27). We can therefore use all the technologywe have for decoupled harmonic oscillators for this problem. In particular we are interested in the operatorsthat create or destroy phonons in the system. The annihilation and creation operators for harmonic modesare defined as,

as(k) =

√

Mωs(k)

2h

(

qs(k) +i

Mωs(k)ps(−k)

)

a†s(k) =

√

Mωs(k)

2h

(

qs(−k) − i

Mωs(k)ps(k)

)

(4.42)

respectively. Notice that the displacements can now be written with help of (4.17), (4.24) and (4.42) as

uα(R) =∑

s

∑

k

es,α(k)eik·R

√

h

2MNωs(k)

(

as(k) + a†s(−k))

. (4.43)

Notice thate∗s,α(−k) = es,α(k).

In terms of these operators the Hamiltonian (4.27) can be rewritten as

H =∑

s,k

hωs(k)

(

a†s(k)as(k) +1

2

)

. (4.44)

where[

as(k), a†s′(k′)]

= δs,s′ δk,k′ . (4.45)

Thus the states of the system can be labeled by the occupationin each phonon state,|ns(k)〉 and the energy

4.3. PHONONS AT FINITE TEMPERATURE 63

is given by (4.28). The creation and annihilation operatorsact in these states according to

as(k)|ns(k)〉 =√

ns(k)|ns(k) − 1〉a†s(k)|ns(k)〉 =

√

ns(k) + 1|ns(k) + 1〉 . (4.46)

The ground state of the system is therefore the empty state,|0〉 which is defined by

as(k)|0〉 = 0 (4.47)

since there are no phonons to be destroyed. The excitations of the problem are therefore obtained by applyingcreation operatorsa† to |0〉.

4.3 Phonons at finite temperature

We have discussed only the properties of crystals at zero temperatures. At finite temperatures we have toconsider the excitation of the phonons out of the ground state. In order to do that we study the partitionfunction of the problem which is defined as

Z(β) = tr[

e−βH]

=∑

n

〈n|e−βH |n〉 =∑

n

e−βEn (4.48)

wheretr is the trace of the operator and the sum is extended over all eigenstatesEn andβ = 1/(kBT ). Thefree energy,F , of a given statistical mechanical problem is related to thepartition function via

Z = e−βF

F = = − 1

βln(Z) . (4.49)

Observe that the mean energy density of the problem is givingby

E =1

V

tr[

He−βH]

tr [e−βH ]

=1

V

∑

nEne−βEn

∑

n e−βEn

= − 1

V Z

∂Z

∂β. (4.50)

When a given system is made out of independent parts the free energy is an additive quantity, that is,F =

∑

i Fi, and thus from (4.49) the partition function is a product of the partition function of the parts,Z =

∏

i Zi.From the phonon problem we have shown that the system decouples into a set of harmonic oscillators

which are labeled bys andk. Thus, the partition function reads,

Z =∏

s,k

∞∑

ns(k)=0

e−βhωs(k)(ns(k)+1/2) (4.51)

where we have to sum over all possible occupations. The sum over the occupations is just a geometric seriesand one gets

Z =∏

s,k

eβhωs(k)/2

eβhωs(k) − 1=∏

s,k

1

2 sinh(βhωs(k)/2)(4.52)


and, from (4.50), one finds

E =∑

s,k

hωs(k)

(

ns(k) +1

2

)

(4.53)

where

ns(k) =1

eβhωs(k) − 1(4.54)

is the mean number of phonons on a given eigenstate at some temperatureT . This is called the Bose-Einsteindistribution function.h The termE0 =

∑

s,k hωs(k)/2 is the ground state of the problem (take the limit ofT → 0 in (4.53). This energy is not a observable quantity. We can however observe how a system exchangeenergy with a heat bath at some temperatureT . The quantity that characterizes this exchange of energy isthe specific heat at constant volume,CV . The specific heat measures how the energy changes as we vary thetemperature of the system. Since temperature is related to the number of excited states, the specific heat isessentially a counting of the number of available states in the system. This quantity is given by

CV =1

V

∂E

∂T=

∂

∂T

1

V

∑

s,k

hωs(k)

eβhωs(k) − 1. (4.55)

In order to calculate the thermodynamic quantities we definethe concept of density of states. The densityof states is the number of states with a given energyE. Mathematically it is given by,

N(E) =1

V

∑

n

δ(E − En) . (4.56)

For phonons the density of states is just

N(E) =1

V

∑

s,k

δ(E − hωs(k)) . (4.57)

In terms of this quantity the specific heat can be written

CV =∂

∂T

∫ ∞

0dE

N(E)E

eβE − 1(4.58)

which can be proved by direct substitution. Thus, all the thermodynamic function can be obtained directlyfrom the knowledge of the density of states.

In general, the density of states is a very complicated function of the energy due to the shape of thephonon dispersion relation. We have seen, however, that we haved acoustic modes andd(Nb − 1) opticalmodes. Moreover, these modes are characterized by the fact that the acoustical modes have a dispersionproportional tok while the optical modes are dispersionless at smallk. What is the effect of temperatureon the phonons? At low temperatures only acoustical modes are excited since to excite an optical mode onehas to pay an energyhω0. Therefore forkBT ≪ hω0 we do not expect to excite any optical mode. Theacoustical modes, however, can always be excited at low temperatures since the dispersion vanishes withk,that is, it is always possible to find a modek such thatkBT ≈ hck. Indeed, these two types of excitationsare the most basic ones in condensed matter physics: gaplessand gapfull. Most of physical properties ofsolids can be understood on the basis of this classification.Without doing any calculation we can predict thebehavior of the thermodynamic functions quite well.

4.3. PHONONS AT FINITE TEMPERATURE 65

k

k

x

y

k(a)

0

n(E)

0 E/(k T) B

hck=k T B

(b)

Figure 4.3: (a) Phase space for phonons in two dimensions; (b) Shape of the Bose-Einstein distribution.

For an acoustical mode at some given temperatureT the number of modes available is given by thespherical volume ink-space of radiusk at energykBT (see Fig.4.3)(a)). Ind dimensions this volumeis proportional toT d. Naturally the number of states available at this temperature is proportional toT d.Thus the specific heat has to behave likeT d. The energy of the states is of orderkBT . Therefore thethermal energy of the system,∆E, (∆E = E − E0) has to be proportional toT d+1 (in accordance with(4.55) the specific heat is proportional toT d). This argument is valid as far as the Bose-Einstein occupationnumber does not give an important contribution. This requires low temperatures (see Fig.4.3(b)). At hightemperatures the occupation number gives a contribution oforderkBT/E and therefore the mean energy in(4.58) is justE ≈ kBT

∫

dEN(E) which is linear in the temperature. The specific heat, by its turn, has tobe temperature independent. This is known as the Dulong-Petit law. The integral

∫

dEN(E) is essentiallythe total number of states per unit of volume in the system which isdN/V ∝ 1/ad wherea is the latticespacing. Observe that besides the thermal energykBT the only other quantities which can appear in theexpression of the energy is the phonon velocityc, the Planck constanth and the lattice spacinga. The onlyquantity with dimensions of energy that we can construct from these constants ishc/a. Since we know thatthe mean thermal energy is proportional to(kBT )d+1 the only form allowed for the mean energy per unit ofvolume is

∆E =(kBT )d+1

(hc/a)dΦd

(

hc/a

kBT

)

(4.59)

whereΦd(x) is a dimensionless universal function such that forx ≫ 1 it is constant and whenx ≪ 1it behaves likexd−1 accordingly to the Dulong-Petit law. Eq.(4.59) is called a scaling form of the meanenergy because it was only based on dimensional analysis andthe study of trivial limits of the problem.This scaling form of the energy is possible because acousticmodes are critical, that is, the frequency andmomentum scale in a well defined way. In this case we can definedwhat is called a dynamical exponent,z. This can be understood in the following way: suppose the lattice constant of the problem is changedby a constantb, that is,a → ba. Since the momentum is proportional to1/a then it has to change ask → k/b. But the frequency is also proportional to the momentum and thereforeω → ω/b. Thus the ratio


k/ω is unchanged by any change in the scale of the problem. In thiscase we say thatz = 1. In a systemwhere the frequency does not depend linearly with momentum achange in the lattice spacing can lead to achange in the frequency such thatω → ω/bz and thus the invariant ratio is notk/ω butkz/ω which definesa generic dynamical exponent. The physical reason for this is due to the “relativistic” invariance of theacoustic problem, that is, the system of acoustic phonons isinvariant under a Lorentz transformation.

k

k

x

y

k D

Figure 4.4: Geometry of the Debye approximation in two dimensions: the square Brillouin zone is replacedby a circle of radiuskD.

Now that we know the result of the calculation let us considerhow a serious calculation would lead tothe form (4.59). We approximate the actual frequency by the acoustic relation (4.39). We have to remember,however, that the dispersion is a periodic function of momentum and by replacing the exact dispersion bya linear one we have disregarded wavevectors close to the zone boundary (k ≈ G/2). We have thereforeto cut-off the integrals at some maximum vector,kD. In order to estimate this wavevector we replace theoriginal unit cell by a spherical one with radiuskD (see Fig.4.4). The number of states in the original cellisN , the number of atoms in the crystal. Moreover, the volume in k-space per wave vector is(2π)d/V andtherefore the number of states is(2π)dN/V = (2π)dn. In order to preserve the number of states for thespherical problem we require that

(2π)dn =

∫

ddk θ(kD − k)

kD =

(

d(2π)dn

Sd

)1/d

(4.60)

whereSd =∫

dΩ is the total solid angle ind dimensions (S1 = 2, S2 = 2π andS3 = 4π) andθ(x) = 1if x > 0 and vanishes otherwise is a step function. This is called theDebye model. Observe thatkD isproportional to1/a and therefore acts as a high energy cut-off for the model. Forthis model the density ofstates for acoustical modes is written as

ND(E) =∑

s

∫

ddk

(2π)dδ (E − hcs(nk)k)

=∑

s

∫

dΩ

(2π)d

∫ kD

0dkkd−1δ (E − hcs(nk)k) . (4.61)

4.4. INELASTIC SCATTERING 67

We now change the variables asx = hcs(nk)k and write

ND(E) =dSd

(2π)dEd−1

(hc)dθ (kBΘD − E) (4.62)

wherekBΘD = hωD = hckD ∝ hc/a is the Debye energy and is the scale of energy that should appear in(4.59). Moreover,

1

cd=

1

d

∑

s

∫

dΩ

Sd

1

cds(nk)(4.63)

gives an estimate of the average phonon velocity. Notice that the density of states is finite only for energiessmaller than the Debye energy. By substitution of (4.62) into (4.58) one finds (4.59) with

Φd(x) =dSd

(2π)d

∫ x

0dz

zd

ez − 1. (4.64)

The characteristic energy scale in solids is defined by the Debye temperature. The sound velocity is of orderof 104 m/s in most solids and since the lattice spacing is of order ofthe typical Debye temperatures are oforder of hundreds of Kelvin.

The optical modes can be treated in a much simpler way since they are dispersionless and have a gapfor excitation of energyhω0. It is clear from (4.58) that in this case the physical properties are going tobe completely determined by the Bose-Einstein factor whichis exponentially small at low temperatures.Thus one expect all the thermal properties of optical modes to decay exponentially with temperature, thatis, e−hω0/(kBT ). Since we haveNb(d− 1) optical branches in the spectrum we can write that the density ofstates is simply

NE(E) =Nb(d− 1)Ns

Vδ(E − hω0) (4.65)

where the prefactor is chosen in such way that the total number of states is correct. Thus, using (4.58), onefinds

∆E =(d− 1)nhω0

eβhω0 − 1(4.66)

where we have used thatN = NsNb. At low temperatures, as expected, the mean energy vanishesexponen-tially and at high temperatures one recovers the Dulong-Petit result. This simple model for optical phononsis called Einstein model.

4.4 Inelastic scattering

Consider the problem of the scattering of a solid taking intoaccount lattice oscillations, that is, phonons.We assume that the interaction between the probe and the solid is weak and since this is a time dependentproblem (the atoms undergo harmonic motion) we are going to use time dependent perturbation theory. Theprobe (which can be light, neutrons or whatever) is assumed to be initially at some distant timet0 → −∞in a plane wave state (eip·r/

√V ) with wave-vectorp and the solid in a eigenstate of the HamiltonianH0

given by|I〉 with energyEI . The probe interact with the solid via a potentialV and is scattered into a newplane wave state with wave-vectorp′ (see Fig.4.5) and the solid is left in another eigenstate|F 〉 with energy


EF at some distant timet→ ∞. Observe that the energy must be conserved in the scatteringand therefore

EI − EF = E , (4.67)

whereE is the change in energy of the probe during scattering. The scattering is inelastic because the probedoes not conserve its energy during scattering. The rate of transition is simply given by Fermi’s golden rule

T (q, E) =2π

h

∑

F

δ (E − EI + EF ) |〈p′, F |V |p, I〉|2

=2π

h

∑

F

δ (E − EI + EF ) ×

× | 1V

∫

ddreiq·r〈F |V (r)|I〉|2 (4.68)

whereq = p′ − p is the momentum transfer during scattering.

p

p’

|I>

|F>

E

E

I

F

t 0

t

Figure 4.5: Scattering process.

We will assume as before that the interaction of the probe andthe solid is short ranged and can be writtenas

V (r) = V0

∑

R

δ (r− r(R))

V (r) = V0

∑

R

δ (r− R − u(R)) (4.69)

where we have used (4.12) and for simplicity of notation we assume one atom per unit cell. Substitution in(4.68) leads to

T (q, E) =2πV 2

0

V 2h

∑

F

δ (E − EI +EF ) |∑

R

eiq·R〈F |eiq·u(R)|I〉|2 . (4.70)

The cross-section of the crystal is proportional to the transition rate and this allows us to define the so called

4.4. INELASTIC SCATTERING 69

dynamical form factor,S(q, ω), which is given by

S(q, ω) =1

N

∑

F

δ (ω + (EF − EI)/h) |∑

R

eiq·R〈F |eiq·u(R)|I〉|2 . (4.71)

which can be Fourier transformed in the frequency domain,

S(q, ω) =

∫

dt

2πeiωtS(q, t) (4.72)

where

S(q, t) =1

N

∑

F

ei(EF−EI)t/h∑

R,R′

eiq·(R−R′)〈I|eiq·u(R′)|F 〉〈F |e−iq·u(R)|I〉

=1

N

∑

F

∑

R,R′

eiq·(R−R′)〈I|eiq·u(R′)|F 〉〈F |eiH0t/he−iq·u(R)e−iH0t/h|I〉

=1

N

∑

F

∑

R,R′

eiq·(R−R′)〈I|eiq·u(R′)|F 〉〈F |e−iq·u(R,t)|I〉

=1

N

∑

R,R′

eiq·(R−R′)〈I|eiq·u(R′)e−iq·u(R,t)|I〉 . (4.73)

Where we have used that: (1)|I〉 and |F 〉 are eigenstates ofH0; (2) the operator identityUeAU−1 =eUAU

−1; (3) changed from the Schrödinger representation to the Heisenberg representation, that is,eiH0t/hAe−iH0t/h =

A(t); and (4) that the final states form a complete set∑

F |F 〉〈F | = 1. We will assume further that initiallythe solid is an equilibrium situation at temperatureT so that〈I|...|I〉 becomes a thermal average. Noticethat the dynamical form factor only depends on the properties of the crystal since the probe is already out ofthe problem. Notice that we can write (4.73) in a simpler formif we use (2.50)

S(q, t) = N〈I|ρ(−q, 0)ρ(q, t)|I〉 (4.74)

is the so-called density-density correlation function. This result is a natural result since the probe is coupledto the ion density in (4.69) and therefore the response of thesolid occurs by creating density fluctuations.

In order to calculate the thermal average in (4.73) we are going to use a result from the appendix whichsays that if the ground state of the problem is the one for harmonic oscillators and the operatorsA andBare linear combination of destruction and creation operators we have

〈eAeB〉 = e12〈A2+2AB+B2〉 (4.75)

where in our caseA = iq · u(R′) andB = −iq · u(R, t). Thus we have

〈eiq·u(R′)e−iq·u(R,t)〉 = e12〈2q·u(R′)q·u(R,t)−(q·u(R′))2−(q·u(R,t))2〉

= e12

P

µ,ν qµqνCµ,ν(R−R′,t) (4.76)

where

Cµ,ν(R − R′, t) = 〈(

uµ(R, t) − uµ(R′, 0)) (

uν(R, t) − uν(R′, 0))

〉 (4.77)

is the phonon-phonon correlation function which depends only on the relative position of the ions. Thus,


the dynamic form factor reads

S(q, t) =∑

R

eiq·ReP

µ,ν qµqνCµ,ν(R,t) . (4.78)

Most of the phenomena in physics can be expressed in terms of correlation function such as (4.77). This isone important example since most of the excitations in solids (even electronic ones) have bosonic character.

The calculation of the correlation function (4.77) is very simple because theu operators are just linearcombinations of creation and annihilation operators. We can easily show using (4.43) and the commutationrelations between the creation and annihilation operatorsthat

Cµ,ν(R, t) =∑

s

∫

ddk

(2π)deµs (k)eνs (−k)

h

2Mωs(k)[

(1 + ns(k))(

e−i(k·R−ωs(k)t) − 1)

+ ns(k)(

ei(k·R−ωs(k)t) − 1)]

. (4.79)

Notice that the complete evaluation ofS(q, ω) is very complicated. One can however look at somesimple limits. Let us study the limit of elastic scattering,that is, when the energy is conserved in thescattering. In this case we have to take the limit ofω → 0 in (4.72). This is the so-called static limit and isequivalent of taking the limit of long timest→ ∞ in (4.72) since the exponential term oscillates strongly inthis limit and the integral is dominated by small values of the frequency. In this limit the exponential termsin (4.79) also oscillate strongly and give vanishing contribution. The only term left is

limt→∞1

2

∑

µ,ν

qµqνCµ,ν(R, t) = −2W (T )

= −∑

s

∫

ddk

(2π)d|q · es(k)|2 h

2Mωs(k)(1 + 2ns(k)) (4.80)

which is the so-called Debye-Waller factor. The physical meaning of this factor is clear if we calculate thedynamical form factor from (4.78) and (4.72):

S(0)(q, ω) = e−2W (T )δ(ω)Nδq,G (4.81)

which is the result of elastic scattering obtained before except by the Debye-Waller factor which tells us thatvirtual transitions induced by the collision of the probe with the crystal at finite temperatures decreases theintensity of the scattering. We have been able, therefore, to reproduce the results of elastic scattering.

The next correction to elastic scattering is obtained by retaining the time dependence on correlationfunction (4.77). In order to do that we replace the exponent in (4.78) by

S(q, t) =∑

R

eiq·R exp

1

2

∑

µ,ν

qµqν [limt→∞Cµ,ν(R, t) + ∆Cµ,ν(R, t)]

=∑

R

eiq·Re−2W (T )∞∑

n=0

[

1

2

∑

µ,ν

qµqν∆Cµ,ν(R, t)

]n

(4.82)

where

∆Cµ,ν(R, t) = Cµ,ν(R, t) − limt→∞Cµ,ν(R, t) . (4.83)

4.5. PROBLEMS 71

It is possible to show that each power inn in (4.82) is due to the scattering of the probe ton phonons. Thefirst order correction to the static term produces a correction in the dynamic form factor which is given by

S(1)(q, ω) = e−2W (T )∑

s

|q · es(q)|2 h

2Mωs(q)[(1 + ns(q)) δ (ω + ωs(q))

+ ns(q)δ (ω − ωs(q))] (4.84)

observe that the first term increases the number of phonons byone and therefore is related with the emissionof one phonon by the probe. The second process is associated with the absorption of one phonon by theprobe. Theδ functions guarantee the conservation of energy for each oneof these process. Other correctionscan be calculated immediately from (4.82). An important property of (4.84) is that the scattering happenswhenω = E/h = ωs(q). Thus, by varying the energy of the probe one can map the dispersion relation ofthe phonons.

4.5 Problems

1. Calculate the commutation relations betweenuα(k) andPα(k) starting from (4.16).

2. The group velocity of an excitation is defined as the velocity at which a wave-packet made out of asuperposition of plane waves propagate and is given by

vg =dω(k)

dk.

What is the group velocity of an acoustical wave whenk → 0? What is the group velocity of anoptical wave whenk → 0?

3. Calculate the phonon dispersion and polarizations for anone-dimensional lattice with two atoms withmassesM1 andM2 as in Fig.4.1(b).

4. What is the sound velocity for the problem in which the chain has two different masses?

5. Assume that a chain is made out of3 different atoms with different masses. What can you say aboutthe spectrum of the system? Where are gaps located?Hint: no calculations needed.

6. Using the commutation relations betweenqs(k) andps(k) show that the creation and annihilationoperators obey the commutation relations (4.45).

7. Here we are going to consider the long wavelength theory ofacoustic phonons. Define two new fields,

Πs(r) =1√V

∑

k

ps(k)e−ik·r

φs(r) =1√V

∑

k

qs(k)eik·r (4.85)

which obey canonical commutation relations[φs(r),Πs′(r′)] = ihδ(r − r′)δs,s′ . By linearizing the

phonon dispersion,ωs(k) = csk show that Hamiltonian density associated with the acousticcase canbe written in terms of these fields as

H =∑

s

[

Π2s(r)

2ρ+c2sρ

2(∇φs(r))2

]

(4.86)


whereρ = M/a is the lattice mass density anda is the lattice spacing. Eq.(4.86) is the Hamiltoniandensity of a field theory for particles moving with ”light" velocity cs. Notice that in this field theoryall the knowledge of the lattice is lost since we assumed thatk ≪ 1/a or equivalently that we arelooking at wavelengths such thatλ ≫ a. Show that classical equation of motion for (4.86) is thewell-known wave equation:

d2φsdt2

= c2s∇2φs . (4.87)

8. Prove that the equation (4.64) is correct and obey the scaling properties. Evaluate the low temperaturespecific heat for the Debye model ind = 1, 2, 3.

9. Calculate the specific heat for the Einstein model for phonons with frequencyω0.

10. Show that the field theory associated with the Einstein model is given by the following Hamiltoniandensity:

H =∑

s

[

Π2s(r)

2ρ+aMω2

0

2φ2s(r)

]

. (4.88)

11. Prove eq.(4.79).

12. Calculate the Debye-Waller factor in 1, 2 and 3 dimensions at zero temperature for acoustic phonons.What is the physical meaning of your results? This result is related to the famous Mermim-Wagnertheorem.

13. Prove that eq.(4.84) is correct.

4.5.1 Appendix: Thermal averages for bosons

Notice that in (4.73) we are interested in calculating an average of exponential of operators. But from (4.43)

these operators are linear combinations of harmonic creation and annihilation operators,∑

n

(

αnan + γna†n

)

,

where each mode is independent from each other, that is,[an, a†m] = δn,m (for the phonon casen = (s,k)).

Since these modes are independent we can look at each mode individually. Let

A = αAa+ γAa†

B = αBa+ γBa† (4.89)

we want to evaluate an average of the form〈eAeB〉. First we use an operator identity which is valid whenthe commutator[A,B] is a c-number:

eAeB = eA+Be[A,B]/2 . (4.90)

Notice thatA+B can now be written as

A+B = αa+ γa† (4.91)

where

α = αA + αB

γ = γA + γB . (4.92)

4.5. PROBLEMS 73

Thus we are interested in calculating

〈eαa+γa†〉 = e−αγ/2〈eαaeγa†〉= eαγ/2〈eγa†eαa〉 (4.93)

where we have used (4.90). Now we define the function

f(α) = 〈eαaeγa†〉= eαγ〈eγa†eαa〉 . (4.94)

Now we use the cyclic property of the trace in order to write

〈eγa†eαa〉 = tr[

e−βHeγa†

eαa]

/Z

= tr[

eαae−βHeγa†]

/Z

= tr[

e−βHeβHeαae−βHeγa†]

/Z

= 〈eβHeαae−βHeγa†〉 . (4.95)

The Hamiltonian for the problem is simplyH = hω(a†a + 1/2) and if we use the operator identityUeAU−1 = eUAU

−1and

exa†aae−xa

†a = exa (4.96)

we obtain from (4.95):

〈eγa†eαa〉 = 〈eαe−hωβaeγa†〉

= f(αe−hωβ) (4.97)

where we have used the definition off . Now, from (4.94) we find

f(α) = eγαf(αe−hωβ) (4.98)

which can be solved by iteration, that is,

f(α) = eγα(1+e−hωβ)f(αe−2hωβ) = eγα(1+e−hωβ+e−2hωβ)f(αe−3hωβ) (4.99)

and so on. If we do this process an infinite number of times we find

f(α) = eγα/(1−e−hωβ)f(0) (4.100)

where we have used the result of a geometric series∑∞

0 xn = 1/(1 − x). Observe, however, that

f(0) = 〈eγa†〉 =∑

n

e−βhωn〈n|eγa† |n〉 = 1 (4.101)

and therefore

f(α) = eγα/(1−e−hωβ) . (4.102)


The Bose-Einstein factor is

n =1

ehωβ − 1(4.103)

and therefore

f(α) = eγα(1+n) . (4.104)

With this result at hand we can rewrite

〈eαa+γa†〉 = e12γα(1+2n) (4.105)

and finally

〈eAeB〉 = exp

1

2[αAγB − γAαB + (γA + γB)(αA + αB)(1 + 2n)]

= exp

1

2〈A2 + 2AB +B2〉

(4.106)

which can be obtained directly from the definition ofA andB, the commutation relation[a, a†] = 1, thedefinition n = 〈a†a〉 and that〈a2〉 = 〈

(

a†)2〉 = 0.

Chapter 5

Electrons in Solids

We have seen that the difference in the mass of the electrons and ions allow us to separate their time (orenergy) scales. This introduces a major simplification since we can treat electrons and ions separately. Wehave investigated the problem of the coupling between ions and showed that the ion problem has elementaryexcitations which are called phonons. The short range character of the interaction between ions and thefact that they form a lattice makes their theoretical treatment quite simple. The same thing does not happenwith electrons. This can be seen clearly from the experimental fact that electronic properties of solids canvary wildly, that is, we can have metals, insulators, magnetic systems, superconductors, etc, with propertieswhich depend only on the way the electrons interact among themselves.

In this chapter we are going to study the interaction betweenthe electrons and the atoms in a crystal.We are going to explore the fact that in a crystal the potential felt by the electrons has to be periodic dueto the periodic arrangement of the atoms. We are going to consider first the case of a one-dimensionalchain of atoms since the mathematical treatment is quite simple and it is relevant for the experimentalcase of one-dimensional systems such as organic conductors, polymers and other low dimensional systems.The treatment of the higher dimensional problem is completely analogous and we will generalize the one-dimensional results to higher dimensions.

5.1 The one-dimensional chain

Let us consider now a problem where a symmetry helps us to solve the problem entirely. Consider a ringof N identical atoms which are separated from each other by an angle 2π/N as shown on Fig.natom. Welabel the states of the electron localized in each one of these atoms byψn with n = 1, ...,N . Observe thatthis problem has rotational invariance, that is, if we rotate the system by2π/N we should not notice anydifference. In the limit whereR → ∞ andN → ∞ butN/R is a constant (which is just the linear densityof the chain) the problem goes into a linear chain of atoms with periodic boundary conditions, that is, theatomN + 1 is simply atom1. If we had no atoms on the ring and we allow an electron to move alongthe ring then the system has translational invariance, thatis, we could rotate the ring by any angle and theproblem should look like the same. The operator that produces translations (or in this case, rotations) is thetranslation operator

R(l) = eipl/h (5.1)

wherep is the momentum operator. In order to understand why this operator generates translations considerthe transformation

R(l)xR−1(l) = eipl/hxe−ipl/h = x+ l (5.2)

75

76 CHAPTER 5. ELECTRONS IN SOLIDS

as you can easily shown by using the commutation relations[x, p] = ih. In the absence of the atoms theHamiltonian of the problem has to commute withR(l) for any value ofl by translation (rotation) invariance.

t

1N 2

3

t

R

N−1

Figure 5.1: A ring of atoms with radiusR and hopping energyt.

In the presence of the atoms this is not true because the electron feels the potential of each atom. ThepotentialV (x) felt by the electron has to beperiodic, that is, ifx is the coordinate along the ring one musthaveV (x + a) = V (x) wherea = 2πR/N is the lattice spacing. Thus, although the Hamiltonian of theproblem does not commute withR(l) for any value ofl it has to commute forl = a. Thus, the problemdoes not have a continuous symmetry but a discrete one. Sincethe Hamiltonian has to commute withR(a)one knows thatH andR(a) share the same eigenvectors. Thus, if we find the eigenvectors ofR(a) we willalso find the eigenvectors ofH. Moreover, becauseR(a) is written like (5.1) its eigenvalues are related tothe eigenvalues of the momentum operator, that is,

p|k〉 = hk|k〉 . (5.3)

In this case the eigenvalue problem forR(a) reads

R(a)|k〉 = eika|k〉 . (5.4)

We would like, however, to rewrite the states|k〉 in terms of the states|n〉 of the electron localize onatomn. As we used previously this state is essentially a vector withN entries where only thenth entry is1and all the others are zero:

|n〉 =

000...010...0

. (5.5)

5.1. THE ONE-DIMENSIONAL CHAIN 77

We know from basic quantum mechanics that if we have a complete set of orthogonal states any state in theHilbert space has to be written as a linear combination of these states, in other words, the states|n〉 span theHilbert space (

∑

n |n〉〈n| = 1)

|k〉 =∑

n

|n〉〈n|k〉 =∑

n

ck,n|n〉 (5.6)

whereck,n are unknown coefficients. We further know thatR(a) has to move the electron one unit in thelattice and therefore

R(a)|n〉 = |n+ 1〉R−1(a)|n〉 = |n− 1〉 . (5.7)

ApplyingR(a) to |k〉 in (5.6) and using the above relation we find

R(a)|k〉 =∑

n

ck,nR(a)|n〉 =∑

n

ck,n|n+ 1〉 =∑

n

ck,n−1|n〉 . (5.8)

Using (5.4) together with (5.6) and the fact that the states|n〉 are orthogonal to each other (〈n|m〉 = δn,m)we see that

ck,n = ck,n−1e−ika (5.9)

which gives a recursion relation for eachck,n starting from the previous one. Suppose we start fromck,1. Inthis case it is very simple to see that

ck,n = ck,1e−ikan (5.10)

and thus the expansion (5.6) becomes

|k〉 = ck,1∑

n

e−ikan|n〉 (5.11)

and from the fact that|k〉 has to be normalized (〈k|k〉 = 1) we find

|ck,1| =1√N. (5.12)

We have therefore solved the eigenvalue problem in (5.4).

Because the system has the periodicity generated byR(a) it means that the energy of the system doesnot change when we translate the system bya, that is, the Hamiltonian commutes withR(a): [H,R(a)] = 0.From basic quantum mechanics we know that it implies that theeigenstates ofR(a) are also eigenstates ofH. This result, however, does not tell us what is the energy of these states. In order to know that we have touse the Hamiltonian itself. Now, let us go back to our original problem of the tunneling of electrons betweendifferent atoms. The Hamiltonian can be constructed in exactly the same way as for the case of the other


molecules the only difference is that we have aN ×N matrix:

H = −t

0 1 0 0 ... 11 0 1 0 ... 00 1 0 1 ... 00 0 1 0 ... 00 ... 0 1 0 11 0 ... 0 1 0

. (5.13)

Observe that if we applyH to a vector like (5.5) we immediately obtain

H|n〉 = −t (|n+ 1〉 + |n− 1〉) . (5.14)

The meaning is rather obvious: the Hamiltonian has only a hoping term which moves the electron betweennearest neighbor sites on the chain.

Since we know that|k〉 is an eigenstate of the system let us calculate its energy. Inorder to do that wejust have to apply the Hamiltonian the state:

H|k〉 = − t√N

∑

n

e−ikan (|n+ 1〉 + |n− 1〉)

= − t√N

∑

n

(

e−ika(n−1) + e−ika(n+1))

|n〉

= −t(

eika + e−ika)

|k〉 (5.15)

which can be written as

H|k〉 = Ek|k〉 (5.16)

where

Ek = −2t cos(ka) (5.17)

which gives the spectrum of the problem and is shown as a function of k in Fig.5.2. Observe that the stateof lowest energy hask = 0 with energy−2t. From (5.11) this state is simply given by

|k = 0〉 =1√N

∑

n

|n〉 =1√N

111...111...1

(5.18)

which shows that the probability of finding the electron in any atoms is the same and given by1/N , that is,the electron is spread uniformly over the entire chain! Thatis the way the kinetic energy is maximized.

Observe that we started withN states and therefore we have to end withN states. It means that there areonlyN allowed values ofk! To find out what are these values we have to remember that the ring is periodic


-3 -2 -1 1 2 3k

-2

-1

1

2

E

Figure 5.2: Spectrum for an electron moving in a one-dimensional chain as a function ofk.

and therefore|n+N〉 = |n〉. Thus, from (5.10) we must have

eikaN = 1 (5.19)

which implies that the values ofk are quantized and given by

km =2πm

Na(5.20)

wherem = 0,±1, ...,±(N − 1)/2, N/2 for N even andm = 0,±1, ...,±(N − 1)/2 for N odd producesthe requiredN states. Observe that the available states go fromk = −π/a(1 − 1/N) to k = π/a. Inthe limit of N → ∞ the "distance" between two states which is just2π/(Na) goes to zero and we have−π/a < k ≤ π/a which is called the Brillouin zone. Notice that the tunneling between atoms has brokenthe degeneracy of the originalN states intoN non-degenerate states for each value ofk. Thus if the distancebetween the atoms is large enough so thatt → 0 no tunneling takes place and the electron is stuck in oneof the atoms. No motion can occur and the system is an insulator that is highly degenerate. Ast increaseswhen we approach the atoms the tunneling grows and the degeneracy is broken as shown in Fig.5.3. Theargument here is not only valid for the ground state but for any state of the atom. For instance, we couldhave electrons coming from other orbitals and going from atom to atom. Thus, each orbital gives rise towhat is called a band of states which essentially have a dispersion given by (5.17). We also say that eachband has a finite bandwidth,W , which is the energy difference between the highest and lowest energy statewhich for (5.17) is justW = 4t.

Because the electron is spread over the entire lattice it is very easy to move it around. This is true ifthe band is not full. This is simple to understand: from the Pauli principle we can only have one electronfor each value ofk. Suppose we put 1 electron on a lattice ofN atoms. The state of lowest energy hask = 0 (compare this state with the states we found for the H2 and the molecule with 3 atoms!). Oncethis state is occupied we can put another electron on this state but with opposite spin. The next state to beoccupied is the closest tok = 0 since, according to (5.17) (see Fig.5.2) the energy grows with k. Thus, if


E

1/a

N

N

N

N

W

∆

Figure 5.3: Energy of the system as a function of the distancebetween atoms.

we putNe electrons in the system they have to occupy symmetrically the states aroundk = 0. Obviouslythe band will be full whenNe = 2N . In this case there are no more states available! The next state in thesystem corresponds to an atomic state with larger energy, that is, there is a gap,∆, between bands whichin the atomic picture is just the energy between the discretestates. Thus, in order to make an excitation inthe system we have to excite the electron over the gap and thiscosts energy. Thus, at low temperatures noconduction can happen and the system is a band insulator. In real space this is also simple to understand:whenNe = 2N the electrons cannot hop anymore to the same orbital since itis fully occupied (Pauliprinciple) and therefore it has to hop to another orbital at higher energy, as shown in Fig.5.4.

The scenario that we have been describing here is known as tight binding approximation because weassume that the wavefunction on each atom is very close to an atomic wavefunction. This is only true is thetunnelingt is small. From Fig. 5.3 one sees that whenR decreases the bands can start to overlap and thegap between bands disappear. This implies that the electrons just feel a very weak potential created by thelattice. In this case one can just treat the periodic potential created by the lattice as a perturbation. Let usconsider this case in more detail.

In the absence of a periodic potential the electron undergoes free motion in a circle of lengthL. TheSchrödinger equation for this case is simply

− h2

2m

d2

dx2ψ(x) = Eψ(x) (5.21)

which can solved immediately in terms of plane waves:

ψk(x) =1√Leikx

E0k =

h2k2

2m. (5.22)

Observe that the energy is a continuous function ofk in complete contrast with the tight binding picture and


(b)

(a)

t

t

∆

∆

∆

∆

Figure 5.4: Motion of electron in a periodic array of atoms: (a) band less than filled; (b) filled band.

it is shown on Fig.5.5(a). The question that remains to be answered is what happens when we introduce theperiodic potential of the lattice. Observe that because thepotential is periodic we haveV (x) = V (x + a)and if one expandsV (x) in Fourier series

V (x) =∑

k

Vke−ikx (5.23)

the condition of periodicity implies that

k =2πn

a(5.24)

wheren is an integer. Thus, even without computing the effects of the potential we already know that themomentum is only defined in the Brillouin zone, that is,−π/a < k ≤ π/a. Thus, one has to shift theportions of the energyEk inside of the Brillouin zone as shown in Fig.5.5(b). This is the effect of the pureperiodicity of the problem even ifV (x) → 0!

WhenV (x) is finite but small we have to take into account the effect of the potential in the Schrödingerequation

(

− h2

2m

d2

dx2+ V (x)

)

ψE(x) = EψE(x) . (5.25)

Let us treat the problem in perturbation theory. Since we know that the problem whenV (x) = 0 is solvedin terms of plane waves we look for a problem of (5.25) which isa linear combination of plane waves

ψE(x) =1√L

∑

k

ck,Eeikx (5.26)

where we would like to know the values ofck,E. Because of the periodic boundary conditions, that is,


E

k

k

−π/a π/a

(a)

Ek

−π/a π/a

k

(b)

Figure 5.5: Electron dispersion: (a) in the absence of a periodic potential; (b) in the presence of a periodicpotential.

ψE(x+ L) = ψE(x) one has

k =2πn

L(5.27)

wheren = 0,±1, ..., N/2. We have now to normalize the wavefunction over all space

∫ L/2

−L/2dx|ψ(x)|2 = 1 (5.28)

which can be written as

1

L

∑

k,p

c∗k,Ecp,E

∫ L/2

−L/2dxei(k−p)x = 1 . (5.29)

In order to perform the integral one has to remember thatk andp are given by (5.27) which implies that theabove integral has the form:

∫ L/2

−L/2dxei(p−k)x = 2

sin(p− k)L/2

(p− k)=L

π

sin(n−m)π

n−m. (5.30)

Observe that the above integral is zero unlessn = m (that is,p = k) in which case we write

∫ L/2

−L/2dxei(p−k)x = Lδp,k (5.31)

whereδp,k is the so-called Kronecker delta which has the properties that δp,k = 1 if p = k andδp, k = 0 if


p 6= k. Thus we conclude that

∑

k

|ck,E |2 = 1 (5.32)

gives the normalization of the wavefunction.

If we substitute (5.26) into (5.25) one finds

∑

k

h2k2

2mck,Ee

ikx +∑

k,p

Vke−ikxcp,Ee

ipx =∑

k

Eck,Eeikx (5.33)

in order to solve this equation we multiply the equation above bye−ipx and integrate overx. We are goingto use the following useful integral (5.31) to obtain

(E − E0p)cp,E =

∑

p′

Vp−p′cp′,E (5.34)

where

Vk−p =1

L

∫ L/2

−L/2dxV (x)ei(k−p)x . (5.35)

Observe that (5.34) is a matrix equation whereVk,p are the matrix elements of the potential and the collectionof ck,E for fixedE form a vector. In order to see that let us remember that because of the boundary conditionskn = 2πn/L with n = 0,±1, ..., N/2 there areN allowed values ofk. Thus, let us numberkm withm = 1, ..., N . We will usem to number the eigenstate (which is equivalent to usek in the Brillouin zone).That is, we can rewrite (5.34) as

E −Ek1 0 ... 00 E − Ek2 ... 0... ... ... ...0 ... E − EkN−1

00 ... 0 E −EkN

ck1,Eck2,E...

ckN−1,E

ckN ,E

=

V1,1 V1,2 ... V1,N

V2,1 V2,2 ... V2,N

... ... ... ...VN−1,1 ... VN−1,N−1 VN−1,N

VN,1 ... VN,N−1 VN,N

ck1,Eck2,E...

ckN−1,E

ckN ,E

(5.36)

which is aN ×N diagonalization problem exactly like in the tight binding problem we solved earlier.

We do not wish at this point to solve (5.34) in its full form butfor small values ofV , that is, we wouldlike to solve the problem as a power series expansion of the potential. Observe that forV = 0 the solutionis straightforward, namely, for a given ith eigenvalue we have: ifE = Eki

thencki,E = 1; if E 6= Ekiwe

havecki,E = 0. Another way to rewrite this result is to define a wave numberk such that:

E =h2k2

2m(5.37)


and rewrite (5.34) as

h2

2m(k2 − p2)cp,k =

∑

p′

Vp−p′cp′,k (5.38)

and therefore the solution forV = 0 is: if p = k we haveck,k = 1 and ifp 6= k cp,k = 0. The result can besummarized as

c0p,k = δp,k . (5.39)

The next order correction is obtained if we assumeE = E0k +E

(1)k andcp,k = c0p,k + c

(1)p,k with E(1)

k and

c(1)p,k given in first order inV and substitute on (5.38):

(

h2

2m(k2 − p2) + E

(1)k

)

(δp,k + c(1)p,k) =

∑

p′

Vp−p′(δp′,k + c(1)p′,k)

E(1)k δp,k +

h2

2m(k2 − p2)c

(1)p,k = Vp−k (5.40)

where we have kept only terms to first order inV . The solution of this equation is straightforward again: ifk = p we find

E(1)k = V0 (5.41)

and ifk 6= p we find

c(1)p,k =

Vp−kh2

2m(k2 − p2)(5.42)

which is the first order correction. Notice thatc(1)k,k = 0! Observe that

Vk−k=0 =1

L

∫ L/2

−L/2dxV (x) = V (5.43)

is just the mean value of the potential in the system. Thus, weconclude that first correction to the energy isa simple shift in the overall energy of the system. We observethat the wavefunction of the problem at thisorder in perturbation theory can be written as

ψk(x) ≈1√L

eikx +∑

p 6=k

Vk,pE0p − E0

k

eipx

. (5.44)

The normalization condition (5.32) is obeyed to order inV 2.

The second order correction proceeds in the same way, that is, we writeE = E0k + E

(1)k + E

(2)k and

ck,p = c0k,p + c(1)k,p + c

(2)k,p and substitute on (5.38). The first order terms cancel because of (5.40) and the

second order equation becomes

E(2)k δp,k + E

(1)k c

(1)p,k +

h2

2m(k2 − p2)c

(2)p,k =

∑

p′

Vp−p′c(1)p′,k (5.45)


which has the solution: ifk = p we find

E(2)k =

∑

p′

Vk−p′c(1)p′,k =

∑

p′

|Vp′−k|2h2

2m(k2 − (p′)2)(5.46)

where we used (5.42) and the fact thatVp = V ∗−p. If k 6= p then, from (5.45) we have

c(2)p,k =

∑

p′

Vp−p′Vp′−kh2

2m(k2 − (p′)2)− V0Vp−k

h2

2m(k2 − p2)(5.47)

which proportional toV 2. In order to calculate the change in the energy we have to calculate the matrixelementVk,p. From (5.35) and (5.23) one has

Vk,p =1

L

∑

n

Vn

∫ L/2

−L/2dxei(k−p)xei2πnx/a

=∑

n

Vnδk,p+2πn/a (5.48)

Thus,

E(2)k =

∑

n 6=0

|Vn|2E0k − E0

k−2πn/a

. (5.49)

Observe that ifk = ±πn/a the denominator in the above equation vanishes and perturbation theory fails!Thus, for this particular points something else has to happen and one cannot use non-degenerate perturbationtheory. For the other values ofk the result (5.49) is fine since no divergences occur.

Thus, let us go back to the original equation (5.34) and try tounderstand what is going on. For simplicitywe are going to assume thatV (x) = V0 cos(2πx/a). This implies that we haveV0 = 0 andV1 = V−1 =U0/2. Substituting (5.48) into (5.34) one finds

(E0k − E)ck,E +

∑

n

Vnck−2πn/a,E = 0 . (5.50)

But because of our choice ofV (x) this equation simplifies to

(

hk2

2m− E

)

ck,E +U0

2

(

ck+2π/a,E + ck−2π/a,E

)

= 0 (5.51)

which again is an infinite matrix problem. The first three terms read

(

h(k + 2π/a)2

2m− E

)

ck+2π/a,E +U0

2

(

ck+4π/a,E + ck,E)

= 0

(

hk2

2m− E

)

ck,E +U0

2

(

ck+2π/a,E + ck−2π/a,E

)

= 0

(

h(k − 2π/a)2

2m− E

)

ck−2π/a,E +U0

2

(

ck−4π/a,E + ck,E)

= 0 (5.52)

What do we do now? Well our perturbation theory failed fork = π/(2a). Thus we will retain in this matrixproblem only the values ofk andk − 2π/a and forget about all the other Fourier components. In this case


the problem reduces to[

hk2

2m − Ek U0/2

U0/2h(k−2π/a)2

2m − Ek

]

(

ck,Eck−2π/a,E

)

= 0 . (5.53)

which can be solved at once

Ek,± =1

2

(

h(k − 2π/a)2

2m+hk2

2m

)

±

√

(

h(k − 2π/a)2

2m− hk2

2m

)2

+ U20

. (5.54)

Observe that exactly atk = π/a we have

Eπ/a,± = ±|U0|/2 (5.55)

that is, there is a gap in the spectrum with magnitude

∆π/a = Eπ/a,+ − Eπ/a,− = |U0| (5.56)

which implies that the two branches in Fig.5.5(b) are now split and there is a gap between them as shown inFig.5.6.

Ek

−π/a π/a

kU0U

0

Figure 5.6: Effect of a periodic potential on a free electronsystem.

Moreover, by direct substitution of (5.55) into the matrix equation (5.53) we find

ck,+ = ck−π/a,+

ck,− = −ck−π/a,− (5.57)

which implies from (5.26) that

ψ+,k(x) ∝ cos(πx/a)

ψ−,k(x) ∝ sin(πx/a) (5.58)

which have a very simple meaning if we consider a plot of the potential felt by the atoms and the shape ofthese function as in Fig.5.7. The ground state wavefunctionψG = ψ− has its maximum at the position ofthe ions while the next excited stateψE = ψ+ has its maximum in between the ions. Since the charge of theelectron is opposite to the ions the energy is minimized whenthe electrons have most of the charge on thetop of the positive ions as in Fig.5.7(a).

Thus, the appearance of gaps in the spectrum of a free electron system are related to the breaking ofthe translational symmetry of the system (which now has onlydiscrete translational symmetry). It does notmatter from what limit we start, that is, electrons localized on atoms or free electrons. When the hybridiza-

5.2. DIMENSIONS HIGHER THAN ONE 87

x

x

V(x)

V(x)

(a)

(b)

|ψ |G

2

|ψ |E

2

a

Figure 5.7: Wavefunctions for an electron in a periodic potential of the ions: (a) ground state; (b) first excitedstate.

tion between atoms is small the system behaves more like isolated atoms and the bands reflect the quantumnature of the atomic states. As the hybridization grows the system behaves more like a free electron gas.Observe that the electron plays a dual role in this dance: it binds atoms together and also gives rise to chargeconduction. Depending on the type of system and on the type oforbital that participates on the formation ofa solid the properties of the solid can change dramatically as one sees from the simples arguments we havegiven. In Fig.5.8 we show how the free electron solution compares with the solution close to the Brillouinzone boundaryk = G/2 = π/a. We see that the free electron solution works very well closeto the bottomof the band but fails to describe the system at the Brillouin zone boundary.

5.2 Dimensions higher than one

In higher dimensions the situation is very similar to the onedimensional case and we still have to solve theSchrödinger equation

(

− h2

2m∇2 + V (r)

)

ψ(r) = Eψ(r) (5.59)

where due to the symmetry we haveV (r) = V (r + T) whereT =∑d

i=1 niai is a lattice vector. Becauseof this symmetry we can write

V (r) =∑

G

VGeiG·r (5.60)


0.1 0.2 0.3 0.4 0.5k

0.2

0.4

0.6

0.8

1

E

Figure 5.8: Solution of the problem close tok = G/2 and comparison with the free electron result (dashedline).

whereG is a reciprocal lattice vector (notice thatT · G = 2πn which is the Bragg law). The generalsolution of (5.59) can be written as

ψE(r) =∑

k

ψk,Eeik·r (5.61)

where the Fourier components obey the equation (by direct substitution of (5.60) and (5.61) into (5.59))

(

hk2

2m− E

)

ψk,E +∑

G

VGψk−G,E = 0 (5.62)

which is the higher dimensional analogue of (5.34).

Let us consider now the general properties of (5.62). (5.62)can be thought (as in the case of (5.36)) as amatrix equation for the coefficientsψk−G,E. For a given eigenvalueE the coefficientsψk,E have to be linearcombinations of other coefficients at wavevectors shifted by G. Thus, if we define againE = h2k2/(2m)then we must have accordingly to (5.61):

ψk(r) =∑

G

ψk−Gei(k−G)·r . (5.63)

We can rewrite this wavefunction in a more interesting way,

ψk(r) = eik·ruk(r) (5.64)

where

uk(r) =∑

G

ψk−GeiG·r . (5.65)


Observe thatuk(r) = uk(r + T) is a periodic function. Thus, we have shown that in a periodiclattice thewavefunction can be written in a very special form which is extended over the whole lattice. This is calledBloch’s theorem. The result of this demonstration is that the wavefunction can be defined in the Brillouinzone of the material since we can reach anyk outside of the zone by translating by a reciprocal lattice vector,that is,ψk+G(r) = ψk(r). Moreover, the energy is also a periodic function of the reciprocal lattice sincewe can always shift the momentum in (5.62) by a reciprocal lattice vector,Ek = Ek+G. This result iscompletely general and does not depend on the particular form of the potential!

In the limit where the potential is weak, as we have seen previously, we can try perturbation theory.As before we will find that the perturbation theory fails atk = ±G/2 which is the border of the Brillouinzone. This failure is again related with the opening of a gap in the spectrum due to the Bragg scattering,that is, the energy of the system is only slightly modified in for most wave-vector except whenǫ0k = ǫ0k−G

where perturbation theory gives a divergent contribution.The origin of this divergence can be understoodimmediately since the condition for the divergence is equivalent to

|k| = |k −G| (5.66)

which is the Bragg law (see (2.38)). This result makes a lot ofsense since electrons are as good waves asany other microscopic particles. Thus when the electron wave-vector is such that the Bragg law (5.66) isobeyed the electron undergoes coherent Bragg diffraction.Observe that this condition is equivalent to saythat k has to lie on the zone boundary in order to undergo Bragg diffraction and this is the point whereperturbation theory fails and one has to use the matrix approach described earlier. In this case we obtain theformation of a gap at these points, that is, the gap is a straight consequence of Bragg diffraction. Considerfor instance the one dimensional case studied before. Condition (5.66) implies that perturbation theory failsfor k = ±G/2 = ±nπ/a and one has a sudden jump in the energy at this point. A graphical way to depictthis situation is shown in Fig.5.9 where we see that the only substantial difference between the free electroncase and the actual case occurs whenever a gap opens at the Bragg points. In order to obtain the bandsdiscussed in previous approach one has just to fold back the points back to the Brillouin zone as in Fig.5.8.

Ek

k0 π/a 2 π/a 3 π/a−π/a−2 π/a−3 π/a

|U |π/a

|U |2π/a

Figure 5.9: Modification of the free electron dispersion (dashed line) due to the Bragg reflection of theelectrons.

This construction can be carried out to higher dimensions where the points where gaps open are stillgiven by the Bragg condition (5.66). Consider for instance the problem of a two dimensional square lattice


such as in Fig.5.10. As the momentum varies as shown in Fig.5.10(a) the gaps open as in Fig.5.10(b).

k

k

x

yk

a

bc

π/a

π/a

−π/a

−π/a 2π/a

a b ck

Ek

(a)

(b)

Figure 5.10: (a) Brillouin zone in k-space (shaded area) andneighboring zones; (b) Electron dispersionalong the direction ofk.

At this point one natural question comes out: how are these results related with the tight binding calcu-lation we have done previously? The answer to that is very straightforward. In the tight binding limit theelectrons are tightly bounded to the atoms and therefore thepotential is strong. However, accordingly toBloch’s theorem, (5.64), we should be able to solve the problem in terms of periodic functions. The wave-function of the problem has to obey Block’s theorem which is astatement of the periodicity of the system,nothing more. Observe that from (5.64) and (5.65) the wavefunction of the problem must obey

ψk(r + T) = eik·Tψk(r) . (5.67)

We would like to construct the wavefunction from localized orbitals instead of the plane waves which are sonatural for the weak potential problem. These localized orbitals are the solution of the atomic Schrödingerequation:

(

− h2∇2

2m+ VA(r)

)

ψA(r) = EAψA(r) (5.68)

whereVA(r) is the atomic potential andEA the binding energy of the electron. Let us consider the problemwhere the atoms are far apart so the tunneling is small. We would like to construct the wavefunction fromlocalized orbitals like the ones given in (5.68) which has this property. Remember that in the case ofmolecules this can be done by taking linear combinations of localized orbitals (which we called bondingand anti-bonding). A similar idea here is to choose the linear combination

ψk(r) =1√N

∑

T

eik·TψA(r− T) . (5.69)

Observe that this wavefunction obeys (5.67) Moreover, in this approximation each band carries the atomic


numbers, that is, for each atomic level we have a band with itsquantum numbers.

Observe that (5.69) is a solution of the atomic equation (5.68) but instead we would like to solve theactual Schrödinger equation (5.59). It is clear that this isnot the actual ground state of the problem. However,in the limit of infinite separation between the atoms it is an eigenstate. Moreover, the spectrum of theproblem can be written from (5.59) by using the orthogonality of the wavefunctions as

Ek =

∫

ddr ψ∗k(r)

[

− h2∇2

2m+ V (r)

]

ψk(r) (5.70)

which, as we said before, is periodic on the reciprocal lattice, that is,Ek+G = Ek. Since the energy isperiodic on the reciprocal lattice it can be always expand asa Fourier series of the direct lattice (rememberthat the direct lattice is the reciprocal of the reciprocal lattice!) that is,

Ek =∑

T

tTeik·T . (5.71)

In order to calculate the coefficientstT we are going to use (5.70). If we substitute (5.69) into (5.70) we find

Ek =∑

T,T′

eik·(T−T′)

N

∫

ddr ψ∗A(r− T′)

[

− h2∇2

2m+ V (r)

]

ψA(r − T)

=∑

T,T′

eik·(T−T′)

N

∫

ddr ψ∗A(r + T − T′)

[

− h2∇2

2m+ V (r)

]

ψA(r)

=∑

T

eik·T∫

ddr ψ∗A(r + T)

[

− h2∇2

2m+ V (r)

]

ψA(r) (5.72)

where we have used the periodicity of the potentialU(r + T) = U(r). Thus we have shown that (5.71) iscorrect and that

tT =

∫

ddr ψ∗A(r + T)

[

− h2∇2

2m+ V (r)

]

ψA(r) (5.73)

where the integral runs through the unit cell. We now rewriteV (r) = VA(r) + δV (r) whereδV (r) =V (r) − VA(r). In the limit where the distance of the atoms is very large we expectδU → 0 and thereforecan be treated as a perturbation. In this case, using (5.68) one gets

tT = EA

∫

ddr ψ∗A(r + T)ψA(r) +

∫

ddr ψ∗A(r + T)δV (r)ψA(r) . (5.74)

Observe now that the atomic wavefunction decays exponentially with distance (ψA(r) ∝ e−r/ao wherea0

is the Bohr radius) and therefore the overlap integrals above will be very small forT 6= 0. In this case it isa good approximation to use the on-site (T = 0) and nearest neighbor contributions since the integrals willdecay exponentially with distance. This is known as the tight binding approximation. In this approximationwe write

tT ≈ EAδT,0 − tA∑

~δ

δT,~δ

(5.75)

where~δ represent the nearest neighbors sites andtA is the overlap given in (5.74) for nearest neighbors.


Substitution of the above expression into (5.71) leads to

Ek ≈ EA − tA∑

~δ

eik·~δ . (5.76)

Observe that corrections to the tight binding approximation can be computed directly from (5.73).

As an example consider the one-dimensional case. In this case we have just two nearest neighbors whichcorrespond toδ = ±a wherea is the lattice spacing. In this case the energy becomes,

Ek = EA − 2tA cos(ka) (5.77)

and has the shape shown in Fig.5.2. For a cubic system ind dimensions the tight binding model predicts adispersion

Ek = EA − 2tA

d∑

i=1

cos(kia) (5.78)

which has its minimum value atEA − 2dtA and its maximum atEA + 2dtA. Thus, we say that the systemhas a bandwidthW = 4dtA. In the atomic limittA → 0 and the bandwidth vanishes while as the tunnelingbetween atoms increases the bandwidth increases proportionally to the hoppingtA as in Fig.5.3. Observethat gaps that we have computed in the free electron gas just represent the difference in energy betweenthe atomic levels that decreases as the tunneling between atoms increases. When the energy levels start tooverlap we are back to the nearly free electron gas.

Notice that at long wavelengthska≪ 1 we can approximate the problem by expanding the exponentialin (5.76) to second order inka,

Ek ≈ EA − ZtA +tA2

∑

~δ

(k · ~δ)2 (5.79)

whereZ is the number of nearest neighbor sites (Z = 2d for a cubic system) and we have used that∑

~δk · ~δ = 0 by inversion symmetry. We can now use that

~δ =

d∑

i=1

ani (5.80)

wherea is the lattice spacing andni is the unit vector along the main crystal axis. In this case werewrite(5.79) as

Ek ≈ EA − ZtA +

d∑

i=1

h2k2i

2m∗A

(5.81)

which has the form of a free electron dispersion where

m∗A =

h2

2tAa2(5.82)

is the so-called effective mass of the electrons. Observe that this mass can be very different from the freeelectron mass since its physical meaning is completely different of that one. The effective mass measures thestatic dragging of the lattice by the electron motion since the electron interacts with the atomic cores. This


can be easily understood if we take a semi-classical approach. As we have seen the electrons are describedby a dispersionǫk. From this dispersion we can define the velocity by

vk =1

h∇ǫk (5.83)

which can be checked to be correct for the free electron case whereǫk = h2k2/(2m) andvk = hk/m.Suppose an electrical fieldE is applied to the system. This electric field will cause the electrons to accelerate.It turns out that the electrons deep inside the Fermi sea cannot accelerate since they would have to maketransitions to already occupied states. This is forbidden by Pauli’s principle. The electrons at the Fermisurface, however, can be accelerated freely. If we imagine this electrons to behave like wavepackets (that is,like classical particles) then the acceleration is given by

dvk

dt=

1

h

d

dt

d∑

i=1

∂ǫk∂ki

ni

=1

h

d∑

i,j=1

∂2ǫk∂ki∂kj

dkjdt

ni (5.84)

which is valid fork ≈ kF and where we used the chain rule andni is the unit vector in thei-direction.Observe that the momentum of the electron is simplyhk and therefore by Newton’s equation of motion inan external field we have

hdk

dt= eE (5.85)

which substituted in (5.84) leads to

dvk

dt=

e

h2

d∑

i,j=1

∂2ǫk∂ki∂kj

Ejni . (5.86)

Observe that (5.84) is very similar to the classical equation

dv

dt=eE

m. (5.87)

This similarity allows us to define the effective mass tensor

m−1ij =

1

h2

∂2ǫk∂ki∂kj

. (5.88)

Observe that for a cubic system with dispersion given by (5.78) in the limit wherekFa≪ 1 we obtain (5.82)as expected.

In this picture the difference between metals and band insulators is very simple to understand. At bandfillings smaller than one there is in average less than two electron per site as in Fig.5.4(a). Thus the electronscan move freely since electrons with opposite spin can hop from site to site. When the band becomes full(that is we have two electrons per site) the Pauli principle does not allow motion as shown in Fig.5.4(b). Inorder to move an electron one has to excite the electron to an unfilled atomic state and therefore one has topay an energy price which is the gap energy.


5.3 Problems

1. Prove eq.(5.62).

2. Show that (5.69) obeys (5.67).

3. Show that a band insulator always has an even number of electrons per atom. Also argue that theopposite is not true, that is, solids with an even number of electrons per atom can be metallic.

4. In the free electron gas the mass of the electron is justm. In the tight binding case the mass of theelectron is defined as

1

m∗=

1

h2

d2Ekdk2

.

(1) What is the mass of the electron as a function of the momentum? Why is this mass different fromm? (2) What is the mass of the electron when the band is empty? What happens to this mass whenthe tunneling between atoms goes to zero? Give a physical interpretation for your result.

5. In these notes we have used the so-called reduced zone scheme in which the momentum is plottedonly in the first Brillouin zone−π/a < k ≤ π/a. Another possible scheme is to plot the energy-momentum relation over all momentum space in which case we donot fold the energy in the Brillouinzone. Starting from the free electron problem (parabolic band) make a qualitative plot of the energyof a particle in a periodic potential as a function ofk from −4π/a to 4π/a. What are the positions ofthe gaps in momentum space? What are the values of the gaps in terms of the Fourier components ofthe periodic potential?Hint: no calculations needed.

6. Solve exactly the matrix problem in (1.52) for the case where ck±4pi/a,E = 0. Make a plot of theenergy bands you get from the solution. What is the value of the gaps in the spectrum?

7. Show that for a cubic system we have:

m−1ij =

cos(kia)

m∗δi,j .

Chapter 6

The Free Electron Gas

In the last chapter we have seen that the main difference between metals and band insulators is that ininsulators the bands are full and therefore one has to pay an energy equal to the gap in order to make anexcitation. In a metal an electric field can excite electronsfrom one momentum state to another with almostno energy cost while in an insulator this is not possible because of the gap. In this chapter we are going todiscuss the physics of metals based on the picture of the lastsection, that is, the picture that the electrons donot interact among each other but only fill energy states accordingly to the Pauli principle. This is called theelectron gas approximation.

The Pauli principle asserts that the ground state of the problem is obtained filling up all the states withlowest energy. We observe that the properties of the system depend strongly on the way we fill up the bands,that is, the so called filling factor. Suppose we assume periodic boundary conditions. The allowed vectorskare

ki =2πniNia

(6.1)

wherei = x, y, z where we have written the sizeLi = Nia. Observe that maximum allowed value ofniis Ni/2 since at this point we reach the boundary of the Brillouin zone,Gi/2 = π/a for the cubic lattice.Thus, the number of states available isN = NxNyNz which is the number of primitive cells. If we takeinto account the spin we have2N independent states in each energy band. If the atom contributes with justone electron the band is half-filled with electrons. If each atom contributes with two electrons the band iscompletely filled. The state of lowest energy is obtained by filling up the spectrum starting with thek = 0as in Fig.6.1(a). For eachk we can have two electrons with opposite spins. In momentum space this isequivalent of filling up all the states symmetrically up to the largest value which depend on the total numberof electrons as in Fig.6.1(b).

We are interested in the thermodynamic limit, that is, the limit thatN → ∞ andL → ∞ but ρ =N/Ld is finite. In this limit the distance between the states goes to zero and therefore one fills densely themomentum space. It is clear that the number of electrons in the system can be written as

N = 2∑

k

θ(kF − k) (6.2)

wherekF is the largest possible momentum and the factor2 comes from the two spin projections. In thethermodynamic limit we use our old trick

∑

k

→ Ld

(2π)d

∫

ddk (6.3)

95

96 CHAPTER 6. THE FREE ELECTRON GAS

Ek

k2π/L

(a)

k

kx

y

(b)

Figure 6.1: (a) Ground state of a free Fermi gas as a function of energy; (b) Ground state in momentumspace showing the lines of equal energy (circles).

and find

ρ =N

Ld= 2

∫

ddk

(2π)dθ(kF − k) =

2Sd(2π)d

∫ kF

0dkkd−1

=2Sdk

dF

d(2π)d(6.4)

which defines the so-called Fermi momentum,

kF =

(

d(2π)dρ

2Sd

)1/d

(6.5)

that depends only on the electron density and the dimensionality of the system. Associated with the Fermimomentum there is the so-called Fermi energy which is the energy of the highest occupied state,

EF =h2k2

F

2m. (6.6)

Thus, in this case, the finite size spectrum of Fig.6.1 becomes the one in Fig.6.2. The points in momentumspace,kF , such that

ǫkF= EF (6.7)

defines the Fermi surface. Observe that the system is highly degenerated since we have a large number ofstates for a given energy. The magnitude of this degeneracy is again defined in terms of a density of stateswhich is defined exactly like in the phonon problem, that is, Eq.(4.56). For the free electron problem one

97

ky

(b)

k

(a)

kx

k−kF F

EF

kF

kE

Figure 6.2: (a) Ground state of the free Fermi gas; (b) Groundstate in momentum space.

has

N(E) =2

V

∑

k

δ(E − ǫk) = 2

∫

ddk

(2π)dδ(E − ǫk)

=2Sd

(2π)d

∫ ∞

0dkkd−1δ

(

E − h2k2

2m

)

=Sd

(2π)d

(

2m

h2

)d/2

Ed−22

=dρ

2EF

(

E

EF

)d−22

(6.8)

where we have used (6.5) and (6.6).

The concept of a Fermi surface is one of the most important concepts in the physics of electrons insolids. In a solid of densityρ the volume occupied by one electron is in averageSdr

d0/d and therefore the

mean distance between electrons is of order of

r0 =

(

d

Sdρ

)1/d

. (6.9)

Since we are talking about atomic scales it is convenient to use the Bohr radius,a0 = h2/(me2), as adistance scale. Thus, the number,

rs =r0a0

(6.10)

gives us an idea of the density of the solid relative to the atomic volume of the atom. In this units the Fermi


energy (6.6) can be rewritten using (6.5) as

EF =

(

d2(2π)d

2S2d

)2/dme4

2h2

1

r2s. (6.11)

Observe that apart from the geometric factors the Fermi energy is proportional to the ionization energy of theHydrogen atomEI = me4/(2h2) ≈ 13.6 eV. In most metalsrs is of order of one or smaller (the distancebetween electrons is of the order of the distance between atoms). In this case it is easy to see that the Fermienergy is of order of eV, that is, of order of10, 000 K. The ground state energy can now be easily calculated:

E0 =∑

k

h2k2

2mθ(kF − k)

E0

N=

d

d+ 2EF . (6.12)

6.1 Elementary excitations

ky

(b)

k

(a)

kx

k−kF F

Ek

EF

k

k’

q

Figure 6.3: (a) Particle-hole excitation across the Fermi surface; (b) Particle-hole excitation in momentumspace.

The lowest energy excitation corresponds to take an electron with momentumk < kF and transfer itto a state with momentumk′ > kF . In this case a hole is left behind (see Fig.6.3). The momentum of thisparticle-hole excitation is

q = k′ − k (6.13)

6.1. ELEMENTARY EXCITATIONS 99

and has a characteristic energy

hω(q,k) =h2(k − q)2

2m− h2k2

2m=

h2

2m(2q · k + q2) . (6.14)

At some temperatureT all particle-hole pairs such thathω(q,k) ≤ kBT can be excited. At low temper-atures only a region of energykBT can be excited around the Fermi surface. This is clear contrast withthe phonon problem where any phonon state can be excited withany number of phonons. The fact that thephase space for excitations in the electronic system is muchsmaller than in the phonon problem is a pureconsequence of Pauli’s principle since one cannot occupy states that are already occupied. Obviously theexcitation with lowest energy has|k| = kF andq → 0. In this limit the excitation energy reduces to

ω(q||) ≈ vF q|| (6.15)

whereq|| is the component ofq in the direction ofkF and

vF =hkFm

(6.16)

is the Fermi velocity. The striking discovery here is that the dispersion relation for particle-hole excitationsis very similar to the one of acoustic phonons (see 4.39)). Observe that what we actually have done isthe linearization of the dispersion of the electrons close to kF as shown in Fig.6.4. The fact that at lowenergies particle-hole excitations behave as acoustic phonons is not a mere coincidence. It turns out that theelementary excitations of Fermi systems have bosonic character. We will come back to this interesting issuelater but now we are going to consider another important excitation of Fermi systems.

Ek

EFK TB

k

k−kF F

Figure 6.4: Linearization of the electron dispersion closeto the Fermi surface.

So far we have considered the situation where we only have excitations with smallq, that is, whenq ≪ kF . However, we can also have low energy excitations with largemomentumq. Consider the casewherek is close to the Fermi surface butk′ = k + q is on the opposite side of the Fermi sphere. In thiscase we can make an excitation withq ≈ 2kF that also has a low energy excitation sincehω(q = 2kF , k ≈kF ) ≈ h2/(2m)(−2(2kF )kF + (2kF )2) ≈ 0. Thus,2kF excitations with low energy are clearly possible(later we will see that these excitations are rather important). Thus, the excitation spectrum of the electron


gas is bounded by two energy scales, namely,

hω−(q) =h2

2m(2kF q + q2)

hω+(q) =h2

2m(−2kF q + q2) (6.17)

which correspond, respectively, to the small momentum and large momentum processes. Dimensionalityalso plays a role here. Notice that in one dimension the Fermisea becomes a line that runs from−kF to+kF and there are no excitations with zero energy at finite momentum since the only possible excitationshave eitherq = 0 or q = 2kF (see Fig.6.5). In higher dimensions there is a macroscopic number ofexcitations at low energies because the Fermi surface is a continuum. It is easy to see that in one-dimensionit is not possible to generate any particle-hole excitationwith energy smaller than

hωmin = −hω+(q) =h2

2m(2kF q − q2) (6.18)

while this is possible in 2 and 3 dimensions. If one draws the curves generated by (6.17) and (6.18) in theω × q diagram we get the situation shown in Fig.6.6.

F

−k +kF F

q ~ 0

(a)

(b)−kF +kF

q ~ 2 k

Figure 6.5: Fermi points in one dimension and the possible particle hole excitations: (a) small momentumtransfer; (b)2kF momentum transfer.

(b)

ω

0 2k F

q

ω

q0 2k F

(a)

Figure 6.6: (a)Particle-hole continuum in 1D; (b)Particle-hole continuum in 2D and 3D.

Let us go back to our original problem of electrons on a box of size L in three dimensions. Supposewe displace the whole electron gas through a distancex with respect to background of charge producedby the ions. In this case we will have an excess of positive charge on one side of the box and an excessof negative charge on the opposite side of the box (see Fig.6.7). This effect will create an electric fieldacross the box. By Gauss law the electric field is simplyE = 4πσ whereσ is the charge density on thewall of the box. This density is simplyσ = eρx. Of course this field will act on the electrons in such a

6.2. FREE ELECTRON GAS IN A MAGNETIC FIELD 101

way to move the displaced charge density back to its originalplace so charge neutrality is obtained again.Mathematically this come from the fact that the force applied on the charge on the opposite side of the boxis F = −NeE = −Ne4πσ = −4πρe2Nx. Thus the equation of motion for the displaced density is

(Nm)d2x

dt2= −4πρe2Nx(t)

d2x

dt2= −ω2

px(t) (6.19)

where

ω2p =

4πρe2

m(6.20)

is the so-called plasmon frequency of the electron system. Observe that (6.19) predicts that the displacedcharge undergoes harmonic motion with the plasmon frequency. Since the plasmon frequency is associatedwith the displacement of the whole electron system it has a characteristic wave-vectorq = 0. Thus thissimple argument shows that the electron gas, besides the particle-hole excitations described by (6.15), alsohas a collective mode called the plasmon that has a dispersion

ω(q) ≈ ωp (6.21)

at small wavelengths. Again, we have a similarity between the optical phonon problem and the electrongas. These similarities do not stop here. Observe that the plasmon is a high energy excitation since we needtemperatures of the order ofhωp in order to excite this mode. The plasmon is a massive excitation. Usingthe equations given before we can show that

hωp ≈ 41.7r−3/2s (6.22)

in eV. Thus, for metallic densities the plasma frequency is of order of the order of40 eV or40, 000 K. Thus,in order to excite a plasmon in a metal one has to use high energy probes such as X-rays or high energyelectrons. As in the phonon problem where acoustic phonons describe the the low temperature propertiesof ion vibration, the particle-hole excitations dominate the behavior of Fermi liquids at low temperatures.Observe that all the elementary excitations we are discussing here have a counterpart in the physics ofelementary particles. In order to create a particle from thevacuum one has to pay an energy price which istwice the rest energy of the particle. Thus optical phonons and plasmons behave like actual particles althoughthey are collective modes which are born out of the interaction between ions and electrons. The result forthe plasmon frequency is only valid for three dimensional solids. In lower dimensions the situation is morecomplicated because the electric field is not the one createdby a capacitor plate but by a line of charge. Inthis case we can show that the plasmon dispersion and the plasmon mode becomes gapless.

6.2 Free electron gas in a magnetic field

A simple way to probe the properties of an electron gas is to subject it to external fields. One way is toattach it to a battery and look at the current produced by the application of difference of potential. For a freeelectron gas the field will accelerate the electron freely producing an infinite current. This, of course, neverhappens since the electrons collide with impurities so thatthe current is always finite. We will discuss theseissues in detail later. Important properties of an electrongas can be measured by applying a magnetic field.Since the electron has spin and charge the magnetic field couples to the electron in two different ways. Thefirst and most simple coupling is to the electronic spin sincethe magnetic field will try to align the electron


+ + + + + + + + +

+ + + + + + + + +

+++++++++

+ + + + + + + + +

+++++++++

+ + + + + + + ++

+ + + + + + + + +

+ + + + + + + + +

+ + + + + + + + +

+++++++++

+ + + + + + + + +

+++++++++

+ + + + + + + ++

+ + + + + + + + + −

−

−

−

−

−

−

E = 4 π σ

dd

+σ −σ

(a)

(b)

Figure 6.7: Formation of a plasmon in a non-interacting electron gas: (a) uniform positive background pluselectron gas; (b) displaced electron gas relative to background.

spin. This is the Zeeman effect that you probably have seen inthe study of the Hydrogen atom. The secondeffect is due to the Lorentz force that the electron feels when is in motion. These are relativistic effects butthey have different origins. While the Lorentz force can be applied to classical particles, the Zeeman effectis a pure quantum mechanical effect. We are going to study each effect separately.

6.2.1 Zeeman effect and Pauli susceptibility

Consider a magnetic fieldB applied to a free electron gas. The Zeeman shift in energy dueto the magneticfield is just

HZ = −me · B (6.23)

whereme is the electron magnetic moment. For the electron spinS the magnetic moment is written as

me = µeS

h(6.24)

where

µe = gsµB ≈ 1.16 × 10−8 (6.25)

in units of eV/Gauss. Heregs ≈ 2 is the g-factor andµB = eh/(2mc) is the Bohr magneton. In order tostudy the Zeeman effect we have to add the new term (6.23) to the free electron Hamiltonian. Without anyloss of generality we choose the magnetic field to be along thez axis. Thus, it is clear that the problem canbe completely diagonalized in the basis|k, σ〉 whereσ =↑, ↓ where↑ means that the spin is parallel to thefield and↓ that the spin is anti-parallel to the field (Sz| ↑〉 = (h/2)| ↑〉 andSz| ↓〉 = −(h/2)| ↓〉). The


Ek

EF

n n

B

2µ BB

kF

k k

kF

Figure 6.8: Magnetization in a free Fermi gas.

eigenenergies of the problem can be written as:

Ek,↑ =h2k2

2m− µBB

Ek,↓ =h2k2

2m+ µBB . (6.26)

Thus, the dispersion of each mode is shifted relative to eachother by a quantity2µBB. The ground state ofthe system is obtained by filling up the states starting from the lowest energy state up to the Fermi energy,EF , as shown in Fig.6.8. Since the up and down spins are in equilibrium with each other it means that thechemical potential is the same, that is, the field does not change the Fermi energy of the system but only thenumber of species of electron as depicted in Fig.6.8. Thus, in the presence of a magnetic field the numberof ↑ spins per unit of volume,ρ↑, is different from the number of↓ spins,ρ↓. This implies the system hasbeen magnetized with a magnetizationM per unit of volume given by

M = µB(ρ↑ − ρ↓) . (6.27)

Observe that the change in the number of the two species of electrons is equivalent to a change in theFermi momentum of them. Thus we can define two new Fermi momenta,kF,↓ andkF,↑ which are related tothe number of fermions as in (6.4):

ρσ =Sd

d(2π)dkdF,σ . (6.28)

Since the total number of electrons is conserved in the presence of the magnetic field we have a constraintthat

ρ↑ + ρ↓ = ρ

kdF,↑ + kdF,↓ =d(2π)dρ

Sd. (6.29)

The other constraint that we discussed before is that the chemical potential is the same for each species of


electrons:

EF =hk2

F,↑

2m− µBB =

hk2F,↓

2m+ µBB

k2F,↑ − k2

F,↓ =4mµBB

h2 . (6.30)

Equations (6.29) and (6.30) have to be solved at the same time. It is useful to parameterize the unknowns as

kF,↑ = k0 cosh θ

kF,↓ = k0 sinh θ (6.31)

so that (6.30) is solved at once:

k0 =

√

4mµBB

h2 (6.32)

sincecosh2 θ − sinh2 θ = 1. Substituting (6.31) on (6.29) one gets

coshd θ + sinhd θ =d(2π)dρ

kd0Sd(6.33)

which is a complicated equation forθ. The quantity we are after is the magnetization of the systemwhich isgiven in (6.27):

M =SdµBk

d0

d(2π)d

(

coshd θ − sinhd θ)

. (6.34)

Instead of the general solution of this problem we are going to look in the limit of weak magnetic field.In this limit (B → 0) we see that from (6.32) thatk0 → 0 and therefore the right hand side of (6.33)diverges. It implies thatθ → ∞. In this case we can approximatecosh θ ≈ sinh θ ≈ eθ/2 and we find

eθ =

(

2d−1d(2π)dρ

Sd

)1/d1

k0(6.35)

In this approximation we can write (6.34) as

M ≈ µB21−dSdkd0

(2π)de(d−2)θ

≈ µ2BN(0)B (6.36)

whereN(0) is the density of states at the Fermi energy which is given in (6.8) for E = EF . Noticethat the magnetization is directly proportional to the density of states at the Fermi surface. The magneticsusceptibility of the system is given by

χP (T ) = limB→0

∂M

∂B= µ2

BN(0) (6.37)

is called the Pauli susceptibility and it is an exact result since only the linear term in the magnetic field con-tributes. Thus, the measurement of the magnetic susceptibility of a free electron gas is a direct measurementof its density of states at the Fermi energy.


Let us consider a simple case of the general problem discussed here. Let us considerd = 2. In this caseone can calculate the magnetization exactly from (6.34) and(6.32),

M =mµ2

BB

h2π(6.38)

which is strictly linear with the magnetic field. In this case(6.36) is not approximate but exact.

Finally we observe that by increasing the field one can fully polarize the electron gas (that is,ρ↑ = ρ).As we can see from Fig.6.8 this happens at a critical field,Bc, given by

Bc =EF2µB

. (6.39)

For a metallic system this will happen at fields of the order of108 G or 104 Tesla (1 Tesla=104 Gauss).These are very large fields.

6.2.2 Landau levels and de Hass-van Alphen effect

After discussing the spin part of the magnetic field problem we will consider the orbital part, that is, theeffect of the Lorentz force on the electron which is given byev × B/c. Classically, it is easy to showthat in the plane perpendicular to the field the electrons undergo circular motion. This motion is a result ofthe centripetal forcemv2/R whereR is the classical radius of the orbit. It is clear that the velocity of theelectron on its orbit isv = 2πR/T = ωcR whereT is the period of the orbital motion andωc = 2π/T isthe cyclotron frequency. The cyclotron frequency can be obtained immediately from the classical equationof motion

evB/c = mv2/R ,

ωc =eB

mc. (6.40)

Classically the radius of the orbit depends on the velocity of the electron. This is not entirely correct inquantum mechanics. In quantum mechanics the angular momentum is always quantized in units ofh. Forthe circular motion the angular momentum is justpR. Thus, from the quantization of the angular momentumwe find

pR = nh ,

Rn =√nl0 , (6.41)

where

l0 =

√

ch

eB, (6.42)

is the so-called cyclotron radius that gives a measure of thesize of the electron wavefunction in the presenceof an external magnetic field. Observe that apart from the magnetic field the cyclotron radius only dependson universal quantities. If we choose the magnetic field to begiven in Tesla, the cyclotron radius can bewritten as

l0 =250√B

. (6.43)


We see, therefore, that the cyclotron radius is usually muchlarger than the typical separation between atomsin a solid. In order forl0 to be of order of1 Åone needs fields of the order of6×104 T. These are incrediblylarge fields. As we saw previously at these fields the electrongas is completely spin polarized.

As you already know the Hamiltonian for a charged particle ina magnetic field is

H =1

2m

(

p − e

cA)2

(6.44)

whereA(r) is the vector potential which is related to the magnetic fieldvia B = ∇ × A. Assume forsimplicity that the field points in thez direction, that isB = Bz. As you know the problem of a particle inan electromagnetic field has what is called gauge invariance, that is, we can replace the vector potential byA → A +∇φ (whereφ(r) is an arbitrary function) without changing the value of the actual magnetic fieldsince∇×∇φ = 0 is a mathematical identity. This gauge invariance allows usto choose any form ofA(r)such thatB = Bz holds. Here we will choose the so-called Landau gauge:

A = −Byx . (6.45)

Substitution of (6.45) directly into (6.44) leads to

H =1

2m

(

px +eB

cy

)2

+p2y

2m+

p2z

2m. (6.46)

Observe that the Hamiltonian does not depend explicitly onx andz. Therefore, in thex andz directionsone has free motion. In this case the wavefunction of the problem can be written as

Ψkx,kz(r) = ei(kxx+kzz)ψ(y) (6.47)

whereψ(y) is unknown. In this case the Hamiltonian of the problem reduces to

H =1

2m

(

hkx +eB

cy

)2

+p2y

2m+h2k2

z

2m. (6.48)

A simple redefinition of variables helps a lot

δy = y + y0

y0 =hkxmωc

(6.49)

and the Hamiltonian of the problem becomes

H =p2y

2m+mω2

c

2(δy)2 +

hk2z

2m(6.50)

which is the Hamiltonian for a harmonic oscillator plus a free particle part. We have therefore solved theproblem entirely. The wavefunction is given by (6.47) withψ(y) ∝ Hn(y − y0) is a Hermite polynomial ofordern.

Since we have mapped the problem into the harmonic oscillator the spectrum is giving by

En(kz) = hωc

(

n+1

2

)

+h2k2

z

2m(6.51)

and is shown in Fig.6.9. The physics here is straightforward: the electron undergoes circular motion in


E

E

k

k

hωc

hω c

µ

µ

z

z

n=0

n=0

(a)

(b)

Figure 6.9: Landau levels of a free Fermi gas: in (b) the field is larger than in (a).

the plane perpendicular to the field (due to the Lorentz force) and free motion along the direction of thefield (since the Lorentz force vanishes). Circular motion isessentially harmonic motion in thex and ycoordinates. We are interested in the physical quantities such as the magnetization of the system. Noticethat when we change the magnetic field by an infinitesimal amount δB the internal energy of the systemchanges byδU = −MδB. Thus, the magnetization is simply

M(B) = −∂U∂B

. (6.52)

Observe, however, that the energy does not depend onkx and therefore the problem is highly degenerate.The reason for this degeneracy can be understood by looking into (6.49). Observe that the wavefunction ofthe electron is centered aty0 which is called thecenter of orbit. The energy of the problem is invariant if weshift the center of the orbit in the plane perpendicular to the field because the field is homogeneous in theperpendicular plane. Physically one has to impose the condition that0 ≤ y0 ≤ Ly. It implies from (6.49)that the momentum in thex direction is bounded by

0 ≤ kx ≤ mωcLy/h . (6.53)

But from the periodic boundary conditions we require thatkx = 2πnx/Lx and therefore the number of


states ink space for fixedkz is simply

Nφ =mωcLxLy

2πh=

Φ

φ0(6.54)

whereΦ = BLxLy is the total magnetic flux through the plane perpendicular toB and

φ0 =ch

e≈ 4 × 10−7 (6.55)

in Gauss-cm2 is the so-called flux quanta.

As we have seen so far the interesting part of the electron motion happens at the plane perpendicularto the field. It is possible to build artificial semiconducting structures such that the electrons are confinedto a very thin layer of thicknessLz. In this casekz = 2πnz/Lz is quantized. From now on we focus onthe two-dimensional electron gas and forget about the motion in the third direction. It is easy to generalizeour discussion here to the three dimensional situation in ordinary solids. From now on we just forget thekzdependence in (6.51) (as you can easily convince yourself all we are going to do here is equivalent to workwith fixed kz). The energy levels are shown in Fig.6.10. Each level can comportNφ electrons (we forgetspin for the moment being). If there are a total ofNe electrons in the system the ground state is obtained byfilling up the lowest energy states. Observe that the number of states per electron is simply

1

ν=Nφ

Ne=

B

σφ0(6.56)

where

σ =Ne

LxLy(6.57)

is the planar density of electrons.

Consider initially the case whereNφ > Ne, that is, only the first Landau level has electrons (0 < ν < 1).Clearly it implies very strong magnetic fields or small electronic densities since we must have from (6.56)thatB > σφ0. In semiconductors we have ordinarilyσ ≈ 1011 cm−2 which impliesB > 4 × 104 G (orB > 4 Tesla). In this case the total energy of the system is simply

E0 = Nehωc2

E0

Ne=

heB

2mc. (6.58)

From (6.52) we find

M(B)

Ne= −µe = − he

2mc(6.59)

is the electronic magnetic moment (notice that2µeB = hωc). Observe that this magnetization is constantas far asNφ > Ne or ν < 1. Suppose we decrease the magnetic field in the system with fixed numberof electrons. Then the fractional occupationν increases and the distance between Landau levels decreases.Whenν = 1 the first Landau level is fully occupied if we reduce the magnetic field even furtherNφ < Ne <2Nφ (or 1 < ν < 2) and some of the electrons, say,δN = Ne − Nφ have to occupy the second Landau


E

n=0

n=1

n=2

n=3

n

n−1

Nφ

N

N

N

N

N

φ

φ

φ

φ

Nφ −n Nφ

φ

hωc/2

ωh c/2

ωh c/2

ωh /2

3

5

7

ω /2h(n−1/2)

(n−1/2) ω /2h

Figure 6.10: Landau levels of a two dimensional free Fermi gas.

level. In this situation the energy of the system is

E1 = Nφhωc2

+ δN3hωc

2E1

Ne= hωc

(

3

2− 1

ν

)

(6.60)

and the magnetization is

M1

Ne= µe

(

−3 +4

ν

)

. (6.61)

Observe that something drastic happened because atν = 1 there is a jump in the magnetization from−µe to+µe. This jump occurred because the electrons in the lowest levels move to higher Landau levels as the fielddecreases. Observe that this effect occurs every timeν is an integer number, that is, each time a Landau levelis fully occupied. The generalization of the above argumentis simple. Assume that the last fully occupiedLandau level isn−1 so thatn is occupied byN −nNφ electrons (see Fig.6.10). In this casen < ν < n+1and the total energy of the system is

En = Nφ

n−1∑

m=0

hωc

(

m+1

2

)

+ (Ne − nNφ)hωc

(

n+1

2

)

EnNe

= hωc

(

n+1

2− n(n+ 1)

2ν

)

(6.62)


1 2 3 4 5n

-1

-0.5

0.5

1

M

N µ

Figure 6.11: Magnetization of a Fermi gas as a function ofν.

and the magnetization

Mn

Ne= 2µe

(

n(n+ 1)

ν− n− 1

2

)

(6.63)

which is shown in Fig.6.11 as a function ofν. Notice that the magnetization of the system is a periodicfunction ofν for ν > 1. Thus the magnetization of the electron gas is a periodic function of1/B as we seefrom (6.56). This effect is observed experimentally in the magnetization of solids and it is known as the deHaas-van Alphen effect.

6.2.3 Flux quantization

In our discussion of the Landau levels we defined a quantity called the flux quantaφ0 = hc/e but we gaveno interpretation for it or even why it is called a quanta. In order to understand why the magnetic flux isquantized we have to study the effect of the magnetic field in the electron wave function. Consider theproblem of a two dimensional electron gas in a plane where an infinite solenoid is perpendicular to it. Thesolenoid has areaS and an internal magnetic fieldB. The magnetic field outside the solenoid is zero. Themagnetic field is given by the usual relationB = ∇×A. Although the magnetic field is zero outside of thesolenoid the vector potential is still finite. As a consequence we must have

A(r) = ∇Ω(r) (6.64)

outside of the solenoid since∇ × A = ∇ × ∇Ω = 0. Moreover, the magnetic flux through the plane isgiven by

Φ =

∫

AB · dS =

∫

CA · dl =

∫

C∇Ω · dl = ∆Ω (6.65)


B

C

Figure 6.12: Geometry of the Bohm-Aharonov effect.

whereA is the area of the solenoid,C is a closed curve outside of the solenoid andl is a vector thatparameterizes this curve. Observe that the flux through the plane depends only on the change of the functionΩ around a closed curve.

The Schrödinger equation for this problem is

1

2m

(

−ih∇− e

cA(r)

)2ψ(r) = Eψ(r)

− h2

2m

(

∇ +ie

chA(r)

)2

ψ(r) = Eψ(r) . (6.66)

For the region outside the solenoid we can use (6.64) and rewrite the above equation as

− h2

2m

(

∇ +ie

ch∇Ω(r)

)2

ψ(r) = Eψ(r) (6.67)

and a simple inspection of this equation leads to the conclusion that the wavefunction has the form

ψ(r) = e−iech

Ω(r)φ(r) (6.68)

where

− h2

2m∇2φ(r) = Eφ(r) (6.69)

obeys the free particle problem. As expected bygauge invariancethe spectrum of the problem does notdepend on the magnetic field (since there is none) but the wavefunction now has a new phase factor thatdepends on the magnetic field. This result has amazing consequences and one of them is the so-calledBohm-Aharonov effect. Since there is no field outside the solenoid we must require that the wavefunctionis invariant under closed curves that circulate the solenoid. Suppose the electron starts at a pointR with awavefunctionΨB(R) and returns to the same point after going around the solenoidas in Fig.6.12

From (6.68) the wavefunction of the electron after the closed curve is

ΨA(R) = e−iech

∆ΩΨB(R) (6.70)


y

x

δy0

φ0

Figure 6.13: Quantization of the magnetic flux in the Landau gauge.

but sinceΨA(R) = ΨB(R) we must require

e

ch∆Ω = 2πn (6.71)

wheren is an integer. Using (6.65) we obtain that

Φ = nφ0 (6.72)

whereφ0 = hc/e is the flux quantum. Our argument shows therefore that for a free electron moving in aclosed orbit around the solenoid the magnetic flux through the orbit must be quantized. The relevance ofthis result for the Landau level problem comes from the fact that the degeneracy of each Landau level isthe total magnetic flux measured in units of the flux quantum, (6.54). Physically, this can be understoodif we recall that in the Landau gauge the electron motion in the x direction is free while in they directionwe have harmonic motion as shown in Fig.6.13. Observe that the wavefunction in they direction is peakedaroundy0 and becausekx = 2πnx/Lx is quantized the values ofy0 from (6.49) are also quantized asy0(nx) = (φ0/BLx)nx. Thus, the magnetic flux through a strip of sizeδy0 is simply,Bδy0Lx = φ0. Thus,exactly one flux quantum goes through the strip.

6.3 Free electrons at finite temperature

We have talked only about what happens to the electron gas at zero temperature. As we said before if weincrease the temperature we can excite electrons to states out of the Fermi surface producing particle-holeexcitations. In order to talk about finite temperature effect one has to discuss first the concept of chemicalpotential. The chemical potential is the energy required inorder to change the number of particles in thesystem by one, that is, ifE(N) is the free energy of the system withN electrons than the chemical potential,µ, is defined as

µ = E(N + 1) − E(N)

≈ ∂E

∂N(6.73)

where the last line is valid ifN ≫ 1 which is the case under consideration here. At zero temperature, in thefree Fermi gas, the energy required to put one extra electronin the system is essentially the Fermi energy,

6.3. FREE ELECTRONS AT FINITE TEMPERATURE 113

EF since as we put one extra electron it has to go to an unoccupiedstate at the Fermi surface. In the phononproblem the chemical potential is zero since phonons can be produced with infinitesimally small energies.We observe that in the Fermi gas the excitations occur close to the Fermi surface, or chemical potential, thusit is convenient to measure all the energies relative to the Fermi energy (the natural energy scale in metals).Observe that one usually calculates the total energy of the system assuming a fixed number of particles. Itturns out, however, that for most calculations at finite temperature it is more convenient to work with fixedchemical potential. Moreover, one actually have the systems of interest in contact with reservoir of particles(such as batteries, for instance) which keep the chemical potential constant and allow the number of particlesfluctuate. In the thermodynamic limit,N → ∞, these fluctuations are of order

√N and therefore irrelevant

compared toN itself. Thus we define the free energy,F , of the system as

F = E − µN , (6.74)

which is known as the Legendre transformation of the thermodynamic potential. Thus, using (6.73), we finddF = dE − µdN −Ndµ = −Ndµ, that is

N = −∂F∂µ

(6.75)

which gives the average number of particles as a function of the chemical potential.

Let us consider here the simple problem of theN non-interacting electrons described by a HamiltonianH with single energiesǫα. By the Pauli principle we know that the occupationnα of each state can be only0 or 1. A stateΨn of the system is described by the occupation of each single particle state, that is, eachstate is represented by a set of numbersΨn = (n1, n2, n3, ....). The total number of electrons in the systemis

N =∑

α

nα , (6.76)

while the total energy of a particular stateΨn is

En =∑

α

nαǫα . (6.77)

The partition function of the problem is defined as

Z = e−βF =∑

n

〈n|e−β(H−µN)|n〉

=

1∑

nα=0

e−βP

α nα(ǫα−µ)

=∏

α

1∑

nα=0

e−β(ǫα−µ)nα

=∏

α

(

1 + e−β(ǫα−µ))

(6.78)

and the free energy is

F = − 1

β

∑

α

ln(

1 + e−β(ǫα−µ))

. (6.79)


Therefore, using (6.75), one finds

N =∑

α

1

eβ(ǫα−µ) + 1(6.80)

which by comparison with (6.76) allows us to define the mean occupation per stateα as

nα =1

eβ(ǫα−µ) + 1(6.81)

which is the so-called Fermi-Dirac occupation number. Naturally the average energy density of the systemin simply

E =1

V

∑

α

ǫαnα (6.82)

and other physical quantities can be calculated as well. As in the phonon case we define the density of states

N(E) =1

V

∑

α

δ (E − ǫα) (6.83)

and transform all the sums into integrals, in particular, the mean energy density is

E =

∫

dE N(E)E n(E) . (6.84)

Observe that the Fermi-Dirac distribution has the properties one expects for electrons. Consider forinstance the zero temperature limit, that is,β → ∞. If E > µ the exponent in the exponential in (6.81)is positive and therefore whenβ → ∞ the Fermi-Dirac occupation vanishes. IfE < µ the exponent isnegative and at zero temperature it vanishes leading ton = 1. The conclusion is that atT = 0 we have astep function

n(E) = θ(µ− E) (6.85)

as in Fig.6.14(a) and only states below the chemical potential are occupied, as expected. At low temperatureskBT ≪ µ only states in a region of widthkBT with energy close toµ are thermally excited, as shown inFig.6.14(b).

These properties of the Fermi-Dirac distribution are very useful if we use the fact that the derivative ofa theta function is a Dirac delta function, that is,

∂n

∂E= −δ(µ− E) . (6.86)

At finite temperatures the delta function is broadened bykBT . In order to explore the usefulness of theproperties of the Fermi-Dirac distribution consider an integral of the distribution with an arbitrary function


n

n

E

E

µ

µ

1

1

0

0

(a)

(b)

K TB

Figure 6.14: Fermi-Dirac distribution: (a) zero temperature; (b) finite temperature.

f(E)

I(µ, T ) =

∫ +∞

−∞dEf(E)n(E)

=

∫ +∞

−∞dE

dF

dEn(E)

= −∫ +∞

−∞dEF (E)

∂n

∂E, (6.87)

where we have defined as

F (E) =

∫ E

−∞dE′f(E′) . (6.88)

Observe that the last line was obtained by integration by parts. We assume that the integrand vanishes in theE → −∞ limit. Observe that the derivative of the Fermi-Dirac distribution at low temperatures is highlypeaked aroundE = µ and assuming thatF (E) is a smooth function close to this point we make a Taylorseries expansion forF (E):

F (E) = F (µ) +∞∑

n=1

F (n)(µ)

n!(E − µ)n , (6.89)

whereF (n)(µ) is thenth derivative of the function atE = µ. We further observe that

∫ +∞

−∞dE

(

− ∂n

∂E

)

= 1 , (6.90)

and that the Fermi-Dirac function is an even function ofE − µ. By direct substitution of (6.89) into (6.87)


we obtain

I(µ, T ) =

∫ µ

−∞f(E)dE +

∞∑

n=1

f (2n−1)(µ)

(2n)!

∫ +∞

−∞dE(E − µ)2n

(

− ∂n

∂E

)

, (6.91)

which can be further simplified if we change variablesx = β(E − µ) in order to obtain

I(µ, T ) =

∫ µ

−∞f(E)dE +

∞∑

n=1

fn(µ)(kBT )2n , (6.92)

where

fn =f (2n−1)(µ)

(2n)!

(

2 − 22(1−n))

ζ(2n) , (6.93)

whereζ(n) is a Riemann zeta function. Equation (6.92) gives the finite temperature correction to the zerotemperature result in powers ofT . This is called the Sommerfeld expansion. We also notice that thecorrections to the zero temperature result is given in even powers of the temperature. In particular, at lowtemperatures the first correction will always be of orderT 2, thus, the first term in (6.92) can be replaced by

∫ µ

−∞f(E)dE =

∫ EF

−∞f(E)dE +

∫ µ

EF

f(E)dE ,

≈∫ EF

−∞f(E)dE + (µ−EF )f(EF ) . (6.94)

6.3.1 Specific heat

Let us apply this result for the case of the free electron gas.From (6.82) we havef(E) = EN(E) and

E ≈∫ EF

0EN(E)dE

+ EF (µ− EF )N(EF ) +π2

6

(

EFN′(EF ) +N(EF )

)

(kBT )2 (6.95)

and the mean density is

ρ ≈∫ EF

0N(E)dE + (µ− EF )N(EF ) +

π2

6N ′(E)(kBT )2 . (6.96)

Notice that number of particles at finiteT is the same as at zero temperature and therefore from (6.96) weget

µ ≈ EF − π2

6

N ′(EF )

N(0)(kBT )2 , (6.97)

which when substituted in (6.95) gives

E ≈ E0

V+π2

6N(0)(kBT )2 . (6.98)


Finally, from (6.98), we obtain the electronic contribution for the specific heat

CVV

=∂E

∂T=π2

3N(0)k2

BT , (6.99)

which is independent of the dimensionality of the problem (that only modifies the density of states whichfor the case of free electrons is given in (6.8)).

Eq. (6.99) could have been obtained without all this mathematical workout if we have used the physicalmeaning of the specific heat which is the number of excitations in the system. Obviously in the case of asystem with a Fermi surface the number of excitations available is proportional to thekBT times the numberof states per unit of energy which is simplyN(0) which gives the correct order of magnitude of the specificheat. Since the average energy of the particle-holes iskBT the excitation energy will beN(0)(kBT )2 whichmisses the correct value by a factorπ2/6. Another way to calculate the specific heat is to use the analogywe made before between the particle-hole excitations and acoustic phonons. As we showed in (6.15) theparticle-hole excitations have a dispersion relation of one-dimensional acoustic phonons since it dependsonly on the component of the momentum perpendicular to the Fermi surface. The specific heat of onedimensional phonons, according to (4.59), indeed behave like T . In order to show that this is more thana coincidence let us calculate the specific heat of these particle-hole excitations. First we observe that thedensity of states at the Fermi surface of a Fermi gas can be written as in (6.8)

N(0) = 2

∫

ddk

(2π)dδ(EF − ǫk) . (6.100)

Observe that only states at the Fermi surface, that is, withk suchEF = ǫk contribute to the integral. Thuswe make a simple change of variables

k = kF + q (6.101)

and therefore

ǫk ≈ EF + q · ∇ǫkF

= EF + q||vF (6.102)

where we have used (6.16). As expected only the component ofq normal to the Fermi surface appears inthe energy in agreement with (6.15). Therefore it is naturalto rewrite

q = q||kF

kF+ q⊥ (6.103)

whereq⊥ is tangent to the Fermi surface. Observe that the density of states at the Fermi energy can bewritten as

N(0) = 2

∫

dq||dd−1q⊥

(2π)dδ(q||)

vF

= 2

∫

dd−1q⊥(2π)d

1

vF(6.104)

involves only the component ofq perpendicular to the Fermi surface. Now assume for the moment beingthat the particle-hole excitations have bosonic character. Then the partition function for these bosons is the


same one as the acoustic phonons case, (4.52)

Z = e−βF =∏

q

[

sinh(βvF q||/2)]−1

. (6.105)

The specific heat is simply

CVV

=β2

4

∫

dq||dd−1q⊥

(2π)d(vF q||)

2

sinh2(βvF q||/2)

=N(0)vFβ

2

4

∫ ∞

−∞dq||

(vF q||)2

sinh2(βvF q||/2)

= 2π2

3N(0)T , (6.106)

where we have used (6.104). Observe that this is twice than the expected result (6.99). This happenedbecause we have double counted the number of particle-hole excitations. The mapping of the particle-holeproblem into bosons has an extra requirement thatq|| ≥ 0, that is, we count only the excitations outsidethe Fermi surface. If we take this into account we get the correct result. The connection between theparticle-hole excitations of an electron system goes beyond what has been discussed here. We can show thatin dense Fermi systems there is an operator identity betweenfermions and bosons. This identity is calledbosonization for obvious reasons.

6.3.2 Correction to the Pauli susceptibility

Another interesting application of the Sommerfeld approximation is the calculation of the correction to thePauli susceptibility. In the presence of a magnetic field theZeeman energy shifts the energy of the up spinsby−µBB while the down spins the energy shifted by+µBB. Therefore the density of states of each speciescan be directly obtained by the density of states of the unpolarized case,N(E). Indeed, by definition,

N↑(E) =1

V

∑

k

δ(E − ǫk + µBB) ,

N↓(E) =1

V

∑

k

δ(E − ǫk − µBB) , (6.107)

but from definition (6.8)

N↑(E) =1

2N(E + µBB) ,

N↓(E) =1

2N(E − µBB) . (6.108)

The magnetization is given by

M =µB2

∫ ∞

−∞dE n(E) (N(E + µBB) −N(E − µBB)) , (6.109)

and the magnetic susceptibility is simply

χ(T ) = limB→0∂M

∂B= µ2

B

∫

dEn(E)dN

dE. (6.110)


Using the Sommerfeld expansion one finds

χ(T ) = µ2BN(0)

(

1 +π2

6[ln(N(0))](2)(kBT )2

)

(6.111)

which gives the correction to the Pauli susceptibility.

6.3.3 Landau diamagnetism

Finally let us consider the problem of a two dimensional electron gas in a magnetic field. Since the levelsare quantized the density of states is not a smooth function of the energy but has delta functions at the energyof each Landau level

N(E) = 2Nφ

∞∑

n=0

δ (E − hωc(n + 1/2)) , (6.112)

whereNφ gives the degeneracy of each level and the factor of2 account for the two spin orientations. Thefree energy of the system is given in (6.79)

F = −2Nφ

β

∞∑

n=0

ln(

1 + eβ(µ−hωc(n+1/2)))

, (6.113)

It is very hard to perform this sum. Since we are interested onthe susceptibility of the electron gas we justhave to evaluate the free energy at very small fields, that is,kBT ≫ hωc. In this case we rewrite the sum as

F

S= − 2B

βφ0

∞∑

n=0

ln(

1 + eβ(µ−hωc/2)e−xn

)

, (6.114)

whereS = LxLy is the total area and

xn =n

kBT/(hωc)= (βhωc)n . (6.115)

Thus, whenkBT ≫ hωc(βhωc ≪ 1) the separation between the allowed values ofxn is very small and wecan replace the sum by an integral. However, we are not interested in strictly zero magnetic field and wehave to keep the first correction. This is done with the help ofthe so-called Euler-MacLaurin formula,

r∑

n=0

f(xn) =1

h

∫ xr

x0

dxf(x) +1

2(f(x0) + f(xr))

+

∞∑

n=1

B2nh2n−1

(2n)!

(

f (2n−1)(xr) − f (2n−1)(x0))

(6.116)

wherexn = x0 + nh andBn is a Bernoulli number. Application to this formula to (6.114) leads to

F

S= − 2B

βφ0

∫ ∞

0dx ln

(

1 + z e−βhωcx)

+e2B2

24πmc2z

z + 1, (6.117)

where

z = eβµ , (6.118)


is called the fugacity of the system. Observe that all the dependence of the free energy on the chemicalpotential can be obtained from the fugacity. Thus any derivative in respect to the chemical potential can bereplaced by a derivative with respect to the fugacity, in particular, from (6.75) one has

N = −∂F∂µ

= −βz∂F∂z

. (6.119)

A simple change of variablesy = −ze−βhωcx in (6.117) leads to

F

S=

m

πβ2h2Li2(−z) +e2B2

24πmc2z

z + 1, (6.120)

where

Li2(x) =

∫ 0

xdy

ln(1 − y)

y=

∞∑

n=1

xn

n2, (6.121)

is a polylogarithmic function of second degree.The total number of electrons is obtained from (6.75)

σ =N

S=

m

πβh2 ln(1 + z) +e2B2β

24πmc2z

(z + 1)2, (6.122)

which is a transcendental equation which givesz as a function ofσ. Let us first consider theB = 0 case,then

1 + z0 = eβEF , (6.123)

where we have used (6.4). Since we are interested in the low density limit it is clear that corrections toz0are going to be very small. We therefore writez ≈ z0 ≫ 1. Substituting this result in (6.120) we find

F

S≈ E0 +

e2B2

24πmc2(6.124)

whereE0 is the ground state energy of the system in the absence of an external field. Finally the magneti-zation density is given by (6.52),

M

S= − e2B

12πmc2= −N(0)µ2

BB

3(6.125)

whereN(0) = m/(πh2) is the density of states of a two dimensional electron gas (see (6.8) withd = 2)andµB = he/(2mc). Therefore, the susceptibility is

χL = −N(0)µ2B

3= −χP

3(6.126)

where we have used the result (6.37). Observe that the sign isnegative implying that the magnetizationis contrary to the direction of the field in accord with Lenz’slaw of electromagnetism. This susceptibilitywas originally derived by Landau and carries his name. Observe therefore that the total susceptibility of anelectron gas is

χT = χP + χL =2

3χP . (6.127)

6.4. PROBLEMS 121

6.4 Problems

1. Prove eq.(6.11) and calculate for Al (rs ≈ 2) and Cs (rs ≈ 5.5) their respective Fermi energies. Showthat in 3 dimensions we have:

EF =50.1

r2s

where the energy is measured in eV.

2. Prove (6.12).

3. Calculate the Fermi velocity ind dimensions and show that in 3 dimensions we can write:

vF =4.2 × 106

rs

where the velocity is measured in m/s. What is the value of theFermi velocity for Al and Cs?

4. Prove (6.22).

5. Prove eq.(6.36).

6. Calculate the exact expression for the magnetization of aelectron gas ford = 1 andd = 2.

7. What happens with the magnetization of a free electron gasfor B > Bc ?

8. Prove (6.41).

9. Show that the mean square deviation〈(δy)2〉 is proportional to the cyclotron length squared, that is,l20.

10. Solve the problem of an electron in a magnetic field in the symmetric gauge:A = B(−yx + xy)/2.

11. Prove eq. (6.62).

12. Calculate the jump in the magnetizationν andν + 1 for a two dimensional electron gas.

13. Prove (6.111).

Chapter 7

Disordered electronic systems

Up to this point we have only discussed the physics of perfectly ordered systems, that is, crystals. We haveseen that in a periodic potential the momentumk is a good quantum number and that the wavefunctionof the electrons looks very much like a plane wave, that is, itis delocalized over the entire crystal. As aconsequence, a crystal can conduct current without any lossof energy since the electrons, accordingly to(5.86) can be accelerated indefinitely. Real materials, however, always contain imperfections like impurities,dislocations, etc. We have seen, for instance, that even forentropic reasons there are always vacancies insolids. Thus, in the presence of an electrical field the electrons will be scattered by these defects. Aselectrons move through the system they will exchange momentum with the lattice and undergo Brownianmotion as shown in Fig.7.1.

Figure 7.1: Motion of an electron in the presence of scattering centers.

When the number of imperfections is small the electrons willmove ballistically between impurities.This is called the weak disorder limit. If however the numberof impurities is large then new effects canarise since there can be a destructive interference of the electron wavefunction due to disorder. We are goingto briefly discuss these two extreme limits.

7.1 Weak scattering

Let us consider the influence of an electrical fieldE in the motion of electrons in a solid when the electronscollide with impurities. We are going to assume a simple model of elastic scattering (no phonons arecreated). In this case the electrons, when they collide withthe impurities, transfer momentum to the solid.After many collisions the electron velocity decreases as a consequence of the collisions. A simple modelfor this process is based on the assumption that there is a mean timeτ between collisions so that in betweencollisions the motion is purely ballistic. The ordered caseis obtained whenτ → ∞, that is, it takes alarge amount of time between collisions. Let us rewrite the semi-classical equation of motion (5.86) for an

123

124 CHAPTER 7. DISORDERED ELECTRONIC SYSTEMS

isotropic solid as

dvk

dt=eE

m∗− 1

τvk (7.1)

that is known as the Langevin equation and describes the motion of a particle in a viscous environment.Observe thatτ can depend on the momentumk but as explained before we are going to consider momentaclose to the Fermi surface where the momentum dependence ofτ is negligible when the scattering is weak.Observe that equation (7.1) leads to an exponential decay ofthe velocity with time in a time scale given byτ . If we wait long enough we obtain the steady state condition,that is, the electron reaches a final velocity

〈vk〉 =eτ

m∗E (7.2)

which allows us to define the electron mobility,µ, as

〈vk〉 = µE (7.3)

where

µ =eτ

m∗. (7.4)

Notice that〈vk〉 has nothing to do with the actual velocity (that is, the Fermivelocity) of the electron. Thisvelocity is the net effect of the collisions. In between impurities indeed the electron moves with the Fermivelocity but because the direction of motion of the electronis changing all the time its velocity along theelectric field is smaller than its velocity in between collisions.

The steady state current in the system is just

J = eρ〈vk〉 (7.5)

whereρ is the average electronic density. Using (7.2) and (7.5) we can write

J = σE (7.6)

where

σ =e2ρτ

m∗(7.7)

is the electric conductivity of the system. The resistivity, ρR, is defined as its inverse, namely,

ρR =1

σ=

m∗

e2ρτ. (7.8)

Since we are talking about electrons at the Fermi surface with a Fermi velocity it is clear that in a time periodτ the electrons will move a distanceℓ such that

ℓ = vF τ (7.9)

which is called the mean free path of the electrons. Equation(7.7) is known as the relaxation time approxi-mation.

At this point the main element of the theory, namely, the relaxation timeτ is a phenomenological pa-rameter. We can estimate its value from a microscopic model if we assume that the number of impurities is

7.1. WEAK SCATTERING 125

very small so that single impurity scattering dominates therelaxation mechanism. Consider the scatteringof an electron from a state|k〉 to a state|k′〉 in the presence of an impurity potentialV . In this case we justuse Fermi Golden’s rule tells us that the scattering rate is giving by

Wk,k′ =2πNi

hδ(ǫk − ǫk′)|〈k|V |k′〉|2 (7.10)

whereNi is the number of impurities. The total scattering rate is obtained by integrating (7.10) over all finalstates|k′〉. We have to remember, however, that not all the states contribute. In Fig.7.2 we show the actualsituation in which the electric field is applies in some arbitrary direction. Observe that only the componentof the momentum in the direction out of the field contributes to the relaxation. Forward scattering events inthe direction of the field cannot contribute since the electron retains its momentum in this direction. Thus,the important fact is not that the electron is scattered but that the amount of momentum along the electricfield is changed in the scattering process. If we take this effect into account we can write the expression forthe relaxation time as

1

τ=∑

k′

Wk,k′

[

1 − cos(

θk,k′

)]

(7.11)

whereθk,k′ is the angle between the two states.

E

k

k’

θ

Figure 7.2: Scattering of an electron wave by an impurity in the presence of the field.

Equation (7.10) is valid only in the Born approximation. A better approximation can be obtained if weobserve that in the Born approximation the scattering amplitudefk(θk,k′) is given by

fk(θk,k′) ≈ −m∗V

2πh2 〈k|V |k′〉 . (7.12)

In general the differential cross-section is given by

σ(θ) = |fk(θ)|2 (7.13)

which allows us to rewrite (7.10) in a more general way, that is,

Wk,k′ =(2πh)3Ni

(V m∗)2δ(ǫk − ǫk′)σ(θk,k′) . (7.14)


It is a simple calculation to show that the (7.11) reduces to

1

τ= 2πcvF

∫ π

0dθ sin(θ)[1 − cos(θ)]σ(θ) (7.15)

wherec = Ni/V is the impurity concentration. From the knowledge of the differential cross section of theimpurity we can calculate the relaxation time from (7.15).

We can get further insight into (7.15) if use the partial waveexpansion of the wavefunction in thepresence of the impurity. As we know from elementary quantummechanics the asymptotic value of thewavefunction of the electron due a central potential at the origin can be written as

ψk(r, θ) ≈∞∑

l=0

(2l + 1)ileiδl(k)sin(kr − lπ/2 + δl(k))

krPl(cos(θ)) (7.16)

whereδl(k) is the phase shift of the electron wavefunction in the presence of the potential (obviously,δl(k) = 0 if V = 0). From (7.16) one can obtain the scattering amplitude in terms of the phase shifts,

fk(θ) =1

k

∞∑

l=0

(2l + 1)eiδl(k) sin(δl(k))Pl(cos θ) (7.17)

and the differential cross-section is given in (7.13). By direct substitution of (7.17) into (7.15) one finds

1

τ=

4πcvFk2F

∞∑

l=1

l sin2(δl−1(kF ) − δl(kF )) . (7.18)

It is now interesting to investigate the effect of the impurities in the spectrum of the system. Assumethat an impurity is at the center of an sphere of radiusR which is of the order of the size of the system. Weimpose the condition that the wavefunction vanishes at the surface, that is,ψk(R, θ) = 0. Then, from (7.16),we find

kn,l =π(

n− δl(k)π + l

2

)

R(7.19)

wheren is an integer. In the absence of the impurity (δl(k) = 0) we easily see thatkn,l = π(n + l/2)/Rand therefore, for fixedl, the distance between two allowed values ofk is π/R. In the presence of theimpurity this energy levels are shifted byδl/R. Consider first the case without impurities. Suppose we fillup all the states up to the Fermi momentumkF = π(Nmax + l/2)/R whereNmax is the label of the lastoccupied state. In the presence of the impurity we will havekF = π(N ′

max − δl(kF )/π + l/2)/R. Sincethe density of the system is kept constant the Fermi momentumcannot change. Thus we must require thatN ′max = Nmax + δl(kF )/π which implies that the number of available states has changed. Taking into

account that the degeneracy of each state is2(2l + 1) (the factor of2 comes from the spin component) thetotal number of new electrons required to fill up the levels tothe same wave-vectorkF is

Z =

∞∑

l=0

2(2l + 1)δl(kF )

π. (7.20)

Observe that in the presence of the impurity we have an extra amount of chargeZ in the system. So, in orderfor the system to be neutral we have to require thatZ equals the amount of electrons given by the impurityto the system, that is, the valence difference between the impurity and the host metal. Of course an impurity

7.2. STRONG SCATTERING 127

with Z = 0 cannot be distinguished from the metal itself. This is the so-called Friedel sum rule and reflectsthe charge neutrality of the solid: the electric charge of the impurity must be neutralized by an excess ofelectrons in its vicinity. Of course, far away from the impurity the the system relaxes to the case withoutimpurities wherekF is given by the value of the average density.

Observe that Friedel sum rule is very useful since it relatesthe phase shift to the valence of the impurity.Suppose for simplicity that only s-wave scattering is important (l = 0) andZ = 1 (notice that ifZ > 1 thens-wave scattering alone cannot account for the phase shift). Then from (7.20) one has

δ0 =π

2(7.21)

which is the value of the phase shift at a resonance. That is, the impurity is resonant with the Fermi system.From (7.18) we have

1

τ=

4πcvFk2F

. (7.22)

Thus under these conditions the resistivity is given by

ρR =4πc

e2ρkF

= 4(π

3

)1/3 c

e2ρ(7.23)

where in the last line we used the free electron result (6.5).For typical electronic densities one hasρ/c ≈4µΩ cm.

7.2 Strong scattering

Let us now consider the validity of (7.18). The condition forits validity is that the electron moves freelybetween collisions. If the mean free time between collisions becomes too short we cannot describe theelectron motion as free. In order to estimate the validity ofthe theory we have to remember thatτ gives anestimate of the mean free time that the electron stays in a state with well defined momentumk and energyapproximately given byEF . That is,τ is the lifetime of the state. Thus, from the uncertainty principle, thedescription below is only valid if

τ ≫ h

EF(7.24)

which, by usingEF = h2k2F /(2m

∗) andτ = ℓ/vF implies that

kF ℓ≫ 1 (7.25)

which implies that the wavelength of the electronλF = 2π/kF has to be larger than the mean free path.This is nothing but the condition of ballistic motion. Condition (7.25) can also be written in a differentform if we realize that the Fermi wave-vector, given in (6.5)is approximately given byπ/a wherea is thelattice spacing. Therefore, we have the condition thatℓ ≫ a which implies that the mean distance betweenimpurities has to be much larger than the lattice spacing.

The question that comes to mind is: what happens whenkF ℓ ≈ 1 or ℓ ≈ a? It is clear that in this casewe have, in average, one impurity per unit cell, that is,c ≈ 1/a3. Let us assume that the impurity potential


is short ranged and can be described by

V (r) = V0e−r/a

r/a. (7.26)

Using the simplest Born approximation (7.10) we find that thedifferential cross section can be written as

σ(θ) ≈ 4(m∗)2a6V 20

h4 (7.27)

in the limit wherea → 0 (that iska << 1). Substituting (7.27) into (7.18) and using (5.82), (7.9) andc = 1/a3 one finds

1

ℓ=

16πa3V 20 (m∗)2

h4 =4π

a

(

V0

t

)2

(7.28)

and observe that the condition ofℓ < a impliesV0 ≈ t, that is, the strength of the disordered potential is oforder of the kinetic energy. It is clear that in this limit it is wrong to do perturbation theory around the planewave state!

So, if the plane wave state is incorrect what is the nature of the ground state? In order to solve theproblem let us consider the one-dimensional potential of Fig.7.3(a). As we have seen the solution for thisproblem is a Bloch wavefunction which in the atomic limit canbe well described by (5.69) and is depictedon Fig.7.3(b). Suppose we now have a potential like the in Fig.7.4(a) where the heights of the potentialwells fluctuate around some mean valueV and with varianceV0. What we are going to argue is that thewavefunction looks like Fig.7.4(b), that is, the wavefunction is localized around the site in the lattice withlowest energy. That is, the wavefunction, instead of be given by (5.69), looks like

ψ(r) =e−r/ξ√N

∑

T

eik·TψA(r − T) . (7.29)

whereξ is the so-called localization length and gives the envelopeof the function in Fig.7.4(b).

E

V(x)

|ψ|2

(a)

(b)

Figure 7.3: (a) Ordered potential; (b) Module square of the Bloch wavefunction.

In order to understand how localization arises in this problem consider our familiar problem of themolecule with two atoms. We now have only two sites1 and2 and the state of the electron localized on site

7.2. STRONG SCATTERING 129

E

V(x)(a)

(b)

V0

|ψ|2

ξ

Figure 7.4: (a) Disordered potential; (b) Module square of the wave function.

1 (2) is |1〉 (|2〉). We now generalize the Hamiltonian (1.43) in order to allowthe two states to have differentenergies and the tunneling is still described by (1.45), that is, the complete Hamiltonian reads

H = V |1〉〈1| + (V + V0)|2〉〈2| + t (|1〉〈2| + |2〉〈1|) . (7.30)

which is our toy model for the disordered problem. SinceV is just a shift in the energy of the problem wejust set it to zero. This problem can be solved exactly as before by writing linear combinations of the states|1〉 and|2〉

|B〉 = α|1〉 + β|2〉|A〉 = α|2〉 − β|1〉 (7.31)

which are the analogue of the bonding and anti-bonding states. In this new basis the Hamiltonian is diago-nalized and it looks like

H = EA|A〉〈A| + EB |B〉〈B| . (7.32)

The coefficientsα, β,EA andEB depend on the parameterst andV0 and are left for you to calculate.

Let us consider the extreme cases where our physical intuition can work for us. The perfect latticehappens when the two potential wells are the same, that is,V0 = 0. If V0 << t we haveα ≈ β ≈ 1/

√2

andEB − EA ≈ 2t that is the case we discussed previously. In this case the twowavefunctions for thebonding and anti-bonding states have equal weight in the twowells. This is the case equivalent to the Blochstate where the electron, by tunneling between atoms, is extended over the crystal. Consider however theopposite case in whichV0 ≫ t. In this case you can show that

α

β≈ V0

2t≫ 1 (7.33)

which means that the probability of finding the particle in the site1 is much larger than the probability offinding the particle in site2, that is, the particle is trapped inside of the well. If we nowturn to the infinitecrystal this argument suggests that there should be not onlyfluctuations in the phase of the wavefunction


as we go from site to site but also fluctuations in the amplitude! These fluctuations become larger as theratioV0/t increases. IfV0/t is very large we expect the wavefunction of the electron on that particular siteto be very little affected by the presence of the other sites in its neighborhood and therefore should decayexponentially away from the site as the wavefunction (7.29)implies. Thus, our expectation is that for agiven electron energy there must be a minimum value ofV0/t for which the electron becomes localizedThis is the so-called Anderson localization transition. The critical value ofV0/t is a non-universal numberwhich depends on the type of lattice, dimensionality and type of disorder in the material. In particular, ithas been shown that in one dimension all the electronic states are localized. The fact that the localizationdepends on the energy of the electron is clear from Fig.7.4 since electrons with large energies do not care forthe disordered potential but electrons with low energies will feel the bottom of the wells. Thus, for a givenvalue ofV0/t above the critical value where localization happens, thereis also a minimal energyEC , belowwhich all the states are localized and above which the statesare extended. This is the so-called mobilityedge. If we make a plot of the density of states of a such material it will look like the one on Fig.7.5. Belowthe mobility edge the states are localized and above it they are extended.

E

N(E)

Ec

Delocalize

Localized

Figure 7.5: Density of states for a disordered system.

7.3 Scaling Theory of Localization

In order to understand the problem of localization considerthe expression of the conductivity given in (7.7).It can be rewritten in terms of the electron mean free path (7.9) and the Fermi momentum (6.4) as:

σ =

(

2Sd(2π)dd

)

e2

hkd−2F (kF ℓ) . (7.34)

It was originally proposed by Mott that the transition between metallic to insulating behavior due to disorderwould occur whenkF ℓ ≈ 1. The reason for this is that on a lattice the smallest distance between impuritiesis one lattice spacinga. Therefore,ℓ = a is the smallest value possible for the mean free path. SincekF ≈ 1/a we must conclude thatkF ℓ ≈ 1 at the transition. This argument would imply that the smallestconductivity in a system, according to (7.34), would be given by

σmin ≈(

2Sd(2π)dd

)

e2

hkd−2F . (7.35)

7.3. SCALING THEORY OF LOCALIZATION 131

Thus, according to Mott the conductivity of a disordered system, as a function of the disorder strength ofkF ℓ would look like as in Fig.7.6 with a discontinuous drop of theresistivity at the transition between metaland insulator.

F

σ

k lF

σmin

k a

Figure 7.6: Behavior of the conductivity at the metal insulator transition according to the Mott minimum.

The prediction of the Mott minimum of conductivity was not fully confirmed experimentally and wasalso questioned theoretically because it does not take intoaccount the statistical nature of the disorderedstate. It turns out that the metal insulator transition is a continuous transition in terms of the disorder strengthand not discontinuous as Mott’s argument would make it. In order to understand this transition consider thedifference between a localized and delocalized state. A localized state, having a characteristic sizeξ shouldbe insensitive to boundary conditions in a system of sizeL > ξ. On the other hand, in a metallic system,the wave-function is extended over the entire volume of the system and therefore it should be sensitive tothe boundary conditions. Now imagine taking a system of volumeV and breaking it into pieces of sizeLd

so thatL ≫ ℓ. It does not make a lot of sense to talk about the conductivityof a piece of a material of sizeLd < V since the conductivity is an extensive quantity. Nevertheless, we can talk about the dimensionlessconductance of this piece that is defined as:

g(L) = σLd−2 . (7.36)

Notice that, according to (7.34) this quantity is proportional to(kFL)d−2. In a metallic system the conduc-tivity in one piece of a material does not change from piece topiece and therefore the dependence ofg(L)with L is essentiallyg(L) ∝ Ld−2 ≫ 1 in the limit of ξ ≫ L ≫ ℓ. However, in the insulating case oneexpectsg(L) ∝ e−L/ξ ≪ 1 since the wavefunction is localized in whenL ≫ ξ, ℓ. These two limits are theasymptotic limits of the conductance. However, we are interested in the limit of the conductance close tothe metal-insulator transition whereg(L) ≈ 1.

In order to understand the behavior of the conductance closeto the metal-insulator transition one has tounderstand the behavior of the conductance for all values ofg. This seems to be an impossible task sincewe need to know details about the disorder strength, distribution of disordered sites, etc. Nevertheless, wecan make some progress if we make a few assumptions about the behavior ofg. We first notice that thelogarithmic derivative ofg with L must be weakly dependent onL sinced ln(g)/d ln(L) ∝ (d − 2) wheng ≫ 1 andd ln(g)/d ln(L) ∝ ln(g) wheng ≪ 1. The scaling theory of localization makes the assumptionthat the function

β(g) =d ln(g)

d ln(L)(7.37)

is only dependent ong but notL. This assumption has strong consequences since the asymptotic behaviorof g(L) imposes constraints in the behavior ofβ(g). In fact, if one plotsβ(g) as a function ofln(g) one getsFig.7.7.


c

β (g)

ln(g)

D=3

D=2

D=1

0

−1

1

g

Figure 7.7: Behavior ofβ(g) according to scaling theory of localization.

Notice that whileβ(g) vanishes atg = gc in d = 3 it is always negative ford = 2 andd = 1. Themeaning of this result becomes obvious when one looks carefully at what happens close tog = gc. Let usexpandβ(g) to first order close togc:

β(g) ≈ 1

ν

g − gcgc

(7.38)

where we have assumed that(g−gc)/gc ≪ 1 andν = (gcdβ(gc)/d ln(g))−1. Using (7.37) we can calculateg(L) explicitly:

g(L) ≈ gc + (g0 − gc)(L/L0)1/ν (7.39)

whereg0 is the value of the conductance at the scaleL = L0. Let us consider first the case ofg0 > gc. Inthis case equation (7.39) predicts thatg(L) should grow asL grows indicating a metallic state. However,g(L) cannot grow indefinitely as indicated in (7.39) because we have assumed thatg ≈ g0 ≈ gc and one hasto stop (7.39) at some scaleL = ξM where(g − gc)/gc ≈ 1. It is easy to see that

ξM (g0) ≈ AL0

[(g0 − gc)/gc]ν(7.40)

whereA is a number of order of unit and indicating thatξ(g) diverges atg = gc with a critical exponentν.The physical meaning ofξ is clear: for regions of sizeL < ξM the system behaves as an insulator while forL > ξM it looks like a conductor. Atg = gc the system becomes an insulator sinceξM becomes of the sizeof the system.

Forg0 < gc we rewrite (7.39) as:

g(L) ≈ gc(1 − (gc − g0)/gc(L/L0)1/ν) (7.41)

indicating thatg(L) decreases asL grows indicating insulating behavior. Notice that this expression wouldpredict thatg(L) would vanish atL = ξloc defined by:

ξloc(g0) ≈L0

[(gc − g0)/gc]ν. (7.42)

However, by definitiong(L) > 0 and the vanishing ofg(ξloc) is just a consequence of the approximation ofexpanding closing tog = gc. In fact one realizes that (7.41) can be rewritten as:

g(L) ≈ gce−(L/ξloc)

1/ν(7.43)

7.4. PROBLEMS 133

that has the same asymptotic behavior of (7.41). Notice thatfor L≫ ξloc the conductance vanishes indicat-ing thatξloc is the localization length of the system: forL > ξloc the system looks like an insulator and forL < ξloc the system looks like a metal.

From the definition (7.36) one can find the actual conductivity of the material: forg < gc the system isan insulator andσ = 0 while for g > gc the system is a metal and forL > ξM in d = 3 the conductivity canbe written as:

σ(g) ≈ g(ξM )

ξM= B

gcL0

[(g − gc)/gc]ν (7.44)

whereB is a number of order unit, indicating that the conductivity vanishes continuously at the metalinsulator transition with an exponentν as shown in Fig.7.8.

INSULATOR

σ

k lFMETAL

Figure 7.8: Actual behavior of the conductivity at the metalinsulator transition.

Notice that there are various ways to cross the metal-insulator transition. One of them is to change thestrength of the disorder and therefore to changeℓ. However, we can also cross the metal insulator transitionby changing the electronic density, that is, changingkF . In the later case one has to change the Fermi energyacross the mobility edge, namely, we can write, to first approximation close to the transition that

g(EF ) ≈ gc +dg(Ec)

dEF(EF − Ec) (7.45)

so that forEF > Ec the system is metallic (g > gc) and forEF < Ec the system is an insulator (g < gc).Finally we comment about the problem in dimensions smaller than3. In those cases, as shown in Fig.7.7,

we haveβ(g) < 0 for all values ofg indicating thatσ(L) → 0 asL grows. Thus, the scaling theory oflocalization predicts that ford < 3 the electronic system should be localized even in the presence of aninfinitesimal amount of disorder. This result has been challenged by recent experiments in two-dimensionalelectron systems confined in heterostructures.

7.4 Problems

1. Using (7.11) and (7.14) and assumingǫk = h2k2/(2m∗) prove (7.15).

2. Prove that in the Born approximation the scattering amplitude for the potential (7.26) is given by:

fk(θ) = − 2m∗V0a3/h2

4(ka)2 sin2(θ/2) + 1.

Calculate the differential scattering cross-section and compare with (7.27).


3. Diagonalize the Hamiltonian (7.30) exactly and calculateα, β, EA andEB .

4. Consider the case the problem of localization in two dimensions. In this case we find thatβ(g) = 0in the limit of g ≫ 1. Corrections to this extreme limit can be written as power series in1/g as:

β(g) ≈ a/g + O(1/g2) . (7.46)

(i) What is the sign ofa? Why? (ii) Calculate the localization lengthξloc in this case and show that itis an exponential function of the conductance.

Chapter 8

Semiconductors

As we have seen in the previous chapters we can classify cleannon-interacting electron systems either asmetals or insulators depending whether they have a partially filled or completely filled band, respectively.Metals do not have a gap in the charge spectrum and can conductelectricity very well, insulators, however,cannot conduct charge if the temperature is much smaller than the gap energy,Eg. When the thermal energy,KBT , at room temperature is of order ofEg then one can promote electrons from the valence band to theconduction band, making electrical conduction possible. Systems whereEg ≈ KBT at room temperatureare called semiconductors. It is quite amazing that the electronic revolution of the20th century was notdriven by the development of metals but by the progress in theunderstanding of semiconductors. These arethe basic elements for the production of transistors and other electronic elements in modern computers. It istherefore of interest to understand their physics.

Here we are going only to consider the case where the conduction band and the valence band electronicenergies can be written as:

Eck = Ec +h2k2

2mc

Evk = Ev −h2k2

2mv(8.1)

wheremc andmv represent the different curvatures of the conduction and valence band, respectively. Noticethat in terms of our definition (5.88) of the effective mass, the valence band has negative mass while theconduction band has positive mass. The bands described in (8.1) are not generic because many times thetop of the valence band is not at the same point as the bottom ofthe conduction band (these are calledindirect gap semiconductors). Many times the bands are shifted by a finite momentumQ. However, thereare some systems where the two edges are at the same point in momentum space (these are called direct gapsemiconductors). Another simplification in (8.1) is that weare using infinite parabolic bands. As we knowthis approximation is only good if we are considering statesclose to the edge of the bands. If the temperaturebecomes too high one starts to excite particles at higher energy states and the deviation from parabolicitybecomes an important factor. Here, however, we are going to consider cases wherekBT ≤ Eg = Ec − Evin which case we constrain ourselves to states close to the edges of the bands. For simplicity, from now onwe are going to setEv = 0 so thatEg = Ec.

It is a very simple exercise to calculate the density of states of a semiconductor as a function of energy.Using the results of the previous sections we find that forE ≥ Eg:

Nc(E) =1

2π2

(

2mc

h2

)3/2√

E − Eg (8.2)

135

136 CHAPTER 8. SEMICONDUCTORS

is the density of states for the conduction band and forE ≤ 0 we have

Nv(E) =1

2π2

(

2mv

h2

)3/2 √−E (8.3)

is the density of states for the valence band. Observe that the density of states is zero everywhere in the gap,0 < E < Eg.

When an electron with momentumke is excited from the valence band to the conduction band it leavesa hole behind in the valence band. Since the total number of electrons do not change we can either de-scribe the problem in terms of electrons or holes. The total momentum of the system before and after thetransition is the same which implies that the momentum of thehole,kh, has to be−ke. Moreover, in de-scribing the system in terms of holes the energy of the holes is obtained by inverting the valence band so thatEh(kh) = −E(ke) which implies that the mass of the hole is given bymh = mv. Because the properties ofsemiconductors depend strongly on the number of electrons and holes we will be always talking about thenumber of such carriers in the system.

Let us calculate the number of electrons in the conduction band,nc, and holes in the valence band,pv,at a fixed temperatureT . It is clear that

nc =

∫ ∞

Eg

dENc(E)f(E) (8.4)

and

pv =

∫ 0

−∞dENv(E)f(E) (8.5)

where

f(E) =1

eβ(E−µ) + 1(8.6)

is the Fermi-Dirac distribution function (β = 1/(kBT )). In what follows we are going to assume that thetemperature is such that

Eg − µ ≫ kBT

µ ≫ kBT (8.7)

so that the Fermi-Dirac distribution function (8.6) can be replaced by

f(E) ≈ e−β(E−µ) . (8.8)

Because the chemical potentialµ depends on the amount of charge in the system these conditions can onlybe checked a posteriori. We are going to see that this is indeed the case of interest.

Using (8.8) we can write (8.4) as

nc ≈ 1

2π2

(

2mc

h2

)3/2 ∫ ∞

Eg

dE√

E − Ege−β(E−µ)

= Nc(T )e−β(Eg−µ) (8.9)

8.1. INTRINSIC SEMICONDUCTORS 137

where

Nc(T ) =1

2π2

(

2mckBT

h2

)3/2

(8.10)

that should not be confused with the density of states (8.2).By the same token we can show that

pv = Pv(T )e−βµ (8.11)

where

Pv(T ) =1

2π2

(

2mvkBT

h2

)3/2

. (8.12)

Equations (8.9) and (8.11) give the density of electrons at the conduction band and holes in the valenceband. Observe, however, that these expressions still depend on the chemical potentialµ which is undefined.However, if we multiply (8.9) and (8.11) we find that

ncpv = NcPve−βEg (8.13)

which is independent on the chemical potential. This expression is very important in the theory of semicon-ductors.

8.1 Intrinsic Semiconductors

In intrinsic semiconductors the number of impurities is vanishing small. This implies that all the holes inthe valence band are due to the excitation of electrons from the valence band to the conduction band. In thiscase it is obvious that:

nc = pv = ni (8.14)

whereni is the intrinsic density of a semiconductor and can be obtained directly from (8.13):

ni =√

NcPve−βEg/2 . (8.15)

This result implies that at finite temperatures the number ofelectrons in the conduction band or holes in thevalence band is exponentially small with temperature.

Using (8.14) and (8.15) in (8.9) we can immediately obtain the value of the chemical potential;

µ =Eg2

+3kBT

4ln

(

mv

mc

)

(8.16)

which shows that at low temperatures the chemical potentialis essentially in the center of the energy gap.Observe that in this case the conditions (8.7) are obeyed as long as

kBT ≪ Eg/2 (8.17)

which is obeyed in most semiconductors where the energy gap is of order of a few electron volts.


8.2 Extrinsic Semiconductors

In the presence of external atoms the physics of a semiconductor can change profoundly. Imagine that wedope a semiconductor with another atom that has a valence+1 larger than the host atom (it gives up oneelectron more to the conduction band). In this case we can think of the external atom as simply the hostatom with one electron and one proton more. These atoms are called donors. At first sight this is equivalentto add a Hydrogen atom to the host material. If this was the case the extra electron would be strongly boundto the impurity atom with an energy of the order ofEb = −me4/h2 ≈ −13.6 eV. One has to remember,however, that a Hydrogen atom on a host material is not the same as a Hydrogen atom in vacuum. Theproperties of the motion of the electron are changed by the presence of other atoms and other electrons.A first consequence of the fact that there are other atoms in the system implies that the electron is movingin the external periodic potential of the lattice. Thus, themass of the electron is not its bare mass but itseffective mass,mc, which can be much smaller than the electron mass (usually,mc ≈ 0.1me). This impliesthat the binding energy is decreased already by one order of magnitude. However, the biggest effect is notdue to electron-ion interactions but due to the electron-electron interactions.

In a metal (a system with no gap) charges can move freely and therefore an external charge can be easilyshielded (or screened) by a local change in the electron density. Thus, in a metal the potential seen by theextra electron would be essentially zero far away from the impurity and the electron would be free to go tothe conduction band. In an perfect insulator at zero temperature charges cannot move freely because thereis a gap in the charge spectrum and it requires a large amount of energy to move electrons around. Thisimplies that in an insulator the potential seen by the electron is essentially the bare potential (Hydrogenatom) and it would be strongly bounded as discussed previously. However, in a semiconductor the situationis middle way between a metal and an insulator: there is charge motion because there areni electrons in theconduction band (see (8.15)) that can screen partially the potential generated by the external impurity. Sothe binding energy is neither as small as in a metal or large asin a insulator.

A way to quantify the amount of screening in a semiconductor is via the dielectric constantǫ. Theeffective potential seen by an electron due to an external charge+e is given by

V (r) = −1

ǫ

e2

r. (8.18)

In semiconductors we can findǫ ≈ 20 − 100. The consequence of (8.18) is that we can replacee2 by e2/ǫ.Putting together the effects of the electron-ion interaction and the electron-electron interactions we see thatthe binding energy of an electron by an impurity is of order:

Eb = −mc

m

1

ǫ2me4

h2 (8.19)

which can be10−4 times smaller than the biding energy of an electron in the Hydrogen atom! In fact fora host of Ga atoms (valence+4) with As impurities (valence+5) it is found that the binding energy is oforder of0.013 eV. For this reason, the radius of motion of the electrons is also increased from a few Bohrradius,a0, to

r0 =m

m∗ǫa0 . (8.20)

Thus, the conclusion is that in a semiconductor the impuritylevels are weakly bound to the impurities andcan be easily ionized. In other words, atT = 0 the impurity level is occupied by a single electron which canbe promoted to the conduction band at finite temperatures. Because these are bound states of electrons theyare localized just below the bottom of the conduction band byan amount given byEb. We callEg = Ed+Eb

8.2. EXTRINSIC SEMICONDUCTORS 139

the donor energy in a semiconductor and to a donor doped semiconductor we call it as n-type.

Analogously to the electron doped case, we can also dope holes in a semiconductor. We just have tochoose an impurity atom that has a valence−1 relative to the host. In this case we are taking one protonand one electron from the host. The situation here is one of a localized negative charge with a hole boundedto it. For the same reasons described above this will lead to abound state of holes above the top of thevalence band. These are called acceptor states and the semiconductor is said to be of p-type. Observe thatthe ionization of a acceptor state is equivalent to a hole going to the valence band or an electron moving fromthe valence band to the impurity level. AtT = 0 the acceptor state has one hole but at finite temperaturessuch thatkBT = Ea = −Eb (wheremc is replaced bymv in (8.19)) an electron can be promoted to theacceptor state while a hole goes to the valence band.

Because the donor and acceptor states are such that|Eb|, Ea ≪ Eg one can easily dope the conductionor the valence bands with external charges. In this case thenc andpv as defined previously will change byan amountδn, that, is:

δn = nc − pv (8.21)

which is equivalent to a change in the chemical potential of the system. Independent of this change thecondition (8.13) is still valid. Using the definition ofni given in (8.15) we can rewrite (8.13) as

ncpv = n2i . (8.22)

Solving (8.21) and (8.22) fornc andpv we find

nc =

√

(

δn

2

)2

+ n2i +

δn

2

pv =

√

(

δn

2

)2

+ n2i −

δn

2(8.23)

which gives the number of electrons and holes as functions ofthe intrinsic density and the extrinsic densityof impurities. Whenδn > 0 there are more donors than acceptors and the system is n-type. If δn < 0 wehave the opposite situation and the system is p-type.

In order to calculate the value ofδn we need to know how many electrons (holes) are ionized and goto the conduction (valence) band and how many remain boundedto the impurity atoms. Let us call thedensity of donorsNd, the density of acceptorsNa, density of electrons bounded in the donor statesnd andthe density of holes bounded in acceptor statespa. From the fact that the total number of electrons and holeshas to be conserved it is easy to see that:

δn = nc − pv = Nd −Na − (nd − pa) . (8.24)

SinceNd andNa are fixed for a given semiconductor we are left with the calculation ofnd andna.

Let us firstly consider the problem of the donors. The occupation Nj of the donor level can beN0 = 0(no electrons),N1 = 1 (spin up or down) orN2 = 2 (one electron with spin up and another with spindown). The doubly occupied state requires the donor state tobe charged relative to the host and thereforeis energetically not stable. We therefore disregard doubleoccupancy. The donor state with zero electronsmakes no contribution to the energy and thereforeE0 = 0. The donor state with one electron has energy


E1 = Ed. Thus, in thermal equilibrium the average number of electrons in donor sites will be

nd = Nd

∑

j Nje−β(Ej−µNj)

∑

j e−β(Ej−µNj)

= Nd2e−β(Ed−µ)

1 + 2e−β(Ed−µ)

=Nd

eβ(Ed−µ)

2 + 1. (8.25)

Observe that becauseEd ≈ Eg we expect that forkBT ≫ Eg −Ed we must havend ≪ Nd, that is, almostall the donor levels are ionized the electrons are in the conduction band.

In the case of acceptors the situation changes. The occupation Nj can be0 holes (or two electronsNj = 2), 1 hole (or one electron with either spin up or down,Nj = 1), or 2 holes (that is, zero electronsNj = 0). The state with two holes is again unstable because it is positively charged relative to the host. Theenergy of the state with one hole is zero,E1 = 0, while the state with no holes has energyEa (this is theenergy cost to take one electron from the valence band and putin the acceptor level, in this case, the holethat was bound in the acceptor level goes into the valence band). In this case, the number of electrons in theacceptor level can be calculated as before:

< n > =2e−β(Ea−2µ) + 2eβµ

e−β(Ea−2µ) + 2eβµ

=eβ(µ−Ea) + 1eβ(µ−Ea)

2 + 1. (8.26)

Since the maximum occupation of the acceptor level is2, the average number of holes in an acceptor levelis:

pa = Na(2− < n >) =Na

eβ(µ−Ea)

2 + 1. (8.27)

Since we expectEa ≪ kBT ≪ Eg the acceptor levels to be fully ionized at room temperature.In this casethe holes will be all essentially in the valence band.

The conclusion of these calculations is that we can approximate (8.24) by

δn ≈ Nd −Na (8.28)

at room temperature. Substitution of (8.28) in (8.23) allows us to give a better definition to intrinsic andextrinsic semiconductor. The intrinsic semiconductor is the one in whichni ≫ |Nd −Na| and therefore

nc ≈ ni +Nd −Na

2

pv ≈ ni −Nd −Na

2(8.29)

and the number of electrons in the conduction band and holes in the valence band only change by a small

8.3. THE P-N JUNCTION 141

amount. An extrinsic semiconductor is one such thatni ≪ |Nd −Na| so that

nc ≈ Nd −Na

pv ≈ n2i

Nd −Na(8.30)

for Nd > Na and

pv ≈ Na −Nd

nc ≈ n2i

Na −Nd(8.31)

for Nd < Na. Thus, in an extrinsic semiconductor there is large difference between the number of chargecarriers. It is exactly the flexibility in controlling the number of charge carriers that makes semiconductorsso attractive for making devices.

8.3 The p-n junction

Let us consider the problem of a junction between two identical semiconductors with energy gapEg so thatone of them hasNa acceptors and the other hasNd donors. While the p-type semiconductor has free holes,the n-type has free electrons. Thus, when these two semiconductors are put in contact through an interface(or junction) we expect electrons and holes to flow between them and establish chemical equilibrium acrossthe junction, that is, the chemical potential becomes the same in both sides of the junction. In order words,when these two different systems are put in contact a potential φ(x) appears in the system so the systemremains in equilibrium. Here we are going to assume that the junction is an infinite plane and that thepotential only varies in thex direction.

The first assumption of our description of the junction is that the potential varies smoothly across thejunction and is such that very far away from the junction it vanishes. Consider, for instance, the p-typesemiconductor to be atx < 0 and the n-type to be atx > 0. Forx = −∞ the p-type semiconductor canbe described as in the previous sections and there areNa acceptors which neutralize the extra holes in thesystem. Atx = +∞, deep inside of the n-type semiconductor, there areNd donors that neutralize the extraelectrons in the system. Consider the potentialφ(x) built up by the junction in the system. The energy ofelectrons (holes) can be rewritten as

Ed(x) = Ed − eφ(x)

Ea(x) = Ea − eφ(x) (8.32)

by the same token the edges of the conduction band,Ec, and the valence band,Ev, also change withφ(x)so that the chemical potentialµ is the same in both sides of the junction:

Ec(x) = Eg − eφ(x)

Ev(x) = −eφ(x) . (8.33)

Because of this change in the energies of electrons and holes, their densities will also change. In fact, wecan write:

nc(x) = Nce−β(Eg−eφ(x)−µ)

pv(x) = Pve−β(µ+eφ(x)) . (8.34)


Thus, far away from the junction we can write:

nc(x→ +∞) = Nd = Nce−β(Eg−eφ(+∞)−µ)

pv(x→ −∞) = Na = Pve−β(µ+eφ(−∞)) (8.35)

and if we multiply the two equations we get:

NaNd = NcPve−β(Eg−eδφ)

δφ = φ(+∞) − φ(−∞) (8.36)

which can also be written as:

e δφ = Eg + kBT ln

(

NdNa

NcPv

)

(8.37)

which gives the total drop of the potential across the junction. Observe that because usuallyEg ≫ kBT thedrop is essentially given by the gap in the semiconductor. Notice also that the actual values ofφ(±∞) donot have especial physical significance because we can fix oneof them to be zero. Only the difference ofpotential has physical meaning. Using (8.35) we can rewrite(8.34) as

nc(x) = Nde−βe(φ(+∞)−φ(x))

pv(x) = Nae−βe(φ(x)−φ(−∞)) . (8.38)

The last equation gives the total drop in the potential across the junction but does not tell us how doesthe potential changes as a function ofx. In order to know that we have to remember that the potential obeysthe Poisson equation:

−∇2φ = −d2φ

dx2=

4π

ǫρ(x) (8.39)

whereǫ is the dielectric constant of the semiconductor andρ(x) is the local charge density in the systemthat can be written asρ = eδn whereδn is given in (8.24):

ρ(x) = e(Nd(x) −Na(x) − nc(x) + pv(x)) (8.40)

whereNd(x) = NdΘ(x) andNa(x) = NaΘ(−x) with Θ(x) = 1 (−1) if x > 0 (x < 0). If we use (8.40)and (8.38) into (8.39) it is clear that we will find a highly non-linear problem to solve. Instead we are goingto simplify the problem and assume the thatφ(x) varies only considerably in a region of size−dp < x < dnacross the junction (dp anddn have to be calculated), that is, forx < −dp we have

φ(x) = φ(−∞) , (8.41)

and forx > dn we have

φ(x) = φ(+∞) . (8.42)

In this case, using (8.38) it is clear that forx > dn we havenc(x) = Nd and forx < −dp we havepv(x) = Na. In this case we can break the (8.39) into different parts: for x > dn or x < −dp we have

d2φ

dx2= 0 (8.43)

8.3. THE P-N JUNCTION 143

while for−dp < x < 0 we have

d2φ

dx2=

4πe

ǫNa (8.44)

and for0 < x < dn we have

d2φ

dx2= −4πe

ǫNd . (8.45)

We have to solve these equations using the boundary conditions (8.41) and (8.42) and impose that thepotential and its derivative are continuous across the interfaces atx = −dp, x = 0 andx = dn. A simplecalculation shows that for−dp < x < 0 we have

φ(x) = φ(−∞) +2πeNa

ǫ(x+ dp)

2 (8.46)

and for0 < x < dn we have

φ(x) = φ(+∞) − 2πeNd

ǫ(x− dn)

2 (8.47)

where

dn =

√

Na/Nd

Na +Nd

ǫδφ

2πe

dp =

√

Nd/Na

Na +Nd

ǫδφ

2πe. (8.48)

These equations provide the full profile of the potential generated by the junction between two semiconduc-tors.

Suppose that an external potentialV is applied to the junction. It is clear that this case the difference ofpotential can be either increased or decreased and that in general we must have

δφ = δφ0 − V (8.49)

whereδφ0 is the potential difference in the absence of the applied potential and is given in (8.37). Becauseof this new potential we see that the regions of variation of potential given bydp anddn change to:

dn,p(V ) = dn,p(0)

√

1 − V

δφ0(8.50)

whereδn,p is given in (8.48). Thus, by applying an external potential one can change the region where theelectronic density varies because of the junction. This effect can be used to make a device called a rectifier.

In the absence of an external voltageV the number of electrons (or holes) crossing the junction in onedirection is the same as the number of electrons crossing thejunction in the opposite directions since thesystem is in equilibrium. Thus, if a hole current flowing fromthe p-type to the n-type is calledIh,0, thereis an equal but opposite current from the n-type to the p-type. The same is true for the equilibrium electroncurrentIe,0. When a voltageV is applied to the junction we see that the total potential across the junctionis modified according to (8.49). Sinceeδφ0 is the barrier height between the two sides of the junction whenV = 0, it becomes clear that when a potential−V < 0 is applied to the n-type side of the barrier, the barrier


height for holes is decreased relative to the other side. In this case, holes coming from p-type side of thejunction see a smaller potential barrier while the holes on n-type side of the junction see essentially the samepotential as before. In a tunneling junction the current increases exponentially with the barrier height sinceit depends on the tail of the wave-function below the barrier. Thus, while the current for holes going fromthe n-type to the p-type isI0, the current for holes going from p-type to n-type isI0eβeV and thus, the nethole current in the system is:

If = Ih,0(eβeV − 1) . (8.51)

This is called the forward current. For electrons the situation is the same. When a potentialV > 0 is appliedto the n-type side of the junction the situation is reversed:holes moving from the p-type to the n-type sideof the junction see a larger potential and its current is decrease doIh,0e−βeV while holes moving from then-type toward the p-type side of the junction have essentially the same currentIh,0. In this case the netcurrent is simply

Ir = Ih,0(1 − e−βeV ) . (8.52)

This is called the reverse current. Thus, at zero temperature (β → ∞) one seesIf → ∞ while Ir = Ih,0.This implies, that for positive voltages the current through the junction is very high while for negativevoltages the current is essentially the equilibrium current. Because of this highly non-linear behavior of thecurrent in regards to the sign of the voltage this device is called a rectifier.

8.4 Problems

1. CalculateNc(T ) in (8.9) explicitly.

2. Show (8.29), (8.30) and (8.31).

3. Solve (8.39) explicitly and show that (8.46), (8.47) and (8.48) are indeed correct.

4. Consider a potential barrier between two metals1 and2. The potential profile of the barrier can bewritten as

V (x) = Θ(x)Θ(a− x)[

V1 + (V2 − V1)x

a

]

(8.53)

wherea is the thickness of the barrier andV1 > V2. In the absence of any applied external potentialthe electric current flowing from left to right,I1→2 = I0, is equal but opposite to the current flowingfrom right to left (I2→1 = −I0) because the two sides of the junction are in equilibrium, that is,I1→2 + I2→1 = 0.

(i) Make a plot of the potential as a function ofx.

(ii ) If a potentialV > 0 is applied to the left side of the junction, what is the net current flowing in thesystem, that is, what is the value ofI = I1→2 + I2→1?

(iii ) If a negative potential−V < 0 is applied to the left side of the junction what is the net current inthe system?

(iv) Make a schematic plot of the current versus the voltageV at a finite temperatureT . Can you usethis device as a rectifier?

Chapter 9

Electron-phonon interactions

So far we have studied the problem of non-interacting systems such as free phonons and free electrons. Inthis chapter we are going to start the study of interacting systems. It is very useful, however, to introduce anew language for the problem which is usually called "secondquantization” and allows us to work directlywith operators that act on states with well defined occupation number. In the next two sections we are goingto discuss this new representation for bosons and fermions,respectively. We are going to see that this newrepresentation makes our life much easier when calculatingphysical quantities of experimental relevance.

9.1 Boson operators

As you know from the harmonic oscillator and the phonon problem discussed previously, it is possible todefine creation,b†i , and annihilation,bi, operators which create and destroy bosons in a quantum state |i〉with occupationni = 0, 1, 2, ...,∞. These operators commute among themselves

[

bi, b†j

]

= δi,j ,

[bi, bj ] = 0 , (9.1)

and when applied to a state with a well defined occupation,|ni〉, give

bi|ni〉 =√ni|ni − 1〉 ,

b†i |ni〉 =√ni + 1|ni + 1〉 , (9.2)

that implies

b†ibi|ni〉 = ni|ni〉 . (9.3)

For obvious reasonsNi = b†i bi is called the number operator. A many-body state of bosons can be describedin terms of the occupation of each state|i〉 with i = 0, ...,∞ as

|n0, n1, n2, ...〉 =(b†0)

n0

√n0!

(b†1)n1

√n1!

(b†2)n2

√n2!

...|0〉 (9.4)

where|0〉 is the vacuum state with no bosons. The Hilbert space of all states which can be represented inthis way is called the Fock space.

A simple example is a problem of free bosons inside of a box of volumeLd (you can convince yourself

145

146 CHAPTER 9. ELECTRON-PHONON INTERACTIONS

that this is actually the case of phonons). In this case the eigenstates of the problem are plane waves

φp(r) =1

Ld/2eip·r , (9.5)

where, due to periodic boundary conditions, we have

pα =2πnαL

, (9.6)

with α = x, y, z. Thus the many-body state can be constructed from operatorsbp in the way describedbefore providing the occupationnp to each state|p〈. We can invert this picture to real space by defining theoperator

ψ(r) =1

Ld/2

∑

p

e−ip·rbp , (9.7)

which annihilates bosons in a superposition of momentum states with amplitudee−ip·r/√Ld that is just the

amplitude of having a particle annihilated at positionr. Since a superposition of plane waves is an envelopefunction in real spaceψ(r) actually destroys a particle at positionr. Similarly we can define the creationoperator of a particle at positionr asψ†(r). Another way to understand the physical meaning ofψ†(r)comes from usual space representation of a state in Dirac notation, namely,|r〉. In this representation wedefine the operation

|r〉 = ψ†(r)|0〉 , (9.8)

that is, the operatorψ†(r) creates the state|r〉. This allows us to change representations using closurerelations. By definition the state|i〉 is created by acting withb†i on the vacuum

|i〉 = b†i |0〉 , (9.9)

where|i〉 forms a complete set, that is,

∑

i

|i〉〈i| = 1 . (9.10)

Using the closure relation above, (9.8) and (9.9) we find

ψ†(r)|0〉 =∑

i

|i〉〈i|r〉

=∑

i

〈i|r〉b†i |0〉 , (9.11)

and since the vacuum is unique we have

ψ†(r) =∑

i

〈i|r〉b†i . (9.12)

Observe now that〈i|r〉 is nothing but the complex conjugate of the projection of thestate|i〉 into the space

9.2. FERMION OPERATORS 147

representation, that is, the single-particle wavefunction φi(r). Thus,

ψ†(r) =∑

i

φ∗i (r)b†i . (9.13)

This result is the generalization of (9.7) to any quantum state |i〉. Thus, we can always expand an operatorin a complete basis of single-particle states. Observe thatthe wavefunction is a coefficient of the expansion.For this reason this process is called second quantization.

The commutation rules of these new operators defined in real space can be obtained immediately from(9.1),

[

ψ(r), ψ†(r′)]

=1

Ld

∑

p,p′

e−ip·r+ip′·r′[

bp, b†p′

]

= δ(r − r′) ,

[

ψ†(r), ψ†(r′)]

= 0 . (9.14)

Thus, we can represent the many-body state ofN bosons in real space as

|r1, r2, ..., rN 〉 =1√N !ψ†(rN )...ψ†(r2)ψ

†(r1)|0〉 . (9.15)

Observe that because of the commutation rule (9.14) we have the following identity

|r1, r2, ..., rN 〉 = |r2, r1, ..., rN 〉 , (9.16)

which implies that if we exchange two bosons the wavefunction does not change. Furthermore,

ψ†(r)|r1, r2, ..., rN 〉 =√N + 1|r1, r2, ..., rN , r〉 , (9.17)

so that adding a particle produces the state with the correctsymmetry. We further notice that for a singleparticle one can recover the usual one-particle picture quite easily. Sinceψ†(r) creates a particle at positionr the wavefunction of the particle can be obtained by (look at (9.8)):

〈φ|ψ†(r)|0〉 = φ∗(r) ,

〈0|ψ(r)|φ〉 = φ(r) , (9.18)

whereφ(r) is the usual wavefunction of the particle.

9.2 Fermion operators

Let us now go back to the problem of isolated atoms and ask if wecan define creation and annihilationoperators for fermions and what properties these operatorsshould have. For the moment being we are goingto consider the problem without spin. Consider a single state of an atom which by the Pauli principle canonly be unoccupied|0〉 or occupied by a single electron|1〉 (which is the case we studied for the formationof the H+

2 molecule). We define the creation and annihilation operators for fermions in the most intuitive


way,

c|0〉 = 0 ,

c|1〉 = |0〉 ,c†|0〉 = |1〉 ,c†|1〉 = 0 . (9.19)

The conditionc†|1〉 = 0 is the Pauli principle, that is, we cannot put two electrons in the same quantumstate. Moreover, from (9.19) we can prove that

(cc† + c†c)|0〉 = |0〉 ,(cc† + c†c)|1〉 = |1〉 ,c2|1〉 = c|0〉 = 0 ,

(c†)2|0〉 = c†|1〉 = 0 . (9.20)

Since the states|0〉 and|1〉 generate the whole Hilbert space of the problem we can immediately write

cc† + c†c = c, c† = 1 ,

c, c = c†, c† = 0 , (9.21)

which are valid for all states of the problem. Observe that instead of commuting with each other we say thatthe fermions anti-commute. As for the case of the bosons we can define a number operator

N = c†c , (9.22)

which has the required properties

N |0〉 = c†c|0〉 = 0 ,

N |1〉 = c†c|1〉 = c†|0〉 = |1〉 . (9.23)

Thus we can define the state by its occupation,|N〉, as in the case of bosons.

Let us now consider the situation where we have two states with occupationN0 andN1 and the statescan be labeled by their occupation, that is,|N0,N1〉. In this case we have four different states in the Hilbertspace, namely,|0, 0〉, |1, 0〉, |0, 1〉 and|1, 1〉. We now define the creation and annihilation operators exactlyas before, sayc0 andc1. In analogy with the single level case we have:

c0|0, 0〉 = 0 ,

c0|1, 0〉 = |0, 0〉 ,c†0|0, 0〉 = |1, 0〉 ,c†0|1, 0〉 = 0 ,

c1|0, 0〉 = 0 ,

c1|0, 1〉 = |0, 0〉 ,c†1|0, 0〉 = |0, 1〉 ,c†1|0, 1〉 = 0 . (9.24)

Observe that we have not defined how the operators behave whenwe apply them to a state which is already


occupied, say,c0|1, 1〉. The reason is that we need to build into the operator language the correct symmetryof the problem, that is, the Pauli principle: the wavefunction has to be anti-symmetric when we exchangetwo electrons. Suppose we want to exchange two particles in the state|1, 1〉. We have to perform thefollowing operations: (i) we destroy, say, the particle in the state1,

c1|1, 1〉 = |1, 0〉 , (9.25)

which defines the wayc1 acts on this state; (ii ) we destroy a particle in the state0 following (9.24):c0|1, 0〉 =

|0, 0〉; (iii ) we create a particle in state1 again using (9.24):c†1|0, 0〉 = |0, 1〉; (iv) and finally put a particleon state0, that is,c†0|0, 1〉. This series of operations can be written as

c†0c†1c0c1|1, 1〉 = c†0|0, 1〉 , (9.26)

which by Pauli’s principle must have the opposite sign of thestate|1, 1〉. Observe that this is possible if weimpose the condition

c†1c0 = −c0c†1 , (9.27)

and use the number operatorsN0 = c†0c0 andN1 = c†1c1 as

N0N1|1, 1〉 = |1, 1〉 . (9.28)

Thus we have,

c†0|0, 1〉 = −|1, 1〉 . (9.29)

By the same token we can show that

c†0|1, 1〉 = c0|0, 1〉 = 0 ,

c†1|1, 0〉 = |1, 1〉 ,c†1|1, 1〉 = c1|1, 0〉 = 0 ,

c0|1, 1〉 = −|0, 1〉 . (9.30)

The last equation requires that

c1c0 = −c0c1 . (9.31)

From these considerations we see that the correct way to express the Pauli principle with creation andannihilation operators for different states is by imposinganti-commutation relations

ci, c†j = δi,j ,

ci, cj = c†i , c†j = 0 . (9.32)

It is clear from the argument given above that the procedure can be extended to as many states as wewish. In this case the many-body state represented by electron occupationn0, n1, n2, ... with ni = 0, 1 canbe written as

|n0, n1, n2, ...〉 = (c†0)n0(c†1)

n1(c†2)n2 ...|0〉 , (9.33)


with the anti-commutation rules given in (9.32). Observe that the total number of particles in this state iswell defined and it is an eigenstate of the operator

N =

∞∑

i=0

c†i ci . (9.34)

9.2.1 Free electrons in a box

Consider now, as in the case of bosons, the problem of free electrons in a box of volumeLd. The one-bodyeigenstates are plane waves and the momentump and the spinσ =↑, ↓ which are the quantum numbers ofthe problem. The operator that annihilates an electron in real space at positionr and spinσ is written as:

ψσ(r) =1

Ld/2

∑

p

e−ip·rcp,σ . (9.35)

Using (9.32) it is simple to show that

ψσ(r), ψ†σ′(r

′)

= δ(r − r′)δσ,σ′ ,

ψ(r), ψσ′ (r′)

= 0 , (9.36)

and the many-body state ofN electrons is written as

|r1, σ1; r2, σ2; ...; rN , σN 〉 =1√N !ψ†σN

(rN )...ψ†σ2

(r2)ψ†σ1

(r1) , |0〉 (9.37)

in complete analogy with the boson many-body state. Moreover, from the anti-commutation rules we obtainimmediately

|r1, σ1; r2, σ2; ...; rN , σN 〉 = −|r2, σ2; r1, σ1, ..., rN , σN 〉 , (9.38)

in accordance with Pauli’s principle.We can represent any operator in terms of creation and annihilation operators. Consider, for instance,

number operator given in (9.34). Using (9.35) we can write

cp,σ =1

Ld/2

∫

ddreip·rψσ(r) , (9.39)

and from (9.34) one gets,

N =∑

p,σ

c†p,σcp,σ ,

=∑

σ

∫

ddrψ†σ(r)ψσ(r) , (9.40)

which allows us to define the density operator

ρ(r) =

N∑

i=1

δ(r − ri) =∑

σ

ψ†σ(r)ψσ(r) , (9.41)

sinceN =∫

ddrρ(r). The kinetic energy,K, of the electrons can also be obtained in the new operator


language if we recall for each momentump the kinetic energy is simplyh2p2/(2m) and therefore one justhas to count how many electrons with momentump there are, that is,

K =∑

p,σ

h2p2

2mc†p,σcp,σ . (9.42)

Using the identity (9.39) one can rewrite the kinetic energyas

K =∑

σ

∫

ddrψ†σ(r)

(

− h2∇2

2m

)

ψσ(r) . (9.43)

Other operators like the current operator,

J(r) = − ih

2m

(

ψ†σ(r)∇ψσ(r) − (∇ψ†

σ(r))ψσ(r))

, (9.44)

and the spin operator at positionr

S(r) =∑

σ,σ′

ψ†σ(r)Sσ,σ′ψσ′(r) , (9.45)

whereSα = hσα/2 with α = x, y, z are written in terms of Pauli matricesσα, can be easily written in thisnew language.

9.2.2 Free electrons in a lattice

So far we have discussed the problem of free electrons in a box. We can, however, use any basis of states.For instance, in systems where the electrons are confined to be very close to the atoms (such as in thetight-binding approximation) we can use the atomic orbitals instead of the plane waves as a starting pointsince the atomic quantum numbers describe more accurately the physical situation. For simplicity let us goback to the H+2 molecule problem discussed in Chapter 1. In the second quantized language we have thestates|1〉 and|2〉 which correspond to a state of the electron localized on atom1 or atom2, respectively. Interms of our new language the state of the whole system is given by |n1, n2〉 whereni with i = 1, 2 givesthe occupation of each atom. Acting on these states we have creation,c†i , and annihilation,ci, operatorsthat obey the anti-commutation rules (9.32). The Hamiltonian of the system is given by (1.43) and (1.45).Observe that the states of the electron can be described in terms of the creation and annihilation operatorsas in (9.24). Thus, in our new language, the Hamiltonian is easily seen to be

H = E0

(

c†1c1 + c†2c2

)

− t(

c†1c2 + c†2c1

)

. (9.46)

Observe that the tunneling term makes the Hamiltonian non-diagonal. In order to diagonalize this Hamilto-nian we define, in analogy with (1.55), the anti-bonding and bonding operators,

cA =1√2(c1 − c2) ,

cB =1√2(c1 + c2) , (9.47)


which obey the anti-commutation rules. In terms of this new operators the Hamiltonian (9.46) becomes

H = (E0 + t)c†AcA + (E0 − t)c†BcB (9.48)

as expected. Observe that the state of the system is now givenby |nA, nB〉 and the ground state is simply|0, 1〉 in this new basis.

The molecule problem just described can be easily generalized to the case of the solid. Here we assumethat the electrons to be in the tight binding approximation and only one atomic orbitalA is involved. Incomplete analogy to the physics discussed at the end of Chapter 5 we only allow the electrons to hopbetween its nearest neighbor atoms. LetRi be the atom site in a static lattice andci,σ the electron operatorthat destroys an electron in a certain orbital at sitei with spinσ. The Hamiltonian of the system as

H = EA∑

i,σ

c†i,σci,σ − tA∑

<i,j>,σ

(

c†i,σcj,σ + c†j,σci,σ

)

(9.49)

where< i, j > means that sitesi andj must be nearest neighbor sites. The problem imposed by (9.49) canbe easily solved by Fourier transforming the electron operator

cp,σ =1√N

∑

j

eip·Rjcj,σ (9.50)

observe that by the translation symmetry the momentump is defined only in the first Brillouin zone. Forsimplicity we will assume a cubic lattice. You can easily show that in this case the Hamiltonian (9.49) canbe written in terms of the operators in (9.50) as

H =∑

p,σ

Ekc†k,σck,σ (9.51)

with Ek given in (5.76).

9.3 Electron-phonon interaction

Let us go back to the physical problem on how to treat the problem of lattice vibrations coupled to theelectrons. Again we are assuming the Born-Oppenheimer approximation. The basic Hamiltonian can besplit into,H = He +Hp +He−p whereHe involves electrons only;Hp is given by (4.44) andHe−p is theCoulomb interaction between electrons and the ions

He−i =∑

n,m

Ve−i(re,n − ri,m) , (9.52)

wherere,n is the position of thenth electron andri,m the position of themth ion. As previously we aregoing to assume that the departures of the ions from their equilibrium positions are very small so, as inChapter 4, we writeri,m = Rm + um, and expand the interaction (9.52) to leading order in the deviationsu. Observe that to zeroth order inu we have only

∑

n,m Ve−i(re,n − Ri,m) which is the potential of thestatic periodic lattice which can be readily incorporated into the electron partHe which in this case is givenby (9.51). The fluctuation part of interest is,

He−p ≈ −∑

n,m

um · ∇Ve−i(re,n − Ri,m) . (9.53)

9.3. ELECTRON-PHONON INTERACTION 153

If we Fourier transform the potentialVe−i to momentum space

Ve−i(r) =1

N

∑

p

e−ip·rVe−i(p) , (9.54)

equation (9.53) becomes

He−p ≈i

N

∑

n,m,p

e−ip·re,neip·RmVe−i(p)p · um . (9.55)

Observe, however, that,∑

n

e−ip·re,n = ρ(p) , (9.56)

where

ρ(p) =

∫

ddr

N∑

n=1

δ(r − re,n)e−ip·r =

∫

ddrρ(r)e−ip·r , (9.57)

is the Fourier transform of the electronic density operator. Moreover, from (4.43), one has

∑

m

eip·Rmum =∑

k

ek

∑

m

e−i(k−p)·Rm

√

h

2MNωk

(

ak + a†−k

)

,

=∑

G

ep+G

√

Nh

2Mωp+G

(

ap+G + a†−p−G

)

. (9.58)

Using (9.56) and shifting the sum over the momentum we rewrite the electron-phonon interaction as

He−p ≈1

Ld/2

∑

p,G

V (p + G)ρ(p + G)(

ap + a†−p

)

(9.59)

where

V (p) = i

√

h

2ρsMω(p)Ve−i(p)p · ep (9.60)

andρs = N/Ld is the density of the lattice. Using (9.41) and (9.35) we have

ρ(r) =∑

σ

ψ†σ(r)ψσ(r) =

1

Ld

∑

p,k

ei(k−p)·rc†k,σcp,σ ,

=1

Ld

∑

q

eiq·r∑

p

c†p+q,σcp,σ , (9.61)

where in the last line we have change variables toq = k− p. Comparing (9.61) with (9.57) one has

ρ(q) =∑

p,σ

c†p+q,σcp,σ . (9.62)


Thus the Hamiltonian, written in terms of creation and annihilation operators for electrons and phonons,becomes

He−p ≈1

Ld/2

∑

p,G,k,σ

V (p + G)c†k+p+G,σck,σ

(

ap + a†−p

)

. (9.63)

Let us consider first the case whenVe−i = 0 which is well-known: the ground state of the problem hasthe electrons in a Fermi sea with ground state energy,E0 and Fermi momentumkF which depends on thedensity as discussed in Chapter 5, and there are no phonons (since we are talking about zero temperature).Let us label this state by|FS〉. Observe that the perturbation has actually two different types of operators(consider only the case whereG = 0), namely,c†k+p,σck,σap andc†k+p,σck,σa

†−p. In the first case a phonon

with momentump is destroyed, an electron state with momentumk is destroyed and an electron state withmomentumk + p is created. There is a simple way to represent this process graphically as in Fig.9.1 (a).Let us represent the momentum of the electron by a continuousline and the momentum of the phonon bya dashed line. Physically the process is one in which an electron absorbs a phonon and it is scattered to anew momentum state. In the second process an electron of momentumk emits a phonon of momentum−p

and is scattered to a new momentum statek + p as shown in Fig.9.1 (b). The diagrams which are shown inFig.9.1 are known as Feynman diagrams and they are very useful since they allow a clear representation ofthe physical process of scattering. Furthermore, they allow us to rewrite perturbation theory in very simpleterms.

p

k

k+p(a)

k

k+p

−p(b)

Figure 9.1: Scattering process: (a) phonon absorption; (b)phonon creation.

9.3.1 The optical case

An important type of coupling is the one between optical phonons and electrons. In the case of ioniccrystals or polar semi-conductors the optical phonons produce a dipolar field when it oscillates. This dipolarfield polarizes the electrons and leads to the coupling discussed here. To be more specific consider the


displacement vector,D, which is written in terms of the electric field,E, and polarization vector,P, as

D = E + 4πP . (9.64)

In the absence of free charges Maxwell’s equations tell us that,

∇ ·D = 0 , (9.65)

which, when Fourier transformed together with (9.64), leads to

∑

p

p · (Ep + 4πPp) = 0 . (9.66)

At long wavelengths (G = 0) and for longitudinal phonons both the electric field and thepolarization fieldare parallel top and we must have from (9.66) that

Ep = −4πPp . (9.67)

Observe now that the electric field will create a potential due to the electric dipole term which is simplygiven by

E(r) = −∇φ(r) = −i∑

p

peip·rφp . (9.68)

φ(r) is the potential felt by the electrons due to the phonons. Comparing (9.67) with (9.68) we obtain

ipφp = 4πPp , (9.69)

that allows us to compute the potential once the polarization is known.

Since the displacement of the ions create a dipole field one must also have

Pp = κeup = κeepqp . (9.70)

whereκ is a real constant andqp is the phonon displacement given in (4.24). Usingep = ip/|p| (sincee∗p = e−p), together with (9.69) and (9.70) we find:

φp =4πeκ

|p| qp , (9.71)

or, in terms of the boson creation and annihilation operators, we find

φ(r) =∑

p

4πeκ

|p| eip·r

√

h

2ρsLdω0

(

ap + a†−p

)

. (9.72)

Thus, the energy associated with this term is simply

He−p =∑

n

eφ(re,n) =

∫

ddrρ(r)eφ(r) ,

=α

Ld/2

∑

p

1

|p|ρ(p)(

ap + a†−p

)

, (9.73)


where

α = 4πe2κ

√

h

2Mρsω0(9.74)

is the electron-phonon coupling constant. Observe that this result could be obtained from (9.60) withVe−p(p) ∝ 1/p2 which is the case of the Coulomb potential in three dimensions. Substituting (9.62) into(9.73) and rewriting the whole Hamiltonian of the electron phonon problem we have

H =∑

k,σ

Ekc†k,σck,σ +

∑

p

hω0a†pap

+α

Ld/2

∑

p,k,σ

1

|p|c†k+p,σck,σ

(

ap + a†−p

)

(9.75)

which has the form described in (9.63).

Let us consider the effect of the phonons on the electron ground state in perturbation theory, that is, wewriteH = H0 +He−p whereH0 is the free electron-phonon problem and treatHe−p as a perturbation. Theground state ofH0, as we said before, is a filled Fermi sea and no phonons are present. In first order wehave〈FS|He−p|FS〉 which vanishes since the perturbation does not conserve thenumber of phonons andtherefore cannot have a finite expectation value. We have to rely on second order perturbation theory. Foreach value ofk the shift in the energy is given by

δE =∑

n 6=0

|〈n|He−p|0〉|2E0

0 − E0n

, (9.76)

whereH0|n〉 = E0n|n〉. SinceHe−p has one creation and one annihilation operator the excited states that

participate in the sum (9.76) must have at least one excited phonon. Moreover, sinceap|0〉 = 0 becausethe ground state has no phonons we are left witha†−p|0〉 = | − p〉. Moreover, by applying the operator

c†k+p,σck,σ to the ground state of the electron system we create a particle-hole pair exactly as in Fig.6.3.Thus we have to ensure that|k| < kF and |p + k| > kF in order to get a finite result. Observe that theenergy of such excited state isEn = E0

0 − Ek + Ek+p + hω0. Thus, for each value ofk one can write,

δEk = −2α2

Ld

∑

p

1

p2

Θ(kF − |k|)Θ(|p + k| − kF )

Ek+p − Ek + hω0(9.77)

and the total energy is, of course, given byδE0 =∑

k δEk (the factor of2 comes from the two spinprojections).

9.3.2 The polaron problem

Instead of evaluating the change in the energy for any value of kF we are going to look at the case wherethere is just only one electron in the system. This is the so-called polaron problem. Of course this is onlypossible is the electron density is very low andkF → 0. In this case one does not have to worry aboutthe step functions in (9.77). The ground state, of course, has one electron at the bottom of the band withenergyEk = h2k2/(2m). We will consider the three dimensional case (d = 3). Transforming the sums into


integrals we have

δEk = −4mα2

h2

∫

d3p

(2π)31

p2

1

(p + k)2 − k2 + P 20

= − 4mα2

(2π)2h2

∫ π

0dθ sin θ

∫ +∞

0dp

1

p2 + 2pk cos θ + P 20

= − 4mα2

(2π)2h2

∫ 1

−1du

∫ +∞

0dp

1

p2 + 2pku+ P 20

= − 4mα2

(2π)2h2P0

∫ 1

−1du

∫ +∞

0dx

1

x2 + 1 + 2(k/P0)xu(9.78)

whereP0 =√

2mω0/h. Since at low energies the momentum of the electron is small we perform anexpansion in terms ofk/P0. We rewrite

δEk ≈ − 4mα2

(2π)2h2P0

∫ +∞

0dx

∫ 1

−1du

(

1

x2 + 1+

8

3(k/P0)

2u2 x2

(1 + x2)3

)

= − 4mα2

(2π)2h2P0

(

π +π

6(k/P0)

2)

. (9.79)

Observe, therefore, that the total electron energy to second order in perturbation theory can be written as

Ek ≈ −E0 +h2k2

2m∗(9.80)

where

E0 =mα2

2πh2P0

m∗ = m

(

1 +2mE0

3h2P 20

)

. (9.81)

So we see that up to second order in perturbation theory the electron-phonon interaction only changes themass of the electron and replaces it by an effective massm∗. This is a result of the dragging of the latticedue to the electron motion.

9.3.3 The acoustic case

In many cases of interest we neglect all the reciprocal lattice vectors in (9.59) exceptG = 0. This is thelong wavelength limit of the theory since theG 6= 0 represent short wavelengths. For acoustic phononsthe polarization vector at long wavelengths is in the direction of the propagation of the wave, that is,ep =ip/|p|. Moreover, in most solids the interactions between the ionsand the electrons is very short ranged(since the electrons essentially feel only the atomic interaction). In this case we can replaceVe−i(p) = Dby a constant. The electron-acoustic phonon interaction reduces to

δHe−p ≈ D∑

p

√

h

2ρsLdωp

|p|ρ(p)(

a(p) + a†(−p))

. (9.82)


Another elegant way to re-express this part of the Hamiltonian is to use (9.41), (4.85) and (4.43) and write

δHe−p ≈D√ns

∑

σ

∫

ddr

∫

ddr′K(r− r′)ψ†σ(r)ψσ(r)φ(r′) , (9.83)

where

K(r) =i

Ld

∑

p

|p|eip·r . (9.84)

The form of the kernel depends strongly on the dimensionality.

9.3.4 Solitons in one dimension

Consider the one-dimensional system, for instance. In thiscase the polarization vector is just1 and from(9.60) we have

K(x) = i

∫ ∞

−∞

dp

2πpeipx =

d

dx

∫ ∞

−∞

dp

2πeipx

=dδ(x)

dx. (9.85)

Substituting the kernel into (9.83) and doing integration by parts one finds,

δHe−p ≈ D∑

σ

∫

dxψ†σ(x)ψσ(x)

∂φ(x)

∂x. (9.86)

Observe that in this new formulation the electron-phonon problem is described by the Hamiltonian

H =∑

σ

∫

dxψ†σ(x)

(

− h2

2m

∂2

∂x2

)

ψσ(x) +

∫

dx

[

Π2

2ρs+c2sρs2

(

∂φ

∂x

)2]

+ D∑

σ

∫

dxψ†σ(x)ψσ(x)

∂φ(x)

∂x. (9.87)

Let us study the equations of motion for this Hamiltonian in the Heisenberg representation. You can easilyshow that the equations of motion are

ih∂ψσ(x, t)

∂t+h2

2m

∂2ψσ(x, t)

∂x2= Dψσ(x)

∂φ(x)

∂x∂2φ(x, t)

∂t2− c2s

∂2φ(x, t)

∂x2=

D

ρs

∑

σ

∂

∂x

(

ψ†σ(x, t)ψσ(x, t)

)

. (9.88)

Now observe that although we are dealing with equations of motion for operators we can project theseequations into single particle states (since we are dealingwith the problem of a single electron) as in (9.18).In this case instead of the adjoint of operators we are going to have the complex conjugate of the functions.Observe that the equation of motion for the phonon field has the form of a relativistic wave equation with alight velocity cs. As we know from classical mechanics this equation has a traveling wave solution, that is,can be written as

φ(x, t) = φ(x∓ cst) . (9.89)


In order to explore this type of solution we change variablesto χ = x− cst. Moreover, we are interested inthe stationary states of the electron and therefore we look for a solution of the type:

ψ(x, t) = ψ0(x− cst)eih(mvx−E0t) , (9.90)

whereψ0 is assumed to be real andv andE0 are the velocity and energy of the electron. In terms of the newvariables (9.88) becomes

− h2

2m

d2ψ0

dχ2+D

dφ

dχψ0(χ) =

(

E0 −mv2

2

)

ψ0(χ) ,

dφ

dχ= − D

ρs(c2s − v2)ψ2

0(χ) , (9.91)

where we have integrated the second equation in (9.88) once.Observe that the first of these equations isvery similar to the Schrödinger equation for a particle moving under the influence of a potential given by

V = Ddφ

dχ, (9.92)

but the second equation allows to rewrite this potential as

V = − D2

ρs(c2s − v2)ψ2

0 , (9.93)

which shows that the potential itself depends on the wave-function. We say that the potential is determined ina self-consistent way. Observe that this potential is attractive if v < cs and repulsive ifv > cs. Substitutingthe second equation of (9.91) into the first one finds

− h2

2m

d2ψ0

dχ2− D2

ρs(c2s − v2)ψ3

0(χ) =

(

E0 −mv2

2

)

ψ0(χ) (9.94)

which is the so-called non-linear Schrödinger equation which appears in many different fields of physics.You can show that the solution for this equation reads

ψ0(χ) =

√g

2sech

(gχ

2

)

(9.95)

where

g =m

ρs(c2s − v2)

(

D

hcs/a

)2

(9.96)

and

E0 =mv2

2− h2g2

8m. (9.97)

Observe that this solution exists only forv < cs and is unstable otherwise. In this case the potential felt bythe electron is given by (9.93) and (9.95) as

V = − h2g2

4msech2

(gχ

2

)

. (9.98)


Physically what is happening is that the electron is deforming the lattice around it creating a phonon cloud.This phonon cloud responds to the electron by creating a trapping potential which is given by (9.98). Inthis way, electron and phonon-distortion move together as asingle object called a soliton. This situation isdepicted on Fig.9.2.

-10 -5 5 10g x2

-0.4

-0.2

0.2

0.4

0.6

0.8

1

Figure 9.2: Polaron formation: electron is trapped in the self-consistent potential created by the phonons.

9.3.5 Retarded Interactions

We have seen that in the limit of empty band the main effect of the phonons on the electrons is a change inthe electron mass. In a system with finite density this is no longer true. The reason being that phonons, likephotons (which produce the Coulomb interaction), can induce interactions among the electrons. Phonons,however, are scalar fields while photons are vector fields. But besides this difference, from the point of viewof the electrons, they have very similar effect. Here we are going to consider the dynamical problem ofelectrons interacting with phonons. So let us go back to the original electron-phonon Hamiltonian (9.63) inthe long wavelength limit (G = 0) together with the free theory,

H =∑

k,σ

Ekc†k,σck,σ +

∑

p

hωpa†pap

+1

Ld/2

∑

p

V (p)(

ap + a†−p

)

ρ(p) . (9.99)

In the Heisenberg representation the equation of motion forthe phonon operators can be easily obtained

ihdapdt

= [ap,H]

= hωpap +1

Ld/2V ∗(p)ρ†(p) , (9.100)


where we have used thatV (−p) = V ∗(p) and

ρ(−p) =∑

k,σ

c†k−pck =∑

k,σ

c†kck+p = ρ†(p) . (9.101)

A similar equation can be obtained fora†p Equation (9.100) has the form of the equation for a forced har-monic oscillator and the solution is straightforward:

ap(t) = ap(t0)e−iωp(t−t0) − i

hLd/2V ∗(p)

∫ t

t0

dt′ρ†(p, t′)e−iωp(t−t′) , (9.102)

wheret0 is some arbitrary time. We are going to assume that the interaction between the electrons andphonons is switched on very slowly (that is, adiabaticly) sothat in the infinitely distant past the interactionis off and one has the non-interaction problem. In this case one can pickt0 = −∞. Observe that in thiscase the first term in left hand side of (9.102) is strongly oscillating and can be neglected. The second term,however, has the electronic density on it and can produce some new effect.

k’ k+p

k’+p k

p

Figure 9.3: Scattering process mediated by phonons.

If we substitute (9.102) into (9.99) we see that the Hamiltonian of the problem gets an extra piece whichreads

Hextra = − 1

hLd

∑

p

|V (p)|2∫ t

−∞dt′ρ†(p, t′)ρ(p, t) sin

[

ωp(t− t′)]

. (9.103)

Observe that this extra term only carries electronic degrees of freedom and moreover, it is retarded. Thephysical meaning should be clear by now: the phonons generate a retarded density-density interaction be-tween the electrons. This is exactly what photons do indeed.The only difference is that for photons wecan choose the Coulomb gauge where the interaction is instantaneous. But if instead one decides to usethe Lorentz gauge one would generate an interaction which very much looks like (9.103). Moreover, theHamiltonian is a conserved quantity and therefore can be defined at any timet from now on we pickt = 0.It is convenient to rewrite the problem in terms of Fourier components, that is, we define

ρ(p, t) =

∫ +∞

−∞

dω

2πe−iωtρ(p, ω) , (9.104)

and use the identity

∫ t

−∞dt′eiΩt

′

= −i eiΩt

Ω − iǫ, (9.105)


whereǫ→ 0. In this case (9.103) can be written as

Hextra =

∫

dω

2π

∫

dω′

2π

∑

p

|V (p)|2ωp/(hLd)

(ω′ − iǫ)2 − ω2p

ρ†(p, ω′)ρ(p, ω) . (9.106)

The problem is not solved yet because we do not know what the electrons are doing. It could be very wellthat in (9.103) averages to zero. In order to understand whatis going on one needs to look at the electronicproblem as well. In order to do that we are going to assume thatthe Fermi energyEF of the electrons ismuch larger than any phonon energy (like the Debye energy, for instance) in this case the electron operatorobeys the equation

ihdck,σdt

= [ck,σ,H] ≈ Ekck,σ , (9.107)

since the corrections to this equation are of orderV /EF . Thus we write

ck,σ(t) ≈ ck,σ(0)e−iEkt , (9.108)

and therefore

ρ(p, t) ≈∑

k,σ

c†k+p,σ(0)ck,σ(0)e

i(Ek+p−Ek)t . (9.109)

Thus we get from (9.104),

ρ(p, ω) ≈∑

k,σ

c†k+p,σ(0)ck,σ(0) 2πδ(Ek+p − Ek − ω) , (9.110)

and finally in (9.106)

Hextra =∑

p,k,k′,σ,σ′

|V (p)|2ωp/(hLd)

(Ek+p − Ek)2 − ω2p

c†k,σck+p,σc

†k′+p,σ′ck′,σ′ , (9.111)

which is the final result. Observe that the interaction appears in terms of four fermion operators. The reasonfor that is simple to understand in terms of the Feynman diagrams we used before. The process describedin (9.111) is one in which two electrons, one with momentumk′ and another with momentumk + p, aredestroyed and two other electrons with momentumk′ + p andk are created. This is a scattering processwhere a momentump is transferred between electrons. It is depicted on Fig.9.3.

The important point about the process described by Hamiltonian (9.111) is the fact that if|Ek+p−Ek| <ωp it has a negative sign which implies attraction. Consider the case of optical phonons, for instance, whereωp = ω0. In this case electrons states sufficiently close to the Fermi surface can feel an attraction betweenthem. The constraint is thatEF − hω0 < Ek+p, Ek < EF + hω0 which implies that there is a region ofthickness of size2hω0 around the Fermi surface where the electrons attract each other due to the interactionwith phonons (see Fig.9.4). For acoustic phonons we just replaceω0 by the Debye frequencyωD but theresult is essentially the same.

The physical reason for which the attraction appear can be traced back to the basic physics of the prob-lem: as the electron moves it attracts the ions in the lattice. The lattice is an elastic medium which canpropagate this information and therefore other electrons,which will also attracted by the lattice, will feelthis information as if the electrons were attracting each other as depicted on Fig.9.5. The net effect is theattraction between electrons.


2hω0

Fk

Figure 9.4: Region of attraction around the Fermi surface due to the electron-phonon coupling.

9.3.6 Cooper pairs

Although phonons can indeed induce attractive interactions among the electrons one has to remember thatthe electrons interact directly via the Coulomb repulsion.Thus the total interaction can be either attractiveor repulsive, depending on the net interaction between the electrons. The possibility of having electronsattracting each other is quite “attractive" and we are goingto explore this situation here. Suppose we havetwo electrons close enough to the Fermi surface such as they attract each other (as in Fig.9.4). The only roleplayed by the other electrons in the system will be via the exclusion principle. In this case we are back to theproblem of two particles with central attractive forces (exactly like in a H atom). We can therefore just dothe simplest calculation which is to solve the problem in thecenter of mass of the system which we supposeto be at rest (so that the two electrons have opposite momentum). In this frame the problem reduces to thesolution of the Schrödinger equation for the wavefunction of the pair, which we, callψ(r) wherer is therelative coordinate between the electrons. The Hamiltonian of the problem is simply

H =p2

1 + p22

2m+ V (r) =

p2

m+ V (r) , (9.112)

wherep is the relative momentum of the pair andV (r) is the attractive interaction given in (9.111). Inmomentum space the Schrödinger equation becomes

(

k2

m− E0

)

ψ(k) +1

Ld

∑

k′

V (k − k′)ψ(k′) = 0 . (9.113)

We will assume that the interaction is attractive and existsonly when the electron is inside of the shell inFig.9.4. Sincek andk′ are very close tokF we will assume that the matrix element is essentially constantclose to the Fermi surface, that is,

V (k − k′) = −V0Θ(ω0 − |Ek′ |) . (9.114)

In this case the solution of (9.113) reads

ψ(k) =κ

k2

m − E0

, (9.115)

where

κ =V0

Ld

∑

k′

Θ(ω0 − |Ek′ |)ψ(k′) . (9.116)


Substituting (9.115) into (9.115) we find

1 =V0

Ld

∑

k′

Θ(ω0 − |Ek′ |)(k′)2

m −E0

, (9.117)

which is the equation forE0. Using the density of statesN(ǫ) =∑

k δ(ǫ−k2/2m)/Ld the (9.117) becomes

1 = V0

∫ EF +hω0

EF

dǫN(ǫ)

2ǫ− E0. (9.118)

We will also assume that the density of states is constant over the interval of integration we have

1 =V0N(0)

2ln

(

2(EF + hω0) − E0

2EF − E0

)

, (9.119)

which can be easily solved

E0 = 2EF +2hω0

1 − e2

N(0)V0

, (9.120)

that is the binding energy of the two electrons. Observe thatwhen the attraction is turned offV0 → 0 oneobtains the expected result, that is2EF , which is energy of two electrons at the Fermi surface. Observehowever that whenV0 6= 0 we haveE0 < EF , that is, the energy of the system is lowered by the formationof a bound state. What this implies is that the Fermi surface is unstable with attractive interactions since theelectrons will lower the energy of the system by forming pairs. The binding energy of these pairs is simply

∆ = 2EF − E0 =2hω0

e2

N(0)V0 − 1≈ 2hω0e

− 2N(0)V0 , (9.121)

which is a non-analytic expression in terms of the interaction and therefore cannot be obtained in pertur-bation theory (which always gives a power series expansion). These pairs are called Cooper pairs. Cooperpairs are the heart of a phenomena called superconductivitywhere pairs condense to form a macroscopicquantum state in clear violation of Pauli’s principle. The reason for that is that Cooper pairs, being formedof two fermions, behave very much like free bosons and therefore can Bose-Einstein condense. This is acompletely new state of matter. Observe, moreover, that theformation of Cooper pairs only requires annet attractive interaction between electrons and therefore does not care very much about the mechanismfor the attraction. The electron-phonon interaction is only one of many mechanisms that can make theelectron-electron interaction attractive. More on this inthe future...

+ + + +

+ + + +

− −

p

Figure 9.5: Physical coupling between phonons due to the lattice.

9.4. PROBLEMS 165

9.4 Problems

1. Using the orthogonality and closure of the states|i〉 and the creation operator in real space defined in(9.12) show that the operatorsΨ(r) andΨ†(r) obey bosonic commutation relations.

2. Show that for fermion operators we have

cs|n0, n1, n2, ...〉 = (−1)Ss√ns|n0, n1, ...ns − 1, ...〉

c†s|n0, n1, n2, ...〉 = (−1)Ss√ns + 1|n0, n1, ...ns + 1, ...〉

where

Ss = n1 + n2 + ...+ ns−1 .

3. Show that for fermion operators

Sz(r) =h

2

(

ψ†↑(r)ψ↑(r) − ψ†

↓(r)ψ↓(r))

Sx(r) =h

2

(

ψ†↑(r)ψ↓(r) + ψ†

↓(r)ψ↑(r))

and using the anti-commutation rules between electrons show that

[

Sx(r), Sy(r′)]

= ihδ(r − r′)Sz(r) . (9.122)

4. Show that (9.47) obeys the anti-commutation relations and prove that (9.48) is indeed correct. Writedown all the states of the system from the new basis in terms ofthe old basis.

5. Using (9.50) prove that (9.51) is indeed correct.

6. Prove (9.79).

7. Expand (9.78) to all orders ink/P0 and calculate the exactly value of the integral. What happenswhenk becomes large? What is the physical meaning of your result?

8. Prove (9.88).

9. Show that (9.95) is a solution of (9.94) by direct substitution.

10. (i) Find the displacement field profile,φ(x), for the static polaron (v = 0) by solving (9.91). What doyou conclude about the lattice distortion around the electron?

(ii ) Show that due to translational invariance of the problem the polaron mass is given by

M0 = ρs

∫ +∞

−∞dx

(

∂φ

∂x

)2

(9.123)

and calculate the ratio between the bare electron massm andM0.

11. (i) In the polaron problem the potential felt by the electron isgiven in eq. (9.98). Consider the casewhere the polaron is at rest (v = 0) and solve the Schrödinger equation in this case (9.91). Show thatone of the eigenstates is given by (9.95) with eigenenergy given in (9.97). (Hint: The solution can be


found in the book by P. M. Morse and H. Feshbach,Methods of Theoretical Physics(McGraw-Hill,New York, 1953)).

(ii ) Find all the other eigenstates of the problem and their respective eigenenergies. Show that thesestates are scattering states with finite phase shiftδ(k). Give an expression forδ(k).

12. Obtain the time evolution ofa†p for the electron-phonon problem described by Hamiltonian (9.99).


14. Prove that the Heavyside Theta function can be written as

Θ(t) =

∫ +∞

−∞

dω

2πi

eiωt

ω − iǫ

and from this result prove (9.105).

Chapter 10

Magnetism

One of the most important problems in condensed matter physics is magnetism. Magnetism is a quan-tum mechanical phenomenon since it has its origins on the spin of the electron, its orbital motion and theCoulomb interaction between electrons. Naturally, magnetism is associated with the response of the elec-trons to a magnetic field. We have seen that for a free non-interacting electron gas we have two main typesof response, paramagnetic when the effect of the magnetic field is only the magnetic polarization of themedium, and diamagnetic when the response of the medium is against the magnetic field. In isolated atomsand molecules a similar type of response appears, as we have seen in Chapter 1. It is however the interactionbetween different atoms in a solid that leads to the phenomenon of magnetic ordering that is observed innatural magnets like iron.

As we have seen in Chapter 1 the magnetic response of isolatedatoms can be summarized in the threeHund’s rules that are phenomenological rules based on simple energetic arguments. In short, the Hund’srules provide statements about the magnetic ground state ofmany electron atoms when the Coulomb energyis minimized. In a solid, however, the atoms are localized inwell defined positions that are dictated by thesymmetry of the lattice. Thus, the atoms are subject to electromagnetic fields that are called crystal fields.We are going to discuss the effect of crystal fields and the Coulomb interaction among electrons in differentatoms lead to the phenomena of magnetism in solids.

10.1 Crystal fields

As we have seen in Chapter 1 the degeneracy of a magnetic atom is lifted by the internal electron-electroninteraction in an isolated atom. Moreover, spin-orbit effects lift the degeneracy even further. The thirdHund’s rule states that the final degeneracy of an isolated atom with total angular momentumJ = L + S is2J+1 due to the different projections of theJz component ofJ . The spin-orbit interaction is very importantin 4f-electron systems where the partially filled 4f shells lie deep inside of the ion and therefore are screenedfrom the outside charges. In other systems with 3d- or p-shells the effect of the interaction between theelectrons and other nuclei is very important because these orbitals are exposed to the electrostatic potentialof other atoms in the lattice. Thus, in the case of transitionmetals with d shells the lattice potentials aremore important than the spin-orbit effects and should be considered first (that is, Hund’s third rule is notapplicable). In this case the degeneracy of the2l + 1 states can be lifted by the Coulomb interaction as weshow in Fig.10.1. In some systems the crystal field interaction is more important than the third Hund’s ruleand one has to take the effect into account first. Observe thatin this case the two first Hund’s rules stillapply and therefore there is a degeneracy of(2S + 1)(2L + 1). Since the electrostatic field of the latticecouples only to the charge degrees of freedom only the angular momentum degrees of freedom are affected.If the crystal field is very asymmetric then the degeneracy ofthe angular momentum is lifted and as a result

167

168 CHAPTER 10. MAGNETISM

the expectation value of the angular momentum has to vanish,that is,〈L〉 = 0, even though we still have〈L2〉 = L(L+ 1). Consider, for instance the case of p-orbitals in a free atom. These orbitals can be writtenasxf(r), yf(r) andzf(r) wherex, y, z refers to the first Legendre polynomial (l = 1) andf(r) is thesolution of the radial equation. In a free atom these orbitals are degenerate. It is easy to show that theseorbitals do not carry well defined angular momentum projection which are obtained as linear combinationsof them: (x + iy)f(r) with mz = +1, zf(r) with mz = 0 and(x − iy)f(r) with mz = −1. But sincethe orbitals are degenerate it really does not matter how we represent them. In the presence of an externalelectric field the energy of the orbitals is split as in Fig. 10.1. If the electric field is very asymmetric one cansplit the energy of all the states. Consider the case where the orbitalyf(r) has the lower energy. In this casethe angular momentum of this state can be computed immediately

〈Lz〉 =

∫

dryf(r)1

i

(

x∂

∂y− y

∂

∂x

)

yf(r) (10.1)

but observe that

(〈Lz〉)∗ = −〈Lz〉〈(Lz)†〉 = −〈Lz〉

〈Lz〉 = −〈Lz〉 (10.2)

and therefore〈Lz〉 = 0. Where in the first line above we have integrated by parts and in the second line weused the fact thatLz is a hermitian. From the classical point of view this can be understood as the precessionof the angular momentum in the crystal field such that its magnitude is fixed but it has zero average. In thiscase we have〈J〉 = 〈S〉 which is known as quenching of the orbital angular momentum.

−

−

− −

−

−

+

+

+

z

x

y

z

z

x

x

y

y

py

pz

py

px

p

px

z

field

Crystal

Figure 10.1: Crystal field effects for a p-orbital.

Thus, in 3d electron systems the spin-orbit interaction is aperturbation on the crystal field effects. Letus consider this problem more closely. The spin-orbit effect is described by a Hamiltonian (1.25). Let us

10.1. CRYSTAL FIELDS 169

consider the case in which an external magnetic field is applied:

HP = λL · S + µBB · (L + 2S) , (10.3)

that we consider as a perturbation of the problem in the presence of a crystal field.Moreover, we are only interested in the spin degrees of freedom since〈J〉 = 〈S〉. Therefore we perform

perturbation theory in the orbital degrees of freedom alone. Let |0〉 be the ground state of the problem in acrystal field. Observe that first order perturbation theory gives

δE1 = 2µBB · S (10.4)

due to the quenching of the orbital angular momentum. Let us consider now the second order perturbationtheory in the absence of the field. If|n〉 is an orbital eigenstate of the system in the presence of crystal fieldeffects then second order perturbation theory leads to a change in energy given by

δE2 = λ2∑

n 6=0

|〈n|∑

a SaLa|0〉|2E0 − En

= λ2∑

n 6=0

∑

a,b

〈n|La|0〉〈n|Lb|0〉E0 − En

SaSb

= −∑

a,b

Γa,bSaSb (10.5)

where

Γa,b = λ2∑

n 6=0

〈n|La|0〉〈n|Lb|0〉En − E0

. (10.6)

Observe that in this case the spin-orbit effect induces an interaction between different components of thespin. This is called single ion anisotropy. If one takes the principal axis of the crystal asx, y andz andexpress the components ofΓa,b asΓx, Γy andΓz, one can show that (10.5) can be written as

δEso = −S(S + 1)

3(Γx + Γy) +

(Γx + Γy − 2Γz)

2S2z

+(Γy − Γx)

2

(

S2x − S2

y

)

. (10.7)

The first term in (10.7) is just an overall shift of the energy of the problem. The second term, because itappears as a sum of twoΓ is usually more important than the last term which appears with the difference.For integerS the term withS2

z splits the energy levels into doubly degenerate levels withSz = ±S,±(S −1), ...,±1 and a non-degenerate level withSz = 0. For half-odd integerS there are doubly degeneratelevels withSz = ±S,±(S−1), ...,±1/2. The term proportional toS2

x−S2y ∝ S2

+ +S2− induces transitions

between states that haveδSz = ±2. Thus, for integerS this term lifts the degeneracy of the degeneratestates which can be linked by±2. For instance, forS = 1 the Sz = 0 state is non-degenerate and theSz = ±1 states have their degeneracy lifted by the spin-orbit. For ahalf-odd integerS the degeneracypersists becauseδSz is always an odd integer, thus, the degenerate ground state has alwaysδSz = ±1 andtherefore are not linked by the perturbation. Thus the degeneracy persists even with spin-orbit coupling.The case with integerS corresponds to an even number of electrons in the atom while ahalf-odd integerS corresponds to an odd number of electrons. So for atoms with an odd number of electrons even in thepresence of crystal fields and spin-orbit interactions the ground state is a doublet. This degeneracy has its


origin on the time reversal symmetry.

Observe that although spin-orbit and crystal fields reduce considerably the symmetry of the atom, thereis one symmetry that persists in the absence of external magnetic fields: time reversal symmetry. Timereversal symmetry changest → −t and therefore changesp → −p but keeps the position invariant, thatis, r → r. It implies that by inversion of the time direction the angular momentum is reversed, that is,L → −L. Classically it just means that we reverse the direction of rotation. Time reversal also inverts thespin,S → −S, because spin and angular momentum are indistinguishable in quantum mechanics. In theabsence of magnetic fields the Hamiltonian is invariant under time reversal. Indeed, consider the generalproblem of an electron moving under the influence of an external potential and the spin-orbit effect,

H = − h2

2m∇2 + V (r) + λS · L (10.8)

and it is clear that under time reversal the Hamiltonian is invariant. Notice that the addition of a magneticfield leads to a new term of the formB · (L + 2S) which is not invariant under time reversal since this termchanges sign. We can obviously define a quantum mechanical operatorK such that

K|r〉 = |r〉K|p〉 = | − p〉K|S〉 = | − S〉 (10.9)

and it is clear that

[H,K] = 0 (10.10)

which means that if|ψ〉 is an eigenstate of the Hamiltonian with energyE (H|ψ〉 = E|ψ〉) thenK|ψ〉 isalso an eigenstate of the system with the same energy. This simple exercise in group theory proves that evenif all the symmetries in a quantum mechanical systems are broken and no magnetic fields are present theground state of the system has to be doubly degenerate, that is, it must be a doublet. This is the so-calledKramers theorem.

In the presence of an external magnetic field the energy shiftof the ground state is given by

δE = δEso +∑

a,b

[

µBga,bBaSb −(µBλ

)2BaΓa,bBb

]

(10.11)

where

ga,b = g0

(

δa,b −1

λΓa,b

)

(10.12)

is the so-called g-tensor. Observe that (10.11) has a deep physical meaning: when a magnetic field is appliedto an atom the spin does not necessarily responds in the direction of the field because of the spin-orbitcoupling. The magnetization of the atom is

Ma = − ∂E

∂Ba

= −∑

b

(

µBga,b〈Sb〉 − 2(µBλ

)2Γa,bBb

)

(10.13)

10.2. THE HEITLER-LONDON THEORY 171

and it depends on the matrixΓa,b. Moreover, the susceptibility is now a tensor and it is defined as

χa,b =∂Ma

∂Bb= 2

(µBλ

)2Γa,b (10.14)

is the induced orbital momentum which arises from the spin-orbit coupling. Observe that in this case anapplied field in one direction can have a response in another different directions. This is a direct consequenceof the crystal fields. Observe that in this case the magnetic moment of the atom can be written as

ma = µB∑

b

ga,b〈Sb〉 (10.15)

as a consequence of spatial anisotropy. Thus, in this case the application of the magnetic field in onedirection can lead to a magnetization in another direction.

10.2 The Heitler-London theory

The next step after the isolated atom problem is to study the case of a few atoms, that is, the molecule. Thiskind of study help us to understand more complicated cases later.

We have seen in Chapter 1 that in a H2 atom the electrons that form the bonding orbital need to havetheir spin anti-parallel to each other due to the Pauli principle. Therefore the system does not have any netmagnetic moment. As we have seen in Chapter 7 this can be described by the Hamiltonian (9.46)

H0 = −t∑

σ=↑,↓

(

c†1,σc2,σ + c†2,σc1,σ

)

. (10.16)

whereci,σ (c†i,σ) destroys (creates) an electron in a localized state at atomi (i = 1, 2) with spinσ (σ =↑, ↓).As before we can diagonalize the problem by a simple linear combination of electron operators

cA,σ =1√2(c1,σ − c2,σ)

cB,σ =1√2(c1,σ + c2,σ) . (10.17)

In terms of this new operators the Hamiltonian (10.16) becomes

H = t∑

σ=↑,↓

[

c†A,σcA,σ − c†B,σcB,σ

]

(10.18)

The ground state for the case where two electrons are presentis

|ΨB〉 = c†B,↑c†B,↓|0〉

=1

2

(

c†1,↑c†1,↓ + c†1,↑c

†2,↓ + c†2,↑c

†1,↓ + c†2,↑c

†2,↓

)

|0〉

=1

2

(

c†1,↑c†1,↓ + c†1,↑c

†2,↓ − c†1,↓c

†2,↑ + c†2,↑c

†2,↓

)

|0〉 (10.19)

where |0〉 is the empty state. Notice that in the last line of (10.19) we used the anti-commutation rulesbetween the electrons. Observe that the state represented in (10.19) is anti-symmetric as required by thePauli principle. Another way to represent this state is to use the occupation of each atom (say|n1,σ, n2,σ′〉)


where

|ΨB〉 =1

2(| ↓↑, 0〉 + | ↑, ↓〉 − | ↓, ↑〉 + |0, ↑↓〉) . (10.20)

In this new notation one has to be very careful about the orderof the spins because of Fermi statistics.Observe, however, that the Hamiltonian (10.16) cannot be a good approximation for the real problem

of the molecule since it completely neglects the Coulomb repulsion between the electrons. In particularit neglects the large Coulomb repulsion for electrons in thesame atom. If this Coulomb energy is verylarge then the double occupancy of the atom costs a lot of energy. But we see from (10.19) that for thenon-interacting problem this states are present. In the Heitler-London theory it is assumed that the Coulombenergy is very large and that double occupancy is avoided. Inthis case we see that a good approximationfor the wave-function for the problem can be obtained from (10.19) by suppressing the operators that createtwo electrons in the same atom. In this case we see that the approximate ground state can be written as

|Ψs〉 =1√2

(

c†1,↑c†2,↓ − c†1,↓c

†2,↑

)

|0〉 (10.21)

which in the other notation is simply

|Ψs〉 =1√2

(| ↑, ↓〉 − | ↓, ↑〉) (10.22)

which is the so-called singlet state we discussed before. Thus the singlet state has the best of two worlds: itminimizes the Coulomb repulsion and maximizes the kinetic energy and therefore is a good starting pointfor the calculation of the true ground state of the problem.

10.3 Coulomb interactions

Let us be more quantitative and consider the Coulomb repulsion of two electrons in the H2 molecule in moredetail. The Coulomb energy of interaction between electrons is simply

HC =e2

2

∫

ddr

∫

ddr′ρ(r)ρ(r′)

|r − r′| (10.23)

whereρ(r) is the electronic density. If we write the density in terms ofthe electron operators (9.41), wehave

HC =

∫

ddr

∫

ddr′∑

σ,σ′

e2

2|r − r′|Ψ†σ(r)Ψσ(r)Ψ

†σ′(r

′)Ψσ′(r′) . (10.24)

Since the electron operators create or destroy electrons atthe position of the atoms we can rewrite them as

Ψσ(r) =2∑

i=1

φi(r)ci,σ (10.25)

whereφi(r) is the wavefunction of the electron localized at the atomi.We consider the case where the number of electrons in each atom is constant. If the number of electrons

do not change it means that electronic motion is completely disregarded. This approximation is only possiblewhen we are dealing with a very large Coulomb repulsion in theatom. Actually this Coulomb repulsion

10.3. COULOMB INTERACTIONS 173

has to be much larger than the kinetic energy gain for an electron to jump from one atom to another sothat in first approximation one can neglect the kinetic energy. This is the case of insulating systems andtherefore can only describe what is called localized magnetism which should be contrasted with the case ofitinerant magnetism, as we shall see later. For the H2 molecule the condition of single occupancy impliesni,↑ + ni,↓ = 1. Direct substitution of (10.25) directly into (10.24) leadto 16 different terms. We consideronly the ones that do not lead to hopping of the electron from one atom to another. There are, therefore,three different types of terms that interest us. The first oneis

HU =U

2

∑

σ,σ′

2∑

i=1

ni,σni,σ′

= U

2∑

i=1

ni,↑ni,↓ +U

2

∑

σ

2∑

i=1

ni,σ (10.26)

where in the last line we used thatn2i,σ = ni,σ for a fermion and

U = e2∫

ddr

∫

ddr′|φi(r)|2|φi(r′)|2

|r − r′| (10.27)

represents the Coulomb interaction in the same atom and it isthe type of interaction we talked about whendiscussing the Hund’s rules. Observe that the last term in the right hand side of (10.26) only shifts the localenergy of the electrons and can be included in the non-interacting part of the Hamiltonian.

The second type of term is the one that involves interaction between electrons in different atoms. Al-though this term is not important if we constraint the atoms to single occupation we shall see later that thisterm is very important for virtual processes. It is written as:

HV = V∑

σ,σ′

n1,σn2,σ′ (10.28)

where

V = e2∫

ddr

∫

ddr′|φ1(r)|2|φ2(r

′)|2|r− r′| . (10.29)

And finally the third term is called the exchange term and is given by

HJ = −JC∑

σ,σ′

c†1,σc1,σ′c†2,σ′c2,σ (10.30)

where

JC = e2∫

ddr

∫

ddr′φ∗1(r)φ1(r

′)φ∗2(r′)φ2(r)

|r− r′| . (10.31)

The exchange term can be written in a more illuminating way ifwe use the representation of the spinoperators in terms of electron operators (9.45). We defineSi = hsi/2 where

sai =∑

α,γ

c†i,ασaα,γci,γ (10.32)


whereσa is a Pauli matrix (a = x, y, z). We need to calculate

s1 · s2 =∑

a

∑

α,β,γ,δ

σaα,βσaγ,δ c

†1,αc1,βc

†2,γc2,δ (10.33)

and for that we can show that∑

a

σaα,βσaγ,δ = 2δα,δδβ,γ − δα,βδγ,δ . (10.34)

Using the above relation we find

s1 · s2 =∑

α,β

(

2c†1,αc1,βc†2,βc2,α − c†1,αc1,αc

†2,βc2,β

)

(10.35)

And therefore we have from (10.30) we have

HJ = −JC2

∑

σ,σ′

n1,σn2,σ′ − 2JCs1 · s2 (10.36)

where the first term can be absorbed into (10.28). We are more interested in the last term which represents aspin-spin interaction between spins in the two atoms. Moreover, sinceJC > 0 the energy is minimized if thespins are such that they are aligned pointing in the same direction. This is called a ferromagnetic coupling.

It is very tempting now to conclude that if the electrons are localized in the atoms, that is, the tunnelingis very small then the electrons should have their spins aligned. This of course contradicts our previousconclusion that in a bonding state the electrons should havetheir spins pointing in the opposite direction ina singlet state. What is wrong here?

What is wrong is that even if the electrons cannot actually hop from one atom to another, they can stillvirtually hop! This is a pure quantum mechanical effect since second order perturbation allows a quantumsystems to undergo transitions over excited states. Indeedlet us consider the case where only interactionin the same ion are taking into account. In the limit where theions are far apart these are the terms thatdominate the physics and they are giving by (10.26). Let us treat now the tunneling Hamiltonian (10.16)as a perturbation. The ground state of (10.26) is given by oneelectron in each atom since the Coulombrepulsion is minimized, that is, in the ground state hasni,↑ + ni,↓ = 1. The energy of this state is zero. Firstorder perturbation theory in the hopping gives a null resultbecause the occupations of the atoms change.In second order perturbation theory one electron can hop back and forth from one atom to another. In thisprocess it has to go through an excited state with double occupancy and energyU . By the Pauli principlethis can only occur if the spins in different atoms are pointing in opposite direction. Indeed, in second orderperturbation theory this process can be written as

HA = −4t2

U

∑

σ,σ′

c†1,σc2,σc†2,σ′c1,σ′ (10.37)

and using again (10.35) we can rewrite the above interactionas

HA =2t2

Us1 · s2 (10.38)

which has the same form of (10.36) but with the opposite sign implying that the spins tend to becomeanti-aligned. This is called a kinetic exchange interaction.

10.4. MAGNETIC ANISOTROPY 175

If we now put put together (10.36) and (10.38) together we getthe total exchange interaction betweenthe electrons in the atoms is given by

HE = Js1 · s2 (10.39)

where

J =2t2

U− 2JC (10.40)

which can be ferromagnetic or antiferromagnetic dependingon the relative values oft2/U andJC . For theH2 molecule it turns out that the antiferromagnetic coupling is larger and therefore the ground state is asinglet.

10.4 Magnetic anisotropy

Observe that the Hamiltonian in (10.39) is symmetric under rotations of the spins if is the product of twovectors. It turns out that in real systems there are magneticanisotropies that appear due to spin-orbit andcrystal fields as we discussed in the previous section. Theseanisotropies also affect the interaction betweenspins in a solid. Consider, for instance, the classical dipolar interaction between two magnetic momentsm1

andm2 separated by a distanceR. The dipolar interaction can be written as

Hdip =m1 ·m2

|R|3 − 3(m1 ·R)(m2 ·R)

|R|5 . (10.41)

Consider now the case of a 3d electron atom which has a quenched orbital momentum. The magneticmoment of the atom is only given by the spin degrees of freedomand the magnetic moment, due to crystalfield effects, is given in (10.15). Direct substitution of (10.15) into (10.41) leads to

Hdip =∑

a,b

S1,aCa,b(R)S2,b (10.42)

where

Ca,b(R) = µ2B

(∑

c gc,agc,b|R|3 − 3

∑

c,d gc,agd,bRcRd

|R|5)

. (10.43)

Thus the total interaction between two atoms can be written as the sum of (10.39) and (10.42)

H =∑

a,b

S1,aJa,b(R)S2,b (10.44)

where

Ja,b = Jδa,b + Ca,b . (10.45)

We can diagonalizeJa,b and find the principal magnetic axis of the system. SinceJa,b is a3 × 3 matrix wecan have3 different eigenvalues. It the problem has symmetry around an axis then two of this eigenvaluesmust be degenerate and so on. These anisotropies lead to a Hamiltonian that is more general than (10.39)


and reads

H = JzS1,zS2,z + JyS1,yS2,y + JxS1,xS2,x , (10.46)

whereJz, Jy andJx can be different from each other.

10.5 Localized Magnetism

All the discussion here can be generalized for a crystal. If the system is insulating and the interaction onlyaffects nearest neighbor atoms the generalization of the Hamiltonian (10.46) is simply

H =∑

〈i,j〉

(JzSi,zSj,z + JySi,ySj,y + JxSi,xSj,x) . (10.47)

The extreme limits of this Hamiltonian can be obtained by varying the parameters in (10.47). For instance,whenJz ≫ Jy, Jx we have

HI = Jz∑

〈i,j〉

Si,zSj,z (10.48)

which is the so-called Ising model. WhenJy = Jx = Jxy >> Jz we have

Hxy = Jxy∑

〈i,j〉

(Si,ySj,y + Si,xSj,x) (10.49)

is the so-called XY model. And finally whenJx = Jy = Jz = J we are back to (10.39)

HH = J∑

〈i,j〉

Si · Sj (10.50)

is the Heisenberg model. Each one of these models have different symmetries. While the Heisenberg modelhas full rotation symmetry in spin space, the XY model only has rotation on a plane (we say that it has aneasy plane) and finally the Ising model is the one with lowest symmetry since the spins are constrained tobe on a fixed axis (we say that the system has an easy axis). The interesting thing about these systems is thatbecause they have different symmetries they order magnetically in different ways leading to what is calleduniversality classes. In nature, however, it is difficult tofind a magnetic system that can be classified as apure Ising, XY or Heisenberg.

10.5.1 Problems

1. Prove (10.7).

2. Prove (10.11).

3. Consider the H2 molecule and obtain the expression for the anti-bonding state in terms of the localizedstates of the electrons.

10.6. MAGNETIC INTERACTIONS IN METALLIC ALLOYS 177

4. Using the representation of (10.20) show that

H0| ↑↓, 0〉 = −t (| ↑, ↓〉 − | ↓, ↑〉)H0| ↑, ↓〉 = −t (|0, ↑↓〉 + |0, ↑↓〉)H0| ↓, ↑〉 = +t (| ↑↓, 0〉 + |0, ↑↓〉)H0|0, ↑↓〉 = −t (| ↑, ↓〉 − | ↓, ↑〉)H0|σ, σ〉 = 0 . (10.51)

5. Use (10.25) and write down all the 16 terms that appear in (10.24).

6. Using the Fourier transform of the Coulomb potential showthatJC > 0.

7. Prove (10.34).

8. Prove (10.38).

9. Prove (10.42).

10.6 Magnetic interactions in metallic alloys

In the last section we considered the problem of magnetic interactions in insulators, that is, systems wheredue to the strong Coulomb interactions the electrons are localized in the atoms. This is not always the case.There are a large number of atoms that show metallic behavior, that is, they can be well described by bandtheory discussed earlier. Metals are usually paramagnetic(a counter example of this assertion is Fe thatcan be metallic and ferromagnetic) and show no particular magnetism. In many compounds atoms withmagnetic moments can be introduced into a metallic host. This is the case for instance of alloys with f- or d-electron atoms. Magnetic atoms can interact with the conduction electrons that can propagate the magneticinteraction between the magnetic moments. This effect happens because the spin of the magnetic momentcan interact with the spin of the conduction electron via a dipolar interaction.

Consider an atom with magnetic momentmS interacting with an electron with magnetic momentme

via (10.41). Eq. (10.41) is valid wheneverR 6= 0 otherwise it becomes singular. Let us consider whathappens whenR = 0. The magnetic moment of the atom,mS , creates a magnetic field,BS , that coupleswith the electron via the Zeeman term,BS · s, wheres is the spin of the electron. The magnetic field can becalculated from electrodynamics starting from the vector potentialA as:

BS = ∇× A (10.52)

whereA is the vector potential for a magnetic dipole that is given by

A =1

|R|3 mS × R = −mS ×∇(

1

|R|

)

. (10.53)

Direct substitution of (10.53) into (10.52) leads to

BS = ∇×∇×(

mS

|R|

)

. (10.54)


Using the identities∇×∇× = ∇(∇·) −∇2 and∇2(1/r) = −4πδ(r) we find

BS = ∇[

∇ ·(

mS

|R|

)]

+ 4πmSδ(R)

=

[

∇∇− ∇2

3

]

mN

|R| +8π

3mNδ(R) . (10.55)

It is easy to show the first term in (10.55) is (10.41) and the second term is a contact interaction at the atom.Thus, the Zeeman energy for the electron interacting with the magnetic moment of the atom is

HZ = −8π

3me · mSδ(R) (10.56)

that is known as the Fermi contact interaction.

In a lattice withN magnetic atoms we can generalize (10.56) to

HZ =∑

j,n,a,b

Ja,b(rj − Rn)sa(rj)Sb(Rn) (10.57)

whereRn is the position of thenth atom,rj the position of thejth electron, andmS = µSS whereµS isthe atom magnetic moment. From (10.15) we have,

Ja,b(rj − Rn) =8π

3µBµSga,bδ(rj − Rn) (10.58)

is the exchange between the spin of the electron and the magnetic moment of the nucleus in the presence ofcrystal fields and spin-orbit effect.

For simplicity we assume that the metal that surrounds the magnetic nucleus is described by plane waves(although the same can be done with Bloch states) and is described by

He =∑

k,σ


whereEk = h2k2/(2m). We want to represent the new term in (10.57) in terms of planewaves as well. Forthat we need the representation of the electron spin in termsof field operators as in (9.45). It is easy to showthat the Hamiltonian can be written as

HJ =∑

k,k′,n

ei(k−k′)·Rn

JzSz(Rn)[

c†k′,↑ck,↑ − c†

k′,↓ck,↓

]

+J⊥2

[

S−(Rn)c†k′,↑ck,↓ + S+(Rn)c

†k′,↓ck,↑

]

(10.60)

where we have usedS± = Sx ± iSy. Observe that the effect of the interaction between the conductionelectrons and the nucleus is a scattering process where a plane wave with wavevectork scatters tok′. In thisprocess the electron can flip its spin.

In order to simplify our problem let us consider first the highly anisotropic limit of (10.60) in whichJ⊥ = 0. Moreover, we are going to assume that there is just one impurity at the origin (Rn = 0). In this


case (10.57) is rewritten as

HZ =∑

j

Jzδ(rj)sz(rj)Sz

=∑

j

Jzδ(rj)Sz(n↑(rj) − n↓(rj)) (10.61)

In this limits it is easier to write the Hamiltonian of the problem in real space in terms of the field operatorsdefined in (9.45):

H =

∫

ddr∑

σ,σ′

Ψσ(r)

[

− h2∇2

2mδσ,σ′ + V (r)σzσ,σ′

]

Ψσ′(r) (10.62)

where

V (r) = JzSzδ(r) (10.63)

and σz is a Pauli matrix andSz is the eigenstate of the operatorSz (that is, Sz|Sz〉 = Sz|Sz〉 whereSz = −S, .., S).

Observe that we can diagonalize the problem entirely if we expand the electron field operator as

Ψσ(r) =∑

E,σ

φE,σ(r)aE,σ (10.64)

whereaE,σ is a fermion operator (a†E,σ, aE′,σ′ = δE,E′δσ,σ′ ) and

− h2∇2

2mφE,σ(r) + σV (r)φE,σ(r) = EφE,σ(r) (10.65)

is a Schrödinger equation for the magnetic scattering of a particle by an impurity since it depends on theelectron spin (here we have used thatσzσ,σ′ = σδσ,σ′ ). In this case the Hamiltonian of the problem in thenew basis reads

H =∑

n,σ

Ena†En,σ

aEn,σ (10.66)

whereEn are the solutions of (10.65).

Let us consider, for simplicity, the case where the electrons interact only weakly with the magneticimpurity and we can use perturbation theory. In (10.65) define E = h2k2/(2m) andU = 2mV/h2 andrewrite (10.65) as

∇2φk,σ(r) +[

k2 − σU(r)]

φk,σ(r) = 0 (10.67)

and transform it to Fourier space

φk,σ(r) =1√N

∑

q

eiq·rφk,σ(q) (10.68)


in order to get

(k2 − q2)φk,σ(q) =∑

p

σU(p− q)φk,σ(p) (10.69)

whereU(p) is the Fourier transform ofU(r). In order to solve (10.69) we have to impose the conditionthat whenU = 0 the solution of the problem is a plane wave, that is,φk,σ(q) = δq,k. Thus, the solution of(10.69) is

φk,σ(q) = δq,k + σ∑

p

U(p− q)

k2 − q2φk,σ(p) (10.70)

which is an integral equation forφk,σ(q). We first observe that because of (10.63) we haveU(p) = U =2mJzSz/h

2 is independent of momentum. Moreover, in first order the solution of the problem is given bythe substitution of the delta function in the first term on thel.h.s. of (10.70) into the second term on the l.h.s.which gives,

φk,σ(q) ≈ 1√N

(

δq,k + σU

k2 − q2

)

(10.71)

which when substituted into (10.68) leads to

φk,σ(r) ≈1√N

eik·r + σU∑

q 6=k

eiq·r

k2 − q2

(10.72)

which is correct to first order inU and it is known as Born approximation.

In order to evaluate (10.72) we have to avoid the point wherep = k where the sum diverges. Moreover,in the thermodynamic limit we replace the sum by an integral over q. We have the problem of avoidingthe points where the integral diverge. The best way to do it isto add an infinitesimal imaginary part to theintegral and write

∑

q 6=k

eiq·r

k2 − q2= ℜ

[∫

ddq

(2π)2eiq·r

k2 − q2 − iǫ

]

(10.73)

whereǫ → 0 at the end of the calculation. This is the principal value of an integral. So let us evaluate thisintegral

∫

ddq

(2π)2eiq·r

k2 − q2 − iǫ=

1

(2π)2

∫ π

0dθ sin θ

∫ ∞

0dqq2

eiqr cos θ

k2 − q2 − iǫ

=1

(2π)2

∫ +1

−1du

∫ ∞

0dqq2

eiqru

k2 − q2 − iǫ

=1

(2π)2r

∫ +∞

−∞dq

q sin(qr)

k2 − q2 − iǫ(10.74)

where in the last line we used that the integrand is even inq. The integrals can be performed by contourintegration. For instance,

∫ +∞

−∞dq

qeiqr

q2 − k2 + iǫ= 2πi

(

e−ikr

2

)

= πie−ikr (10.75)


since we have to close the contour in the upper half-plane in order for the integral to converge. The finalresult is

∑

q 6=k

eiq·r

k2 − q2= 2π2 cos(kr)

r(10.76)

and from (10.72) we have

φk,σ(r) ≈ eik·r + 2π2σUcos(kr)

r. (10.77)

Observe that we can now calculate various properties using (10.64) and (10.77). In the approximationused, the electrons are only scattered by the magnetic impurity but their energy remains the same. Thus theground state is obtained by filling up the momentum states as usual. Observe, however, that the presence ofthe impurity modifies the charge density for different spins. Indeed,

〈nσ(r)〉 = 〈Ψ†σ(r)Ψσ(r)〉

=∑

k,k′

φ∗k,σ(r)φk′,σ(r)〈a†k,σak′,σ〉

=∑

k

|φk,σ|2Θ(kF − k) (10.78)

wherekF is the Fermi momentum which is related to the electron density by kF = (3π2n)1/3. Using(10.77) to leading order inU we find

〈nσ(r)〉 =n

2+ 4π2σU

∫

d3k

(2π)3cos(kr)

rcos(k · r)Θ(kF − k)

=n

2+σU

r2

∫ kF

0dkk sin(2kr)

=n

2+ σ

8k4FmJz

h2 SzF (2kF r) (10.79)

where

F (x) =sinx− x cos x

x4. (10.80)

Notice that the electron density far away from the impurity reaches is bulk valuen/2 but close to the impurityit oscillates strongly. These are called Friedel oscillations. Moreover, the oscillations depend on the spin ofthe electron. This happens because the interaction betweenthe electron and the impurity atom is magnetic.

Consider now the problem of two magnetic impurities at positionsRn andRm. The spinSn at positionRn polarizes the electron gas in its vicinity in the way given by(10.79). Another spinSm atRm feels thispolarization that allows the two moments to interact. In order to get the interaction Hamiltonian betweenthese magnetic moments we use (10.61) and (10.79) to find

HRKKY = −∑

n,m

JzSz(Rn) (n↑(Rm) − n↓(Rm))

∝ − J2z

EF

∑

n,m

F (2kF |Rn − Rm|)Sz(Rn)Sz(Rm) (10.81)


whereEF is the Fermi energy. This interaction between magnetic moments is called RKKY interaction dueto Ruderman, Kittel, Kasuya and Yosida. Observe that the interaction is oscillatory and decays like1/r3 asshown in Fig.10.2.

0 5 10 15 20x

-0.004

-0.002

0

0.002

0.004

F

Figure 10.2: Plot ofF (x)

Notice that at short distances (r ≪ 1/(2kF )) the interaction is ferromagnetic but at large distancesit changes sign and can be antiferromagnetic. Thus, depending on the position of the spins in the latticethis interaction can be effectively ferromagnetic or antiferromagnetic. In the case of disordered alloys bothferromagnetic and antiferromagnetic couplings are possible. At zero temperature this lead to a frustratedmagnetic state called spin glass.

In the generic case whereJ⊥ is not null the form of the RKKY is given by:

HRKKY = −∑

n,m,a,b

Γa,bF (2kF |Rn − Rm|)Sa(Rn)Sb(Rm) (10.82)

whereΓa,b depends on the spin-orbit and crystal field effects.

10.7 The Double Exchange problem

There are some magnetic systems like Mn that, when forming a solid, donate one electron to the conductionband but the remaining core electrons do not form a closed shell. In this case the atom in the lattice has aspinS that can interact with the electrons of the conduction band via an exchange interaction of the typegiven in (10.60). This interaction is usually ferromagnetic because of Hund’s rules coupling, that is, thesystem wants to have locally the largest value of the spin. Inthis case the Hamiltonian of the problem canbe written as:

H = −t∑

<i,j>,σ

(

c†i,σcj,σ + h.c.)

− J∑

i

si · Si (10.83)

whereh.c. means hermitian conjugate andJ > 0 is the Hund’s coupling between localized and itinerantelectrons. This is a problem of great complexity since, as wehave seen before, the electrons mediate theinteraction between spins via the RKKY interaction.

Let us consider the case where we have a large localized spinS (S ≫ 1/2). In this case the spin can beconsidered as a classical variable that can be parameterized by anglesθi andφi relative to some fixed axis(for instance,Sz = S cos(θ), Sx = sin(θ) cos(φ), Sy = S sin(θ) sin(φ)). Note that the Hund’s couplingforces the spin of the electron to orient along the directionof S. If the local spins are all aligned witheach other (ferromagnetic situation) then the second term in (10.83) gives a negative energy contribution,reducing the energy of the system. If the spins are orientedπ degrees from each other (antiferromagneticsituation) the electrons cannot hop from site to site because the hopping term in (10.83) does not flip the spin

10.7. THE DOUBLE EXCHANGE PROBLEM 183

and therefore it costs an energyJ to frustrate the Hund’s coupling. Consider the situation onFig.10.3(a):the electron spin (empty circle) is oriented with all the localized spins (filled circles) and can move freelyover the lattice. In Fig.10.3(b) the antiferromagnetic orientation of the spins makes the hopping from oneatom to the other difficult since it costs an energyJ for the electron with spin up to move to an atom withspin down. WhenJ ≫ t the hopping is completely suppressed.

a)

b)

J

Figure 10.3: (a) Hopping in a ferromagnetic situation; (b) Hopping in a anti-ferromagnetic situation.

This argument shows that in order to gain the electron kinetic energy the spins tend to form a ferro-magnetic state where all the spins are aligned to allow for the electron motion. Notice that because of thesymmetry of the interaction in (10.83) the change in energy can only depend on the relative angle,θi,j,between the localized neighboring spins. Whenθi,j = π the effective hopping between spins,ti,j, is zero,whenθi,j = 0 the effective hopping is simplyti,j = t.

In order to study this formally Let us consider the case of twomagnetic atoms with spinsS1 andS2.The Hamiltonian of the problem (in accord with (10.83) is:

H = −t∑

σ

(c†1,σc2,σ + h.c) − J(S1 · s1 + S2 · s2) (10.84)

In the reference frame of the local spins the energies of the states are+JS or−JS depending if the electronspin is anti-parallel or parallel to the local spin, respectively. Let us call these states|i,−〉 and |i,+〉,respectively (i = 1, 2). Thus, the whole problem reduces to four states, namely,|1,+〉, |1,−〉, |2,+〉, and|2,−〉. The spinsS1 andS2 form an angleθ between them, that is,S1 · S2 = S2 cos(θ). If we choose aquantization axis alongS1 (we writeS1 = Sz), for instance, we have to rotate to project the states ofS2

into this axis. This can be done by rotating the states of in atom 2 via a spin rotation operator around theYaxis:

U = e−iθSy/h = e−iθσy/2

= cos(θ/2) − iσy sin(θ/2) (10.85)

whereσy is theY -Pauli matrix. The states of atom2 relative to the quantization axis of atom1 can bewritten as:

|2, α〉 =∑

γ=±1

Uα,γ |2′, γ〉 (10.86)

where|2′, α〉 refer to the states in the rotated frame. The rotation can be written more explicitly as:

|2,+〉 = cos(θ/2)|2′,+〉 + sin(θ/2)|2′,−〉|2,−〉 = − sin(θ/2)|2′,+〉 + cos(θ/2)|2′,−〉 . (10.87)


Moreover, from (10.84) and (10.87) we see that:

〈i, α|H|i, γ〉 = −αJSδα,γ〈1, α|H|2, γ〉 = −tUα,γ . (10.88)

Thus, the Hamiltonian can be written as:

[H] =

−JS 0 −t cos(θ/2) −t sin(θ/2)0 JS t sin(θ/2) −t cos(θ/2)

−t cos(θ/2) t sin(θ/2) −JS 0t sin(θ/2) −t cos(θ/2) 0 JS

which can be diagonalized in the usual way. The eigenenergies are:

E = ±√

(JS)2 + t2 ± 2JSt cos(θ/2) . (10.89)

ForJ ≫ t the lowest energy eigenvalue is:

E0 ≈ −JS − t cos(θ/2) (10.90)

where the first term corresponds to the Zeeman energy of the electron in the field of the localized momentand the second term corresponds to the effective hopping energy for the spins between the different atoms.Notice that forθ = 0 it gives−t and forθ = π it gives zero, in agreement with the previous discussion.Notice further that while in the RKKY mechanism the first contribution to the energy is of orderJ2/EF (see(10.81)) the current mechanism produces an effect of orderJ and therefore should be dominant at smallJ .The above argument shows that, when generalized to a lattice, the effective Hamiltonian of the problem canbe written, in the limit ofJ ≫ t, as:

Heff = −t∑

〈i,j,〉,σ

cos(θi,j/2)(

c†i,σcj,σ + h.c.)

(10.91)

whereθi,j is the angle between neighboring spins. For a given lattice one has to find out the spin configu-ration that minimizes the energy in (10.91). The model described here was originally proposed by Zener inorder to describe the physics of certain oxides with Mn and the term double exchange comes from the factthat the exchange between Mn atoms occurs via the filled p-orbitals of the O atoms.

10.7.1 Problems

1. Show that in two dimensions the RKKY interaction decays like1/r2 at long distances.

2. Consider the problem of an electron jumping between two atoms with large spins in the limit ofJ ≪ t. (i) What is the ground state energy in this case? (ii ) What is the physical interpretation of thesolution in this limit? (iii ) Will the electron travel over the system for any spin configuration?

3. Consider the problem of electrons moving in a d-dimensional hyper-cubic lattice with a densityn ofelectrons in the limit ofJ ≫ t. Assume that the angle between adjacent spins is always the same, thatis, θi,j = θ independent ofi, j. (i) What is the spin configuration in this case? (ii ) What is the groundstate energy as a function of angle? (iii ) Assume that the angle between adjacent spins is very smalland that the distance between atoms isa. Definek = θ/a and show that the energy of the systembehaves likek2. How does the electron spin behave in this case ?

10.8. THE ANDERSON HAMILTONIAN 185

10.8 The Anderson Hamiltonian

In the last sections we discussed Fermi contact interactionbetween conduction electrons and localized mag-netic moments. The contact interaction is important for s-wave states where the electron has large proba-bility of being at the atomic site. For p- or d-like orbitals that vanish at the origin, the contact interactionalso vanishes. In this case the interaction between a conduction electron and a localized moment has a dif-ferent origin which is related with the hybridization between conduction electrons and localized electrons.Consider the problem of a localized impurity at a positionr = 0 in a Fermi gas as shown in Fig.10.4.

U

U

Figure 10.4: Impurity immersed in a Fermi gas.

A conduction electron can hop back and forth from the electron gas into the impurity site. This hoppingcan be seen as a tunneling term like in (10.16). However, in order to do so, the electron has to pay an energyU if the impurity site is doubly occupied (which was studied in(10.26)). The simplest Hamiltonian thatdescribes this situation is the Anderson Hamiltonian:

HA =∑

k,σ

Ekc†k,σck,σ + Ef

∑

σ

f †σfσ + V∑

σ

(

c†σ(r = 0)fσ + f †σcσ(r = 0))

+ Unf,↑nf,↓ (10.92)


which in momentum space can also be written as:

HA =∑

k,σ

Ekc†k,σck,σ +Ef

∑

σ

f †σfσ +V√N

∑

k,σ

(

c†k,σfσ + f †σck,σ

)

+ Unf,↑nf,↓ (10.93)

whereck,σ andfσ are electron operators for electrons on the conduction bandand the localized impuritystate, respectively (N is the number of sites).Ef is the atomic energy at the impurity,V is the tunnelingenergy between the conduction band and the impurity andU is the Coulomb energy to put two electrons onthe impurity (nf,σ = f †σfσ).

The Hamiltonian describes a true many-body problem. In trying to understand the physics of this Hamil-tonian we have to make approximations. Let us consider first the problem whereU = 0, that is, the Coulombinteraction is absent. One expects from the beginning that no magnetism can be described in this case sincea magnetic moment can only exist in a isolated energy level ifdouble occupancy of the level is not allowed.We will see later how the Coulomb energy can be introduced into the problem. WhenU = 0 the problem isquadratic in the operators and therefore can be diagonalized by an unitary transformation:

fσ = αfσ +∑

k

βkck,σ

ck,σ = γkfσ +∑

k

ιk,k′ck′,σ (10.94)

wherefσ andck,σ are the new electron operators which diagonalize the problem, that is, after the unitarytransformation the Hamiltonian (10.93) withU = 0 can be written as

H =∑

σ

Eff†σfσ +

∑

k


with Ef andEk the new eigenvalues (still to be calculated). Notice that the coefficientsα, βk, γk andιk,k′

depend on the parameters in the Hamiltonian.

One of the ways to solve this problem is to look at the equations of motion for the operators. Using theanti-commutation relations between the fermion operatorsit is easy to show that

[fσ,H] = Effσ +V√N

∑

k

ck,σ

[ck,σ,H] = Ekck,σ +V√Nfσ . (10.96)

On the other hand, using (10.94) and (10.95) one has

[fσ,H] = αEffσ +∑

k

βkEkck,σ

[ck,σ,H] = γkEffσ +∑

k′

ιk,k′Ek′ck′,σ (10.97)


and by direct substitution of (10.94) into (10.96) we find

[fσ,H] = (αEf +V√N

∑

k

γk)fσ +∑

k

(Efβk +V√N

∑

k′

ιk′,k)ck,σ

[ck,σ,H] = (V√N

∑

k

βk +∑

k′

Ekιk′,k)ck,σ + (Ekγk +V√Nα)fσ . (10.98)

Direct comparison of (10.97) and (10.98) leads to

Ef = Ef +V

α√N

∑

k

γk

(Ef − Ek)γk = αV√N. (10.99)

In solving the above equations we have to exercise some care.We are interested in the case whereEf

is within the electron band which implies that the second equation in (10.99) has a singularity atEf = Ek.This singularity can be avoided, however, if we interpret the second equation as the real part of

γk =1√N

V

Ef − Ek − iǫ, (10.100)

whereǫ → 0. Moreover, we are interested in the case of a single impurityin 1023 electrons. Thus, in thiscaseEk ≈ Ek plus corrections of order10−12. Substitution of (10.100) into the first equation of (10.99)leads to the desired final result

Ef = Ef +1

Nℜ

∑

k

V 2

Ef − Ek − iǫ

(10.101)

which is anequationfor Ef . Equation (10.101) shows that the energy of the f-state is shifted by an amountproportional toV 2. One has to remember, however, that the original f-state is not an eigenstate of theHamiltonian since it is actually coupled to the conduction band. Thus, when an f-electron is put into themany-body system it has to decay intof andc states which are the true eigenstates of the problem. There-fore, the f-electron has a finite lifetime. In order to estimate the life-time we go back to (10.101) and interpretthe imaginary part of the f-electron energy as its decay rate(this step can be formally proved with the use ofGreen’s function method). Thus, we have:

h

τf=

1

Nℑ

∑

k

V 2

Ef − Ek − iǫ

=πV 2

N

∑

k

δ(Ef − Ek) (10.102)

where we have used that:

limitǫ→0

[

1

x− iǫ

]

= P(

1

x

)

+ πiδ(x) (10.103)

whereP means the principal value of the function. Notice that (10.102) makes a lot of sense. WhenV → 0the f-state is decoupled from the conduction band and therefore has infinite lifetime. But as the coupling tothe conduction band increases the lifetime of the f-state becomes shorter.

It is interesting to compare the behavior of the density of states in both cases. WhenV = 0 the density


of states of the problem reduces to the density of states of the f-state and the conduction band

Nf (E) = δ(E − Ef ) = limitǫ→01

π

ǫ

(E − Ef )2 + ǫ2= ℑ

1

E − Ef − iǫ

Nc(E) =1

N

∑

k

δ(E −Ek) . (10.104)

In the presence of the couplingV we replaceEf in the first equation in (10.104) byEf − ih/τf in order toget

Nf (E) = ℑ

1

E − Ef − ih/τf

=1

π

h/τf

(E −Ef )2 + h2/τ2f

(10.105)

which is not a Dirac delta function but a Lorentzian with width 1/τf with is given by

h

τf= πV 2Nc(Ef ) (10.106)

as we can readily see from (10.102) and (10.104). Thus, as a result the f-level is not sharp. This reflectsthe fact that when an f-level hybridizes with a conduction band it is not an exact eigenstate of the system(indeed, one sees from (10.94) that the eigenstate is a linear combination of f and conduction band states).This situation is depicted in Fig.10.5.

N(E)

EEf

(a)

N(E)

EEf

(b)1/τf

Figure 10.5: Density of states for theU = 0 Anderson model: (a)V = 0, (b) V 6= 0.

In order to understand what happens in the presence ofU one has to treat the interaction term in (10.93).It is obvious that this term cannot be treated in the same way we treated theV term because it contain fourfermion operators. The problem is highly non-linear. Instead of trying to solve the problem exactly wesearch for an approximate solution. The approximation is called mean field or Hartree-Fock approximation.


In this approximation we replace

Unf,↑nf,↓ → U〈nf,↑〉nf,↓ + Unf,↑〈nf,↓〉 (10.107)

which corresponds to replace the actual value of the occupation at the f-level state by its average. If onemake the substitution (10.107) into (10.93) one sees that the only modification is in the energy of the f-levelstate, that is,

Hf =∑

σ

Efnf,σ → Hf,MF =∑

σ

Ef,σnf,σ (10.108)

where

Ef,σ = Ef + U〈nf,−σ〉 (10.109)

that is, the energy of the f-level with spin, say,↑ is increased byU times the average occupation of the statewith spin↓ and vice-versa. After this approximation is done we can proceed as earlier (in the case ofU = 0)but taking into account that the local energy of the f-level depends also on the spin projection. In particularthe density of states for each spin projection is given by:

Nf,σ(E) =1

π

h/τf,σ

(E − Ef,σ)2 + h2/τ2f,σ

(10.110)

where

Ef,σ = Ef + U〈nf,−σ〉 +1

NRe

∑

k

V 2

Ef,σ − Ek − iǫ

1

τf,σ= πV 2Nc(Ef,σ) . (10.111)

For simplicity we assume that the electron density of statesis essentially constant and thereforeEf,σ =Ef +U〈nf,−σ〉 andτf,σ = τf . In this case, the average occupation of the f-level with spin σ is given by theusual expression

〈nf,σ〉 =

∫ µ

−∞dENf,σ(E)

=

∫ µ

−∞

dE

π

h/τf

(E − Ef − U〈nf,−σ〉)2 + h2/τ2f

=1

πarccot

(

Ef + U〈nf,−σ〉 − µ

h/τf

)

(10.112)

which is a transcendental equation for the occupation. Thuswe have traded a non-linear problem in termsof operators for a non-linear problem in terms of ordinary equations. It is common to rewrite the equationabove in terms of new variables defined as:

y = τfU/h

x =µ−Ef

U(10.113)


so that

cot(π〈nf,σ〉) = y(〈nf,−σ〉 − x) . (10.114)

Let us consider two simple limits of these equations. WhenU = 0 it is obvious that

〈nf,σ〉 = 〈nf,−σ〉 =1

2(10.115)

since〈nf,σ〉 + 〈nf,−σ〉 = 1. This corresponds to the non-magnetic solution of the problem. Thus, as onecould have guessed the Coulomb interaction is fundamental for the appearance of magnetic moments. Inthe limit ofU → ∞ thearccot(x) can giveπ or 0 depending on the sign of the argument. It is obvious thatthe solution close toπ corresponds to the full occupation of the level and the one close to0 corresponds toan empty level. Thus, let us assume that〈nf,↑〉 ≈ 1 and〈nf,↓〉 ≈ 0 and calculate

〈nf,↑〉 ≈ 1 − 1

πy(x− 〈nf,↓〉)

〈nf,↓〉 ≈ 1

πy(〈nf,↑〉) − x(10.116)

which implies that the local magnetization in the magnetic impurity is

mf = 〈nf,↑〉 − 〈nf,↓〉

≈ 1 − 1

πy(1 − x)

≈ 1 − V 2N(0)

U(10.117)

where we have used (10.106) and (10.113). Notice that the above approximation is valid forV ≪ EF , U .That is, in the limit in which the impurity is weakly coupled to the electron gas and the Coulomb energy islarge. In this case the charge fluctuations in the impurity site are very small and only the spin of the impurityhas dynamics.

The transition between the magnetic states at largeU and the non-magnetic states at smallU can bestudied within the mean-field approach by solving (10.114) in the case of〈n↑〉 = 〈n↓〉 = nc. Taking thederivative of this equation with respect ton we find another equation, namely,

y =π

sin2(πnc). (10.118)

Using (10.118) in (10.114) we find that, at the transition,x has to be such that:

x = nc −sin(2πnc)

2π. (10.119)

Further notice that because〈n↑〉 + 〈n↓〉 ≤ 2 we must havenc ≤ 1. A plot of x as a function ofnc as givenby (10.119) andπ/y as a functionnc is shown in Fig.10.6. From these plots we can draw the phase diagramof the problem as a function ofx andπ/y as shown in Fig.10.7. This phase diagram shows that momentformation in an atom embedded in a metallic environment is only possible ifx < 1, that is the impuritylevel is close to the Fermi energy,µ − Ef < U , andπ/y < 1, that is the Coulomb interaction is large,U > πh/τf .

Therefore, the conclusion of this calculation is that for the existence of local moments the Coulomb

10.9. THE KONDO PROBLEM 191

0 0.2 0.4 0.6 0.8 1n

0

0.2

0.4

0.6

0.8

1

x,py

Figure 10.6: Plot of (10.118): continuous line; and (10.119): dashed line.

Non−Magnetic

x

0

1

1/2

1

π/y

Magnetic

Figure 10.7: Phase diagram of the Anderson impurity problemat the mean-field level.

energy is fundamental. The physics of this problem is relatively simple: in the absence of the Coulombinteraction,U = 0, the singly occupied states are degenerate with energyEf and the doubly occupied statehas energy2Ef . In the presence ofU the energy of doubly occupied state isU + Ef > µ and thereforeout of the Fermi surface. Thus, this state is always empty. Itis exactly because the doubly occupied state ishigher in energy that moment formation is stabilized.

10.9 The Kondo problem

Although (10.93) has very interesting physics it describesa large number of effects that are not necessarilyrelated with magnetic scattering. In this section we will study the processes in which the number of electronsin the impurity is constant, that is,

∑

σ nf,σ = 1, so that the impurity has a magnetic moment. This is onlypossible ifV ≪ Ef , U,EF andEf < EF so that the electron on the impurity does not disappear on theconduction band andU + Ef > EF so that the impurity is never doubly occupied. Instead, we are onlyinterested in the virtual process in which an electron can hop from the conduction band into the impurityand back. When an electron does that it goes through and excited state of energyEf + U if the impurity issingly occupied. In perturbation theory this leads to a termwhich is proportionalV 2/(Ek − U − Ef ). Onthe other hand an electron localized at the impurity can jumpfrom the impurity into the conduction bandand back where the intermediate state now has energyEk. Again in perturbation theory this leads to a termof orderV 2/(Ef − Ek). Observe that in both cases we haveEk ≈ EF . These process are not differentfrom the ones discussed previously for the H2 molecule. Thus, second order perturbation theory generates


an interaction between the conduction band and the electronimpurity of the form

Hexch| = Js(r = 0) · S (10.120)

where

J ≈ NV 2

(

− 1

U +Ef+

1

Ef

)

(10.121)

which is always positive (we have fixed the energy so thatEF = 0). The Hamiltonian (10.120) is called theKondo Hamiltonian and has exactly the same form as the Fermi contact interaction (10.57). Thus all ourresults about the RKKY interaction are valid for this model if we take the impurity spin to be1/2. However,there is new physics in this model that we have not discussed yet.

As before we consider the anisotropic Kondo problem where the exchanges in the Z and X-Y directionsare different from each other (happen in the presence of crystal fields and spin-orbit effects). The completeHamiltonian reads:

H =∑

k,σ

ǫkc†k,σck,σ + JzS

z∑

k,k′,σ

σc†k,σck′,σ

+ J⊥(

S+s−(r = 0) + s+(r = 0)S−)

(10.122)

whereJz is the longitudinal or Ising coupling andJ⊥ is the transverse of XY coupling. Notice that becauseJz, J⊥ > 0 there is an effective attraction between the impurity and the electrons when they have oppositespins. Attraction between particles implies the possibility of forming bound states. In fact, forJ⊥ = 0 wecan use the results of the previous section, in particular, we write (10.69) as:

(E − h2q2

2m)φE(q) = −Jz

∑

p

φE(p) (10.123)

where we reestablished the usual units. This equation can besolved as

φE(q) =C(E)

E − h2q2

2m

(10.124)

where

C(E) = −Jz∑

p

φE(p) . (10.125)

Substitution of (10.124) into (10.125) leads to

1 = −Jz∑

p

1

E − h2q2

2m

(10.126)

which gives the equation forE. If we now use the definition of the density of states

N(E) =∑

p

δ

(

E − h2p2

2m

)

(10.127)


we can rewrite (10.126) as

1 = Jz

∫

dE′ N(E′)

E′ −E, (10.128)

that has to be solved forE(Jz), the bound state energy. Observe that (10.128) does not haveany informationabout the occupation of the states in the system, that is, it does not carry a Fermi occupation number.However, it predicts the possibility of forming bound states that can be written as:| ⇑, ↓〉 or | ⇓, ↑〉, where⇑,⇓ represent the impurity spin and↑, ↓ represent the electron spin. In the above problem the two states aredegenerate in energy. In this localized picture we immediately see that the couplingJ⊥ in (10.122) lifts thedegeneracy between these two states so that the final states of the problem are:

|s〉 =1√2

(| ⇓, ↑〉 − | ⇓, ↑〉)

|t〉 =1√2

(| ⇓, ↑〉 + | ⇓, ↑〉) (10.129)

which are the singlet and triplet states. Notice that due to the transverse term in the Hamiltonian the singlestate is lower in energy byJ⊥ relative to the triplet state.

For the reasons given below the approach given above has a bigflaw associated with the fact that weare treating theJz andJ⊥ in a very non-perturbative way and we will show that this is incorrect physically.However, we borrow the ideas of the naive calculation above in order to simplify the problem to a treatableform. Let us first define new states by:

|+〉 = | ⇑, ↓〉|−〉 = | ⇓, ↑〉 . (10.130)

These states are not “pure” spin states since they involve the impurity spin and the localized electron spin.However, we see that the operator associated withJ⊥ has a very simple effect when acting on those states,namely:

(

S+s−(r = 0) + s+(r = 0)S−)

|+〉 = |−〉(

S+s−(r = 0) + s+(r = 0)S−)

|−〉 = |+〉 (10.131)

that is, theJ⊥ operator acts as a Pauli matrixσx on these states. By the same token, the operatorSz has avery simple action on those states:

Sz|+〉 = |+〉Sz|−〉 = −|−〉 (10.132)

and thereforeSz acts on these states as the Pauli matrixσz. Thus, by changing the basis we can rewrite theKondo Hamiltonian as:

H =∑

k,σ

ǫkc†k,σck,σ + Jzσ

z∑

k,k′,σ

σc†k,σck′,σ + J⊥σ

x . (10.133)

This problem still has too many degrees of freedom. As we knowfrom our study of metals perturbations inmetals can only affect states close to the Fermi energy and unfortunately (10.133) still contains states withhigh energy close to the bottom of the Fermi sea. If we consider the momentak andk′ in (10.133) closeto the Fermi surface we see the effect of the impurity is to scatter and electron with momentumk from an


occupied state inside of the Fermi sea (leaving a hole behind) and putting this electron outside the Fermi seain a unoccupied state with momentumk′. That is, it creates particle-hole pairs. Thus, it is more convenientto work in the language of particle-hole pairs instead of thelanguage of the original fermions.

As we showed in Chapter 6 (see Eq.(6.106)) the low energy excitations of the Fermi gas, the particle-hole pairs, have bosonic character and at low temperatures provide a natural description of the specific heat,for instance. The energy of a particle-hole pair close to theFermi surface is

ωq = vF q (10.134)

wherevF is the Fermi velocity andq > 0 is the module of the momentum transfer for the production of theparticle-hole pair (k′ = k + q). Thus, particle-hole pairs behave like harmonic oscillators in momentumspace (like acoustic phonons). Thus, at low energies we can replace the free electron Hamiltonian by:

H0 =∑

k,σ

(ǫk − µ)c†k,σck,σ → Hph =

∑

q>0

[

P 2q

2M+Mω2

q

Q2q

2

]

(10.135)

whereµ is the chemical potential. Notice that the mass of the oscillatorsM is undefined. The reason is givenbelow. Furthermore, in the scattering process the electrontransfer momentum to the impurity (momentumis not conserved in the scattering problem) and this leads tothe creation of particle-hole pairs with finitemomentumq. Since the momentum operator associated with the pair isPq we see that theJz operator in(10.133) can be replaced by:

Jzσz∑

k,k′,σ

σc†k,σck′,σ → Jzσz∑

q>0

Pq . (10.136)

Defining now creation and annihilation operators for the harmonic oscillators in the usual way:

aq =

√

Mωq2

(

Qq −i

√

MωqPq

)

a†q =

√

Mωq2

(

Qq +i

√

MωqPq

)

(10.137)

we find that the Hamiltonian in terms of bosons can be written as:

H =∑

q>0

ωqa†qaq + λσz

∑

q>0

√qi(aq − a†q) + J⊥σ

x . (10.138)

We rewrite the Hamiltonian by redefining the creation and annihilation operators by a phase factor:

bq = iaq = eiπ/2aq (10.139)

so that:

H =∑

q>0

ωqb†qbq + λσz

∑

q>0

√q(bq + b†q) + J⊥σ

x , (10.140)

where

λ = Jz

√

MvF2

. (10.141)


The Hamiltonian in (10.140) is called the dissipative two-level system model since it describes a problemof a two-state system (the eigenstates ofσz) coupled to a heat bath of harmonic oscillators. Notice thattheparameterM remains undefined. The reason for that is that the oscillators are produced by the couplingof the impurity to the particle-hole excitations and therefore their mass is also determined by this coupling.The procedure presented here cannot tell howM behaves as a function ofJz and thus we can think ofM(or λ) as parameters of the problem that have to be obtained by different means. What is guaranteed fromour construction is that the number of degrees of freedom in the problem is correct and that the low energyphysics of Hamiltonian (10.140) is the same as the original fermionic problem (10.122). One way to obtainλ is to solve the problem via Bethe ansatz or numerical methodsand obtainλ as a function ofJz. From nowon we think ofλ as a parameter of the problem in its own right.

The Hamiltonian in (10.140) has another free parameter thatis not explicit in the Hamiltonian. Noticethat the bosons are a result of the linearization of the electronic dispersion close to the Fermi surface (seeFig.6.4) and therefore our procedure is only valid if the momentum of the particle-hole pair is smaller thana cut-off,Λ. The size of this cut-off can be estimated to be of the order ofthe Fermi momentum,kF , which,in most metals, is proportional to the inverse of the latticespacing,a. That is, our approximation assumesthat:

q ≪ Λ ≈ kF ≈ 1

a(10.142)

which implies that our theory only describes wavelengths much larger thana. Moreover, the sums in(10.140) are bounded from above byΛ.

We have in our hands the tools to discuss the reason why perturbation theory fails for this problem.Notice that the operatorJ⊥ flips the pseudo-spin|+〉 to |−〉 and vice-versa. In this case the sign of theinteraction in theJz term changes from+ to −, that is, there is a sudden switch on of the interaction! Aslong as the system is in one of the eigenstates ofσz not much happens (there is a simple shift of the bosonmomentum) but once the pseudo-spin flips a potential is switched on and the ground state changes! Whenwe study time dependent perturbation theory we learn that the perturbation potential has to be switched onvery slowly so that states of the system do not change drastically. This is not the case here and this is whyperturbation theory fails.

In what follows We describe a way to treat this problem with a renormalization group (RG) method thatgives results beyond perturbation theory. The main idea is to use the fact that the harmonic oscillators withenergy close tovFΛ = W ≈ EF are very fast compared with the impurity spin. In fact, the flipping time ofthe spin can be estimated by the uncertainty principle to be:

tflip ≈h

J⊥≫ h

W(10.143)

since we assume thatJ⊥ ≪W ≈ EF . In this case, the fast (high energy) harmonic modes always “see” theimpurity spin as static (in the same sense that in the Born-Oppenheimer approximation the electrons “see”a static lattice of ions). Thus, the fast oscillators oscillate many periods before the spin flips and their onlyeffect is to “dress” the spin (in the same way that electrons “glue” the ions). In order to understand howthese high energy degrees of freedom “dress”, or more elegantly, renormalize the spin flips, let us considerthe effect of changing the cut-offΛ to Λ − dΛ (see Fig.10.8). Observe that this is equivalent to change theenergy scale fromW toW − dW (W = vFΛ).


W-dW

W

0

ω

ΛΛ−δΛ q

Figure 10.8: Dispersion of the particle-hole excitations and the RG procedure of reducing the cut-offΛ.

The state of the pseudo-spin can be now calculated in first order perturbation theory inλ as:

|+, 0〉 → |+〉R =1√N

|+, 0〉 + λσzΛ∑

q=Λ−dΛ

√q

ωqb†q|+, 0〉

=1√N

|+, 0〉 +λ√vF

Λ∑

q=Λ−dΛ

1√q|+, q〉

|−, 0〉 → |−〉R =1√N

|−, 0〉 − λ√vF

Λ∑

q=Λ−dΛ

1√q|−, q〉

(10.144)

where|±, 0〉 is the state without bosons and|±, q〉 is the state with one boson with momentumq such thatΛ − dΛ < q < Λ. The subscriptR indicates that this is the renormalized state andN is the normalizationfactor that can be easily calculated from the condition that〈±|±〉R = 1, that is,

N ≈ 1 +

(

λ

vF

)2 Λ∑

q=Λ−dΛ

1

q. (10.145)

In order to calculate the renormalized value of the couplingconstants, that is, the value of the couplingconstant when we reduce the cut-off we realize that the bare value of the coupling is given by:

J⊥ = 〈+, 0|H|0,+〉 (10.146)

as you can easily prove from (10.140). The renormalized value of the coupling,JR⊥ is then given by:

JR⊥ =1

N〈+|H|+〉R

=1 −

(

λvF

)2∑Λ

q=Λ−dΛ1q

1 +(

λvF

)2∑Λ

q=Λ−dΛ1q

≈ 1 − 2

(

λ

vF

)2 Λ∑

q=Λ−dΛ

1

q(10.147)


where in the last line we have used the fact that we are doing perturbation theory. The sum overq can beeasily performed:

Λ∑

q=Λ−dΛ

1

q= ln

(

Λ

Λ − dΛ

)

≈ dΛ

Λ. (10.148)

Furthermore, the value of the renormalized cut-off is (see Fig.10.8):WR = W −dW = W (1−dΛ/Λ) andtherefore we can write

JR⊥WR

=J⊥W

1 − αdΛΛ1 − dΛ

Λ

JR⊥WR

=J⊥W

(

1 + (1 − α)dΛ

Λ

)

(10.149)

where we have defined the constantα given by:

α = 2

(

λ

vF

)2

(10.150)

which, likeM andλ can be viewed as a undetermined function ofJz . Notice that (10.149) tells us how thedimensionless coupling constant

g⊥ =J⊥W

(10.151)

changes as we change the scale of the problem, that is,g⊥ = g⊥(Λ). Eq.(10.149) can be rewritten in theform:

g⊥(Λ − dΛ) = g⊥(Λ) (1 − (1 − α)d ln(1/Λ))

dg⊥dℓ

= (1 − α)g⊥(ℓ) (10.152)

where we have defined

ℓ = ln(Λ0/Λ) = ln(W0/W ) (10.153)

whereΛ0 is the bare value of the cut-off. Eq.(10.152) is the RG equation for the couplingJ⊥. This equationcan be solved as:

g⊥(ℓ) = g⊥(0)e(1−α)ℓ (10.154)

whereg⊥(0) = J⊥/W is the coupling constant of the problem at the original scaleΛ while g⊥(ℓ) is thecoupling at an arbitrary scale. From this result we immediately can conclude the following: ifα > 1 wehaveg⊥(ℓ → ∞) → 0 indicating that at longer and longer length scales (smallerΛ) the coupling vanishesand this term is irrelevant for the long wavelength physics;if α < 1 we haveg(ℓ → ∞) → ∞ and thecoupling constant becomes arbitrarily large as we reduce the cut-off indicating that the coupling is relevantand dominates the physics in the long wavelength limit. The case ofα = 1 is called marginal and the RGequations only have a contribution to orderg2

⊥.

Thus, the conclusion of this RG calculation is that perturbation theory is only reliable in the case ofα > 1 since the perturbation becomes weaker and weaker at low energies. In this case we can effectively


makeJ⊥ = 0 in the Hamiltonian and we see that the problem simply reducesto the problem discussedpreviously of the formation of the bound state (or scattering) out of a impurity. The state of the impurity isan eigenstate of theσz operator.

Forα < 1 the coupling becomes arbitrarily large and perturbation theory fails miserably and in this casewe have to find a better way to solve the problem (which usuallymeans that we have to solve the problemexactly!). We can make an educated guess of what happens in this case. Observe that the fact thatJ⊥becomes relevant in this limit indicates that this is the dominant term in the Hamiltonian. This implies thatthe state of the impurity is an eigenstate of theσx operator. Thus, we expect the ground state of the problemto be:

|s〉 =1√2

(|+〉R − |−〉R) (10.155)

which is the singlet state. This is the so-called Kondo singlet. Notice that our perturbation theory is onlywell defined forJ⊥ ≤ W , that is,g⊥ ≤ 1. Forα < 1 the couplingg⊥ grows under the RG until it reachesg⊥(ℓ∗) = 1 where, from (10.154):

ℓ∗ ≈ 1

1 − αln(1/g⊥(0)) (10.156)

Using (10.153) we see that this scale is associated with an energy scalekBTK such that:

ln(W0/kBTK) =1

1 − αln(W0/J⊥)

kBTK = W0

(

J⊥W0

)1/(1−α)

. (10.157)

For energies or temperatures aboveTK the coupling is small and the problem can be treated perturbativelybut for temperatures smaller thanTK the system flows to strong coupling and the Kondo singlet is formed.This temperature scale is called the Kondo temperature of the problem. Observe that forJ⊥ = 0 we haveTK = 0 and therefore no Kondo effect.

Thus we have found that the behavior of the problem changes dramatically when we go fromα > 1to α < 1. We have argued thatα has to be a function ofJz and therefore there must be a critical value ofJz for which the physics of the problem changes completely. Thequestion is: can we look at our originalHamiltonian (10.122) and understand howJz changes the physics of the Kondo problem? The answer isquite simple, in fact. Notice that the formation of a singletstate requiresJz > 0, that is, an antiferromagneticcoupling between the impurity and the electron gas. IfJz < 0, that is, if we had ferromagnetic couplingthe state of the system would be a triplet, that is,| ⇑, ↑〉 or | ⇓, ↓〉 (degenerate). In this case notice thattheJ⊥ term does not connect these two terms and therefore is not able to lift this degeneracy in first orderperturbation theory. Thus,J⊥ is irrelevant in this case. Thus, we can easily assign the case ofα > 1 withthe case ofJz < 0 while α < 1 for Jz > 0. Thus, the quantity1 − α is an odd function ofJz, that is,1−α(Jz) can be written as a power series expansion inJz with odd powers only. ForJz ≪W0 we expect,based on this argument that:

α(Jz) ≈ 1 − CJzW0

(10.158)

whereC is a constant independent ofJz. Thus, in this limit the Kondo temperature can be written, from


(10.157) as:

kBTK ≈W0

(

J⊥W0

)W0/(CJz)

= W0 exp

− ln(W0/J⊥)

CJz/W0

(10.159)

Notice that forJz = 0 we haveα = 1 and therefore from (10.157) we haveTK = 0 indicating that there isno Kondo effect in this case. These results show that in the Kondo effect both the Ising,Jz , as well as thetransverse,J⊥, couplings are fundamental for the occurrence of the Kondo effect.

In the case of the isotropic Kondo effect (J⊥ = Jz = J) the RG flow can be calculated by differenttechniques and one can show that:

dg

dℓ= g2 (10.160)

whereg = J/W . In this case the Kondo coupling is marginal (very similar tothe previous case withα = 1).We can solve this RG flow as:

g(ℓ) =g(0)

1 + g(0)ℓ(10.161)

which shows that whenℓ∗ = −1/g(0) the coupling constant diverges under the RG, indicating, aspreviouslythe failure of perturbation theory. From this result we obtain the Kondo temperature of the problem as:

kBTK = We−1/g(0) ≈We−W/J (10.162)

and therefore the Kondo temperature is exponentially smallwith the Kondo couplingJ .

The experimental consequences of the formation of the boundstates are immediate: the magnetic sus-ceptibility of the impurity, instead of following the Curielaw, χ ∝ 1/T has to saturate atT = TK sothat

χimp(T ) ∝ 1

TK; (10.163)

the entropy of the system, instead of beingS = kBN ln(2) as it would be for a spin1/2 atom has to goto zero belowTK because of the formation of the bound state, this implies that the specific heat,CV =TdS/dT , has to behave like

CV,imp ∝T

TK(10.164)

for T < TK . All these effects are observed at low temperatures (TK ≈ 1 − 5 K) in magnetic alloys.

Moreover, the Kondo effect leads to the so-called Kondo minimum in the electric resistance of thesesystems. On the one hand, formation of the virtual bound states or scattering resonances implies strongscattering (with the phase shifts close toπ/2 also known as the unitarity limit). The scattering increasesbelowTK and therefore one expects the resistivity to increase with decreasing temperatures. On the otherhand, phonon scattering decreases with decreasing temperature because as the system gets cooler there areless phonons in the system. The final result is therefore a minimum in the resistivity.

Another interesting property of the Kondo effect is its non-locality. SinceEF ≫ kBTK for TK ∼ T ≪EF /kB the only energy scale in the problem is the Kondo temperatureTK . The only information of theelectron gas that is important for this problem is the Fermi velocity,vF , since the states at the bottom of theFermi sea are irrelevant for the physics discussed here. Using the Fermi velocity and the Kondo temperature


we can construct a quantity with dimensions of length,ℓK , that is called the Kondo screening length:

ℓK =hvFkBTK

. (10.165)

The physical meaning of this quantity can be understood if wecalculate the average value of the correlationfunction between the local spinS and the electron spins(r), namely,

〈S · s(r)〉 ∼ e−r/ℓK (10.166)

which decays exponentially with distance from the impurity. Although this result has not been demonstratedhere it seems almost obvious when we realize that the only length scale in this problem is given by (10.165).This result shows that for distances larger thanℓK the electrons and the impurity are uncorrelated. Onlyelectrons in a region of sizeℓ3K around the impurity site participate in the Kondo screeningprocess. We canestimate the number of electrons,NK , as:

NK ∼ nℓ3K ∼ (kF ℓK)3 ∝(

EFkBTK

)3

(10.167)

where we used that the electronic densityn is essentially proportional tok3F and used (10.165) withEF ≈

hvF kF . SinceEF ≈ 1 eV andkBTK ≈ 10−4 eV we see thatNK ≈ 1012 electrons. That is, a huge numberof electrons participate in the screening process.

10.9.1 Problems

1. Consider the Hamiltonian for the Kondo problem (10.120) in its isotropic form (Jz = J⊥) and assumethatJ ≪ EF and that the electrons have a spherical Fermi surface.

(i) Using the first Born approximation, calculate the scattering amplitudet(1)(k, k′) for the scatteringof an electron from an state|k, ↑ to a state|k′, ↑ as a function ofJ andSz.

(ii ) Show that in the second Born approximation the scattering amplitude,t(2) is given by:

t(2)(k, k′) =∑

k′′,σ

1

Ek − Ek′

[

(1 − nk′′)〈k, ↑ |HK |k′′, σ〉〈k′′, σ|HK |k′, ↑〉

+ nk〈k′′, σ|HK |k′, ↑ 〈k, ↑ |HK |k′′, σ〉 . (10.168)

Herenk is the Fermi-Dirac occupation number. Notice that the orderof the matrix elements matters!Explain why.

(iii ) Show that the above expression can be approximately written as:

t(2)(k) ≈ 2 (J)2 Sz∑

k′′

nk′′

Ek − Ek′′

. (10.169)

Assuming a constant density of statesN(0) for the electrons calculate the sum above and show that itcan be approximately written as:

t(2)(k) ≈ 2 (J)2 SzN(0) ln(W/|EF − Ek|) (10.170)

whereW is the bandwidth. Notice thatt(2)(kF ) is divergent!

(iv) The total scattering by the impurity,〈t(2)〉 can be obtained by integrating (10.170) around the

10.10. ITINERANT MAGNETISM 201

Fermi surface in a region of energy of sizekBT . Show that in this case〈t(2)〉 is logarithmicallydivergent with the temperature.

(v) The total resistance generated by the presence of the impurity is written as: ρ(T ) = ρ0(1 −〈t(2)〉) whereρ0 is the temperature independent part of the resistance (see Chapter 7). If the phononcontribution to the resistance is given byρP (T ) = AT 5 show that this gives rise to a resistanceminimum in the behavior of the total resistivity of the material.

10.10 Itinerant Magnetism

So far we have discussed the problem of magnetism arising from impurities on a metallic host. In thesesystems there are two types of electrons: localized electrons coming from f or d shells and conduction bandelectrons from s or p shells. In many systems such as Fe, magnetism appears due to electrons in a singleatomic shell. In this case one cannot differentiate betweenthe electron that produces the magnetism and theelectron that participates in the conduction. Systems likethis are called itinerant. Our starting point is goingto be the Hubbard model discussed previously:

H =∑

k,σ

(ǫk − µ)c†k,σck,σ + U∑

i

ni,↑ni,↓ . (10.171)

Here we have introduced the chemical potentialµ and work in the canonical ensemble keepingµ fixed andallowing the number of electrons to fluctuate. The complete solution of (10.171) was only obtained in onedimension via the so-called Bethe ansatz. In higher dimensions one has to use approximate solutions. Inthis section we are going to treat the problem in the Hartree-Fock approximation. This is an uncontrolledapproximation. Although this kind of approach has its limitations, it can provide a deep insight on thephysics of the problem.

The complexity of (10.171) is associated with the treatmentthe interaction term. Because it involvesfour electron operators it is a highly non-linear problem. In mean field theory we replace the interactionterm by

ni,↑ni,↓ → 〈ni,↑〉ni,↓ + ni,↑〈ni,↓〉 − 〈ni,↑〉〈ni,↓〉 (10.172)

where the averages are evaluated in the ground state. Noticethat the factorization is done so that〈ni,↑ni,↓〉 =〈ni,↑〉〈ni,↓〉, that is, the fluctuations in the up and down spin channels areindependent.

It is convenient to rewrite the problem in terms of the total number of electrons per site and the magne-tization as

Ni = 〈ni,↑〉 + 〈ni,↓〉Mi = 〈ni,↑〉 − 〈ni,↓〉 (10.173)

or equivalently,

〈ni,↑〉 =Ni +Mi

2

〈ni,↓〉 =Ni −Mi

2(10.174)


in which case the mean-field Hamiltonian is given by

HMF =∑

k,σ

(ǫk − µ)c†k,σck,σ +

U

2

∑

i

[Ni(ni,↑ + ni,↓) −Mi(ni,↑ − ni,↓)]

− U

4

∑

i

(N2i −M2

i ) . (10.175)

In what follows we are going to assume that state is always homogeneous and therefore

Ni = n =Ne

N(10.176)

whereNe is the total number of electrons andN the total number of sites. In this case we can rewrite(10.175) as

HMF =∑

k,σ

(ǫk − µ)c†k,σck,σ +UNn2

4+Un

2

∑

i

(ni,↑ + ni,↓)

+U

4

∑

i

M2i − U

2

∑

i

Mi(ni,↑ − ni,↓) (10.177)

where the second term in the first line on the r.h.s. of (10.177) is a total energy shift that can be absorbed intothe definition of the chemical potential, the third term is just a shift of the electron energy, the first term onthe second line on the r.h.s. is the energy required to magnetize the system and the last term is the interactionbetween the magnetization and the electron themselves. This term has the form of a local applied magneticfield. Observe that in this problemMi is an unknown and has to be calculated self-consistently. Asusual inmean field approaches one has to guess the nature of the groundstate, that is, the form of the magnetization.In a ferromagnetic system all spins point in the same direction and thereforeMi = M0 for all sites. In anantiferromagnetic system the situation is more complicated because one can have different sub-lattices inwhich all the spins are up or all the spins are down. For instance, for a square lattice the magnetization canbe given by

Mi,j = M0(−1)i+j (10.178)

where each site is located at(ai, aj) wherea is the lattice spacing. This classical state of the electronsystemis called the Néel state and it is shown on Fig. 10.9. Observe that although the magnetization changes frompoint to point in space the Fourier transform of (10.178) is quite simple

M(q) ∝ δ(q − Q) (10.179)

whereQ = (π/a, π/a) is the so-called ordering vector. For a ferromagnet we haveQ = 0.

Here we are going to consider only the case of the ferromagnetand leave the antiferromagnet as aproblem for the reader. In the ferromagnetic case we can rewrite (10.177) as

HMF =∑

k,σ

(ǫk − µ)nk,σ +NUM2

0

4+UM0

2

∑

i

(ni,↑ − ni,↓)

=NUM2

0

4+∑

k

[(

ǫk − µ− UM0

2

)

nk,↑ +

(

ǫk − µ+UM0

2

)

nk,↓

]

(10.180)


x

y

a

a

Figure 10.9: Two-dimensional Néel state.

where we have used∑

i ni,σ =∑

k nk,σ. Observe that (10.180) is already in diagonal form since it onlydepends on the occupation in momentum space. The total energy only depends on the occupation of eachspin state, that is, in the ground state

〈nk,σ〉 = Θ(EF,σ − ǫk) (10.181)

whereEF,σ is the Fermi energy of each spin state and at this point it is anunknown. Moreover,M0 in(10.180) and it is also unknown. In order to calculate this parameters one has to minimize the total energyof the system as a function of them. The energy of the system iswritten as:

E =NUM2

0

4+∑

k

[(

ǫk − µ− UM0

2

)

Θ(EF,↑ − ǫk)

+

(

ǫk − µ+UM0

2

)

Θ(EF,↓ − ǫk)

]

(10.182)

As usual in this kind of problems it is useful to introduce theelectronic density of states

N(ǫ) =1

V

∑

k

δ(ǫ− ǫk) (10.183)

in which case the energy densityE = E/V becomes

E =ρsUM

20

4+

∫ EF,↑

−∞dǫ

(

ǫ− µ− UM0

2

)

N(ǫ)

+

∫ EF,↓

−∞dǫ

(

ǫ− µ+UM0

2

)

N(ǫ) (10.184)


whereρs = N/V is the solid density. In order to calculate the unknown parameters in the energy densitywe have to minimize it so that

∂E∂µ

= −ρ

∂E∂EF,σ

= 0

∂E∂M0

= 0 (10.185)

whereρ = Ne/V is the electron density. Furthermore, in order to make sure that the solution is really aminimum and not a maximum one has calculate the second derivatives of the energy with respect to thevariational parameters. The reader can do it as an exercise.From (10.185) one finds

∫ EF,↑

−∞dǫN(ǫ) +

∫ EF,↓

−∞dǫN(ǫ) = ρ

(

EF,σ − σUM0

2− µ

)

N(EF,σ) = 0

ρsM0 −∫ EF,↑

−∞dǫN(ǫ) +

∫ EF,↓

−∞dǫN(ǫ) = 0 . (10.186)

We can first solve forµ andM0 using the two equations for the spins:

µ =EF,↑ + EF,↓

2

M0 =EF,↑ − EF,↓

U(10.187)

which can be now substituted into the other two equations:

∫ EF,↑

−∞dǫN(ǫ) +

∫ EF,↓

−∞dǫN(ǫ) = ρ

∫ EF,↑

EF,↓

dǫN(ǫ) = ρsEF,↑ − EF,↓

U(10.188)

which is the set of equation with defineEF,↑ andEF,↓. After these are calculated one can obtain the chemicalpotential and magnetization from (10.187). Using (10.187)we can rewrite (10.188) as

ρ = 2

∫ EF,↑

−∞dǫN(ǫ) + ρs

EF,↑ − EF,↓U

ρsEF,↑ − EF,↓

U=

∫ EF,↓

EF,↑

dǫN(ǫ) . (10.189)

Observe that the last equation in (10.189) can be rewritten as

ρsEF,↑ − EF,↓

U=

∫ EF,↑−EF,↓

0dǫN(ǫ+ EF,↓) (10.190)

which for fixedEF,↓ is a single equation forEF,↑ − EF,↓. Since the density of states has a finite bandwidth


the integral in (10.190) is a monotonic function ofx = EF,↑ − EF,↓ and saturates at large valuesx asshown in Fig. 10.10. The solution of (10.190) is the intercept of a straight line of slopeρs/U and ther.h.s. of (10.190). Observe that there is a critical value ofU , say,UC , above which one finds a solutionwith x = x∗ 6= 0 and therefore from (10.187) corresponds to a system with finite magnetization, that is, aferromagnetic state. ForU smaller thanUC the only solution isx = 0 which is the paramagnetic state. Inorder to calculateUC one has to linearize the r.h.s. of (10.190) in order to get

UC =ρs

N(EF,↓)(10.191)

whereEF,↑ is determined by the first equation in (10.189):

ρ = 2

∫ EF,↑

−∞dǫN(ǫ) (10.192)

which completes the solution of the problem atU = UC . The condition for the existence of a ferromagneticstate forU larger thanUC as given in (10.191) is called Stoner criterion. This criterion, which is based ona mean field approach, is only a estimate of how largeU has to be (or how large the density of states has tobe) in order for the system to have a spontaneous ferromagnetic magnetization. As we are going to see inthe next chapter the mean field solution is an extreme approximation that has its limitations. It provides us,however, with the information that the Coulomb interactionis fundamental for the creation of an orderedmagnetic state.

xx*

U>UU=UU<U

CC

C

0

Figure 10.10: Diagrammatic solution of (10.190): solid line is the r.h.s. of (10.190).

10.10.1 Problems

1. Consider the problem of itinerant magnetism in an antiferromagnet in a square lattice in two dimen-sions in which the magnetization is given in (10.178). This can be rewritten as

M(r) = M0 cos(Q · r)

whereQ = (π/a, π/a).


(i) Show that for a square lattice the electron dispersion relation in the tight binding approximation isgiven by

ǫk = −2t (cos(kxa) + cos(kya)) .

And therefore show thatǫk = −ǫk+Q.

(ii ) Make a plot of the Fermi surface in momentum space, that is,(kx, ky) at half filling (ǫk = 0).Show that the opposite sides of the Fermi surface can be linked by the vectorQ. Observe that thismeans that the symmetry of the problem has been broken and nowstates which differ by a vectorQ are actually the same state. Explain what symmetry has been broken and show that the magneticBrillouin zone of the problem is the Fermi surface you just plotted.

(iii ) Write down the mean field Hamiltonian for the problem and show that it depends onck,σ andck±Q,σ. Define the following spinor:

[Ψσ(k)] =

(

ck,σck+Q,σ

)

and show that the Hamiltonian of the problem can be written as

H =UM2

s

4+∑

k,σ,σ′

[Ψσ(k)]†[

Hσ,σ′]

[Ψσ′(k)]

where

[

Hσ,σ′]

=

[

ǫkUMs

2 τzUMs

2 τz −ǫk

]

whereMs is the staggered magnetization andτz is a Pauli matrix.

(iv) Diagonalize the above Hamiltonian and find the new electronic spectrum. Show that a gap opensin the spectrum at the Fermi surface. Interpret your result in terms of the symmetry of the problem.

(v) Calculate the total energy of the problem by filling up all the energy states below the gap (sincethe system is at half filling) and by minimizing it find the equation that determinesMs.

(vi) In this item we are going to use the Debye theory for phonons in the context of the antiferromagnet.Expand the spectrum around the(π/a, π/a) point to second order in the momentum and perform theintegral of item

(v). Note that you have to introduce a cut-off in the upper limitof the integral so it converges. Thiscut-off, sayΛ, can be calculated by using Debye’s theory for phonons. Whatis the functional form ofMs as a function ofU andt.

Chapter 11

Magnetic phase transitions

In the last chapter we discussed the origins of magnetic interactions in solids and how they lead to magneticorder. At high temperatures magnetic atoms behave independently from each other leading to the param-agnetic behavior. At low temperatures the atoms can take advantage of magnetic exchange interactions andlower their energy by ordering in a some specific form that depends on the particular character of the inter-actions. The two main types of ordering are ferromagnetic and antiferromagnetic, depending if the systemorders with its magnetic moments parallel or anti-parallelto each other.

In this chapter we focus on the problem of magnetic ordering and its consequences to the magneticbehavior of solids. When the system orders magnetically we say that long range order has been established.When a system attains long range order the internal symmetryof the system is usually lowered. Thus,with the decreasing of the temperature there is a spontaneous lowering of the symmetry and thus a phasetransition: a transition between a highly symmetric phase at high temperatures and a low symmetry phaseat low temperatures. The study of the phase transition has reached a great deal of development with theintroduction of the concept of renormalization group and universality. These are two of the most importantconcepts in modern condensed matter physics.

11.1 Spontaneous symmetry breaking

In order to understand how a phase transition occurs in a magnetic system we consider for instance theferromagnetic Ising model,

H = − 1

N

N∑

i,j=1

JijSzi S

zj (11.1)

whereJij > 0. Observe that the Hamiltonian (11.1) has a symmetry for the reversal of the spins, that is, wecan reverse all the spins and the Hamiltonian remains the same. Mathematically the Hamiltonian is invariantunderSzj → −Szj . The question here is: is the ground state of the (11.1) invariant under this symmetry?Firstly let us assume that the ground state is invariant under the overturn of spins. We can specify any stateof the system in the terms of the eigenstates ofSz, that is,

|0〉 = |Sz1 , Sz2 , ..., SzN 〉 . (11.2)

If this states is invariant under the overturn of spins we certainly have

| − Sz1 ,−Sz2 , ...,−SzN 〉 = |Sz1 , Sz2 , ..., SzN 〉 . (11.3)

207

208 CHAPTER 11. MAGNETIC PHASE TRANSITIONS

It follows immediately that

〈Sz1 , ..., SzN |Szj |Sz1 , ..., SzN 〉 = 〈−Sz1 , ...,−SzN |Szj | − Sz1 , ...,−SzN 〉= −〈Sz1 , ..., SzN |Szj |Sz1 , ..., SzN 〉 (11.4)

from which we have to conclude

〈0|Szj |0〉 = 0 . (11.5)

Thus, if the ground state has the symmetry of the Hamiltonianthen the average magnetic moment is zero. Itis possible, however, that the ground state does not have thesymmetry of the Hamiltonian, that is, it is notinvariant underSzj → −Szj . In this case we cannot conclude (11.5) and we can have that〈0|Szj |0〉 6= 0. Ifthis happens the symmetry of the Hamiltonian is broken and the system acquires finite magnetic moment.In a system without disorder, that is, a system with the translation symmetry of the lattice, all the spinsare equivalent and therefore it does not make sense to speak of a particular spin. In this case we define anaverage quantity over all spins

M =1

N

N∑

j=1

〈0|Szj |0〉 (11.6)

whereM 6= 0 in the phase with broken symmetry.M is the magnetization of the system and it can be finitebecause the spins interact ferromagnetically so they can align along a given direction. If the ground state ofthe system is such thatM 6= 0 then we can say that the symmetry is spontaneously broken since the groundstate does not have the symmetry of the Hamiltonian. Observe, however, that we have two possibilities:either all the spins point in the up direction or they point inthe down direction. For these two states thesymmetry is broken and they are degenerate since they have the same energy. This degeneracy can be liftedby applying a magnetic field to the system. In the presence of an infinitesimal field the degeneracy of thesetwo states is lifted and the system picks the state with lowest energy. We say that the external magnetic fieldis the symmetry breaking field.

At high temperatures we expect the magnetic moments to act independently, that is, paramagnetically. Atthese temperatures we would haveM = 0 since the spins point randomly. If the system at zero temperaturehasM 6= 0 then there is a critical temperatureTc such that aboveTc the system is disordered, that, isM(T > Tc) = 0, and above which the system is ordered, that is,M(T < Tc) 6= 0. At T = Tc we havea phase transition. Since the magnetization changes from zero to a finite value at the transition we say thatM(T ) is the order parameter of the magnetic problem. Moreover we have two possibilities atT = Tc: i)the magnetization can jump discontinuously fromM(T = Tc + ǫ) = 0 toM(T = Tc − ǫ) 6= 0 atT = Tc(ǫ→ 0) in which case we say that the transition is of first order as shown in Fig.11.1(a);ii ) the magnetizationcan go continuously fromM(T = Tc + ǫ) = 0 toM(T = Tc − ǫ) 6= 0 atT = Tc in which case we say thatthe transition is of second order as shown Fig.11.1(b). Observe that the fact that the magnetization changesatT = Tc implies that the magnetic susceptibility, that is,

χ(T ) =

(

∂M

∂B

)

B→0

(11.7)

diverges atT = Tc. In order to understand why this is so, consider the plot of the magnetization as afunction of the magnetic field at fixed temperature as shown inFig.11.2. AtT > Tc the magnetization isjust paramagnetic and is zero atB = 0. If T < Tc one hasM(T ) 6= 0 for B = 0. Thus, atT = Tc, themagnetization has infinite curvature implying a divergent susceptibility.

11.1. SPONTANEOUS SYMMETRY BREAKING 209

M

TTcM

TTc

M0

M0

(a)

(b)

Ordered Phase

Disordered Phase

Ordered Phase

Disordered Phase

Figure 11.1: Magnetization of a system during a phase transition: a) First order; b) Second order.

Although we have talked about ferromagnets the concept of broken symmetry, order parameter andsymmetry breaking field can be generalized to any system withlong range order. As an example considerthe case of an antiferromagnet such as the one shown on Fig. 10.9. Observe that in this case the systemindeed has〈Szj 〉 6= 0 at each site but the total magnetization as defined in (11.6) vanishes even in the orderedphase. This is because the homogeneous magnetization is notthe order parameter of the antiferromagnet.Let us look at the magnetization of an antiferromagnet as theone in Fig. 10.9. It is easy to see that the localmagnetization at positionr = (n,m)a is

M(r) = Ms(−1)n+m (11.8)

whereMs is a constant. Observe that the magnetization oscillates from site to site and therefore vanishes ifwe sum over all sites. However, it is clear from the equation above thatMs has to be finite in the orderedphase. We can obtainMs from (11.8) if we multiply both sides by(−1)n+m and sum over all sites:

Ms =1

N

∑

r

(−1)n+mM(r) =1

N

N∑

j=1

(−1)j〈Szj 〉 (11.9)

which is called the staggered magnetization of the system. Furthermore, observe that the ferromagnetic stateretains the symmetry of the lattice while in the antiferromagnetic case the unit cell doubles (one needs atleast one spin up and one spin down in the same unit cell) and then the translational invariance of the latticeis also broken. Consider Fourier transforming the local magnetization in each case. For a ferromagnet thelocal magnetization is uniform,M(r) = M0, and therefore its Fourier transform has Dirac delta peaks atk = 0 and all other reciprocal lattice vectorsG. Thus the ferromagnet has the symmetry of the lattice. In theantiferromagnetic case as given in (11.8) the Fourier transform has Dirac delta peak atQ = (π/a, π/a) asyou can easily show. Finally, the symmetry breaking field in this case is not an homogeneous magnetic field


M

M

−M

0

0

T>Tc

T<Tc

T<Tc

B0

Figure 11.2: Magnetization as a function of external field atfixed temperature.

since it cannot differentiate between the two degenerate states of the antiferromagnet (which are obtained byflipping all the spins in the system). In the antiferromagnetic case a staggered field is the symmetry breakingfield.

11.1.1 Critical exponents

In the case of second order phase transitions the order parameter goes smoothly to zero at the critical tem-perature which implies that the magnetization has to behavelike

M(T ) ∝ (Tc − T )β (11.10)

for T very close (and smaller) thanTc. β is known as a critical exponent. The magnetic susceptibility isgiven by

χ(T ) ∝ (T − Tc)−γ (11.11)

for T > Tc and

χ(T ) ∝ (Tc − T )−γ′

(11.12)

for T < Tc since the susceptibility has to diverge on both sides of transition but not necessarily with thesame exponents. Another exponents are defined for other physical quantities. For instance, the specific heatat fixed field also diverges atT = Tc and one has

CH(T ) ∝ (T − Tc)−α (11.13)

for T > Tc and

CH(T ) ∝ (Tc − T )−α′

(11.14)

11.1. SPONTANEOUS SYMMETRY BREAKING 211

for T < Tc. Another interesting property is that the magnetization does not have to be linear with themagnetic field atT = Tc and we define another exponentδ by

M ∝ |B|1/δsgn(B) . (11.15)

Besides the critical exponents that define the thermodynamic functions there are exponents that definedynamical correlations. Correlation functions are important for experiments that measure spatial and tem-poral correlations among spins. In the phonon problem we showed that the dynamical form factorS(k, ω)is given in terms of a density-density correlation function. Remember that the neutrons interact with thesystem of interest via the spin-spin interaction. Thus, neutron scattering is very sensitive to magnetic order.Like in the case of scattering of neutrons by phonons we wouldhave Bragg scattering of the neutrons belowTc when the system has long range order. Observe that as in the case of atoms in a solid the Bragg peaksare directly related with the magnetic ordering in the system. In the case of the ferromagnet the magneticordering does not break the symmetry of the lattice and in thecase where the ordering is commensuratewith the lattice the Bragg peaks are at the reciprocal lattice vectorsG. Thus, in a ferromagnet magneticpeaks superimpose to lattice peaks. This is not the case of anantiferromagnet where a Bragg peak appearsatQ which is not a reciprocal lattice vector. As in the case of phonons, perfect order implies the presence ofinfinitely sharp Bragg peaks. In the magnetically disordered phase we do not expect these peaks to disappearimmediately but to become broader due to lack of long range order. In momentum space this broadening isgiven by1/ξ whereξ is called the magnetic correlation length. In order to understand the meaning of thecorrelation length let us assume for simplicity that the shape of the peak (or intensity) in the magneticallydisordered phase is given by a Lorentzian shape

I(k) =I0/ξ

(k − Q)2 + 1/ξ2(11.16)

whereQ is the ordering vector (Q = 0 for the ferromagnet andQ 6= 0 for the antiferromagnet) andI0 isa constant. Observe that whenξ → ∞ the Lorentzian in (11.16) goes to a Dirac delta function atk = Q

as expected in the case of complete order. Thusξ is a function of temperature and diverges atT = Tc.Moreover, it is easy to show that the intensity is proportional to the spin-spin correlation function for spinsat different sites,〈Szi Szj 〉. In order to see how the correlation function behaves one hasto Fourier transform(11.16) to real space. It is a simple exercise to show that

I(ri − rj) ∝ 〈Szi Szj 〉 ∝ exp

−|ri − rj |ξ(T )

. (11.17)

Now the meaning of the term correlation length is clear: it tells what is the characteristic distance abovewhich the spins are uncorrelated. AtT = Tc one expects the spins to be strongly correlated andξ(T )diverges. One then defines the critical exponents forξ(T ) as

ξ(T ) ∝ (T − Tc)−ν (11.18)

whenT > Tc andB = 0 and

ξ(T ) ∝ (Tc − T )−ν′

(11.19)

whenT < Tc andB = 0. In some cases the correlation function decays algebraically with the distance (in


which case the neutron scattering intensity is not a Lorentzian)

〈Szj Szj 〉 ∝1

|ri − rj |d−2+η(11.20)

with a new exponentη for a system ind dimensions.

It turns out that these critical exponents are not independent of each other because of the thermodynamicrelations between the various quantities. This is related with the so-called scaling hypothesis. Consider thefree energyF (T,B) which we parameterize in terms of the reduced temperature

t =T − TcTc

. (11.21)

The scaling hypothesis asserts that, close to the phase transition, it is always possible to find two parametersat andaB such that

F (λatt, λaBB) = λF (t, B) (11.22)

for any value of the real numberλ.

In order to understand how the exponents are relate let us differentiate (11.22) with respect toB in orderto get the magnetization:

λaB∂F (λatt, λaBB)

∂(λaBB)= λ

∂F (t, B)

∂B

λaBM(λatt, λaBB) = λM(t, B) . (11.23)

Consider the case whereB = 0 andt→ 0. From (11.23) we have:

M(t, 0) = λaB−1M(λatt, 0) (11.24)

that is valid for anyλ. In particular for

λ =

(

−1

t

)1/at

(11.25)

we find

M(t, 0) = (−t)1−aB

at M(−1, 0) . (11.26)

But from (11.10) we have in the limit oft→ 0−

β =1 − aBat

. (11.27)

Let us now taket = 0 and letB → 0 in (11.23)

M(0, B) = λaB−1M(0, λaBB) (11.28)

and chooseλ = B−1/aB in order to get

M(0, B) = B1−aB

aB M(0, 1) (11.29)

11.2. MEAN FIELD APPROACH 213

which by direct comparison with (11.15) leads to

δ =aB

1 − aB. (11.30)

Similarly we can derive twice with respect toB in order to obtain

λ2aBχ(λatt, λaBB) = λχ(t, B) . (11.31)

Taking the limit ofB = 0 first and choosingλ = (−t)−1/at we can prove

γ′ =2aB − 1

at(11.32)

and using again (11.31) withλ = t−1/at one finds

γ′ = γ (11.33)

implying that the susceptibility diverges with the same exponents on both sides of the transition. Moreover,combining (11.32) with (11.27), (11.30) and (11.33) one finds

γ = β(δ − 1) (11.34)

which relates three different exponents.

Using exactly the same arguments for derivatives with respect to the temperature we can show that

α′ + β(δ + 1) = 2 (11.35)

and

α = α′ (11.36)

and combining (11.35) with (11.34) one finds

α+ 2β + γ = 2 . (11.37)

Many other relations between the exponents can be obtained in this way. Although the scaling hypothesis isan assumption and cannot be proved, it makes many predictions for critical exponents that can be checkedexperimentally or analytically for specific models. This hypothesis has been very successful in explaincritical behavior in systems with second order phase transitions.

11.2 Mean field approach

In order to calculate the exponents that define the critical behavior of a magnetic systems we have severaloptions. One of them is to solve the problem exactly. This is amajor task since the problems at hand are verycomplicated. The other approach is to use mean field theories. In mean field theories we look at a particularspin and try to replace all the other spins in the problem by a the average field of the other spins. We havediscussed this procedure in the case of itinerant magnetic systems at the end of Chapter 8. Here we are goingto discuss it in the context of localized magnetism. Let us goback to the problem of the ferromagnetic Isingproblem in a magnetic field. We will assume that the interaction occurs only among next nearest neighbors.


The Hamiltonian given by

H = − J

N

∑

〈i,j〉

Szi Szj − gµB

∑

i

Szi B . (11.38)

In the mean field approach we replace the interaction by

Szi Szj → 〈Szi 〉Szj + 〈Szj 〉Szi . (11.39)

Moreover, we assume that the magnetization is uniform so that 〈Szi 〉 = M in which case Hamiltonian(11.38) is replaced by

HMF = −gµB∑

i

BeffSzi (11.40)

where

Beff = B + λM(T,B) (11.41)

is the effective magnetic field,Z is the number of nearest neighbors andM(T,B) is the magnetization givenin (11.6) and

λ =2ZJ

(gµB)2(11.42)

is the so-called molecular field. Observe that (11.40) is theHamiltonian of a set of independent spinsS in amagnetic field. This problem was studied for paramagnets in Chapter 8. In particular the magnetization ofthis problem is given in (1.37),

M(T,B) = M0BS[βgSµBBeff ]

= M0BS[βgSµB (B + λM(T,B))] (11.43)

whereM0 = gµBS. Notice that (11.43) has to be solved self-consistently to give the magnetization. ForB = 0 this equation reduces to

M = M0BS [βgSµBλM ] (11.44)

which is shown graphically on Fig.11.3. Observe that this equation has the trivial solutionM = 0 for all T .This equation has also a non-trivial solutionM 6= 0 if the slope of the straight line is larger than the slopeof Brillouin function. Thus let us consider the Brillouin function close toM = 0. The expansion ofBS(x)for smallx is

BS(x) ≈ S + 1

3Sx− (S + 1)(1 + 2S + 2S2)

90S3x3 (11.45)

and therefore there is a non-trivial solution for temperaturesT < Tc. In order to calculateTc we use (11.44)to find where the slopes in both sides of (11.44) equal each other. That is,

1 = gµBSS + 1

3S

gµBSλ

kBTc(11.46)

11.2. MEAN FIELD APPROACH 215

and therefore (11.42) we have

Tc = λC =2ZJS(S + 1)

3kB(11.47)

where

C =(gµB)2S(S + 1)

3kB(11.48)

is the Curie constant.

-4 -2 2 4M

-1.5

-1

-0.5

0.5

1

1.5

B

Figure 11.3: Graphical solution of eq. (11.44)

For simplicity let us consider the case of aS = 1/2 spin in which the Brillouin function reduces toB1/2(x) = tanh(x/2) and (11.43) reduces to

M = M0 tanh (βgµB(B + λM)/2) . (11.49)

In order to simplify our calculations we define

m =M

M0

τ =T

Tc(11.50)


and (11.49) becomes

m = tanh

(

gµBBβ

2+m

τ

)

(11.51)

which can be rewritten with the help of trigonometric identities as

b =m− tanh(m/τ)

1 −m tanh(m/τ)(11.52)

where

b = tanh

(

gµBBβ

2

)

. (11.53)

Let us consider the case thatT ≈ Tc wherem ≈ 0 and expand (11.52) as

b ≈ m

(

1 − 1

τ

)

+m3

[

1

3τ3+

1

τ

(

1 − 1

τ

)]

. (11.54)

In zero field,b = 0, (11.54) gives

m2 ≈ 3τ2

(

1

τ− 1

)

∝ (Tc − T ) (11.55)

which can be immediately compared with (11.10) to give

βMF =1

2. (11.56)

At the critical temperature, that is,τ = 1, and small fields one finds

gµBB

2kBTc≈ m3

3(11.57)

which, by comparison with (11.15), gives

δMF = 3 . (11.58)

In order to calculate the susceptibility we observe that

χ(T ) =∂M

∂B=∂M

∂m

∂m

∂b

∂b

∂B

=CT

∂m

∂b. (11.59)

If we differentiate (11.54) with respectb we get

∂m

∂b≈[(

1 − 1

τ

)

+m2

τ3

]−1

(11.60)

11.3. MEAN FIELD AND BEYOND 217

and from (11.59) one finds

χ(T ) ≈ CT

[(

1 − 1

τ

)

+m2

τ3

]−1

. (11.61)

In the disordered phase (T > Tc) we havem = 0 and therefore

χ(T ) ≈ CT − Tc

(11.62)

which from (11.11) gives

γMF = 1 (11.63)

while in the ordered phase (T < Tc) we use (11.55) and find

χ(T ) ≈ C2T

1

Tc − T(11.64)

which from (11.12) we find

γ′

MF = 1 . (11.65)

Observe that the scaling hypothesis predicts from (11.35) and (11.36) thatαMF = α′

MF = 0 which impliesthat the specific heat does not diverge at the transition.

11.3 Mean field and beyond

Observe that the mean field theory has no small parameter and cannot be explained in terms of a perturbativeexpansion. In order to understand better the nature of the mean field let us consider the problem in the basisof Szi (Szi |σi〉 = σi|σi〉 whereσi = ±1). In this case the energy of the problem depends on the configurationσi and from (11.38) we have

E(σi, B) = −∑

i,j

Ji,jσiσj −∑

i

σiB . (11.66)

where we assume thatJi,j only depends on the relative distance|ri−rj | between the two spins. The partitionfunction of the problem is given by

Z[B] =∑

σi

e−βE(σi,B) (11.67)

where the sum is over all the spin configurations. Observe that (11.67) is quite useful since the magnetizationat zero field of the system is

M =

∑

σi

∑

j σje−βE(σi,B=0)

∑

σie−βE(σi,B=0)

=1

βZ[B = 0]

(

∂Z[B]

∂B

)

B=0

. (11.68)


We will rewrite (11.67) in a different way using a well known identity

∫ +∞

−∞dxe−x

2/(4a2)+sx =√

2πa ea2s2 . (11.69)

Using (11.66) in zero field (B = 0) and the above identity we can write

Z =∑

σi

eP

i,j Ki,jσiσj

∝∑

σi

∫ ∞

−∞

N∏

j=1

dφje− 1

4

P

i,j φiJ−1i,j φj+

P

i φiσi

∝∫ ∞

−∞

N∏

j=1

dφje− 1

4

P

i,j φiJ−1i,j φj

∑

σi

eP

i φiσi (11.70)

whereKi,j = βJi,j . We now observe that

∑

σi

eP

i φiσi =∏

i

2 cosh(φi) = 2eP

i ln(cosh(φi)) . (11.71)

Moreover, we define,

ψi =∑

j

1

2J−1i,j φj (11.72)

and the partition function becomes

Z ∝∫ ∞

−∞

N∏

j=1

dψje−

P

i,j ψiKi,jψj+P

i ln(cosh(2P

j Ki,jψj)) . (11.73)

Observe that what we have done was to trade a sum over discretevariables for an integral over a continuousvariable. But in doing so we have generated a non-linear termin the partition function.

11.3.1 Interactions with infinite range

Let us now consider the “artificial" case of interactions with infinite range. In this case we have

Ji,j =2J

N

Ki,j =K

N=

2Jβ

N. (11.74)

Observe that in this case the partition function in (11.73) simplifies considerably since we can change vari-ables to a “center of mass" variable

ψ =1

N

∑

i

ψi (11.75)


in which case the partition function can be written as

Z ∝∫ +∞

−∞dψ e−NKψ2−ln(cosh(2Kψ)) (11.76)

which is a simple integral but with a very complicated integrand. Notice that the exponent in (11.76) scaleswith N the number of spins in the system. But we are interested in thelimit of N → ∞ in which case theintegral is dominated by the saddle point value of the integral that is, the minimum of the function

S(ψ) = Kψ2 − ln (cosh(2Kψ)) (11.77)

which is plotted of Fig.11.4. The minimum of this function isgiven by

∂S

∂ψ= 0

ψ = tanh(2Kψ) (11.78)

which gives the value ofψ. Observe that (11.78) is identical to (11.49) withψ ∝ M . In particular, ifK < 1/2 (T > Tc = 2J/kB) the only solution isψ = 0 and forK > 1/2 (T < Tc = 2J/kB) asolution withψ 6= 0 exists. This simple calculation shows that mean field theoryis equivalent to solve theproblem with interactions with infinite range! Thus, we should suspect that in the real case with short rangeinteractions fluctuations around the mean field solution aregoing to be very important.

In order to understand the how the phase transition occurs let us consider the free energy for the orderparameter as given by (11.77). Let us expandS(ψ) to fourth order inψ (ln cosh x ≈ x2/2 − x4/12):

S(ψ) ≈ K(1 − 2K)ψ2 +4K4

3ψ4 . (11.79)

Observe that the quadratic term changes from negative to positive whenK > 1/2 toK < 1/2. Thereforethe curvature of the free energy atψ = 0 changes atK = 1/2. In this case the ground state of the systemfor K < 1/2 which is given byψ = 0 becomes metastable atK = 1/2. ForK > 1/2 the state withψ = 0 becomes unstable and the ground state is given by the two solutions of (11.78) which are degeneratereflecting the symmetry of the problem. It is clear that the full form of (11.77) is not needed. The physics ofthe problem is already present in (11.79). The term proportional toψ4 is only need to keep the free energybounded from below in the ordered phase. Without this term the system would be thermodynamic unstablesince the ground state would be given byψ → ±∞ with infinite negative energy whenK > 1/2 which isclearly unphysical. In the disordered phaseK < 1/2 the free energy is naturally bounded and the term inψ4 is irrelevant for the physics.

11.3.2 Local interactions

Naturally the interactions between spins are short-ranged. In this case the previous approach is not appro-priate. IndeedKi,j is really localized close tori = rj instead. The best way to realize this is to look at theFourier transform ofKi,j,

K(r) =1

N

∑

k

eik·rK(k) . (11.80)

On the one hand, ifK(r) has infinite range thenK(k) is a Dirac delta function atk = 0. On the other hand,if K(r) is localized thenK(k) is a smooth function ofk. Therefore, if we work in momentum space we can


-4 -2 2 4y

0.5

1

1.5

2

2.5

3

3.5

S y

Figure 11.4: Plot of eq. (11.77): dashed line:K < 1/2; continuous line:K > 1/2.

always work with functions that can be expand in power series. So let us introduce the Fourier transform ofψi,

ψ(ri) =1√N

∑

k

eik·riψ(k) . (11.81)

Moreover, becauseK(r) andψ(ri) are real functions we must haveK(−k) = K∗(k) andψ(−k) = ψ∗(k).

As we explained in the case of the interactions with infinite range interactions we do not need to workwith the full free energy (11.73). Instead we are going to usea truncated form of the free energy

F ≈∑

i,j

ψiKi,jψj − 2∑

i

∑

j

Ki,jψj

2

+4

3

∑

i

∑

j

Ki,jψj

4

. (11.82)

For the moment being we are going to focus on the quadratic terms. It is very easy to show with the use ofthese Fourier transforms the following relations

∑

i,j

ψiKi,jψj =∑

k

K(k)ψ(k)ψ(−k)

∑

i

∑

j

Ki,jψj

2

=∑

k

K(k)K(−k)ψ(k)ψ(−k) . (11.83)

Thus the quadratic part of (11.82) can be written as

F0 =∑

k

(

K(k) − 2|K(k)|2)

|ψ(k)|2 . (11.84)


Observe that if the interaction was a Dirac delta function inreal space (local interaction) thenKk) shouldbe a constant. Let us assume thatK(k) is a real function of its argument. Then (11.84) as

F0 =∑

k

K(k) (1 − 2K(k)) |ψ(k)|2 . (11.85)

This equation is very important because it tells us that if there is a wave-vectorQ for which

K(Q) >1

2(11.86)

then the system is unstable at that particular wave-vector.Observe that this condition implies that the phasetransition happens to the componentψ(Q). Sinceψ is essentially the magnetization of the system is impliesthat the order parameter of the system is given by

M(r) = M0ℜ(

eiQ·r)

. (11.87)

This is in complete agreement with our previous discussion.

Let us now go back to the ferromagnetic problem and study whathappens at the unstable wave-vector,that is,k = 0. Close tok = 0 we can expandK(k) as

K(k) ≈ K0

(

1 − (ℓk)2 + O(k4))

(11.88)

whereℓ is a constant with dimensions of length. From now on we are going to disregard all the terms oforderk4 and higher. Observe thatK0 is the Fourier component ofK(k) with zero momentum, that is,

K0 =∑

r

K(r) = ZβJ (11.89)

where we have assumed that the interaction only couples theZ nearest neighbor atoms. Direct substitutionof (11.88) into (11.84) one gets

F0 =∑

k

K0

[

(1 − 2K0) + (4K0 − 1)(ℓk)2]

|ψ(k)|2 . (11.90)

Observe that, exactly like in the case of the mean field theorywhenk = 0 this part of the free energybecomes negative whenK0 > 1/2 which defines the critical temperature

Tc =2ZJ

kB. (11.91)

At this level of approximation we are recovering the mean field approximation.

If we are interested only in temperatures close toTc then we can write

K0 ≈ 1

2+ O(T − Tc)

1 − 2K0 ≈ T − TcTc

+ O(T − Tc)2

4K0 − 1 ≈ 1 + O(T − Tc) (11.92)


and the free energy (11.90) as

F0 ≈ 1

2

∑

k

(

T − TcTc

+ (ℓk)2)

|ψ(k)|2 . (11.93)

It is usual to rewrite (11.93) in a slightly different way by defining

φ(k) = ℓ ψ(k)

µ2 =T − Tcℓ2Tc

(11.94)

in which case (11.93) becomes

F0 ≈ 1

2

∑

k

(

µ2 + k2)

|φ(k)|2 . (11.95)

If one Fourier transform (11.95) back to real space we find

F0 =1

2

∫

dr

(∇φ(r))2 + µ2 (φ(r))2

(11.96)

which actually has a very interesting form. Let us calculate, for instance, the correlation function for for thefieldsφ when they are separated by a distancer,

〈φ(r)φ(0)〉 =

∫∏

r dφ(r)φ(r)φ(0)e−F0

∫∏

i dφ(ri)e−F0(11.97)

Notice however that the integration can be done easier if we Fourier transform since

〈φ(r)φ(0)〉 =∑

k,k′

eik·r〈φ(k)φ(k′)〉 (11.98)

and

〈φ(k)φ(k′)〉 =

∫∏

p dφ(p)dφ(−p)φ(k)φ(k′)e−12

P

q(µ2+q2)|φ(q)|2

∫∏

p dφ(p)dφ(−p)e−12

P

q(µ2+q2)|φ(q)|2(11.99)

and therefore, after a simple Gaussian integral,

〈φ(k)φ(k′)〉 =δk,−k′

k2 + µ2(11.100)

and therefore from (11.98) we have

〈φ(r)φ(0)〉 =∑

k

eik·r

k2 + µ2. (11.101)

Observe that the integral above has poles atk = ±iµwhich indicates that the correlation function will decayexponentially at large distances, that is,

〈φ(r)φ(0)〉 ∝ e−r/ξ (11.102)

11.4. CONTINUOUS SYMMETRIES AND THE GOLDSTONE THEOREM 223

where using (11.94)

ξ ∝ 1

µ∝ 1√

T − Tc(11.103)

is the magnetic correlation length. Observe that the magnetic correlation length diverges at the criticaltemperature when we lower the temperature from the disordered phase. Comparing with (11.18) we find anew exponentνMF = 1/2.

Up to this point we have not talked about the non-linear term in (11.82). Observe that close to the phasetransition the value of the magnetization, orψ, is very small and therefore we can keep only thek = 0 partof K(k). In this case the whole free energy simplifies to a very simpleform which is

F =

∫

dr

1

2(∇φ(r))2 +

µ2

2(φ(r))2 +

λ

4!(φ(r))4

(11.104)

whereλ/4! = K40/ℓ

4 is a positive constant. (11.104) is known as the Ginzburg-Landau free energy whichwas proposed by Ginzburg and Landau only based on the symmetry and the existence of a second orderphase transition. This kind of free energy is the heart of many discussions of the problem of phase transitionsin magnetic and non-magnetic systems.

11.4 Continuous symmetries and the Goldstone theorem

In the previous sections we have considered only the Ising problem which has a discrete symmetry ofspin inversion. As we have seen this discrete symmetry is reflected in the free energy of the system asthe existence of two minima which are separated by an energy barrier as shown in Fig. 11.4. In highersymmetric systems, like the Heisenberg model, the Hamiltonian has the symmetry of rotation in the spinspace since the interaction energy,Si · Sj, only depends on the relative orientation of the spins but not ontheir absolute orientation like in the Ising model. The symmetry of rotation in spin space is a continuoussymmetry and therefore we expect that the physics to be very different from the discrete case. However, thephenomenon of a phase transition is again the same thing: in the ordered phase the symmetry is broken. Thequestion then is: what happens when a continuous symmetry isbroken? Do we get the same type of physicsas in the discrete case?

The simplest way to understand the consequences of a broken continuous symmetry is to consider theGinzburg-Landau free energy for the so called O(N) model. In this model the order parameter hasNcomponents. The Ising case is the particular case whereN = 1! Thus we define aN component vector~φ = (φ1, φ2, ..., φN ) which is governed by a free energy which is the analogue of (11.104)

F =

∫

dr

1

2

(

∇~φ(r))2

+µ2

2

(

~φ(r))2

+λ

4!

(

~φ(r))4

− B · ~φ(r)

(11.105)

whereB is the symmetry breaking field. Observe that by constructionthe free energy (11.105) is invariantunder rotations of the order parameter~φ. This is the so-called O(N) symmetry. The XY model whichinvolves only the X and Y components of the spin is a particular case ofN = 2 while the Heisenberg modelhasN = 3 since it involves all the components, namely, X,Y and Z.

The main difference between (11.105) and (11.104) forN > 1 is the fact that one can rotate continuouslythe vector in theN dimensional space. Let us consider first the case whereN = 2 and the components of thevector~φ are constant in space (in which case we disregard the derivatives in the action). The homogeneous


part of the Ginzburg-Landau free energy is simply

U(φ1, φ2)

V=µ2

2(φ2

1 + φ22) +

λ

4!(φ2

1 + φ22)

2 (11.106)

whereV is the volume of the system. The invariance of the system under rotations becomes obvious if weplot this free energy in the space of(φ1, φ2). If µ2 > 0, that is, in the disordered phase, the minimum ofthe energy is atφ1 = φ2 = 0 but if µ2 < 0 the potential energy looks as in Fig. 11.5. In this case theground state is infinitely degenerate, that is, we can rotatecontinuously the vector around the origin withoutchanging the energy of the system.

-2

0

2

f1 -2

0

2

f2

-20

0

20

U

-2

0

2

f1

Figure 11.5: Energy potential for a O(2) model.

The heart of Goldstone theorem is to state that it requires noenergy to move the order parameter contin-uously along the minima of the potential. Thus, if an infinitesimal field is applied in this particular directionwe can rotate the order parameter by a finite amount. The susceptibility of the system in this particular di-rection is therefore infinity! That is, we get a finite response with an infinitesimal cost. It implies that theremust be excitations of the system which cost no energy in the direction transverse to the ordering direction.This is Goldstone theorem: if a system has a spontaneously broken continuous symmetry there is always anexcitation of the system in the ordered phase that costs no energy to excite.

We have seen already a good example of an application of this theorem: when a set of atoms forms acrystal the translational invariance (which is a continuous symmetry) is lost. This has to be contrasted withthe liquid or gas phase where the translational symmetry persists! In the case of the crystal we know verywell what is the order parameter: it is the Fourier transformof the density,〈ρ(q)〉 which now have Diracdelta peaks at the reciprocal lattice vectorsG. The peaks only exist if the system is crystalline and theydisappear in the gas phase. Thus, Goldstone’s theorem predicts the existence of modes with zero energy inthis phase. But what are these modes? If you go back a few chapters you will find that there are indeedmodes which are gapless in this phase: acoustic phonons! Because acoustic phonons have zero energyat zero wavevector an infinitesimal energy is required to excite them! There are many other examples ofGoldstone modes which we will discuss in the next chapter andthese are very important to understand the

11.5. PROBLEMS 225

behavior of systems in ordered phases with spontaneously broken continuous symmetries.

11.5 Problems

1. Show that the Fourier transform of the local magnetization of a two dimensional antiferromagnet hasDirac delta peaks atQ = (π/a, π/a) andQ + G whereG is a reciprocal lattice vector.

2. Consider the case of an one dimensional magnet with a Lorentzian neutron scattering line shape andshow that the spin-spin correlation function decays exponentially with the correlation functionξ.

3. What is the neutron line shape corresponding to the correlation function given in (11.20).

4. Prove (11.31), (11.32), (11.33) and (11.34).

5. Prove (11.35), (11.36) and (11.37).

6. Prove (11.52) and (11.54).

7. Prove thatαMF = α′

MF = 0 by calculating the specific heat of a ferromagnet in the mean fieldapproximation.

8. Consider the O(N) model where the magnetization in theα direction is:

Mα = 〈φα〉 = Mnα

whereM is the magnitude of the magnetization vectorM andn is an unit vector. Define susceptibilitytensor

χαβ = − ∂2U

∂Bα∂Bβ,

whereα, β = 1, ..., N are the components of the order parameter andBα are components of magneticfield in theN directions. Using the chain rule

∂U

∂Bα=

∂B

∂Bα

∂U

∂B

=BαB

∂U

∂B

whereB =√

∑Nα=1B

2α, prove that

χαβ = nαnβχ||(B) + (δαβ − nαnβ)χ⊥(B)

where

χ||(B) = −∂2U

∂B2

χ⊥(B) = − 1

B

∂U

∂B=M

B.

Observe thatχ||(B) gives the susceptibility of the system in the direction of the order parameterMwhile χ⊥(B) gives the susceptibility in the direction transverse toM. While χ||(B) is finite in the


ordered phaseχ⊥(B) diverges sinceM 6= 0 even whenB = 0. This is again a result of continuoussymmetry and therefore related to Goldstone theorem.

Chapter 12

Magnetic excitations

In the last chapter we have seen that systems with continuousbroken symmetries exhibit gapless modes inthe excitation spectrum. The clearest example of a Goldstone mode is the acoustic phonon that exists incrystals due to the breaking of the translation symmetry. However, crystals also support optical modes thathave a gap. Optical modes are related to the breaking of a symmetry that is discrete: the doubling of unitcell. Therefore, when the system of interest has a particular symmetry that is broken a new mode appearsin the spectrum. If the symmetry is continuous the mode is gapless and when the symmetry is discrete themode has a gap. Since experimentally speaking we are always probing excited states of a given system it isimportant to understand the nature of these excitations. Aswe have seen, the nature of the ground state andthe symmetries characterize it determine the nature of the excitation spectrum.

In this chapter we are going to study the excitation spectrumof magnetic systems and distinguish be-tween the various universality classes: the Ising model that has a discrete symmetry of spin inversion, theHeisenberg model that has the full symmetry of rotation in spin space, and the XY model that has thesymmetry of rotation in a plane.

12.1 The one-dimensional Ising model

Let us consider what is perhaps the simplest magnetic system: the one-dimensional Ising model. Considera chain withN atoms that interact via a Hamiltonian given by:

H = −N∑

i=1

(

Jσzi σzi+1 + hσzi

)

(12.1)

whereJ > 0 is the magnetic exchange andh = µBgB is the magnetic field energy. Since (12.1) iscompletely described in terms of theσz operator we can use the eigenstates of this operator (namely,σzi |σi〉 = σi|σi〉) to write the energy of the system as:

E[σ] = −N∑

i=1

(Jσiσi+1 + hσi) (12.2)

which is a functional of the configurationσ1, σ2, ..., σN (we will assume periodic boundary conditionsso thatσN+1 = σ1). Observe that this problem is classical in the sense that the states of the system aregenerated by classical configurations of the spins. There are no quantum mechanical effects such as spinflips (that is, a tunneling of the spins between different configurations). The ground state of the systemis obviously ferromagnetic and described by the configuration +1,+1, ...,+1 if h > 0. Although the

227

228 CHAPTER 12. MAGNETIC EXCITATIONS

ground state configuration is straightforward the excited states are not so obvious. Consider, for instance aspin flip in the system given by the configuration+1,+1,−1,+1...,+1. This spin flip costs2J in energy(h = 0). Actually this state is highly degenerated because one canflip a spin anywhere along the chain.Now consider a two spin flip which is given by+1,+1,−1,+1...,+1,−1,+1 which costs4J in energyand compare with another two spin flip where the flipped spins are neighbors+1,+1,−1,−1, ...,+1.This last state costs only2J of energy. Actually the state where the flipped spins are all together only cost2J of energy, no matter how long the flipped sequence is. Thus, there is an enormous amount of degeneratestates in the problem.

The question that comes out here is: at what temperatures does the system order magnetically? In orderto answer this question one has to calculate the partition function for the problem,

Z = tr(e−βH)

=∑

σ1,...,σN=±1

e−βE[σ] . (12.3)

The whole problem in calculating this partition function isrelated with the problem of calculating the con-tribution of the interaction term. In order to evaluate thiscontribution we are going to use a trick which isvalid for spin1/2 particles. Consider the following matrix element:

〈σ|eθσx |σ′〉 = 〈σ|(cosh(θ) + σx sinh(θ))|σ′〉= cosh(θ)δσ,σ′ + sinh(θ)δσ,−σ′ (12.4)

where we used that(σx)2 = 1 and thatσx|σ〉 = | − σ〉. The above identity can be written in a moreinteresting way as:

〈σ|eθσx |σ′〉 =

√

sinh(θ) cosh(θ)

(

cosh(θ)

sinh(θ)

)σσ′

=

√

sinh(2θ)

2e

σσ′

2ln(coth(θ)) (12.5)

as you can easily show. Thus, in (12.3) the interaction term in (12.1) can be written as

eβJσiσi+1 =

(

sinh(2θ)

2

)−1/2

〈σi|eθσx |σi+1〉 (12.6)

if we identify 2βJ = ln(coth(θ)) or equivalently

θ =1

2ln(coth(βJ)) . (12.7)

Observe that in this case the partition function can be written as:

Z =∑

σ1,...,σN

N∏

i=1

(

sinh(2θ)

2

)−1/2

〈σi|eβhσzeθσ

x |σi+1〉 . (12.8)

Now using the fact that the states are complete, that is,∑

σi=±1 |σi〉〈σi| = 1 we can rewrite the partition

12.1. THE ONE-DIMENSIONAL ISING MODEL 229

function as:

Z =

(

sinh(2θ)

2

)−N/2∑

σ

〈σ|(

eβhσzeθσ

x)N

|σ〉

=

(

sinh(2θ)

2

)−N/2

tr

[

(

eβhσzeθσ

x)N]

. (12.9)

Let us consider the case without the magnetic field (h = 0). In this case (12.9) is straightforward tocalculate:

Z =

(

sinh(2θ)

2

)−N/2

tr[

eNθσx]

(12.10)

which is the partition function of a completely different problem from the initial one, namely, one which isdescribed by the “Hamiltonian”:

H = −Γσx (12.11)

where

Γ =Nθ

β=N

2βln(coth(βJ)) . (12.12)

Observe that we have mapped our original problem (12.1) of a chain of Ising spins (a classical one-dimensional problem) into a problem of a single spin in a transverse field of strengthΓ ( this is a zerodimensional problem). How this is possible?

What we have shown here is an essential feature of relationship between quantum statistical mechanicsin d dimensions and the classical statistical mechanics ind + 1 dimensions. The eigenstates of problem(12.11) are symmetric (|s〉 = (| + 1〉 + | − 1〉)/

√2) and antisymmetric (|a〉 = (| + 1〉 − | − 1〉)/

√2)

combinations of eigenstates ofσz with energy−Γ and+Γ, respectively. Thus, if we prepare the state of thesystem at timet = 0, say, with spin up (that is,|+ 1〉) quantum mechanics tells us that fort > 0 the state ofthe system will be:

|ψ(t)〉 =1√2

(

e−iΓt/h|s〉 + eiΓt/h|a〉)

= cos (Γt/h) | + 1〉

− i sin (Γt/h) | − 1〉 (12.13)

that is, the spin will oscillate from up to down with time. Thus, if we imagine the time evolution as a linewith N steps we see that along the line the spins are never ordered, except whenΓ = 0 in which case|ψ(t)〉 = | + 1〉 for all time t. In this case the chain is ferromagnetically ordered! Thus,with this argumentone finds that the system is only ordered whenΓ = 0. Going back to (12.12) we see that this is possible inthe limit ofN → ∞ whenβ = ∞, that isT = 0. For any finiteβ, no matter how small,t→ ∞ asN → ∞and the system has to be disordered.

Furthermore, from (12.10) we can compute the partition function and free energy of problem since wehave a problem of a single1/2 spin. We can diagonalize the operator that appears in the exponent of (12.10)which has the eigenvalues:

E± = ±N2

ln(coth(βJ)). (12.14)


Thus, we get immediately:

e−βF = Z = 2

(

sinh(2θ)

2

)−N/2

cosh

(

N

2ln(coth(βJ))

)

(12.15)

and in the limit ofN → ∞ we find

F

N= − 1

βln(2 cosh(βJ)) (12.16)

which is the exact free energy of the 1D Ising model at any temperature. Observe that the free energy is asmooth function of the temperature: forkBT ≪ J we haveF ≈ −JN which is the energy of the fullyaligned ferromagnetic state and forkBT ≫ J we findF ≈ ln(

√2)kBT .

In order to calculate the magnetization as a function of temperature we need to re-examine (12.9) again.Observe that because we are dealing with Pauli matrices we can write:

eβhσzeθσ

x=

[

cosh(θ)eβh sinh(θ)eβh

sinh(θ)e−βh cosh(θ)e−βh

]

(12.17)

and it is easy to show that the partition function in (12.9) becomes:

Z = tr

(

[

eβ(h+J) eβ(h−J)

e−β(h+J) e−β(h−J)

]N)

. (12.18)

Since the trace of a matrix is invariant to an unitary transformation (sincetr(A) = tr(U−1UA) = tr(UAU−1))we rewrite (12.18) as

Z = tr(AN ) = tr(

(

UAU−1)N)

= tr(ΛN ) = λN+ + λN− (12.19)

whereΛ is a diagonal matrix (UAU−1 = Λ) andλ± its eigenvalues. Thus, in order to solve the problems wejust have to diagonalize the matrix that appears in (12.18).It is a simple matter to show that the eigenvaluesare:

λ± = cosh(βh)eβJ ±√

sinh2(βh)e2βJ + e−2βJ (12.20)

which ends with the solution of the problem. Observe thatλ+ > λ− and since we are interested in the limitof N → ∞ we can rewrite (12.19) as:

Z = e−βF = λN+F

N= − 1

βln

(

cosh(βh)eβJ +

√

sinh2(βh)e2βJ + e−2βJ

)

(12.21)

which is the exact expression for the free energy of the problem and reduces to (12.16) whenh → 0.

12.1. THE ONE-DIMENSIONAL ISING MODEL 231

0.1 0.2 0.3 0.4 0.5H

0.2

0.4

0.6

0.8

1

M

Figure 12.1: Magnetization of the one dimensional Ising model as a function of external field for differenttemperatures (temperature increases from top to bottom).

Observe that the magnetization per spin in the problem is given by:

M(h) = −∂F/N∂H

=sinh(βh)

√

sinh2(βh) + e−4βJ. (12.22)

Notice that (12.22) is a quite interesting function. If we take the limit of h → 0 with β finite we findM(h → 0) = 0 that is, there is no magnetization in the system at any finite temperature. If however wetake the limit ofβ → ∞ before we take the limit ofh → 0 we findM(h → 0) = 1 ! That is, the systemis magnetized at zero temperature as we have argued before. Therefore, the magnetization is a singularfunction of the temperature since there is a discontinuous jump in the magnetization atT = 0. A mostamazing effect! The approach to zero temperature is shown onFig.12.1.

The problem of phase transitions in one-dimensional systems can be understood very well by an argu-ment due to Landau-Lifshitz. Consider a state with a finite magnetizationM and free energyF . In theordered state the free energy would be given by:

F = E − TS (12.23)

whereE is the energy of the ordered state andS its entropy. Consider the configurations discussed at the


beginning of this section in which all the spins to the right of a point n are reversed creating a domainwall: (+1,+1,+1, ...,+1,−1,−1,−1). The internal energy of the system will be increased by an amountδE ≈ 2M2J (M = 1 at T = 0). But we can choose the pointn at any point of the lattice ofN sites andtherefore there is an entropy of the orderδS ≈ kB ln(N) associated with the creation of the domain wall.Thus, the net change in free energy due to the creation of domain wall is

δF ≈ 2M2J − kBT ln(N) . (12.24)

Observe, therefore, that if we makeN sufficiently large, that is,

N > N∗ = e2M2JkBT (12.25)

the free energy of the problem becomes negative and the system becomes unstable. Thus, we have toconclude thatM = 0 for any finite temperature. Observe that atT = 0 order is possible becauseN∗ → ∞.

12.2 The Heisenberg-Ising Hamiltonian

Let us consider a system described by the Hamiltonian

H = −∑

〈i,j〉

[

JzSzi S

zj + J⊥

(

Sxi Sxj + Syi S

yj

)]

− gµBH∑

i

Szi (12.26)

which describes an anisotropic magnet in a external magnetic fieldH. We first point out that the total spinof the system,S2

T = (∑

i Si)2 and thez component of the total spin,SzT =

∑

i Szi , commute with the

Hamiltonian. Thus, we can classify all the states with respect to their eigenstates, namely,ST (ST + 1)andSzT . Observe thatST = SN,S(N − 1), ..., 0 while SzT = −ST , ..., ST has2ST + 1 possible values.We will consider two important case of this Hamiltonian: theferromagnetic case withJz > 0 and theantiferromagnetic case withJz < 0. As we are going to see these two cases have completely differentexcitation spectrum and physical properties.

12.2.1 The ferromagnetic case

In the classically ordered state of a ferromagnetic system all the spins are polarized in one direction andtherefore we haveST = SzT = NS whereN is the total number of atoms in the system. This correspondsto the mean value of the operator to be fully polarized that is〈Szi 〉 = S. Observe that the ground state wouldbe described in this way ifJ⊥ = 0 but because of the presence ofJ⊥ this is not completely correct. Let usrewrite:

S±i = Sxi ± iSyi (12.27)

which the raising and lowering operators for spins. In this case Hamiltonian (12.26) is rewritten as

H = −∑

〈i,j〉

[

JzSzi S

zj +

J⊥2

(

S+i S

−j + S−

i S+j

)

]

− gµBH∑

i

Szi (12.28)

which clearly shows thatJ⊥ flips the spins. Classically this term is equivalent to the precession of themagnetic moment around thez axis.

Suppose the magnetic system is in the ordered phase. One expects the total magnetization of the systemto be very close toNS. As in the case of phonons one expects that the fluctuations can be described by

12.2. THE HEISENBERG-ISING HAMILTONIAN 233

some kind of bosonic excitation with operatorsbi andb†i which obey canonical commutation relations:

[

bi, b†j

]

= δi,j . (12.29)

The problem is to find out the relationship between these bosonic operators and the original spin operators.This was done by Holstein and Primakoff. They propose to write the spin operators as:

Szi = S − b†i bi

S+i =

√2S

(

1 − b†ibi2S

)1/2

bi

S−i =

√2Sb†i

(

1 − b†i bi2S

)1/2

(12.30)

which are known as the Holstein-Primakoff transformation.These transformations have nice properties: 1)using (12.29) it can be shown that the operators defined in (12.30) obey the spin algebra; 2) it is easy toshow thatS2

i = S(S + 1) as it should be.

Observe that at zero temperature, because the excitations are bosons, we expect〈b†i bi〉 = 0. Thus it isreasonable to assume that〈b†i bi〉/S ≪ 1 and we can expand the transformation in (12.30) as

Szi = S − b†ibi

S+i ≈

√2Sbi

S−i ≈

√2Sb†i (12.31)

which, of course, is not a true operator relationship but canbe shown to be a good approximation a posteriori.Moreover, the operatorSzi is unchanged. The physical meaning of this transformation becomes clear now.Szi in (12.30) measures the number of overturned spins. This flipping of spins is generated byS±

i . If weimagine that the number of spins equals the number of bosons in the system thenS±

i behave like creationand annihilation operators for bosons.

If we substitute (12.31) into (12.28) and keep only the leading order in the operators we obtain:

H = −JzNZS2 − J⊥NZS − gµBHNS + H (12.32)

where

H = (2JzZ + gµBH)∑

i

b†i bi − J⊥S∑

〈i,j〉

(

b†jbi + b†i bj

)

. (12.33)

Observe thatH has exactly the same form of the tight binding Hamiltonian for electrons moving on a latticewith the difference is that now we have bosons. Thus, like electrons, the ground state is made out of Blochwaves. For the problem where there is just one atom per unit cell solve this problem exactly in the sameway as before, that is, by Fourier transform:

bk =1√N

∑

k

eik·rjbj

bi =1√N

∑

j

e−ik·rjbk (12.34)


and since the system is periodic the wavevector is defined only in the first Brillouin zone (also called mag-netic Brillouin zone). The Fourier transform changes (12.33) into

H =∑

k

Ekb†kbk (12.35)

where

Ek = 2JzZ + gµBH − 2J⊥SZγk (12.36)

and

γk =1

Z

∑

~δ

eik·~δ (12.37)

where~δ is the vector that links two nearest neighbor atoms. The excitations described in Hamiltonian(12.35) are called ferromagnetic magnons. Observe that ifJ⊥ = 0 the magnon spectrum is dispersionlessand therefore there is no propagation! This is the case of theIsing model. If there is a finiteJ⊥ (or finite XYcomponent) the magnons can propagate since there is a finite group velocityck = ∇Ek. Let us considernow the case of long wavelength magnons, that is, in the limitwhere|k · ~δ| ≪ 1. In this case we can writeγk ≈ 1 − (ka)2/2 wherea is the lattice spacing in a cubic system. Therefore from 12.36) we have

Ek ≈ 2(Jz − J⊥)Z + gµBH + J⊥S(ka)2 (12.38)

which can be rewritten as

Ek ≈ ∆ +k2

2m∗(12.39)

where

∆ = 2(Jz − J⊥)Z + gµBH

m∗ =1

4J⊥Sa2. (12.40)

Observe that (12.39) has the form of the dispersion of a free particle with massm∗. This particle is themagnon. Observe that in the limit ofJ⊥ → 0 the mass of the magnon diverges indicating that it localizesin the system. One could ask how these magnons are related to Goldstone’s theorem. Goldstone’s theoremrequires the system to have a continuous symmetry which in the case of this Hamiltonian is obtained in theHeisenberg limit (J⊥ = Jz) and in the absence of fields (H = 0) which leads to∆ = 0, that is, there is nogap in the system atk = 0 implying that the gapless ferromagnetic magnon is the Goldstone mode.

From (12.31) we can calculate the magnetization per atom of the system immediately

M = S − 1

N

∑

k

〈nk〉 (12.41)

wherenk = b†kbk is the occupation number operator for magnons. This number is given by the Bose-Einstein occupation as for the case of phonons. We have

M = S − 1

ρ

∫

ddk1

eβEk − 1(12.42)


whereρ = N/V is the density. Observe that at low temperatures (β → ∞) the integral is dominated by thelong wavelength behavior and we can use (12.39) withd = 3:

M = S − 1

2π2ρ

∫ ∞

0dk

k2

eβ∆eβk2/(2m∗) − 1

= S − 1

2π2ρ(2m∗kBT )3/2

∫ ∞

0dx

x1/2

eβ∆ex − 1. (12.43)

In the absence of external field and in the fully isotropic system, J⊥ = Jz, we have∆ = 0 and thelast integral is a constant and therefore the magnetizationdecreases with a powerT 3/2 of the temperatureshowing that indeed the magnons tend to disorder the classically ordered state. If∆ 6= 0 the magnetizationdecreases exponentially with temperature.

The energy per atom of the system is of course obtained directly from (12.35). Again, at low tempera-tures, we use (12.31):

E =1

N

∑

k

Ek〈nk〉

= ∆ +1

4π2ρm∗

∫ ∞

0dk

k4

eβ∆eβk2/(2m∗) − 1

= ∆ +1

4π2ρm∗(2m∗kBT )5/2

∫ ∞

0dx

x3/2

eβ∆ex − 1. (12.44)

which shows that in the absence of magnetic field and anisotropy the energy of the system goes likeT 5/2

and because of that the specific heat behaves likeCV ∝ T 3/2.

12.2.2 The antiferromagnetic case

Let us now consider the case whenJz < 0 in (12.26). In this case the spins orient anti-parallel to eachother in a structure similar to the one shown in Fig.10.9 which is called the Néel state. We are going toassume a similar structure where only two sublattices,A andB, are present each one withN/2 sites (thatis, two different ferromagnetic lattices which are immersed into each other). This is due to the fact thatan antiferromagnet breaks the translational symmetry of the original lattice and doubles the unit cell. This,of course, does not happen in a ferromagnetic system. In sublatticeA the magnetization is essentially〈Szi 〉 = +S while in sublatticeB we have〈Szi 〉 = −S. Thus, in defining the spin operators in terms ofbosons one has to be careful about each sublattice. The easiest way to proceed is to follow the recipe of the


ferromagnetic problem and define:

SzA,i = S − a†iai

SzB,i = −S + b†ibi

S+A,i =

√2S

(

1 − a†iai2S

)1/2

ai

S−B,i =

√2S

(

1 − b†ibi2S

)1/2

bi

S−A,i =

√2Sa†i

(

1 − a†iai2S

)1/2

S+B,i =

√2Sa†i

(

1 − a†iai2S

)1/2

. (12.45)

Notice that the role played by the lowering and raising spin operators are exchanged between the two sublat-tices because they have different orientation of the spins.Using the same approximations as in the ferromag-netic case and Fourier transforming the bosonic operators you can easily show that the Hamiltonian (12.28)with J replaced by−J and in the presence of a staggered magnetic fieldgµBHs

∑

j(−1)jSzj becomes:

H = −NZJzS2 −NµBHsS + H (12.46)

where

H = ZS∑

k

J⊥γk

(

a†kb†k + akbk

)

+ Jz

(

a†kak + b†kbk

)

+ µBHs

∑

k

(

a†kak + b†kbk

)

(12.47)

which has a structure which is very different from the ferromagnetic case since the operators from differentsublattices mix with each other. The mixing of these operators becomes more obvious if one rewrites (12.47)as:

H = −N(JzZS + µBH) +∑

k

Ψ†k[H(k)]Ψk (12.48)

where

Ψk =

(

akb†k

)

(12.49)

and

[H(k)] =

[

JzZS + µBHs J⊥ZSγkJ⊥ZSγk JzZS + µBHs

]

. (12.50)


In order to diagonalize the Hamiltonian we look for a transformationU such that

Ψk = U(k)Φk

U †(k)[H(k)]U(k) = ǫkI (12.51)

whereǫ is a diagonal matrix and

Φk =

(

αk

β†k

)

(12.52)

with αk andβk are bosonic operators. The simplest transformation has theform

U(k) =

[

uk vkvk uk

]

(12.53)

whereuk andvk are real. Moreover, because the operators are bosons we mustimpose thatu2k − v2

k = 1.Thus,we can write

uk = cosh(θk)

vk = sinh(θk) . (12.54)

This implies that the inverse ofU is

U−1(k) =

[

uk −vk−vk uk

]

(12.55)

and thereforeU is not unitary (U−1 6= U †). Actually it is easy to see thatU † = U and therefore theequation for the energy in (12.51) isUHU = ǫ or HU = ǫU−1 which is not our usual eigenfunctionproblem. Observe moreover that these matrix can be written in terms of Pauli matrices,

H(k) = E0I + E1(k)σx

U(k) = ukI + vkσx

U−1(k) = ukI − vkσx (12.56)

whereE0 = JzZS + µBHs andE1 = J⊥ZSγk. Thus, the equationHU = ǫU−1 leads to the followingset of equations:

(ǫ− E0)uk − E1vk = 0

E1uk + (ǫ+E0)vk = 0 (12.57)

which for a non-trivial solution requires(ǫ− E0)(ǫ+ E0) + E21 = 0, that is,

ǫk =√

E20 − E2

1(k)

=√

(JzZS + µBHs)2 − (J⊥ZS)2γ2k

(12.58)

and moreover we find

tanh(2θk) = − E0

E1(k). (12.59)


Thus, the Hamiltonian becomes

H = −NE0 +∑

k

ǫk

(

α†kαk + β†kβk + 1

)

(12.60)

which has the appropriate diagonal form.

Let us now look for the Goldstone mode. In the absence of a staggered field and in the isotropic limit(Jz = J⊥) the dispersion (12.58) becomes

ǫk = 2JZS√

1 − γ2k (12.61)

and in the long wavelength limit we find

ǫ(k) ≈ csk (12.62)

wherecs = 4√

3JSa is the spin wave velocity in three dimensions. Observe that as indicated by Goldstonetheorem the antiferromagnetic magnon is gapless with linear dispersion relation. This has to be contrastedwith the ferromagnetic case where the dispersion goes likek2. Like the case of acoustic phonons the an-tiferromagnetic magnons have a characteristic velocity. Ferromagnetic magnons, on the other hand, haveGalilean invariance which is characterized by an effectivemass which determines the curvature of the dis-persion.

Let us now consider the sublattice magnetization given by

MA =∑

i

〈SzA,i〉 = NS −∑

k

a†kak

= NS −∑

k

(

u2kα

†kαk + v2

kβkβ†k

)

(12.63)

which atT = 0 becomes

MA = NS −∑

k

v2k = NS −

∑

k

sinh2(θk)

= NS − 1

2

∑

k

[

(1 − γ2k)−1/2 − 1

]

. (12.64)

Thus, the magnetization per unit of volume can be written as

MA

V= nS + n/2 − 1

2

∫

ddk

(2π)d1

√

1 − γ2k

(12.65)

which for a cubic lattice in three dimensions is given by

MA

V= n(S − 0.078) . (12.66)

This is the result of a numeric integration of the equation above. Observe that due to quantum fluctuations(zero point motion of the spins) the staggered magnetization is reduced relative to its classical value.

12.3. PROBLEMS 239

12.3 Problems

1. Using the formalism of subsection (1.1) and the operator identity

limM→∞

(

eAM e

BM

)M= eA+B

show that the one-dimensional Ising model in a transverse field which is described by

Hq =∑

n

(

−J ′σznσzn+1 − Γσxn

)

whereΓ is the strength of the magnetic field has the same partition function as the classical Isingmodel in a square lattice (withN2 sites) which is given by

Hc = −J∑

n,m

(

σzn,mσzn+1,m + σzn,mσ

zn,m+1

)

. (12.67)

Find the relationship betweenJ ′ andΓ to J , β andM .


3. Calculate (12.20) explicitly and show (12.22) to be correct.

4. Consider the one-dimensional Ising model with an interaction which has finite range, that is,

H = −∑

i,j

Ji,jσiσj

where

Ji,j =J

|i− j|n

wheren characterizes the power of decay of the interaction. Using the Landau-Lifshitz argumentshow that it is possible to have long range order at finite temperatures ifn < 2. Show that for thesystem to have a well defined thermodynamic limit one has to requiren > 1. Show that forn = 2 thesame argument requires that the magnetizationM cannot vanish continuously, in order words, eitherthere is zero magnetization at all temperatures or the magnetization is non zero with a discontinuityat the transition.

5. Prove (12.28).

6. Using (12.29) show that[Sxi , Syi ] = 2iδi,jS

zi and prove thatS2

i = S(S + 1) .

7. Prove equations (12.36).

8. Calculate the spin wave velocity for an isotropic spin1/2 Heisenberg model ind space dimensionsfor a hypercubic lattice.

9. In the Debye model for the antiferromagnet one assumes thedispersion of the spin waves to beǫ(k) =csk but cuts-off the dispersion at large wavelengths, say, atΛ. (i) CalculateΛ by assuming that thenumber of states in the spherical zone is the same as in the cubic Brillouin zone. (ii ) Calculate thesublattice magnetization for the Debye model as a function of temperature. (iii ) Find the criticaltemperatureTN above which the sublattice magnetization vanishes.


10. Consider the problem of antiferromagnetic magnons in a uniform magnetic fieldH for the Heisenberg-Ising Hamiltonian. Using the non-unitary transformation find the magnon spectrum in this case. Showthat theαk andβk modes are separated in energy. Calculate theuniform magnetization of the systemat finite temperature assuming a Debye model for the magnons.

11. Using (12.42) and (12.65) find the lower value of the dimensionalitydc above which long range orderis possible (at finite and zero temperature) for a ferromagnet and antiferromagnet, respectively.

Chapter 13

The Interacting Electron Gas

In studying the electron gas in previous chapters we have assumed that the electrons do not interact witheach other. This assumption is not intuitive because we knowthat electrons in free space interact witheach other via a long range Coulomb potential that behaves like 1/r wherer is the distance between thetwo electrons. From the experimental point of view we know that electrons in metals behave very muchlike non-interacting particles. In this chapter we are going to discuss the basic properties of the interactingelectron problem.

Since in a electron system the electrons effectively do not interact via a long range force (as is experi-mentally observed) there is something else in the system that intervenes to modify or weaken the interactionamong electrons. From the microscopic point of view the onlyway an interaction between two particles canbe different from its value in vacuum is if there is a medium. This medium propagates the interaction andtherefore can effectively change the way the particles interact.

The simplest way to understand how this occurs at the microscopic level by the study of the equationsof electromagnetism. When an electric field is applied to a dielectric, the observed field is different fromthe applied field because it polarizes the dielectric. When the system is polarizable we can describe thepolarizationP as the electric dipole moment per unit of volume. In this casethe electric fieldE in thesystem is not simply the applied field butE = D − 4πP whereD is the displacement field that dependsonly on the external distribution of charges and obeys the Maxwell equation

∇ ·D = 4πeρext (13.1)

wheree is the electric charge,ρext is the external (applied) electron density. In many systemsthe polarizationP is linearly related to the electric field throughP = χeE whereχe is the electric susceptibility of themedium. Thus we immediately obtain that

D = ǫE (13.2)

whereǫ = 1 + 4πχe is the dielectric function of the medium. In systems like crystals the dielectric functionis not a constant but a tensor that depends on the direction the field is applied. Here we are going to considerthe case of the isotropic electron gas where the dielectric constant is just a scalar.

As an example consider a single chargeQ at the origin in the presence of the electron gas (in this caseρext(r) = Qδ(r)). If this charge is positive it attracts electrons close to it, if it is negative it repels. Inprinciple one would think that this charge could attract an infinity number of electrons. This is not truenot only because the electrons repel each other and therefore it would lead to an enormous increase of theelectrostatic energy of the system, but also because electrons are fermions and therefore cannot occupy thesame position is space. Thus, basic thinking tells us that the positive charge attracts enough electrons so

241

242 CHAPTER 13. THE INTERACTING ELECTRON GAS

that, as seen from far away, it looks like a neutral object. This is called screening. The same argumentworks for the case of a negative charge. Thus, in the presenceof an external charge the electron gas can notbe uniform. This non-uniformity can be described byδρ(r), the local change in the density of the electrongas. Observe that in the jellium model of the electron gas there is always a background of positive ionsthat neutralize the average densityρ0 otherwise the whole system would be unstable. The main idea behindscreening is that the electric filed in the system is created by the total densityρext + δρ in such a way that

∇ ·E = 4πe (ρext + δρ) . (13.3)

In the general case the external charge can also change in time so that we can Fourier decompose it as

ρext(r, t) =

∫

dk

∫

dωei(ωt−r·k)ρext(q, ω) . (13.4)

One can now Fourier transform (13.1) and (13.3) and use (13.2) to find

ǫ(q, ω) =q · D(q, ω)

q · E(q, ω)=

ρext(q, ω)

ρext(q, ω) + δρ(q, ω)

=1

1 + δρ(q,ω)ρext(q,ω)

(13.5)

gives the dielectric function once the displaced charged can be measured (or calculated). Another way toobtain the dielectric function is to observe that in (13.5) only the longitudinal part of the displacement andelectric field vector enter in the definition of the dielectric function. This is because we assume the system tobe isotropic. Associated with the longitudinal component of the field there we can always define a potentialsuch thatE = −∇φ andD = −∇φext which, accordingly to (13.1) and (13.3), obey the equations

∇2φext(r, t) = −4πeρext(r, t)

∇2φ(r, t) = −4πe (ρext(r, t) + δρ(r, t)) . (13.6)

The potential energies associated with the potentials given above isV = eφ andVext = eφext. From (13.6)and (13.5) we immediately conclude that

ǫ(q, ω) =Vext(q, ω)

V (q, ω)(13.7)

gives the dielectric function in terms of the external potential and the total potential generated by the externalfields. Thus, in order to describe screening one has to calculate the potential function.

13.1 The Thomas-Fermi approach

When one applies an electric field to an electron gas the electrons move from the points of higher potentialto the points of lower potential. In the Thomas-Fermi approach one assumes that, as the electrons, movethey are always in thermodynamic equilibrium with its surroundings. In a three dimensional electron gas thenumber of electrons per unit of volume in the absence of the field can be obtained from (6.4) and is givenby:

ρ0 =N

V=

(2mEF,0)3/2

3h3π2, (13.8)

13.1. THE THOMAS-FERMI APPROACH 243

whereEF,0 = h2k2F /(2m) is the Fermi energy of the system. Because of the electric field the charge density

is not homogeneous and we have a displaced densityδρ(r) that is given by

δρ(r) = ρ(r) − ρ0 . (13.9)

Associated with this charge density there is a local shift inthe Fermi energy of the system,

EF (r) = EF,0 − V (r) , (13.10)

whereV (r) is the total potential felt by the electrons. In the Thomas-Fermi approach we assume that (13.8)gives the change in the charge density due to the shift in the local chemical potential as given by (13.10),that is,

ρ(r) =(2mEF (r))3/2

3h3π2

=(2mEF,0 − V (r))3/2

3h3π2

= ρ0

(

1 − V (r)

EF,0

)3/2

(13.11)

gives the relationship between the potential energy and thedisplaced charge. From (13.6) we see that thepotential energy obeys the Poisson equation

∇2V (r) = −4πe2(ρext(r) + δρ(r))

= −4πe2

[

ρext(r) + ρ0 − ρ0

(

1 − V (r)

EF,0

)3/2]

(13.12)

where we have used (13.9) and (13.11). Notice that (13.12) isa non-linear equation forV (r). It turns out,however, that for practical purposesV ≪ EF,0 and therefore we can linearize (13.12) to give

(

∇2 − q2TF)

V (r) = −4πe2ρext(r) (13.13)

where

q2TF =6πe2ρ0

EF,0=

4me2kF

h2π

=4kFπa0

(13.14)

wherea0 = h2/(me2) the Bohr radius.qTF is the so-called Thomas-Fermi screening length. To understandthe physical meaning of this length scale let us solve (13.13) by Fourier transform

V (k) =4πe2ρext(k)

k2 + q2TF

=Vext(k)

1 +q2TFk2

(13.15)


where

Vext(k) = −4πe2ρext(k)

k2(13.16)

is the external potential created by the charge distribution ρext. Comparing (13.15) with (13.7) one immedi-ately concludes that

ǫTF (k) = 1 +q2TFk2

(13.17)

is the dielectric constant in the Thomas-Fermi approximation. Consider now the case of a single impurity ofchargeQ at the origin. In this caseρext(q) = Q and therefore from (13.15) we find that the total potentialis given by

V (r) =

∫

d3k

(2π)3−4πeQ

k2 + q2TF

= −2eQ

πr

∫ ∞

0dkk sin(kr)

k2 + q2TF

= − eQ

iπr

∫ +∞

−∞dk

keikr

k2 + q2TF

= −eQre−qTF r (13.18)

where the last integral is done by choosing the contour in theupper half complex plane. Observe that inthe Thomas-Fermi approach the effective potential produced by the charge Q at the origin is not the barepotentialVext(r) = −eQ/r but has a Yukawa form and decays exponentially with a characteristic length

λTF =1

qTF=

√

πa0

4kF. (13.19)

In this case we say that the electrons screen the bare potential to a distance of orderλTF . The Thomas-Fermiapproximation gives a simple but powerful picture of the electrostatic behavior of metals. Screening comesfrom the fact that the system has a Fermi surface and therefore it has mobile electrons that can move aroundin order to screen the external charge. Thus, electrons far away from the impurity indeed cannot feel thepotential created by the impurity and moves freely in the system.

13.2 The Random Phase Approximation: RPA

Let us now consider a more involved picture of the screening process that is called Random Phase Approx-imation or simply RPA. In this approach one considers the problem of the response of the electron gas tothe total potentialV (r, t). The assumption, as before, is that the electrons do not couple only to the externalapplied potential but to the total potential produced by allcharges in the system. The energy associated withthis coupling is simply

HV =

∫

drV (r, t)(ρ(r, t) − ρ0) (13.20)

whereρ(r, t) =∑

σ Ψ†σ(r, t)Ψσ(r, t) is the particle density. Observe that we have subtracted from the

energy the energy of the positive background of charge that neutralizes the electron gas. If one Fourier

13.2. THE RANDOM PHASE APPROXIMATION: RPA 245

transforms the above equation we get

HV =∑

q 6=0

Vq(t)ρq(t) (13.21)

where

ρq =∑

k,σ

c†k+q,σck,σ . (13.22)

Notice that the term withq = 0 is excluded from the integral in (13.21) sinceρq=0 = ρ0 is the averageparticle density. The physical meaning of (13.21) is quite straightforward: the potential created by thecharges in the system induces charge fluctuations that are particle-hole pairs described by (13.22) aroundthe Fermi surface.

The total Hamiltonian of the system can be written asH = H0 +HV where

H0 =∑

k,σ

ǫkc†k,σck,σ (13.23)

is the free electron Hamiltonian. In order to calculate the dielectric function one has to calculate the dis-placed charge. In order to do it we are going to look for the equation of motion for the density operator inthe Heisenberg representation, that is,

ihd

dt

[

c†k+q,σck,σ

]

=[

H, c†k+q,σck,σ

]

(13.24)

It is an easy exercise to show that[

H0, c†k+q,σck,σ

]

= (ǫk+q − ǫk) c†k+q,σck,σ[

HV , c†k+q,σck,σ

]

=∑

q′

Vq′

(

c†k+q+q′,σck,σ − c†k+q,σck−q′,σ

)

. (13.25)

Observe that because of the second term in (13.25) the equation for the density fluctuation cannot be solvedin full form. Thus one has to use some approximation.

Consider now the first quantized version of the density operator ρq that is given by

ρq =1

V

∑

n

eiq·rn (13.26)

wherern is the position of thenth electron (you can easily show that (13.26) is correct by the direct Fouriertransform ofρ(r) =

∑

n δ(r − rn))). Therefore, in the r.h.s. of the second equation in (13.25)we see thatρq+q′ appears which, in its first quantized form is:

ρq+q′ =1

V

∑

n

ei(q+q′)·rn . (13.27)

Observe that if the electrons are (statistically) randomlydistributed in the system the above sum only givesa finite result ifq = −q′ because the sum involves random phases that add destructively. In RPA we


approximate (13.25) by taking only theq′ = −q:[

HV , c†k+q,σck,σ

]

≈ V−q

(

c†k,σck,σ − c†

k+q,σck+q,σ

)

(13.28)

that leads to a solution for the equation of motion in (13.24)by Fourier transform in time:

〈c†k+q,σck,σ〉 = V−q(ω)

nk,σ − nk+q,σ

hω − ǫk+q + ǫk(13.29)

wherenk is the Fermi-Dirac occupation number. From (13.22), we get:

〈ρq,σ(ω)〉 = −V−q(ω)∑

k

nk,σ − nk+q,σ

hω − ǫk+q + ǫk. (13.30)

Eq. (13.30) can be rewritten in a slightly different way in terms of the charge density distribution

δρq(ω) = V−q(ω)Π(q, ω) (13.31)

where

Π(q, ω) = −∑

k,σ

nk,σ − nk+q,σ

hω − ǫk+q + ǫk + iη(13.32)

is the so-called polarization function and depends on the electronic system alone. It is important to noticethat the positive background of charge (associated withnq=0) is subtracted to (13.21) and therefore thesum in (13.32) has to be understood as the principal value. This can be accomplished by introducing a smallimaginary partη → 0 in the denominator. Thus, the polarization function has actually a real and a imaginarypart. Indeed from (13.32) we can write

ℜΠ(q, ω) = −P∑

k,σ

nk,σ − nk+q,σ

hω − ǫk+q + ǫk

= −P∑

k,σ

nk,σ

[

1

hω − ǫk+q + ǫk− 1

hω − ǫk + ǫk+q

]

ℑΠ(q, ω) = π∑

k,σ

(nk,σ − nk+q,σ) δ (hω − ǫk+q + ǫk)

= π∑

k

nk,σ [δ (hω − ǫk+q + ǫk) − δ (hω − ǫk + ǫk+q)] (13.33)

where whereP means the principal value of the integral and in the last linewe have changedk → −k − q

and used the fact thatn−k−q,σ = nk+q,σ andǫ−k−q = ǫk+q.

We can now substitute the result (13.31) into the Poisson equation (13.12)

Vq(ω) = −4πe2

q2(ρext(q, ω) + δρ(q, ω))

= −4πe2

q2(ρext(q, ω) + eV−q(ω)Π(q, ω)) (13.34)

13.2. THE RANDOM PHASE APPROXIMATION: RPA 247

assumingVq(ω) = V−q(ω) we find

V (q, ω) =Vext(q, ω)

1 + 4πeq2

Π(q, ω)(13.35)

which, by comparison with (13.7), leads to

ǫRPA(q, ω) = 1 +4πe

q2Π(q, ω) (13.36)

is the RPA expression for the dielectric function. Observe that RPA gives a frequency dependence to thepolarization function that is not present in the Thomas-Fermi result (13.17). This happens because theThomas-Fermi result is purely static, that is, it is valid atω = 0.

In order to check if RPA can reproduce the Thomas-Fermi result let us investigate the static limit of(13.36). From the definition (13.32) atT = 0 and assuming a spherical Fermi surface one has

ℜΠ(q, 0) = −2P∫

d3k

(2π)3Θ(kF − k) − Θ(kF − |k + q|)k2/(2m) − (k + q)2/(2m)

. (13.37)

This integral can be simplified if we take the limit ofq → 0 in that case we can write|k+q| ≈ k+ q cos(θ)whereθ is the angle betweenk andq and thereforeΘ(kF − |k + q|) ≈ Θ(kF − k) − q cos(θ)δ(kF − k)(sinceδ(k) = dΘ(k)/dk). Thus, to leading order inq we have

ℜΠ(q, 0) ≈ 4mP∫

d3k

(2π)3q cos(θ)δ(kF − k)

2qk cos(θ)

≈ mkFπ2

(13.38)

which, when substituted in (13.36), leads to

limq→0

ǫRPA(q, ω = 0) = 1 +4me2kFπq2

, (13.39)

that is identical to the Thomas-Fermi result (13.17). Thus,the Thomas-Fermi approach is a special limit ofthe RPA result whenω = 0 andq → 0. This implies that the RPA result has more information aboutthescreening process in a electron gas than the Thomas-Fermi approach.

It should be clear that the real part of the dielectric function (this is simply related to the real part of thepolarization function in the RPA approach) is related to thescreening processes that go on in the electrongas. The screening process is a coherent process that involves the displacement of electrons in the system.By now you should be curious about the meaning of the imaginary part of the dielectric function and to whatphysical process it describes. In order to understand its meaning it is instructive to look at the frequency-dependent electrical conductivity of the electron gas,σ(k, ω). As we have seen in Chapter 7 in a disorderedenvironment the electron gas has a finite conductivity or resistivity due to the elastic scattering with staticimpurities. But this is not the only source of scattering since the electrons interact with each other. Bydefinition the conductivity of the system is defined by the equation (7.6),J = σE, whereJ is the electriccurrent density. This current is related to the particle current densityj by J = −ej and thus

j(k, ω) = −1

eσ(k, ω)E(k, ω) . (13.40)

On the other hand the number of electrons in the system has to be conserved in the scattering and therefore


the current and the density are related by current conservation equation

∂

∂tn(r, t) + ∇ · j(r, t) = 0 . (13.41)

We are only interested in the fluctuations of the system around the equilibrium particle densityρ0, that is,we writen(r, t) = ρ0 + δρ(r, t) and thus

ω δρk(ω) = k · j(k, ω) . (13.42)

Using thatE(r, t) = −∇φ(r, t) = ∇V (r, t)/e, (13.40), and (13.42) we find that

δρ(k, ω) = iσ(k, ω)k2V (k, ω)

eω. (13.43)

UsingVext(k, ω) = −4πe2ρext(k, ω)/k2, (13.5) and (13.7) it is easy to show that

ǫ(k, ω) = 1 +4πiσ(k, ω)

ω(13.44)

that gives the relationship between the dielectric function and the conductivity. Direct comparison between(13.44) and (13.36) shows that in RPA we have

σRPA(k, ω) = −ie ωk2

Π(k, ω) . (13.45)

Observe therefore that the real part of the conductivity that is related to the dissipative process in the systemis related to the imaginary part of the polarization function. Or more generally, from (13.44) the dissipativeprocesses are related to the imaginary part of the dielectric function. Indeed, from (13.44) one can write

ℜǫ(k, ω) = 1 − 4πℑσ(k, ω)ω

ℑǫ(k, ω) =4πℜσ(k, ω)

ω(13.46)

that relates these two functions. Observe that in experiments we always measure the real part of thesefunctions and through the relations above one can get the imaginary parts as well. Thus the physical meaningof the real and imaginary part are clear now. Moreover, the static conductivity (the one that is measured whena static field is applied to the system) can be obtained from the equations above by taking the limit ofω → 0.However, a word of caution is in place at this point. In calculating that static conductivity one has to takethe limit of q → 0 before we take the limit ofω → 0. The opposite limit, that is,ω → 0 first with q finitedescribes a static electric field that is periodic in space with characteristic wavelength1/q. In this case thecharge distribution is not uniform since it tries to follow the periodicity in the field. Thus, in calculating thestatic conductivity one has always to setq → 0 first.

13.3. RESPONSE OF THE ONE-DIMENSIONAL ELECTRON GAS 249

13.3 Response of the one-dimensional electron gas

Let us apply the methods developed in the last section to study the one-dimensional electron gas. In thiscase the integrals for the polarization function are easierto evaluate. Indeed, using (13.33) we have

ℜΠ(q, ω) =1

π

∫ kF

−kF

dk

(

1

hω − h2kqm − h2q2

2m

− 1

hω + h2kqm + h2q2

2m

)

=m

h2πqln

∣

∣

∣

∣

∣

∣

∣

(

kF + q2

)2 −(

mωhq

)2

(

kF − q2

)2 −(

mωhq

)2

∣

∣

∣

∣

∣

∣

∣

. (13.47)

ℜΠ(q, ω) is depicted in Fig.13.1 forω = 0. Observe thatℜΠ(q, ω) diverges whenq = 2kF for ω = 0.This divergence comes from what is called the nesting of the Fermi surface.

1 2 3 4qkF

0.5

1

1.5

2

2.5

3

3.5

Re@PD

Figure 13.1: Real part of the polarization function for the one dimensional electron gas as a function ofqfor ω = 0.

Nesting it is related to the fact that in some systems it is possible to find a vectork such thatǫq+k = ǫkfor a fixedq. In this case, from (13.32), there is a point in the sum that isdivergent. If this is an isolated pointthen it is not important for the behavior of the polarizationfunction. In one-dimension, however, the Fermisurface is made out of two points,+kF and−kF and therefore the vectorq = ±2kF connects all the pointsin the Fermi surface and leads to the divergence in (13.47). In higher dimensions we need a finite densityof points at the Fermi surface that are nested, otherwise their contribution to (13.32) is a set of measurezero. One of the classic cases where a nesting condition happens is the half-filled tight binding model intwo space dimensions where the Fermi surface has the shape shown in Fig.13.2 and can be connected by awave-vector(π/a, π/a). Because response functions, like the polarization function, have denominators thatinvolve ǫq+k − ǫk, nesting leads to instabilities in the electron gas.


π/a−π/a

π/a

−π/a

k

k

x

y

Nesting vector

Fermi surface

Brillouin zone

Figure 13.2: Nesting of the half filled tight binding model intwo dimensions.

Also from (13.33) we have

ℑΠ(q, ω) =

∫ kF

−kF

dk

[

δ

(

hω − h2kq

m− h2q2

2m

)

− δ

(

hω +h2kq

m+h2q2

2m

)]

=m

h2|q|

[

Θ

(

−ω +hkF q

m− hq2

2m

)

Θ

(

ω +hkF q

m+hq2

2m

)

− Θ

(

−ω +hkF q

m+hq2

2m

)

Θ

(

ω +hkF q

m− hq2

2m

)]

. (13.48)

Observe thatℑΠ(q, ω) has a dependence withq (that is, it behaves like1/q) but it is not finite in thewhole(ω, q) space. Actually it is easy to see that this function is only finite in the shaded area of Fig.13.3.This region is called the particle-hole continuum and it is source of dissipation in the system. Notice thattheℑΠ(q, ω) vanishes atω = 0 for all 0 < q < 2kF . We can show that this behavior is very particular ofone-dimensional systems and in higher dimensionsℑΠ(q, ω) is actually finite in this domain.

13.4 The Kohn anomaly and the Peierls distortion

We have so far discussed the behavior of an electron gas to an external source. The terminology "exter-nal source" is rather loose since the electrons interact with a variety of environmental excitations such asphonons and magnons and also among themselves. Moreover, the electrons interact electrostatically withthe ions in the crystal which, from a broad point of view, are an external source. Indeed, let us consideragain the electron-phonon problem under this new light. In Chapter 9 it was shown that the electron-phononHamiltonian can be written as

H =∑

k,σ

ǫkc†σ,kcσ,k +

∑

q

hωqa†qaq +

∑

q

Uqρq

(

aq + a†−q

)

(13.49)

13.4. THE KOHN ANOMALY AND THE PEIERLS DISTORTION 251

333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

ω

q0 2kF

q + k q2

2m mF

q − k q2

2m mF

q + k q2

2m mF

−

Figure 13.3: Region of the(ω, q) space whereℑΠ(q, ω) is finite.

where the electron-phonon coupling is the same as in (9.59).Observe the close resemblance of (13.49) and(13.21) if we define

Vq = Uq

(

aq + a†−q

)

. (13.50)

This is not just a coincidence since, as we said, the electrons are electrostatically coupled to the lattice.Thus, at least from a perturbative point of view we could see that the electrons respond to the phonons in thesystem and one would suspect that the response should be somewhat related to the polarization function ofthe electron gas. Using the polarization function (13.32) we add and subtractnk,σnk+q,σ in the numerator,regroup the terms together, and changek → −k− q in order to get:

Π(q, ω) = −∑

k,σ

nk,σ (1 − nk+q,σ)

[

1

hω − ǫk+q + ǫk + iη− 1

hω − ǫk + ǫk+q + iη

]

(13.51)

that has the same structure as (9.77). Indeed, let us assume that the coupling between electrons and phononsis weak and use result (13.30) and substitutenq by 〈nq〉 in (13.49) using (13.50) in order to find:

H =∑

q

hωqa†qaq −

∑

q

|Uq|2 Π(q, ωq)(

a−q + a†q

)(

aq + a†−q

)

(13.52)

is a purely bosonic Hamiltonian. The electrons have been traced out of the problem and their effect onlyappears through the polarization function. Notice that in the phonon Hamiltonian we now have operators ofthe formaa anda†a† that do not conserve the number of bosons. This is happening because the electronsare now “hidden" and from the point of view of the bosons everytime an electron emits a boson it is as ifbosons were being produced from vacuum. The full solution ofthe problem involves the diagonalization ofthe Hamiltonian (13.52) that is possible because the problem is quadratic but since we are assuming weakcoupling we neglect the terms that do not conserve the numberof bosons and, apart from a constant factor,


rewrite (13.52) as

H =∑

q

hΩqa†qaq (13.53)

where

Ωq ≈ ωq − 2 |Uq|2 Π(q, ωq) (13.54)

where we have used thatω−q = ωq because of the reflection symmetry of the lattice and that in (13.32)Π(−q, ω) = Π(q, ω). Eq. (13.54) shows that the electrons actually modify the bare frequency of thephonons. This should be expected since what is measured experimentally is not the frequency of the phononsisolated from their environment but the effective frequency that takes into account the interaction of thephonons with their surroundings. We observe two effects: the real part ofΠ(q, ω) renormalizes the phononfrequency while the imaginary part ofΠ(q, ω) produces damping or dissipation of the phonon modes, thatis, when the bare phonon frequency enters a region whereℑΠ(q, ω) is finite the phonon mode becomesunstable and decay into the particle-hole continuum.

Let us consider the application of (13.54) to the one dimensional case. Let us consider the renormaliza-tion of the phonon frequency using (13.47) for the case of theacoustic mode withωq = vsq wherevs is thesound velocity. In this case we find

Ωq = vsq −2m |Uq|2

h2πqln

∣

∣

∣

∣

∣

(

kF + q2

)2 −(

mvsh

)2

(

kF − q2

)2 −(

mvsh

)2

∣

∣

∣

∣

∣

(13.55)

that is plotted in Fig.13.4. Observe that (13.55) has a singularity at q = 2kF (1 + vs/vF ) ≈ 2kF wherevF = hkF /m is the Fermi velocity (notice thatvs ≪ vF ). Because of the logarithmic singularity in (13.55)the phonon frequency vanishes close to2kF in one dimension. In higher dimensions (13.54) predicts a finitefrequency at2kF with a divergent group velocity (dΩq/dq → ∞) that leads to the so-called Kohn anomalyin metals. In one dimension the Kohn anomaly is more severe and leads to the vanishing of the phononfrequency and has strong consequences.

Observe that in first quantized language we can rewrite (13.53) as

H =∑

q

(

PqP−q

2M+MΩ2

q

2XqX−q

)

(13.56)

and therefore the vanishing ofΩq for some wave-vectorq = Q leads to a Hamiltonian of a free particlefor that particular mode. A free particle has an unbounded motion. This leads to the conclusion that thevanishing of the phonon frequency is associated to an instability of the lattice. What this result shows is thatthe lattice becomes “soft" for this particular mode and thatthere is a gain in energy to produce a distortionwith characteristic wavevectorQ, that is, the lattice distorts with a modulation given bycos(Q · r). In thecase of the one dimensional system whereQ = 2kF this distortion is only possible is the electronic densityis commensurate with the lattice. To understand commensuration recall thatQ = 2Nπ/a whereN is anyinteger is a reciprocal lattice vector. Thus, the conditionQ = 2kF implies thatkF = π/(Na). In onedimension one has thatkF = πn/a wheren = N/Ns is the number of electrons per lattice site. Thus, forthe distortion to occur one needsn = 1 that is the condition for commensuration, that is, the number oflattice sites is a multiple of the number of electrons. The simplest case isn = 1 that is, the half-filled band.In this caseQ = π/a and therefore the distortion is of the formcos(πx/a). In this case the unit cell doublessize as shown in Fig.13.5(a,b). Observe that in this case theunit cell doubles its size going froma to 2a.This is called the Peierls distortion and we say that the system is dimerized, that is, transformed to dimers

13.5. ELECTRON-ELECTRON INTERACTIONS 253

1 2 3 4qkF

1

2

3

4

WHqL

Figure 13.4: Dashed line: bare phonon dispersion; Continuous line: result from (13.55).

of atoms. This implies that the Brillouin zone has to shrink from±π/a to ±π/(2a) and a gap has to openat the zone boundary (exactly like in the case of phonons) as shown in Fig.13.5(c,d). It turns out howeverthat the zone boundary is at the Fermi energy and therefore a gap∆ has to open at the Fermi surface andthe system should be insulating. The size of the distortion and the magnitude of the gap cannot be discussedin our picture because these are non-perturbative effects.This effect, however, is seen in one dimensionalsystems such as polymers which, by band structure arguments, should be conductors but by the effect of thePeierls mechanism are actually insulators.

13.5 Electron-electron interactions

So far we have discussed the problem of an electron gas interacting with external charges or the ion latticebut one important question is left out: does our arguments ofscreening apply to the electron gas itself?Do electrons screen their own Coulomb interaction? On the one hand screening requires the electrons tobe indistinguishable from each other and obey the Fermi-Dirac statistics. On the other hand one has to beable to isolate a particular electron in order to apply our argument, that is, we have to be able to distinguishone electron from the others. The problem here is similar to avery familiar puzzle: who was created first,the chicken or the egg? The problem of screening requires oneto separate two aspects of the interactingelectron gas system: the collective electron motion and thesingle electron physics. This is a very delicateand complex problem. One usually has to rely on uncontrolledapproximations. Here we use a mean-field,Hartree-Fock, approach to the electron-electron interaction in three dimensions:

HI =1

2V

∑

q

Uqρqρ−q (13.57)


(a)

(b)

a

2a

kk−kF F

kk−kF F

E

E

∆

(c)

(d)

u

Figure 13.5: (a) Original Lattice; (b) Peierls distortion;(c) Original electron band; (d) Electron band afterPeierls distortion.

whereUq = 4πe2/q2 (where the factor of2 is to avoid double counting). In the mean-field approximationone does the usual substitution:ρqρ−q → 〈ρq〉ρ−q + ρq〈ρ−q〉 that changes (13.57) to

HMF =1

V

∑

q

Vqρq (13.58)

where

Vq = Uq〈ρq〉 . (13.59)

Observe that (13.58) has exactly the form of (13.21) for an external potential. In this picture each electronresponds to the average charge density. Applying the arguments of the previous sections one would arguethat the effective interaction felt by the electrons due to the other electrons would be given (in the staticlimit) by

U effq =Uq

ǫ(q, 0)=

4πe2

q2 + q2TF(13.60)

that is the screened Coulomb interaction. Thus, from this point of view the electrons indeed interact via anscreened (weaker) interaction than the original long rangeCoulomb interaction.

13.6 Landau’s Theory of the Fermi Liquid: An Introduction

In the last sections we have seen that in an electron gas, evenif we start with strong long range Coulombinteractions, screening plays a major role and one ends up with only a mild local interaction among theelectrons. Since the interactions turned out to be weak, theinteracting electron fluid resembles a non-

13.6. LANDAU’S THEORY OF THE FERMI LIQUID: AN INTRODUCTION 255

interacting electrons gas. The main problem here is how to incorporate the electron-electron interactionswithin the properties of the electron gas. Let us imagine that an electron with energyEF is introducedinto the electron fluid as in Fig.13.6(a). As the electron enters the electron gas the other electrons have tomove around to screen its charge. In this case a “correlationhole" with positive charge is formed around theelectron in such way that, as seen from far away, the electronbehaves like a neutral object as in Fig.13.6(b).One imagines that the same picture applies to all the electrons in the system (as we did in the mean fieldtheory described previously). In this case the electrons are dressed by the interactions with all other electronsin the system. Because of this dressing the elementary excitations in the system are not bare electronsbut dressed electrons that we call quasiparticles (as shownin Fig.13.6(c)). Although this picture is quiteappealing we would like to have a more quantitative description. For instance, one would like to know howinteractions affect experimental quantities such as specific heat, magnetic susceptibility and so on.

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

Electron Gas

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

−

−+

(a)

(b)

(c)

Figure 13.6: (a) Electron entering the electron gas; (b) Electron plus its correlation hole; (c) Quasiparticlegas.

A major breakthrough in this field was made by Landau in the 40’s. Landau’s theory of the Fermi liquidis based on two main “intuitive" assumptions:

(1)There is one-to-one correspondence between the quantum numbers of the non-interacting electrongas and the interacting electron gas. Thus, the quasiparticles carry one unit of electric charge, they arespin1/2 excitations and they also carry the momentum quantum numberk. Thus, in Landau’s theory theHilbert space for electrons and quasiparticles is the same and the only possible change in the problem is notdynamic but kinematic. This is also called the adiabatic principle because it assumes that the interactions canbe turned on slowly so that nothing tragic happens in the system (in particular, no phase transition occurs).This principle tells us how the quasiparticles couple to external fields such as the electric and magnetic fields.Since the electric charge is the same as the electrons the electromagnetic field should couple via “minimalcoupling"p− eA/c plus the Zeeman energyS · H.

(2)It is possible to write down an energy functional of the deviations of the quasiparticle occupationrelative to the ground state. As we have seen before we can create excitations in the electron gas by creatingparticle-hole excitations around the Fermi surface. Accordingly to Landau the ground state of an interactingFermi gas is also a Fermi sea of quasiparticles as defined by assumption (1). That is, we can define an


occupation numbernk,σ such that

nk,σ =1

eβ(Ek,σ−µ) + 1(13.61)

whereEk,σ is the quasiparticle energy andµ is the chemical potential of the system. In increasing thetemperature or applying a field to a system what one does is to modify the occupation by an amountδnk,σ.In Landau’s theory the total energy of the system can be written as an expansion in terms of these deviations:

E[δnk,σ ] = E0 +1

V

∑

k,σ

E0k,σδnk,σ +

1

2V

∑

k,σ,k′,σ′

fk,σ,k′,σ′δnk,σδnk′,σ′ (13.62)

plus higher orders inδnk,σ. E0 is just the ground state energy,E0k,σ is the bare dispersion of the quasi-

particles andfk,σ,k′,σ′ is so-called Landau parameter that describes the interaction between quasiparticles.Observe that the actual dispersion of the quasiparticles depend on their interaction. Since we can rewrite(13.62) as

E = E0 +1

V

∑

k,σ

Ek,σδnk,σ

Ek,σ = E0k,σ +

1

V

∑

k,σ

fk,σ,k′,σ′δnk′,σ′ (13.63)

whereEk,σ is the actual quasiparticle dispersion.

In order to understand the physical meaning and consequences of (13.62) let us consider the problemof the specific heat of the electron gas in Landau’s theory. The quasiparticle entropy is given by the usualBoltzmann expression

S = −kBV

∑

k,σ

nk,σ ln(nk,σ) + (1 − nk,σ) ln(1 − nk,σ) . (13.64)

A variationδnk,σ in the distribution function leads to a variationδS in the entropy that is given by

δS = −kBV

∑

k,σ

δnk,σ ln

(

nk,σ

1 − nk,σ

)

=kBβ

V

∑

k,σ

(Ek,σ − µ) δnk,σ (13.65)

where we used (13.61). The specific heat is the variation of the entropy with temperature,δS/δT . Thus,one needs to calculate how the occupationnk,σ changes with temperature. This can be easily obtained from(13.61) as

δnk,σ

δT=Ek,σ − µ

T

(

− ∂nk,σ

∂Ek,σ

)

. (13.66)

Thus, substituting (13.66) into (13.65) we find

δS

δT=

1

V

∑

k,σ

(

− ∂nk,σ

∂Ek,σ

)(

Ek,σ − µ

T

)2

. (13.67)


In order to evaluate the sum in (13.67) we transform it into anintegral by using the usual definition of thequasiparticle density of states:

N(E) =1

V

∑

k,σ

δ(E − Ek,σ) (13.68)

and rewrite (13.67) as

δS

δT=

∫ +∞

−∞dEN(E)

(

E − µ

T

)2(

−∂n(E)

∂E

)

(13.69)

but observe that the derivative in the expression above is highly peaked aroundE = µ (Fermi statistics)while the rest of the argument is a smooth function around this point. Therefore, with great accuracy we canapproximate the above integral as

δS

δT= k2

BN(0)

∫ +∞

−∞dxx2 ex

(ex + 1)2

=π2

3k2BN(0) (13.70)

where we changed variables:x = β(E − µ). The specific heat of the electron gas can be immediatelyobtained from (13.70) since

CV = TδS

δT=π2

3k2BN(0)T (13.71)

looks identical to the free electron gas result (6.99). The temperature dependence of the specific heat isthe same as in the free electron gas but the similarity is onlyapparent because in our definition of thedensity of states for the quasiparticles (13.68) we have used the actual quasiparticle dispersionEk,σ which,accordingly to (13.63), also depends on the Landau parameters fk,σ,k′,σ′ . Thus, in order to be completethe calculation one has to computeN(0) as a function of the interaction parameters. Firstly, we notice thatdue to Landau’s first assumption the number of electrons and the number of quasiparticles have to be thesame. This implies that the Fermi momentum of the quasiparticles is the same as the Fermi momentum ofthe particles,kF = (3πn)3/2, since it depends on the density alone. Let us first go back to the definition ofthe density of states (13.68) and rewrite it as an integral

N(0) =∑

σ

∫

d3k

(2π)3δ(µ− Ek,σ)

=∑

σ

∫

d3k

(2π)3δ(k − kF )

|∇kEk,σ|(13.72)

since by definitionEkF ,σ = µ. Since we are expanding the total energy to second order inδnk,σ we justkeep the quasiparticle energy to first order (take a look at (13.63)). In this case we can replaceEp,σ byE0

p,σ

in (13.72) that depends on the group velocity of the quasiparticles:

vp,σ = ∇pE0p,σ . (13.73)

In a non-interacting Fermi gas the group velocity is simplypF/m wherepF is the Fermi momentum. By


analogy we define the effective mass,m∗, of the quasiparticles as

vp,σ =p

m∗(13.74)

when substituted in (13.72) leads to

N(0) =m∗pF

π2h3 (13.75)

that is the quasiparticle density of states that depends only on the effective mass of the quasiparticles. Thus,what we have done is transferring the problem of calculatingthe density of states to the problem of calculat-ing the effective mass. But life is simpler because the mass is a kinematical quantity that reflects the inertiaof the system. In the case of the interacting electron gas theeffective mass depends on the interaction be-tween the electrons since as one electron moves it has to pushthe other electrons around, that is, the inertiaof the quasiparticle depends on its interaction with the environment.

The most direct way to computem∗ is use an argument due to Landau that assumes Galilean invarianceof the system. Consider observing a Fermi gas from a moving frame with velocityV. If the interactions donot depend on the velocity of the particle (as in the case of the Coulomb interactions) then the only changein the Hamiltonian of the system is in kinetic energy that changes from

K =N∑

i=1

p2i

2m

to

K =

N∑

i=1

(pi −mV)2

2m=

N∑

i=1

p2i

2m− V ·

N∑

i=1

pi +Nm

2V2 (13.76)

since the velocities have to change fromvi to vi−V due to Galilean invariance. Thus, the total energy andmomentum of the system in the moving frame is

Emov = E − P ·V +Nm

2V2

Pmov = P −NmV (13.77)

whereP =∑N

i=1 pi is the total momentum in the rest frame of the electron gas. Consider now adding aquasiparticle with momentump in the rest frame. This leads to an increase in the total mass of the systemby m and the total momentum of the system byp. Thus, from (13.77) we see that momentum of thequasiparticle, as seen in the moving frame is,p−mV while its energy is

Emov,p−mV,σ = Ep,σ − p ·V +mV2

2

Emov,p,σ = Ep+mV,σ − p · V −mV2

2. (13.78)

Now assume thatV is very small and expand (13.78) to first order inV:

Emov,p,σ ≈ Ep,σ + V · (m∇pEp,σ − p) (13.79)


that depends on the quasiparticle energyEp,σ. Using (13.73) and (13.74) in (13.79) leads to

Emov,p,σ ≈ Ep,σ +m−m∗

m∗p · V . (13.80)

On the other hand, from the moving frame the distribution function of the quasiparticles has to change as

nmov,p,σ = np+mV,σ ≈ np,σ +mV · ∇pnp,σ

= np,σ +mV · ∇pEp,σ∂np,σ

∂Ep,σ

= np,σ +m

m∗V · p ∂np,σ

∂Ep,σ. (13.81)

From the change of the occupation we can calculate the energyof the quasiparticle in the moving framefrom (13.63)

Emov,p,σ = Ep,σ +1

V

∑

p′,σ′

fp,σ,p′,σ′m

m∗V · p′ ∂np′,σ

∂Ep′,σ(13.82)

that has to be compared with (13.80). At this point the comparison is not straightforward because we have toevaluate the integral in (13.82). In dealing with an isotropic system (and therefore a spherical Fermi surface)we have an extra symmetry that is the rotations in momentum space. This implies that in the scatteringbetween two quasiparticles with momentump and p′ the interaction parameterfp,σ,p′,σ′ only dependson the angleθ(p,p′) betweenp andp′. In this case we can expandfp,σ,p′,σ′ in Legendre polynomials,Pl(cos(θ)) as

fp,σ,p′,σ′ =

∞∑

l=0

fl,σ,σ′Pl(cos(θ(p,p′))) (13.83)

that can be inverted due to the orthogonality between the Legendre polynomials as

fl,σ,σ′ =2l + 1

2

∫ π

0dθ(p,p′) sin(θ(p,p′))Pl(cos(θ(p,p

′)))fp,σ,p′,σ′

=2l + 1

2

∫ +1

−1du(p,p′)Pl(u(p,p

′))fp,σ,p′,σ′ (13.84)

whereu = cos(θ). Direct substitution of (13.83) into (13.82) leads to

Emov,p,σ = Ep,σ −m

m∗

∑

σ′,l

fl,σ,σ′

∫

d3p′

(2πh)3

(

− ∂np,σ

∂Ep,σ

)

Pl(cos(θ))V · p′

= Ep,σ −m

m∗N(0)

∑

σ′

fl,σ,σ′

6V · p

= Ep,σ −m

m∗

F s13

V · p (13.85)


where we have defined

F sl = N(0)fl,σ,σ + fl,σ,−σ

2

F al = N(0)fl,σ,σ − fl,σ,−σ

2(13.86)

the symmetric and anti-symmetric Landau parameters. Direct comparison of (13.85) with (13.80) we find

m∗

m= 1 +

F s13

(13.87)

that gives the renormalization of the mass in terms of the interactions. The density of states can now beeasily obtained from (13.75).

It is clear that many properties of the interacting electrongas can be directly calculated from Landau’sassumptions. In this case the Landau parameters enter into the theory as phenomenological functions thatdescribe the physics of the interacting Fermi gas. It is possible, therefore, to make many predictions aboutthe physical quantities of interest and check experimentally. The simplicity and power of Landau’s approachshould be clear and its success in explaining the physical properties of many systems such as He3 and metalsmade it the standard approach to the study interacting electron systems.

13.7 Problems

1. Consider a system of electrons moving on a plane. Use a system of coordinates where the threedimensional vectors can be written asR = (r, z) wherer is the vector in the plane. (i) Using theThomas-Fermi approximation show that the Thomas-Fermi screening length is the Bohr radiusa0 anddoes not depend on the density. (ii ) Calculate the effective potential for a chargeQ sitting at the originof the plane (ρext(r) = Qδ(z)δ(r)) and show that at long distances (r ≫ a0) the effective potentialbehaves like

V (r) ≈ −4π2a20eQ

r3.

In this case the potential does not decay exponentially contrary to the problem in three dimensions.

2. Use (13.51) to calculate the polarization function for a three dimensional electron gas with a sphericalFermi surface.Hint: you can find the full solution in Fetter and Walecka, “Quantum Theory of Many-Particle systems" (McGraw-Hill, New York, 1971), pg.158.

3. Calculate the polarization function of a two dimensionalelectron gas in the RPA approximation. (i)PlotℜΠ(q, ω = 0) and show that it does not diverge atq = 2kF . (ii ) In what region of the(ω, q)spaceℑΠ(q, ω) is finite. (iii ) How does the dielectric function compares with the result of question(2)?

4. Consider the Hamiltonian (13.52). (i) Calculate the commutator ofaq with the Hamiltonian and showthat the operatorsaq mixes with the operatora†−q. (ii ) Consider the unitary transformation

bq = uqaq + vqa†−q

what is the condition onuq andvq in order to have canonical commutation relations? (iii ) Showthat the Hamiltonian can be completely diagonalized by the transformation above. What are the newphonon frequencies in this case?

13.7. PROBLEMS 261

5. Consider a one-dimensional electron gas at half filling with hopping matrix elementt0. Considerthe case where the system undergoes a Peierls distortion of size u (that is, the size of the distortionin Fig.13.5). (i) Assume that wavefunction of the electron in each atom has anexponential form,ψ(x) ∼ e−|x|/a0 wherea0 is the lattice spacing. Show that there is a change in hoppingmatrixelement in the system fromt0 to t0 ± 2αu depending on the site (α is a constant).

(ii ) From your result above show that in the tight binding approximation the Hamiltonian for theelectrons of the problem can be written as

He = −∑

n

(t0 + 2αu(−1)n)(

c†ncn+1 + c†n+1cn

)

. (13.88)

(iii ) Show based on general arguments that the elastic energy dueto the Peierls distortion can bewritten as

Hd = 2NKu2 (13.89)

whereN is the number of atoms andK is the spring constant.

(iv) Fourier transform (13.88) and show that it can be written as

He =∑

k

[

ǫk

(

b†kbk − a†kak

)

+ 2αu sin(ka0)(

b†kak + a†kbk

)]

(13.90)

whereǫk = −2t0 cos(ka0), ak = ck andbk = −ick±Q (with Q = π/a) and the sum overk goesfrom −Q/2 toQ/2 (that is, the new Brillouin zone).

(v) Observe that (13.90) is quadratic and can be diagonalized.Define the following unitary transfor-mation

αk = ukak + vkbk

βk = ukbk − vkak (13.91)

whereuk andvk are constants. Show thatu2k + v2

k = 1. Use (13.91) to diagonalize (13.90) and showthat the new Hamiltonian can be written as

He =∑

k

Ek

(

β†kβk − α†kαk

)

(13.92)

whereEk =√

ǫ2k + ∆2k and

∆k = 4αu sin(ka0) . (13.93)

Make a plot of±Ek and show that a gap opens atk = π/(2a). Give a physical meaning to theoperatorsβk andαk.

(vi) Show that

uk =

√

1

2

(

1 +ǫkEk

)

vk =

√

1

2

(

1 − ǫkEk

)

sgn(k)


wheresgn(k) = +1(−1) if k > 0(k < 0).

(v) Assume that the electron system is half filled and show that the total energy of the system (elec-tron+distortion) is given by

E(u)

N= −4t0

πE

[

1 −(

2αu

t0

)2]

+ 2Ku2 (13.94)

whereE[x] =∫ π/20 dt

√

1 − x sin2(t) is an Elliptic integral.

(vi) Show that foru≪ t0/α the energy of the system can be written as

E(u)

N≈ −4t0

π+

[

2K +4α2

πt0− 8α2

αt0ln

(

2t0αu

)]

u2 . (13.95)

What happens whenu→ ∞? From these two results prove that no matter how smallα there is alwaysa value ofu = u0 6= 0 so that the energy is minimum. Thus, you just proved that, no matter howsmall the coupling constantα, the one dimensional system is always dimerized (Peierls theorem).

(vii) Using (13.95) show that

u0 ≈ 2t0αe−1−

πt0K

4α2

for u≪ t0/α. Observe thatu0 is not an analytic function ofα.

6. In this exercise we are going to consider the compressibility of the electron gas in the presence ofinteractions. The physical quantity of interest is the compressibility,κ (or bulk modulus). The com-pressibility is defined as

κ = − 1

V

∂V

∂P=

1

n2

∂n

∂mu

whereV is the volume,P is pressure,n = N/V is the particle density andµ is the chemical potential.Thus, in order to calculateκ one has to consider the change in the quasiparticle occupation, δnk,σ inrespect to the chemical potential.

(i) Using (13.61) show that

δnk,σ = (δEk,σ − δµ)∂nk,σ

∂Ek,σ. (13.96)

(ii ) Show that

δEk,σ =1

V

∑

k′,σ′

fk,σ,k′,σ′δnk′,σ′ (13.97)

and from that prove that

δEk,σ =F s0N(0)

δn

whereδn = 1V

∑

k,σ δnk,σ is the total variation in the number of quasiparticles.

13.7. PROBLEMS 263

(iii ) Using the results of item (i) and (ii ) show that

κ =1

n2

N(0)

1 + F s0.

What is the physical meaning of this result?

7. In the exercise you will calculate the magnetic susceptibility of the interacting electron gas in theLandau approach. Let us consider only the Zeeman energy thatis given by

Hz = −gµBN∑

i=1

Si ·H

whereS is the spin of the quasiparticles which, according to Landau’s assumption, is±h/2.

(i) What is the change in the quasiparticle energy in the presence of the field? What is the differencefrom (13.97)?

(ii ) The change in the occupation is still given by (13.96). Notice however that the change in thechemical potential cannot depend on the direction of the field H and therefore can only be of orderH2. Thus, neglecting the change in the chemical potential due to the field and using the result of item(i) show that

δnσ =N(0)gµB2(1 + F a0 )

σHz

whereδnσ = 1V

∑

k δnk,σ, andHz is the z-component of the magnetic field. From the above resultcalculate the total magnetization of the electron gas and show that the magnetic susceptibility is givenby

χ =N(0)(gµB)2

1 + F a0.

What is the physical meaning of this result ifF a0 < 0?

Introduction to Condensed Matter Physics

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