IV. Vibrational Properties of the Lattice

A. Heat Capacity—Einstein Model

B. The Debye Model — Introduction

C. A Continuous Elastic Solid

D. 1-D Monatomic Lattice

E. Counting Modes and Finding N()

F. The Debye Model — Calculation

G. 1-D Lattice With Diatomic Basis

H. Phonons and Conservation Laws

I. Dispersion Relations and Brillouin Zones

J. Anharmonic Properties of the Lattice

A. Heat Capacity—Einstein Model (1907)

Having studied the structural arrangements of atoms in solids, we now turn to properties of solids that arise from collective vibrations of the atoms about their equilibrium positions.

2212

212

212

212

212

21

1

zkykxkmvmvmv

UKE

zyxzyx

m

kykx

kzFor a vibrating atom:

Classical statistical mechanics — equipartition theorem: in thermal equilibrium each quadratic term in the E has an average energy , so: TkB2

1

TkTkE BB 3)(6 21

1

Classical Heat Capacity

For a solid composed of N such atomic oscillators:

Giving a total energy per mole of sample:

TNkENE B31

RTTkNn

TNk

n

EBA

B 333

So the heat capacity at constant volume per mole is: Kmol

J

V

V Rn

E

dT

dC 253

This law of Dulong and Petit (1819) is approximately obeyed by most solids at high T ( > 300 K). But by the middle of the 19th century it was clear that CV 0 as T 0 for solids.

So…what was happening?

Einstein Uses Planck’s Work

Planck (1900): vibrating oscillators (atoms) in a solid have quantized energies ...,2,1,0 nnEn

[later QM showed is actually correct] 21 nEn

...,2,1,0 nnEn

Einstein (1907): model a solid as a collection of 3N independent 1-D oscillators, all with constant , and use Planck’s equation for energy levels

occupation of energy level n: (probability of oscillator being in level n)

0

/

/

)(

n

kTE

kTE

nn

n

e

eEf classical physics

(Boltzmann factor)

Average total energy of solid:

0

/

0

/

0

3)(3

n

kTE

n

kTEn

nnn

n

n

e

eENEEfNUE

Some Nifty Summing

0

/

0

/

3

n

kTn

n

kTn

e

enNU

Using Planck’s equation: Now let

kTx

0

03

n

nx

n

nx

e

enNU

0

0

0

0 33

n

nx

n

nx

n

nx

n

nx

e

edxd

Ne

edxd

NU Which can be rewritten:

Now we can use the infinite sum:

11

1

0

xforx

xn

n

1

3

1

3

1

13

/

kTx

x

x

x

x

e

N

e

N

ee

ee

dxd

NU

To give: 11

1

0

x

x

xn

nx

e

e

ee

So we obtain:

At last…the Heat Capacity!

Differentiating:

Now it is traditional to define an “Einstein temperature”:

Using our previous definition:

So we obtain the prediction:

1

3/ kTA

VV e

N

dT

d

n

U

dT

dC

2/

/2

2/

/

1

3

1

3 2

kT

kTkT

kT

kT

kTA

Ve

eR

e

eNC

kE

2/

/2

1

3)(

T

TT

VE

EE

e

eRTC

Limiting Behavior of CV(T)

Low T limit:

These predictions are qualitatively correct: CV 3R for large T and CV 0 as T 0:

High T limit: 1TE

RR

TCT

TTV

E

EE

311

13)( 2

2

1TE

TTT

TT

VEE

E

EE

eRe

eRTC /2

2/

/2

33

)(

3R

CV

T/E

But Let’s Take a Closer Look:

High T behavior: Reasonable agreement with experiment

Low T behavior: CV 0 too quickly as T 0 !

B. The Debye Model (1912)

Despite its success in reproducing the approach of CV 0 as T 0, the Einstein model is clearly deficient at very low T. What might be wrong with the assumptions it makes?

• 3N independent oscillators, all with frequency

• Discrete allowed energies: ...,2,1,0 nnEn

Details of the Debye Model

Pieter Debye succeeded Einstein as professor of physics in Zürich, and soon developed a more sophisticated (but still approximate) treatment of atomic vibrations in solids.

Debye’s model of a solid:

• 3N normal modes (patterns) of oscillations

• Spectrum of frequencies from = 0 to max

• Treat solid as continuous elastic medium (ignore details of atomic structure)

This changes the expression for CV because each mode of oscillation contributes a frequency-dependent heat capacity and we now have to integrate over all :

dTCNTC EV ),()()(max

0

# of oscillators per unit

Einstein function for one oscillator

C. The Continuous Elastic Solid

We can describe a propagating vibration of amplitude u along a rod of material with Young’s modulus E and density with the wave equation:

2

2

2

2

x

uE

t

u

for wave propagation along the x-direction

By comparison to the general form of the 1-D wave equation:

2

22

2

2

x

uv

t

u

we find thatE

v So the wave speed is independent of wavelength for an elastic medium!

kvv

f

22

k

)(k is called the dispersion relation of the solid, and here it is linear (no dispersion!)

dk

dvg

group velocity

D. 1-D Monatomic LatticeBy contrast to a continuous solid, a real solid is not uniform on an atomic scale, and thus it will exhibit dispersion. Consider a 1-D chain of atoms:

In equilibrium:

1su

Longitudinal wave:

Ma

su 1su psu

1s s 1s ps

For atom s, p

spsps uucFp = atom labelp = 1 nearest neighborsp = 2 next nearest neighborscp = force constant for atom p

1-D Monatomic Lattice: Equation of Motion

Thus: p

tksaitapskip

tksai ueueceiMu )())(()(2)(

For the expected harmonic traveling waves, we can write xs = sa = position of atom s

)( tkxis

sueu

Now we use Newton’s second law:

p

spsps

s uuct

uMF

2

2

Or:

p

ikpap

tksaitksai eceeM 1)()(2

So: p

ikpap ecM 12 Now since c-p = cp by symmetry,

00

2 1)cos(22p

pp

ikpaikpap kpaceecM

1-D Monatomic Lattice: Solution!

The result is:

0

212

0

2 )(sin4

))cos(1(2

pp

pp kpac

Mkpac

M

The dispersion relation of the monatomic 1-D lattice!

Often it is reasonable to make the nearest-neighbor approximation (p = 1): )(sin

421212 ka

M

c

The result is periodic in k and the only unique solutions that are physically meaningful correspond to values in the range:

ak

a

k

a

a

2

a

a

2 0

M

c14

Dispersion Relations: Theory vs. Experiment

In a 3-D atomic lattice we expect to observe 3 different branches of the dispersion relation, since there are two mutually perpendicular transverse wave patterns in addition to the longitudinal pattern we have considered.

Along different directions in the reciprocal lattice the shape of the dispersion relation is different. But note the resemblance to the simple 1-D result we found.

E. Counting Modes and Finding N()

A vibrational mode is a vibration of a given wave vector (and thus ), frequency , and energy . How many modes are found in the interval between and ?

Ek

),,( kE

),,( kdkdEEd

# modes kdkNdEENdNdN

3)()()(

We will first find N(k) by examining allowed values of k. Then we will be able to calculate N() and evaluate CV in the Debye model.

First step: simplify problem by using periodic boundary conditions for the linear chain of atoms:

x = sa x = (s+N)a

L = Na

s

s+N-1

s+1

s+2

We assume atoms s and s+N have the same displacement—the lattice has periodic behavior, where N is very large.

First: finding N(k)

This sets a condition on allowed k values: ...,3,2,1

22 n

Na

nknkNa

So the separation between allowed solutions (k values) is:

independent of k, so the density of modes in k-space is uniform

Since atoms s and s+N have the same displacement, we can write:

Nss uu ))(()( taNskitksai ueue ikNae1

Nan

Nak

22

Thus, in 1-D: 22

1 LNa

kspacekofinterval

modesof#

Next: finding N()

Now for a 3-D lattice we can apply periodic boundary conditions to a sample of N1 x N2 x N3 atoms:

N1aN2b

N3c

)(8222 3

321 kNVcNbNaN

spacekofvolume

modesof#

Now we know from before that we can write the differential # of modes as:

kdkNdNdN

3)()( kdV

338

We carry out the integration in k-space by using a “volume” element made up of a constant surface with thickness dk:

dkdSdkareasurfacekd )(3

N() at last!

A very similar result holds for N(E) using constant energy surfaces for the density of electron states in a periodic lattice!

dkdSV

dNdN

38)(Rewriting the differential

number of modes in an interval:

We get the result:k

dSV

d

dkdS

VN

1

88)(

33

This equation gives the prescription for calculating the density of modes N() if we know the dispersion relation (k).

We can now set up the Debye’s calculation of the heat capacity of a solid.

F. The Debye Model Calculation

We know that we need to evaluate an upper limit for the heat capacity integral:

dTCNTC EV ),()()(max

0

• 3 independent polarizations (L, T1, T2) with equal propagation speeds vg

• continuous, elastic solid: = vgk

• max given by the value that gives the correct number of modes per polarization (N)

If the dispersion relation is known, the upper limit will be the maximum value. But Debye made several simple assumptions, consistent with a uniform, isotropic, elastic solid:

N() in the Debye Model

First we can evaluate the density of modes:

Next we need to find the upper limit for the integral over the allowed range of frequencies.

dk

dvg

dSv

V

vdS

VN

gg33 8

1

8)(

24 kdS Since the solid is isotropic, all directions in k-space are the same, so the constant surface is a sphere of radius k, and the integral reduces to:

k

32

22

3 24

8)(

gg v

Vk

v

VN

Giving: for one polarization

max in the Debye Model

Since there are N atoms in the solid, there are N unique modes of vibration for each polarization. This requires:

The Debye cutoff frequency

NdN

max

0

)(

Giving: Nv

Vd

v

V

gg

32

3max

0

232 62

max

Dg V

Nv

3/12

max

6

Now the pieces are in place to evaluate the heat capacity using the Debye model! This is the subject of problem 5.2 in Myers’ book. Remember that there are three polarizations, so you should add a factor of 3 in the expression for CV. If you follow the instructions in the problem, you should obtain:

T

z

z

DBV

D

e

dzezTNkTC

/

02

43

)1(9)(

And you should evaluate this expression in the limits of low T (T << D) and high T (T >> D).

Debye Model: Theory vs. Expt.

Universal behavior for all solids!

Debye temperature is related to “stiffness” of solid, as expected

Better agreement than Einstein model at low T

Debye Model at low T: Theory vs.

Expt.

Quite impressive agreement with predicted CV T3 dependence for Ar! (noble gas solid)

(See SSS program debye to make a similar comparison for Al, Cu and Pb)

G. 1-D Lattice with Diatomic Basis

Consider a linear diatomic chain of atoms (1-D model for a crystal like NaCl):

In equilibrium:

M1

a

M2 M1 M2

Applying Newton’s second law and the nearest-neighbor approximation to this system gives a dispersion relation with two “branches”:

2/1

212

21

21

2

21

2121

21

211

2 )(sin4

ka

MM

c

MM

MMc

MM

MMc

-(k) 0 as k 0 acoustic modes (M1 and M2 move in phase)

+(k) max as k 0 optical modes (M1 and M2 move out of phase)

1-D Lattice with Diatomic Basis: Results

These two branches may be sketched schematically as follows:

gap in allowed frequencies

k

a

a

0

optical

acoustic

In a real 3-D solid the dispersion relation will differ along different directions in k-space. In general, for a p atom basis, there are 3 acoustic modes and p-1 groups of 3 optical modes, although for many propagation directions the two transverse modes (T) are degenerate.

Diatomic Basis: Experimental Results

The optical modes generally have frequencies near = 1013 1/s, which is in the infrared part of the electromagnetic spectrum. Thus, when IR radiation is incident upon such a lattice it should be strongly absorbed in this band of frequencies.

At right is a transmission spectrum for IR radiation incident upon a very thin NaCl film. Note the sharp minimum in transmission (maximum in absorption) at a wavelength of about 61 x 10-4 cm, or 61 x 10-6 m. This corresponds to a frequency = 4.9 x 1012 1/s.

If instead we measured this spectrum for LiCl, we would expect the peak to shift to higher frequency (lower wavelength) because MLi < MNa…exactly what happens!

H. Phonons and Conservation Laws

Collective motion of atoms = “vibrational mode”:

)(),( tkxis

suetxu

Quantum harmonic oscillator: ...,2,1,0 nnEn

Energy content of a vibrational mode of frequency is an integral number of energy quanta . We call these quanta “phonons”. While a photon is a quantized unit of electromagnetic energy, a phonon is a quantized unit of vibrational (elastic) energy.

Associated with each mode of frequency is a wavevector , which leads to the definition of a “crystal momentum”:

k

k

Crystal momentum is analogous to but not equivalent to linear momentum. No net mass transport occurs in a propagating lattice vibration, so the linear momentum is actually zero. But phonons interacting with each other or with electrons or photons obey a conservation law similar to the conservation of linear momentum for interacting particles.

Phonons and Conservation Laws

Lattice vibrations (phonons) of many different frequencies can interact in a solid. In all interactions involving phonons, energy must be conserved and crystal momentum must be conserved to within a reciprocal lattice vector:

Gkkk

321

321

22 k

11 k

33 k

Schematically:

Compare this to the special case of elastic scattering of x-rays with a crystal lattice: Gkk

Photon wave vectors

Just a special case of the general conservation law!

I. Brillouin Zones of the Reciprocal Lattice

Remember the dispersion relation of the 1-D monatomic lattice, which repeats with period (in k-space) :a/2

1st Brillouin Zone (BZ)

k

a

a

2

a

a

2 0

M

c14

a

3a

4

a

3

a

4

2nd Brillouin Zone

3rd Brillouin Zone

Each BZ contains identical information about the lattice

Wigner-Seitz Cell--Construction

For any lattice of points, one way to define a unit cell is to connect each lattice point to all its neighboring points with a line segment and then bisect each line segment with a perpendicular plane. The region bounded by all such planes is called the Wigner-Seitz cell and is a primitive unit cell for the lattice.

1-D lattice: Wigner-Seitz cell is the line segment bounded by the two dashed planes

2-D lattice: Wigner-Seitz cell is the shaded rectangle bounded by the dashed planes

1st Brillouin Zone--Definition

Gkk

1k

a

a

2

a

a

2 0

M

c14

a

3a

4

a

3

a

4 k

1k

G

The Wigner-Seitz cell can be defined for any kind of lattice (direct or reciprocal space), but the WS cell of the reciprocal lattice is also called the 1st Brillouin Zone.

The 1st BZ is the region in reciprocal space containing all information about the lattice vibrations of the solid. Only the values in the 1st BZ correspond to unique vibrational modes. Any outside this zone is mathematically equivalent to a value inside the 1st BZ. This is expressed in terms of a general translation vector of the reciprocal lattice:

k

1k k

1st Brillouin Zone for 3-D Lattices

For 3-D lattices, the construction of the 1st Brillouin Zone leads to a polyhedron whose planes bisect the lines connecting a reciprocal lattice point to its neighboring points. We will see these again!

bcc direct lattice fcc reciprocal lattice fcc direct lattice bcc reciprocal lattice

IJ. Anharmonic Properties of Solids

Two important physical properties that ONLY occur because of anharmonicity in the potential energy function:

dxe

dxxe

xkTxU

kTxU

/)(

/)(

1. Thermal expansion2. Thermal resistivity (or finite thermal conductivity)

Thermal expansion

In a 1-D lattice where each atom experiences the same potential energy function U(x), we can calculate the average displacement of an atom from its T=0 equilibrium position:

IThermal Expansion in 1-D

Evaluating this for the harmonic potential energy function U(x) = cx2 gives:

dxe

dxxe

xkTcx

kTcx

/

/

2

2

Thus any nonzero <x> must come from terms in U(x) that go beyond x2. For HW you will evaluate the approximate value of <x> for the model function

Now examine the numerator carefully…what can you conclude?

!0x independent of T !

),0,,()( 43432 kTfxgxandfgcfxgxcxxU

Why this form? On the next slide you can see that this function is a reasonable model for the kind of U(r) we have discussed for molecules and solids.

Potential Energy of Anharmonic Oscillator(c = 1 g = c/10 f = c/100)

0

2

4

6

8

10

12

14

16

-5 -3 -1 1 3 5

Displacement x (arbitrary units)

Pot

ential

Ene

rgy

U (ar

b. u

nits

)

U = cx2 - gx3 - fx4 U = cx2

Do you know what form to expect for <x> based on experiment?

Lattice Constant of Ar Crystal vs. Temperature

Above about 40 K, we see: TxaTa )0()(

Usually we write: 00 1 TTLL = thermal expansion coefficient

Thermal Resistivity

When thermal energy propagates through a solid, it is carried by lattice waves or phonons. If the atomic potential energy function is harmonic, lattice waves obey the superposition principle; that is, they can pass through each other without affecting each other. In such a case, propagating lattice waves would never decay, and thermal energy would be carried with no resistance (infinite conductivity!). So…thermal resistance has its origins in an anharmonic potential energy.

Classical definition of thermal conductivity

vCV3

1

VCwave velocity

heat capacity per unit volume

mean free path of scattering (would be if no anharmonicity)

v

high Tlow T

dx

dTJ

Thermal energy flux (J/m2s)

Phonon Scattering

There are three basic mechanisms to consider:

2. Sample boundaries (surfaces)3. Other phonons (deviation from harmonic behavior)

deviation from perfect crystalline order

1. Impurities or grain boundaries in polycrystalline sample

VC 11 / kT

ph

en

ThighR

TlowT

3

3Thigh

kT

Tlow

VC vTo understand the experimental dependence , consider limiting values of and (since does not vary much with T).

)(T

Temperature-Dependence of

The low and high T limits are summarized in this table:

CV

low T T3

nph 0, so

, but then

D (size)

T3

high T 3R 1/T 1/T

How well does this match experimental results?

Experimental (T)

T3

T-1 ?

(not quite)

Phonon Collisions: (N and U Processes)How exactly do phonon collisions limit the flow of heat?

2-D lattice 1st BZ in k-space:

1q

2q

3q

a2

a2

321 qqq

No resistance to heat flow(N process; phonon momentum conserved)

Predominates at low T << D since and q will be small

Phonon Collisions: (N and U Processes)What if the phonon wavevectors are a bit larger?

2-D lattice 1st BZ in k-space:

1q

2qa

2

a2

Gqqq

321

Two phonons combine to give a net phonon with an opposite momentum! This causes resistance to heat flow.(U process; phonon momentum “lost” in units of ħG.)

More likely at high T >> D since and q will be larger

21 qq

G

3q

Umklapp = “flipping over” of wavevector!