PhononsThe Quantum Mechanics of Lattice Vibrations

What is a Phonon?• We’ve seen that the physics of lattice vibrations in a crystalline solid

reduces to a CLASSICAL normal mode problem.

The goal of the entire discussion has been to

find the normal mode vibrational frequencies of the solid. • In the harmonic approximation, this is achieved by first writing the

solid’s vibrational energy as a system of coupled simple harmonic oscillators & then finding the classical normal mode frequencies & ion displacements for that system.

• Given the results of these classical normal mode calculations for the vibrating solid, in order to treat some properties of the solid,

it is necessary to QUANTIZE these normal modes.

• These quantized normal modes of vibration are called

PHONONS• PHONONS are massless quantum mechanical particles which have no

classical analogue.– They behave like particles in momentum space or k space.

• Phonons are one example of many like this in many areas of physics. Such quantum mechanical particles are often called

“Quasiparticles”Examples of other Quasiparticles:

Photons: Quantized Normal Modes of electromagnetic waves.

Rotons: Quantized Normal Modes of molecular rotational excitations.

Magnons: Quantized Normal Modes of magnetic excitations in magnetic solids

Excitons: Quantized Normal Modes of electron-hole pairs

Polaritons: Quantized Normal Modes of electric polarization excitations in solids

+ Many Others!!!

sphonon

hE

PHONONS • Quantized normal modes of lattice vibrations. The

energies & momenta of phonons are quantized

Phonon wavelength: λphonon ≈ a0 ≈ 10-10 m

phonon

hp

photon

hcE

photon

hp

Comparison of Phonons & PhotonsPHOTONS

• Quantized normal modes of electromagnetic waves. The energies & momenta of photons are quantized

Photon wavelength (visible): λphoton ≈ 10-6 m

2

1nn n = 0,1,2,3,..

E

Quantum Mechanical Simple Harmonic Oscillator

• Quantum mechanical results for a simple harmonic oscillator with classical frequency ω: The energy is quantized

En

• Energy levels are equally spaced!

Often, we consider En as being constructed by adding n excitation quanta of energy to the ground state.

2

10

If the system makes a transition from a lower energy level to a higher energy level, it is always true that the change in energy is an integer multiple of

Phonon absorption or emission

E0 Ground state energy of the oscillator.

ΔE = (n – n΄) n & n ΄ = integers

In complicated processes, such as phonons interacting with electrons or photons, it is known that phonons are not conserved. That is, they can be created and destroyed during such interactions.

Thermal Energy & Lattice Vibrations

As we’ve been discussing in detail, the atoms in a crystal vibrate about their equilibrium positions.

This motion produces vibrational waves.

The amplitude of this vibrational motion increases as the temperature increases.

In a solid, the energy associated with these vibrations is called Thermal

Energy

• A knowledge of the thermal energy is fundamental to obtaining an understanding many of the basic properties of solids. A relevant question is how do we calculate this thermal energy?

• Also, we would like to know how much thermal energy is available to scatter a conduction electron in a metal or semiconductor. This is important; this scattering contributes to electrical resistance in the material.

• Most important, though, this thermal energy plays a fundamental role in determining the

Thermal Properties of a Solid

• A knowledge of how the thermal energy changes with temperature gives an understanding of the heat energy which is necessary to raise the temperature of the material.

• An important, measureable property of a solid is it’s

Specific Heat or Heat Capacity

The thermal energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution. Some other contributions:

Conduction Electrons in metals & semiconductors.

The magnetic ordering in magnetic materials.

Calculation of the vibrational contribution to the thermal energy & heat capacity of a solid has 2 parts:

1. Evaluation of the contribution of a single vibrational mode.

2. Summation over the frequency distribution of the modes.

Lattice Vibrational Contribution to the Heat Capacity

Vibrational Specific Heat of Solids

cp Data at T = 298 K

nn

nP _

Average energy of a harmonic oscillator and hence of a lattice mode of angular frequency at temperature T

Energy of oscillator

1

2n n

The probability of the oscillator being in this level as given by the Boltzman factor

exp( / )n Bk T

Thermal Energy & Heat Capacity Einstein Model

0

/ 2 3 / 2 5 / 2

/ 2 / 2 /

/ 2 / 1

1exp[ ( ) ]

2

.....

(1 .....

(1 )

B B B

B B B

B B

n B

k T k T k T

k T k T k T

k T k T

z nk T

z e e e

z e e e

z e e

_0

0

1 1exp /

2 2

1exp /

2

Bn

Bn

n n k T

n k T

(*)

According to the Binomial expansion for x«1 where

/ Bx k T

nn

nP _

Eqn (*) can be written

_

/

1

2 1Bk Te

'

(ln )x

xx x

_2 2

/ 2_2

/

_/ 2 /2

_/2

/2 2_

22 2 /

1(ln )

ln1

ln ln 1

ln 12

2

4 1

B

B

B B

B

B

B

B B

k T

B k T

k T k TB

k TB

B

k TB

B BB k T

B

zk T k T z

z T T

ek T

T e

k T e eT

k T eT k T T

ke

k k Tk T

k T e

/

/

1

2 1

B

B

k T

k T

e

e

Finally, the result is

_

/

1

2 1Bk Te

This is the Mean Phonon Energy. The first term in the above equation is the zero-point energy. As mentioned before even at 0ºK atoms vibrate in the crystal and have zero-point energy. This is the minimum energy of the system.

The average number of phonons is given by the Bose-Einstein distribution as

1

1)(

TBken

The second term in the mean energy is the phonon contribution to the thermal energy.

(number of phonons) x (energy of phonon) = (second term in )_

Mean energy of a harmonic oscillator as a function of T

Low Temperature Limit

T

2

1

TkB

TkB

12

1_

TBke

2

1_

Zero Point Energy

Since exponential term gets bigger

is independent of frequency of oscillation.

This is the classical limit because the energy steps are now small compared with the energy of the harmonic oscillator.

So that is the thermal energy of the classical 1D harmonic oscillator.

..........!2

12

x

xex

Tke

B

TBk

1

112

1_

TkB

_ 1

2 Bk T

_

Bk T

High Temperature LimitT

2

1

TkB Mean energy of a harmonic oscillator as a function of T

Bk T <<

Heat Capacity C

• Heat capacity C can be found by differentiating the average phonon energy

12

1_

TBke

2

2

1

k TB

k TB

B

Bv

ke

k TdC

dTe

2

2 2

1

k TB

k TB

v B

B

eC k

k T e

k

2

2

1

T

T

v B

eC k

T e

Let

2

2

1

T

T

v B

eC k

T e

Area =2

T

Bk

Bk

Specific heat in this approximation vanishes exponentially at low T and tends to classical value at high temperatures.

These features are common to all quantum systems; the energy tends to the zero-point-energy at low T and to the classical value of Boltzmann constant at high T.

vC

k

where

,T K

3R This range usually includes RT. From the figure it is seen that Cv is equal to 3R at high temperatures regardless of the substance. This fact is known as Dulong-Petit law. This law states that specific heat of a given number of atoms of any solid is independent of temperature and is the same for all materials!

vC

Specific heat at constant volume depends on temperature as shown in figure below. At high temperatures the value of Cv is close to 3R, where R is the universal gas constant. Since R is approximately 2 cal/K-mole, at high temperatures Cv is app. 6 cal/K-mole.

Cv vs T for Diamond

Points:

Experiment

Curve:

Einstein Model

Prediction

Classical Theory of Heat Capacity of Solids The solid is one in which each atom is bound to its side

by a harmonic force. When the solid is heated, the atoms vibrate around their sites like a set of harmonic oscillators. The average energy for a 1D oscillator is kT. Therefore, the averaga energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is

=

where N is Avagadro’s number, kB is Boltzmann constant and R is the gas constant. The differentiation wrt temperature gives;

3 3BNk T RT

23 233 3 6.02 10 ( / ) 1.38 10 ( / )vC R atoms mole J K

24.9 ;1 0.2388 6( ) ( )

J CalCv J Cal Cv

K mole K mole

v

dC

dT

Einstein heat capacity of solids• The theory explained by Einstein is the first quantum theory of solids.

He made the simplifying assumption that all 3N vibrational modes of a

3D solid of N atoms had the same frequency, so that the whole solid had

a heat capacity 3N times

• In this model, the atoms are treated as independent oscillators, but the energy

of the oscillators are taken quantum mechanically as

This refers to an isolated oscillator, but the atomic oscillators in a solid are not

isolated.They are continually exchanging their energy with their surrounding

atoms.

• Even this crude model gave the correct limit at high temperatures, a heat

capacity of the Dulong-Petit law where R is universal gas constant.

2

2

1

T

T

v B

eC k

T e

3 3BNk R

• At high temperatures, all crystalline solids have a specific heat of 6 cal/K per mole; they require 6 calories per mole to raise their temperature 1 K.

•This arrangement between observation and classical theory break down if the temperature is not high.

•Observations show that at room temperatures and below the specific heat of crystalline solids is not a universal constant.

6cal

Kmol

Bk

vC

T3vC R

In each of these materials (Pb,Al, Si,and Diamond) specific heat approaches constant value asymptotically at high T. But at low T’s, the specific heat decreases towards zero which is in a complete contradiction with the above classical result.

• Einstein model also gave correctly a specific heat tending to zero at absolute zero, but the temperature dependence near T= 0 did not agree with experiment.

• Taking into account the actual distribution of vibration frequencies in a solid this discrepancy can be accounted using one dimensional model of monoatomic lattice

Density of States

According to Quantum Mechanics if a particle is constrained;• the energy of particle can only have special discrete energy values.• it cannot increase infinitely from one value to another.• it has to go up in steps.

Thermal Energy & Heat Capacity Debye Model

• These steps can be so small depending on the system that the energy can be considered as continuous.

• This is the case of classical mechanics.• But on atomic scale the energy can only jump by a

discrete amount from one value to another.

Definite energy levels Steps get small Energy is continuous

• In some cases, each particular energy level can be associated with more than one different state (or wavefunction )

• This energy level is said to be degenerate.

• The density of states is the number of discrete states per unit energy interval, and so that the number of states between and will be .

( )

( )d d

There are two sets of waves for solution;

• Running waves

• Standing waves

0

2

L

2

L

4

L

4

L

6

L

k

These allowed k wavenumbers corresponds to the running waves; all positive and negative values of k are allowed. By means of periodic boundary condition

2 2 2NaL Na p k p k p

p k Na L

an integer

Length of the 1D chain

Running waves:

These allowed wavenumbers are uniformly distibuted in k at a density of between k and k+dk. R k

running waves 2R

Lk dk dk

In some cases it is more suitable to use standing waves,i.e. chain with fixed ends. Therefore we will have an integral number of half wavelengths in the chain;

Standing waves:

0L

2

L

6

L

4

L

5

L

2 2;

2 2

n n nL k k k

L L

3

L

7

L

k0

L

2

L

3

L

These are the allowed wavenumbers for standing waves; only positive values are allowed.

2k p

L

for

running wavesk p

L

for

standing waves

These allowed k’s are uniformly distributed between k and k+dk at a density of

( )S

Lk dk dk

2R

Lk dk dk

DOS of standing wave

DOS of running wave

( )S k

•The density of standing wave states is twice that of the running waves.

•However in the case of standing waves only positive values are allowed

•Then the total number of states for both running and standing waves will be the same in a range dk of the magnitude k

•The standing waves have the same dispersion relation as running waves, and for a chain containing N atoms there are exactly N distinct states with k values in the range 0 to ./ a

modes with frequency from to +d corresponds

modes with wavenumber from k to k+dk

The density of states per unit frequency range g():

• The number of modes with frequencies and +d will be g()d.

• g() can be written in terms of S(k) and R(k).

dn

dR

Choose standing waves to obtain ( )g

Let’s remember dispertion relation for 1D monoatomic lattice

2 24sin

2

K ka

m 2 sin

2

K ka

m

dk

ddk

d

d

dk

2cos

2 2

a K ka

m 1

cos2

K kaa

m

;( ) ( )Rdn k dk g d ( ) ( )Sdn k dk g d

( ) ( )Sg k

( ) ( )Sg k

( ) ( )Sg k 1 1

cos / 2

m

a K ka

2cos 1 sin2 2

ka ka

2 2 2sin cos 1 cos 1 sinx x x x

( ) ( )Sg k 2

1 1 4

41 sin

2

m

a K ka

Multibly and divide

( ) ( )Sg k 2

1 2

4 4sin

2

a K K kam m

( )S

Lk dk dk

Let’s remember:

( )gL

2 2max

2 1

a

L Na

2max

4K

m

2 24sin

2

K ka

m

True density of states

constant density of states

N m

K

max 2K

m

K

m

True density of states by means of above equation

1/ 22 2max

2( )

Ng

True DOS(density of states) tends to infinity at ,

since the group velocity goes to zero at this value of .

Constant density of states can be obtained by ignoring the dispersion of sound at wavelengths comparable to atomic spacing.

max 2K

m

/d dk

( )g

The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus

/0

1

2 1kTg d

e

1/ 22 2max

2N

Mean energy of a harmonic oscillator

One can obtain same expression of by means of using running waves.

for 1D

It should be better to find 3D DOS in order to compare the results with experiment.

( )g

3D DOS• Let’s do it first for 2D

• Then for 3D.

• Consider a crystal in the shape of 2D box with crystal lengths of L.

+

+

+ -

-

-

L0

L

y

x

L

Standing wave pattern for a 2D box

Configuration in k-space

xk

yk

L

•Let’s calculate the number of modes within a range of wavevector k.

•Standing waves are choosen but running waves will lead same expressions.

•Standing waves will be of the form

• Assuming the boundary conditions of

•Vibration amplitude should vanish at edges of

Choosing

0 sin sinx yU U k x k y

0; 0; ;x y x L y L

;x y

p qk k

L L

positive integer

+

+

+ -

-

-

L0

L

y

x

•The allowed k values lie on a square lattice of side in the positive quadrant of k-space.

•These values will so be distributed uniformly with a density of per unit area.

• This result can be extended to 3D.

Standing wave pattern for a 2D box

L

L

Configuration in k-space

/ L

2/L

yk

xk

L

L

L

Octant of the crystal:

kx,ky,kz(all have positive values)

The number of standing waves;

3

3 3 33s

L Vk d k d k d k

/L

k

dk

zk

yk

xk

214

8k dk

3 23

14

8s

Vk d k k dk

2

322s

Vkk d k dk

2

22S

Vkk

• is a new density of states defined as the number of

states per unit magnitude of in 3D.This eqn can be obtained by

using running waves as well.

• (frequency) space can be related to k-space:

2

22

Vkk

g d k dk dkg k

d

Let’s find C at low and high temperature by means of using the

expression of . g

B

Tk

High and Low Temperature Limits

• This result is true only if

At low T’s only lattice modes having low frequencies can be excited from their ground states;

3 BNk T Each of the 3N lattice modes of a crystal containing N atoms

dC

dT

3 BC Nk

k

a0

Low frequency long

sound waves

sv k svk

depends on the direction and there are two transverse, one longitudinal acoustic branch:

2

22

Vk dkg

d

1 1

ss s

k dkv

k v d v

and

2

2

2

1

2s

s

Vv

gv

at low T

sv

2 2

2 3 2 3 3

1 1 2

2 2s L T

V Vg g

v v v

Velocities of sound in longitudinal and transverse direction

/0

1

2 1kTg d

e

33

3

/0 01 1

B

BkT x

k Tx

k Td dx

e e

Zero point energy = z

2

/ 2 3 30

1 1 2

2 1 2kTL T

Vd

e v v

3

2 3 3 /0

1 2

2 1z kT

L T

Vd

v v e

B

xk T

Bk Tx

Bk T

d dx

4

43 3

/ 30 0

15

1 1B

kT x

k T xd dx

e e

4 4

2 3 3 3

1 2

2 15B

zL T

k TV

v v

32

3 3

2 1 2

15B

v BL T

k TdC V k

dT v v

at low temperatures

How good is the Debye approximation at low T?

3T

32

3 3

2 1 2

15B

v BL T

k TdC V k

dT v v

The lattice heat capacity of solids thus varies as at low temperatures; this is referred to as the Debye law. Figure illustrates the excellent agreement of this prediction with experiment for a non-magnetic

insulator. The heat capacity vanishes more slowly than the exponential behaviour of a single harmonic oscillator because the vibration spectrum extends down to zero frequency.

3T

The Debye interpolation scheme The calculation of is a very heavy calculation for 3D, so it

must be calculated numerically. Debye obtained a good approximation to the resulting heat

capacity by neglecting the dispersion of the acoustic waves, i.e. assuming

for arbitrary wavenumber. In a one dimensional crystal this is

equivalent to taking as given by the broken line of density of states figure rather than full curve. Debye’s approximation gives the correct answer in either the high and low temperature limits, and the language associated with it is still widely used today.

( )g

sk

( )g

1. Approximate the dispersion relation of any branch by a linear extrapolation of the small k behaviour:

Einstein approximation to the dispersion

Debye approximation to the dispersion

vk

The Debye approximation has two main steps:

23

9( )

D

Ng

2 3 3 3 3

1 2 3 9( ) 3

2 L T D D

V N N

v v

Debye cut-off frequency2. Ensure the correct number of modes by imposing a cut-off

frequency , above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose so that

0

( ) 3D

g d N

2

2 3 3

1 2( ) ( )

2 L T

Vg

v v

22 3 3

0

1 2( ) 3

2

D

L T

Vd N

v v

32 3 3

1 2( ) 3

6 DL T

VN

v v

D

D

D

2( ) /g

The lattice vibration energy of

becomes

and,

First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature.

/0

1( ) ( )2 1Bk T

E g de

3 32

/ /3 30 0 0

9 1 9( )2 1 2 1

D D D

B Bk T k TD D

N NE d d d

e e

3

/30

9 9

8 1

D

BD k TD

N dE N

e

dEC

dT

/2 4

23 2 /0

9

1

D B

B

k T

Dk T

D B

dE N eC d

dT k T e

3

/30

9 9

8 1

D

BD k TD

N dE N

e

Let’s convert this complicated integral into an expression for the specific heat changing variables to

and define the Debye temperature

B

xk T

DD

Bk

The heat capacity is

d kT

dx

kTx

4 /2 4

23 20

9

1

D T xB B

Dx

D B

k T k TdE N x eC dx

dT k T e

3 / 4

20

91

D T x

D Bx

D

T x eC Nk dx

e

The Debye prediction for lattice specific heat

DD

Bk

where

3 /2

0

9 3D T

D D B BD

TT C Nk x dx Nk

How does limit at high and low temperatures?

High temperature

DC

DT

2 3

12! 3!

x x xe x

4 4 42

2 2 2

(1 ) (1 )

1 11

x

x

x e x x x xx

xxe

x is always small

DT

3 / 4

20

91

D T x

D D Bx

D

T x eT C Nk dx

e

3412

5B

DD

Nk TC

How does limit at high and low temperatures?

Low temperature

44 /15

DC

For low temperature the upper limit of the integral is infinite; the integral is then a known integral of .

We obtain the Debye law in the form3T

Lattice heat capacity due to Debye interpolation scheme

Figure shows the heat capacity between the two limits of high and low T as predicted by the Debye interpolation formula.

3 B

C

Nk T

Because it is exact in both high and low T limits the Debye formula gives quite a good

representation of the heat capacity of most solids, even though the actual phonon-density of states curve may differ appreciably from the

Debye assumption.Debye frequency and Debye temperature scale with the velocity of sound in the solid. So solids with low densities and large elastic moduli have high . Values of for various solids is given in table. Debye energy can be used to estimate the maximum phonon energy in a solid.

Lattice heat capacity of a solid as predicted by the Debye interpolation scheme

3 / 4

20

91

D T x

D B xD

T x eC Nk dx

e

/ DT

1

1

Solid Ar Na Cs Fe Cu Pb C KCl

93 158 38 457 343 105 2230 235( )D K

DD

D