Solid state physics 03-lattice vibrations

Solid State PhysicsUNIST, Jungwoo Yoo

1. What holds atoms together - interatomic forces (Ch. 1.6)2. Arrangement of atoms in solid - crystal structure (Ch. 1.1-4) - Elementary crystallography - Typical crystal structures - X-ray Crystallography3. Atomic vibration in solid - lattice vibration (Ch. 2) - Sound waves - Lattice vibrations - Heat capacity from lattice vibration - Thermal conductivity4. Free electron gas - an early look at metals (Ch. 3) - The free electron model, Transport properties of the conduction electrons---------------------------------------------------------------------------------------------------------(Midterm I) 5. Free electron in crystal - the effect of periodic potential (Ch. 4) - Nearly free electron theory - Block's theorem (Ch. 11.3) - The tight binding approach - Insulator, semiconductor, or metal - Band structure and optical properties6. Waves in crystal (Ch. 11) - Elastic scattering of waves by a crystal - Wavelike normal modes - Block's theorem - Normal modes, reciprocal lattice, brillouin zone7. Semiconductors (Ch. 5) - Electrons and holes - Methods of providing electrons and holes - Transport properties - Non-equilibrium carrier densities8. Semiconductor devices (Ch. 6) - The p-n junction - Other devices based on p-n junction - Metal-oxide-semiconductor field-effect transistor (MOSFET)---------------------------------------------------------------------------------------------------------------(Final)

All about atoms

backstage

All about electrons

Main character

Main applications


Lattice Vibration

1. Sound wave

2. Lattice vibrations

3. Heat capacity from lattice vibrations

4. Thermal conductivity

The atomic vibrations in a periodic lattice give rise a wave propagation with en-ergy, which significantly affect heat capacity and thermal conductivity of solids. The quanta of this acoustic wave is named as a “phonon”


Distance between two atoms

V(r)

r

repulsive

attractive

equilibrium distance

Q: Why equilibrium is forbidden ?

pxHeigenberg’s uncertainty principle

At T = 0, the minimum kinetic energy is named as zero point energy

As T is increase, atoms gain more thermal energy and the amplitude of their motions increases.

Interatomic forces


Hook’s law

xKFspring 2

2

1)( KxxV

Simple harmonic motion

dx

xdVF

)(

,sin tAx

Solution for simple harmonic motion

xKmaFspring 2

2 )(

dt

txda

0)()(

2

2

txm

K

dt

txdm

K


...........2

)()()()(

2

22

arar dr

Vdar

dr

dVaraVrV

The potential energy V(r), for a small deviation of r from its equilibrium value a, be expended as a Taylor series about r = a.

No linear term, at equilibrium, 0

ardr

dV Potential energy becomes quadratic


For a small displacement of object, the motion of object can be described with simple harmonic motion. And the restoring force on an object is approximately pro-portional to its displacement (Hooke’s Law).

Now, consider an object on earth

dr

rdVF

)(

r

V(r)


...........2

)()()()(

2

22

arar dr

Vdar

dr

dVaraVrV

The potential energy V(r), for a small deviation of r from its equilibrium value a, be expended as a Taylor series about r = a.

r

V(r)


Now, consider an objects on earth

The elastic limit is where Hook’s law works

The inelastic limit, is where permanent deformation occurs. If the force is applied to the inelastic limit, the sample will not return to its original size and shape, permanent deformation has occurred.


Longitudinal Waves

Transverse Waves

Sound waves

Mechanical waves are waves which propagate through a material medium (solid, liquid, or gas) at a wave speed which depends on the elastic and inertial properties of that medium. There are two basic types of wave motion for mechanical waves: longitudinal waves and transverse waves.

C = Elastic bulk modulusρ = Mass density

CvL

Velocity of sound wave

C = stress/strain

Stress: force per unit areaStrain: ratio of change caused by the stress to the original state

Macroscopic wave: movements of atoms/molecules/parti-cles are much larger than interatomic spacing


The interaction V(r) between nearest neighbor of separation r may, for a small deviation of r from its equilibrium value a, be expended as a Taylor series about r = a.

For a small atomic displacement, the motion of atoms can be described with sim-ple harmonic motion. And the restoring force on each atom is approximately pro-portional to its displacement (Hooke’s Law).

For a large atomic displacement, the anharmonic effects become important.


Now, consider atoms in solids

V(r)

r

...........2

)()()()(

2

22

arar dr

Vdar

dr

dVaraVrV

No linear term, at equilibrium, 0

ardr

dV Potential energy becomes quadratic


Lattice vibrations of one dimensional solids

...........2

)()()(

2

22

ardr

VdaraVrV

ardr

VdK

2

2

We describe 1D crystal as a chain of identical atoms, which connected by springs of

a a a a a a

un-2 un-1 un un+1 un+2

Introduce displacement of nth atom from their equilibrium positions as un(t)

Assume periodic condition nNn uu , so that all atoms are in identical environment


Now, consider only nearest neighbor interactionAnd find equation of motion for nth atom


Force exerted by spring 1:

Force exerted by spring 2:

a a

un-1 un un+1

Spring 1 Spring 2

+u

)( 11 nn uuKF

)( 12 nn uuKF

Therefore, equation of motion for nth atom

)2( 11 nnn uuuKuM


For a normal mode with the same amplitude for all atoms, we can have following trial solution

A normal mode of an oscillating system is a pattern of motion in which all parts of the system move sinusoidally with the same frequency and in phase. The frequencies of the normal modes of a system are known as its natural frequencies or resonant frequencies. Each normal mode oscillates independently of the other modes


naxn 0equilibrium positions of nth atom

k is the reciprocal of the wavelength, k = 2p/l

Q1: What is normal mode ?

Q1: For a matter wave, what does k represent ?

pkmEp

k

,22

k represent momentum

)](exp[ 0 tkxiAu nn

http://en.wikipedia.org/wiki/Oscillation

http://en.wikipedia.org/wiki/Sinusoidal


tkaknaitknaitkaknaitknai AeAeAeKAeM 22

)2( 11 nnn uuuKuM

)](exp[ tknaiAun


ikaika eeKM 22

]1)[cos(2 kaK

kaKM

2

1sin4 22

ka

M

K

2

1sin

4M

K4max



Dispersion relation (relation between frequency w and wavenumber k)

ka

M

K

2

1sin

4frequency dependence ef-fect in wave propagation

In above equation n is cancelled out, this means that the eqn. of motion of all atoms leads to the same algebraic eqn. a our trial function un is indeed a solution of the eqn. of motion of n-th atom.

Our wavelike solutions are uncoupled oscillations called normal modes; each k has a definite w given by above eqn. and oscillates independently of the other modes.

number of modes is expected to be the same as the number of equations N.

0 k

w

a

a

2

a



Dispersion relation (relation between frequency w and wavenumber k)

For a periodic boundary condition,

there should be an integral number of wavelengths in the length of our ring of atoms

Then,

Thus, in a range of 2π/a of k, there are N allowed values of k.

nNn uu

pNa

kkp

Na 22

,pNa p is integer

possible k values are multiple of Na

2

0 k

w

a

a

2

a

Discreet set of many individual points with spacing of

Na

2


0 k

w

ABC

a

a

2

a


ka

M

K

2

1sin

4

ak

a

The restricted range of

includes all possible values of frequency and the group velocityk

What, if anything, is the physical significance of wavenumbers outside this range ?

at k =p/a, w is max

The group velocity of a wave is the velocity with which the overall shape of the waves’s amplitudes propagates through space.



Consider instantaneous atomic displacements for a transverse wave in order to visualize clearly

For a

k

and ,2a

x

un

For a

k7

8 and ,

4

7a (orange)

ak

7

6 and ,

3

7a (blue)

This is the case for the maximum frequency(alternative atom oscillate in antiphase and the waves at this value of k are essentially standing waves) un

x

a

nodes

point A, B, and C correspond to the same instanta-neous atomic displacements as well as the same fre-quency

Behave as if they are held by two springs of 2K giving

M

K4max


0 k

w

ABC

a

a

2

a


At B, the group velocity 0

k

a wave propagating to the right

At A and C, the group velocity 0

k

a wave propagating to the left

At A and C, the atomic displacement, frequency, group velocity are all same.

The k values of point A and C differ by 2p/a

a adding any multiple of 2p/a to k does not alter physical properties of wave.

ak

a

The restricted range of includes all possible modes



Now consider k at the boundary,

The Bragg’s law for the one-dimentional crystal:

sin2d n

For 1D, interatomic spacing a = d, q = 90º

knn

2

a ad 2sin2

Therefore, Bragg’s condition becomesa

nk

Waves with a

k

satisfies Bragg’s reflection Therefore, standing waves occurs at these two k values due to Bragg reflection of running waves.

0k

a

a

2

a


For a long wavelength limit, ,1ka kaka sin

In this limit, the group velocity and phase velocity are equal a k

k

M

Kav

C

vL The speed of sound wave, r is m/a and C is Ka in 1-D


kaKM

2

1sin4 22 222 aKkM a

Therefore, long wavelength limit is basically a sound wave

M

Ka

kkvs

a sL vv

The force required to increase the interatomic distance from a to r

a )(/)( arKaarC

C = stress/strain

Stress: force per unit areaStrain: ratio of change caused by the stress to the original state



0k

a

a

2

a

In real space, a system is periodic with a

In k space, a system is periodic witha

2

The periodic unit (unit cell) in k-space a Brillouin Zone

The first Brillion Zone is a unit cell in k-space centered around the point k=0

The point are known as the Brillouin-Zone boundarya

k

Why dispersion curve should be periodic in a

kk2

)( wtknain Aeu For

akk

2take than

)(2)()( wtknainiwtknaiwtknain AeeAeAeu

x

Reciprocal lattice

a Wigner seitz cell in reciprocal lattice

a

2


Chains of two types of atoms

Now, consider the simplest description of ionic crystal in 1D

Two different types of atoms of masses M and m are connected by identical springs of spring constant K

Un-2Un-1 Un Un+1 Un+2

K K K K

M Mm Mm a)

b)

(n-2) (n-1) (n) (n+1) (n+2)

a

Here, a is repeating distance, the nearest neighbors separations is a/2



Now, consider only nearest neighbor interactionAnd find equation of motion, but now we have two different types of atoms

two equations of motion must be written; One for mass M(nth atom), and One for mass m((n-1)th atom)

M

a a/2

un-1 un un+1

+u

m mM

un-2


M

a a/2

un-1 un un+1

+u

m mM

un-2

For mass M: )()( 11 nnnnn uuKuuKuM

)2( 11 nnn uuuK

For mass m: )()( 2111 nnnnn uuKuuKum

)2( 21 nnn uuuK



)](exp[ 0 tkxiAu nn


Again, use trial solution

20 a

nxn Now, equilibrium position for nth atom is

In order to express relative amplitude and phase of two types of atoms,we introduce complex number a for plane wave function un.

For mass M (nth atom): )](exp[ 0 tkxiAu nn

For mass m (n-1th atom): )](exp[ 0 tkxiAu nn



Eqn. of motion for mass M (nth atom):

t

kaknait

knait

kaknait

knai

AeAeAeKAeM

22222 2

)2( 11 nnnn uuuKuM

)](exp[ 0 tkxiAu nn

222 2ka

ika

ieeKM

)]2

cos(1[22 kaKM

)](exp[ 0 tkxiAu nn

Displacement for mass M:

Displacement for mass m:


)2( 211 nnnn uuuKum


Eqn. of motion for mass m (nth atom):

)](exp[ 0 tkxiAu nn

)](exp[ 0 tkxiAu nn

Displacement for mass M:

Displacement for mass m:

t

kaknait

kaknait

knait

kaknai

AeAeAeKAem

2

2

2222 2

222 2ka

ika

ieeKm

)]2

cos([22 kaKm


)]2

cos([22 kaKm

)]2

cos(1[22 kaKM For mass M:

For mass m:


Now we have a pair of equations for a and w as a function of k

2

2

2

)2/cos(2

)2/cos(2

2

mK

kaK

kaK

MK

a 0)2/(sin4)(2 2224 kaKmMKmM

2/1222 )2/(sin

4)(

)(

ka

MmMm

mMK

Mm

mMK


Now we have two branch of dispersion relation


k

A

BC

0 a

a

2

a

In above equation n is cancelled out, this means that the eqn. of motion of all atoms leads to the same algebraic eqn. a our trial function un is indeed a solution of the eqn. of motion of n-th atom.

The dispersion relations are periodic in k with period 2p/a = 2p/(unit cell length)

For N unit cells, we have 2N equation of motions a 2N normal modes

From periodic boundary condition, possible k values are multiple of

Thus, there are exactly N allowed values of k in the range –p/a < k < p/a

nNn uu 2 Na

2

Optical branch

Acoustic branch


Now examine limiting solutions of the dispersion relations near the points O, A, B, C


k

A

BC

0 a

a

2

a

O



2/1

222

2

)(11

)(ak

Mm

mM

mM

MmK

222)(2

11)(

akMm

mM

mM

MmK

mM

MmK )(2

)(2

22

Mm

aKk

or


Now, for a 1)2/cos( ka ,1kafor mM / or 1


This solution corresponds to point A in dispersion graph. This value of α shows that the two atoms oscillate in an-tiphase with their center of mass(M* = Mm/(M+m)) at rest.

and

m

M

mM

MmK )(22

)(2

222

Mm

aKk

1

This solution represents long-wavelength sound waves in the neighborhood of point 0 in the graph; the two types of atoms oscillate with same amplitude and phase, and the velocity of sound is

and

2/1

)(2

mM

Ka

kvs

2

2

2

)2/cos(2

)2/cos(2

2

mK

kaK

kaK

MK


Acoustic/optical branches

Transverse acoustic mode for diatomic chain

Transverse optical mode for diatomic chain

Acoustic branch: long wavelength limit k g 0, sounds wave

Optical branch: a higher energy vibration need a certain amount of energy to excite this mode

(oscillating charged particle cre-ates electromagnetic wave)


For a k at the boundary p/a


At maximum acoustical point C, M oscillates and m is at rest.

At minimum optical point B, m oscillates and M is at rest.

2/122 4)(

MmMm

mMK

Mm

mMK

Mm

mMKmMK )()(

mK /2 or MK /2

And the corresponding amplitude ratio 0or

for + (optical) for - (acoustic)


Comparison of dispersion relation

0 k

w

a

a

2

a

k

0 a

a

2

a

k0 a

a

2

a

Let m g M, it becomes 1D monoatomic chain with lattice constant a/2 range of k is 2 /(unit cell size)


Lattice vibrations of three-dimentional crystals

For a unit cell containing only one atom, there is three branches of the dispersion relation (One for each cartesian coordinate). And each branch has N normal modes

For a 3D crystal with a primitive unit cell containing two atoms, there is three acoustic and three optical

a total number of modes in crystal is three times the number of atoms in the crystal

Phonon

We consider the lattice vibrations as a collection of quanta of elemental vibrational exci-tation with a mode of angular frequency w

This quanta of lattice vibration is named as Phonon

with energy


Phonon

In classical description: the lattice vibrations have many normal modes of fre-quency w(k), which are independent and harmonic

Describe each normal mode of frequency w as a quantum mechanical harmonic os-cillator, which has discreet energy values with uniform spacing.

Transition to Q. M.

2

1nn

The state en can be considered as a collection of n number of energy units

We introduce the concept of phonon (of energy ) as a quanta for the excitation of the lattice vibration mode of frequency w.

kDispersion relation of wave

kEnergy & Momentum of Phonon

(Crystal momentum)


)2/1(

2

1nn

Phonon

Equally spaced energy level

Transition between neighboring states involvesAbsorption and emission of phonon

Like photons, phonons are bosons and are not conserved

They can be created or destroyed in collisions


•Atoms vibrate about their equilibrium position.

•They produce vibrational waves.

•This motion is increased as the temperature is raised.

In a solid, the energy associated with this vibration and perhaps also with the rotation of atoms and molecules is called as thermal energy.

Note: In a gas, the translational motion of atoms and molecules contribute to this energy.

Thermal Energy and Lattice Vibrations

Q: How the vibrational energy changes with temperature since this gives a measure of the heat energy which is necessary to raise the temperature of the materialRecall that the specific heat or heat capacity is the thermal energy which is required to raise the temperature of unit mass or 1gmole by one Kelvin


Heat Capacity from Lattice Vibrations

The energy given to lattice vibrations is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution.

Other contributions;

i) In metals a from the conduction electrons.

i) In magnetic materials a from magneting ordering.

Atomic vibrations leads to band of normal mode frequencies from zero up to some maximum value.

Calculation of the lattice energy and heat capacity of a solid therefore falls into two parts:

i) the evaluation of the contribution of a single mode

ii) the summation over the frequency distribution of the modes.


Energy and Heat Capacity of a Single Harmonic Oscillator

For a harmonic oscillator of angular frequency w

2

1nn

Then average energy is given by

nn

nP

TkBne /

The probability weight of oscillator being in nth level is given by Boltzman factor

For normalization, 1

nnP

Therefore,

0

// /n

TkTkn

BnBn eeP

0

/

n

TkBneZ

Define Partition function


nn

nP


_0

0

1 1exp /

2 2

1exp /

2

Bn

Bn

n n k T

n k T

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

Geometric series: r

aararara

1...32 when -1<r<1


_

/

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




12

1/

TkBe

Mean energy for a single harmonic oscillator

Zero-point energy contribution of phonons

Contribution of phonons to mean energy for a single harmonic oscillator:

= Energy of phonon ( ) ⅹ mean number of phonons

1

1)(

/

TkBen

Bose-Einstein distribution


Mean energy of a harmonic oscillator as a

function of T

T

2

1

TkB


12

1/

TkBe

low temperature limit TkB



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

12

x

xex

Tke

B

TBk

1

112

1_

TkB

_ 1

2 Bk T

_

Bk T

T

2

1

TkB

12

1/

TkBe

high temperature limit TkB

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

Thermal energy of classical 1D harmonic oscillator is TkB



Heat Capacity: the thermal energy which is required to raise the temperature of unit mass or 1gmole by one Kelvin

,12

1/

TkBe

TC

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

2

2

1

T

T

v B

eC k

T e

Let

Bk



2

2

1

T

T

v B

eC k

T e

Missing area=2

T

Bk

Bk

vC

Specific heat vanishes exponentially at low T’s and tends to classical value at high temperatures.


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

Dulong-Petit law where R is universal gas constant.

2

2

1

T

T

v B

eC k

T e

3 3BNk R


,T K

3R

vC

Einstein Heat Capacity of Solids

Classical limit

Einstein model

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

Cal2388.0J1

moleK

J9.24

vC

a

moleK

Cal6vC

RCv 3 )J/K(1038.1)atoms/mole(10023.633 2323 BNk


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 infinitesimally from one value to another. it has to go up in steps.

Definite energy levels Steps get small Energy is continuous

On atomic scale the energy can only jump by a discrete amount from one value to another.

In classical limit, the steps can be so small a the energy can be considered as continuous.

22

2

L

n

mEn


The density of states( ): number of states per unit energy interval

a the number of states between and will be

Density of States

( )

( )d d

For a one dimentional crystal containing N unit cells of side a, the periodic boundary condition gives the allowed wavenumber

pL

pNa

k 22

,pNa

pNa

kkp

Na 22

Threrefore, allowed wavenumbers are uniformly distibuted in k at a density of )(kR

# of k per unit dis-tance

The number of K in between k and k+dk

dkL

dkkR

2)(


kRunning waves: for periodic boundary condi-tions

Density of States

0

L

2

L

3

L

k

Standing waves: for fixed boundary conditions

2k p

L

k pL

For a one dimentional crystal containing N unit cells of side a with a fixed boundary a integral number of half wavelength in the chain

2 2;

2 2

n n nL k k k

L L

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 p/a

dkL

dkkR

2)( a

dkL

dkkS )(a

0L

2

L

4L

6

L

2

L

4


Density of States

To calculate an energy or heat capacity by summing over normal modes

a Need density of states per unit frequency range )(g

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

12

1/

TkBe

We got average energy of single harmonic oscillator of freq. w

Summing over all normal modes frequencies w give total lattice energy

dge

ETkB

)()12

1(

0 /

( ) ( )Rdn k dk g d

How we get g(w) ? a we know the density of normal modes in k-space

d

dkk

d

dkkg sR )()()( a

dispersion relation

# of modes with freq. from w to w+dw: g(w)dwThere is corresponding # of modes in wavenumber range from k to k+dk


Density of States

d

dkk

d

dkkg sR )()(2)(

Let’s consider one dimention monoatomic chain

ka

M

Ka

dk

d

2

1cos

2/1

kaKM

2

1sin4 22The dispersion relation

)2/(sin1

1

)2/cos(

1)(

2

2/12/1

kaK

MN

kaK

M

a

Lg

KMK

MN

4/1

12

2/1

2/4

12

MK

N

2max

For 1D


( )g

Density of States

For 1D

K

MN

22max

12)(

Ng

If we consider classical limit, a long wavelength limit and ignore the dispersion of sound at wavelengths comparable to atomic spacing, then

DOS(density of states) goes infinity at wmax

Since group velocity becomes zerok /

2/1)/(/ MKakvg

K

MNMKa

Na

d

dkkg S

2/1)/(/)()( constant DOS

M

KM

K2max


Density of States

Now, how we get the DOS for 3D ?

Let’s do it first for 2D

Consider fixed end boundary, 2D box

+

+

+ -

-

-

L0

L

y

x

Standing waves in 2D box has the form )sin()sin()sin(0 tiykxkuu yx

Again, from fixed boundary conditions 2 2;

2 2

n n nL k k k

L L

,pL

kx

q

Lk y

and p, q are positive integer


Density of States

For 1D

k0

L

2

L

3

L

Standing waves: for fixed boundary conditions

For 2D

L

allowed k values

in 2D k-space

xk

yk

L

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

•These values will so be distributed uniformly with a density of (L/p)2 per unit area

• This result can be extended to 3D

Standing waves


Density of StatesFor 3D•The allowed k values lie on a cubic lattice of side p/L in the positive octant of k-space

•These values will so be distributed uniformly with a density of (L/p)3 per unit vol-ume

xk yk

zk

k

dk

zk

yk

xk

kdks

3)( dkk

Vkd

L 23

33

48

1

,)(2 2

2

dkkgdkVk

2

2

2)(

Vk

kg

Standing waves

For running waves, density becomes (L/2p)3

But, sum over all + - k values a 2

2

2)(

Vk

kg


High and Low Temperature Limits

For high temperature limit

3N lattice modes of a crystal containing N atoms contribute kBT to the energy

TNkB3 a BNkT

C 3

For low temperature limit

k

w

0a

Only low energy (long wavelength) modes are excited from their ground states

Linear region (long wavelength acoustic modes): a group velocity = velocity of sound sv

kk


dkVk

dkkg2

2

2)(


For low temperature limit svk

We also got

What we need is g(w)

32

22

2

1

2

1

2)()(

sss v

V

vv

V

d

dkkgg

Average over all direction

We knows that for 3D, there are 3 acoustic modes one longitudinal and two transverse

332

2 21

2)(

TL vv

Vg

dge

ETkB

)()12

1(

0 /

d

evv

VE

TkTL

Z B

0 /

3

332 1

21

2


d

evv

VEE

TkTL

Z B

0 /

3

332 1

21

2


,Tk

xB

a x

TkB

dx

Tkd B

dxe

xTkdx

Tk

e

xTkd

e xBB

xB

TkB

0

3

3

4

0

33

0 /

3

1

)(

1

)/(

1

dxe

xTk

vv

VEE

xB

TLZ

0

3

3

4

332 1

)(21

2 15

4

3

33

2 21

15

2

Tk

vv

kVC B

TL

B



3

33

2 21

15

2

Tk

vv

kVC B

TL

B

Debye’s T3 law

Figure illustrates the excellent aggrement 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.


The Debye interpolation scheme

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


Debye approximation to the dispersionvk

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


Step 1:

Einstein approximation to the dispersion

0

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

Step 2: D

D

0

( ) 3D

g d N


2

2 3 3

1 2( ) ( )

2 L T

Vg

v v


0

( ) 3D

g d N

32 3 3

1 2( ) 3

6 DL T

VN

v v

23

9( )

D

Ng

2 3 3 3 3

1 2 3 9( ) 3

2 L T D D

V N N

v v a

/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

Zero point energy Term gives T dependence



/2 4

23 2 /0

9

1

D B

B

k T

Dk T

D B

dE N eC d

dT k T e

Tkx

B

a ,x

TkB

dx

Tkd B

Define Debye temperature DD

Bk

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


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


For high temperature DT

3 / 4

20

91

D T x

D Bx

D

T x eC Nk dx

e

B

T

DBD Nkdxx

TNkC 39

/

0

2

3


For low temperature DT


3 / 4

20

91

D T x

D Bx

D

T x eC Nk dx

e

15

4 4Infin-ity

3412

5B

DD

Nk TC

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.

Solid Ar Na Cs Fe Cu Pb C KCl

93 158 38 457 343 105 2230 235

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.

D D

D

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

/ DT

1

1

BNk

C

3


Anharmonic Effects

The actual interatomic force is not ideal simple harmonic

...........2

)()()()(

2

22

arar dr

Vdar

dr

dVaraVrV

V(r)

r

a Any real crystal resists compression to a smaller volume than its equilibrium value more strongly than expansion to a larger volume.

The effect of higher order term a anharmonic effects

With harmonic approxiamtion we got independent normal modes, these normal modes do not affect each other and vibrate with their freq. w independently.

Inclusion of higher order terms in Taylor expansion leads to coupling of the modes.

a This coupling can be pictured as collisions between the phonons associated with the modes.

a This collision limit the thermal conductivity associated with the flow of phonons.

Without anharmonic effects, the phonons do not interact with each other. And if there is no boundaries, lattice defects and impurities, the thermal conductivity is infinite.


The coupling of normal modes by the unharmonic terms in the interatomic forces can be pictured as collisions between the phonons associated with the modes.

A typical collision process of

phonon1

phonon2

1 1,k

2 2,k

3 3,k

After collision, another phonon is produced

3 1 2k k k 3 1 2

3 1 2k k k 3 1 2 and

conservation of energy

conservation of momentum

Phonon-phonon collisions


Phonons are represented by wavenumbers with

ka a


0 k

w

C

A

a

a

2

a

a

2

B Normal process !

longitudinal

transverse


Phonons are represented by wavenumbers with

ka a

If lies outside this range add a suitable multible of to bring it back

within the range of . Then, becomes

3k

3 1 2

2nk k k

a

2

a

3 1 2k k k ka a


0 k

w

C

A

a

a

2

a

a

2

The sign of group velocity is reversed

0

k

Umklapp process !B


Phonon3 has ka

Phonon3 has k

a

1 2

k

a

0

a

3'

Umklapp process

3

1 2

k

a

0

a

Normal process

3Longitudinal

Transverse

0n 0n


3 1 2

2nk k k

a


Thermal Conduction by Phonons

A flow of heat takes place from a hotter region to a cooler region when there is a temperature gradient in a solid.

In an electrically insulating solid, the thermal conduction mainly comes from the flow of phonons

In the elementary kinetic theory of gases, the steady state flux of a property P in the z direction is

dz

dPvlflux

3

1Flux: amount that flows

through a unit area per unit time

Angular aver-age

Mean free path

Average velocity of the medium

Gradient of the prop-erty of P in z direction

If P is the energy density, then the flux W is the heat flow per unit area so that

dz

dT

dT

dEvl

dz

dEvlW

3

1

3

1

Heat capac-ity

For a heat flow W

dz

dTKW

Where K is thermal conductivity given by

CvlK3

1

Mean free path: the aver-age distance covered by a moving particle between successive collisions



The essential differences between the processes of heat conduction in a phonon and real gas;

Phonon gas Real gas

•Speed is approximately constant.

•Both the number density and energy density is greater at the hot end.

•Heat flow is primarily due to phonon flow with phonons being created at the hot end and destroyed at the cold end

•No flow of particles

•Average velocity and kinetic energy per particle are greater at the hot end, but the number density is greater at the cold end, and the energy density is uniform due to the uniform pressure.

•Heat flow is solely by transfer of kinetic energy from one particle to another in collisions which is a minor effect in phonon case.

hot cold

hot cold



Temperature dependence of thermal conductivity K

•Temperature dependence of phonon mean free length is determined by phonon-phonon collisions at low temperatures

•Since the heat flow is associated with a flow of phonons, the most effective collisions for limiting the flow are those in which the phonon group velocity is reversed. It is the Umklapp processes that have this property, and these are important in limiting the thermal conductivity

For a heat flow W

dz

dTKW

CvlK3

1

Vanishes at low T with T3 and tends to classi-cal value 3kB at high T

Approximately equal to velocity of sound and so temperature independent.

?

Mean free path



Conduction at high temperature

• At temperatures much greater then the Debye temperature the total lattice energy and heat capacity is given by temperature-independent classical result of

• The rate of collisions of two phonons phonon density.

• At high temperatures the average phonon energy is constant and the total lattice energy T ; phonon number T , so Scattering rate T and mean free length

Then the thermal conductivity of .

D

1T

1

3

1 TCvlK

,3 TNkE B BNkC 3


• Experimental results do tend towards this behaviour at high temperatures as shown in figure 2.19.

1

T

5 10 20 50 100

10

0

10-1

( )T K

2 5 10 20 50 100

( )T K

10

0

10-13T

(a)Thermal conductivity of a quartz crystal

(b)Thermal conductivity of artificial sapphire rods of different diameters




Conduction at intermediate temperature

Referring figure a at ; the conductivity rises more steeply with falling temperature, although the heat capacity is falling in this region. Why?

This is due to the fact that Umklapp processes which will only occur if there are phonons

of sufficient energy to create a phonon with . The relavant phonon that creats Umklapp processes must have an energy compa-

rable to the Debye energy

So, the number of relevant phonons is expected to vary roughly as When , here b is a number of order unity, and emperical values of b are 2

or 3.

So, the mean free path

This exponential factor dominates any low power of T in thermal conductivity, such as a factor of from the heat capacity.

3T

DT

bTDe /

bTDel /

ak /3

DT



Conduction at low temperature

for phonon-phonon collisions becomes very long at low T’s and eventually exceeds the size of the solid,

Because number of high energy phonons necessary for Umklapp processes decay exponentially as

is then limited by collisions with the specimen surface, i.e.

Specimen diameter

T dependence of K comes from which obeys law in this region

Temperature dependence of dominates.

l

3T

l

l

vC

3412

5B

DD

Nk TC

vC

bTDe /



Size effect

• When the mean free path becomes comparable to the dimensions of the sample, transport coefficient depends on the shape and size of the crystal. This is known as a size effect.

• If the specimen is not a perfect crystal and contains imperfections such as dislocations, grain boundaries and impurities, then these will also scatter phonons. At the very lowest T’s the dominant phonon wavelength becomes so long that these imperfections are not effective scatterers, so;

the thermal conductivity has a dependence at these temperatures.

• The maximum conductivity between and region is controlled by imperfections.

• For an impure or polycrystalline specimen the maximum can be broad and low [figure (a) on pg 59], whereas for a carefully prepared single crystal, as illustrated in figure(b) on pg 59, the maximum is quite sharp and conductivity reaches a very high value, of the order that of the metallic copper in which the conductivity is predominantly due to conduction electrons.

3T

3T

/D bTe


Summary

...........2

)()()()(

2

22

arar dr

Vdar

dr

dVaraVrV

V(r)

r

Harmonic approximation for interatomic force

a Force is proportional to the atomic displacement

a a

un-1 un un+1

Spring 1 Spring 2

For 1D monoatomic chain

)2( 11 nnn uuuKuM

)](exp[ 0 tkxiAu nn

By using trial solution

kaKM

2

1sin4 22

We get dispersion relation for a n number of normal mode

0 k

w

a

a

2

a

Periodic with 2p/aPeriodic boundary condition al-low k is multiple of

For long wavelength limit,

Na

2

M

Ka

kkvs


For 1D diatomic chain

Solve eqn. of motion for each type of atom

)2( 11 nnnn uuuKuM

)2( 211 nnnn uuuKuM

Use trial solution for each atoms

)](exp[ 0 tkxiAu nn

)](exp[ 0 tkxiAu nn

Complex number, represent relative amplitude and phase

2/1222 )2/(sin

4)(

)(

ka

MmMm

mMK

Mm

mMK

Then, we get two branch of dispersion relation

k

0 a

a

2

a

Optical branch

Acoustic branch

Periodic with 2p/a

Periodic boundary condition allow k is multiple of Na

2

Long wavelength limit

and)(2

222

Mm

aKk

1

m

M

mM

MmK )(2 and

For a k at the boundary /a

mK /22 MK /22

0


Phonon

Describe each normal mode of frequency w as a quantum mechanical harmonic os-cillator, which has discreet energy values with uniform spacing.

2

1nn

The thermal average en-ergy 1

1)( / TkBe

n

T

2

1

TkB

Einstein model: 3N normal mode with the same freq.

12

13

/ TkBeNE

2

2

1

T

T

v B

eC k

T e

,T K

vC Classical limit

Einstein model

BNk3


Density of States

12

1/

TkBe

We got average energy of single harmonic oscillator of freq. w

Summing over all normal modes frequencies w give total lattice energy

dge

ETkB

)()12

1(

0 /

( ) ( )Rdn k dk g d

d

dkk

d

dkkg sR )()()( a

dispersion relation

For 1D monoatomic chain

kaKM

2

1sin4 22

2/4

12)(

MK

Ng

2max

max 2

K

m

K

m

( )g

K

mN

22max

12)(

Ng


Density of state for 3D

kdks

3)( dkk

L 23

48

1

,)(

2 2

2

dkkgdkVk

Debye model:

0

( ) 3D

g d N

i) Use long wavelength approxi.

ii) Introduce cut-off freq. with constraint

sk

3 / 4

20

91

D T x

D Bx

D

T x eC Nk dx

e

/ DT

1

1

BNk

C

3

High T,

low T, C ~ T3

For 1D, Debye approxi.

For 3D, Debye approxi.


Thermal propertiesFor an insulator, lattice energy mainly account for the material’s thermal propertiesFor a metal, conduction electron provides

If we consider higher order term in interatomic int. a introduce phonon-phonon collision (Anharmonic effect)

Phonon3 has ka

Phonon3 has k

a

1 2

k

a

0

a

3'

Umklapp process

3

1 2

k

a

0

a

Normal process

3Longitudinal

Transverse

0n 0n

Thermal conductivity

CvlK3

1

At high T, ~ T-1

At low T, ~ T3

At intermediate T, ~ bTeT /3

Solid state physics 03-lattice vibrations

Science

Transcript of Solid state physics 03-lattice vibrations