Solid State TheorySolid State TheoryPhysics 545Physics 545

Crystal Vibrations and Phonons

Overview• Ionic motion and the harmonic approximation• Introduction to vibrations and the use of label “k”, the

wave vector, indexing them• Reciprocal space revisited• Vibrations in a finite monatomic lattice, concept of normal

modes• Relationship between frequency of vibration ω and k. • Definition of Brillouin Zone• Vibrations in an infinite monatomic lattice

Ionic MotionInteraction of electrons and ions localises the ions about equilibrium positions in the lattice.

Ions move because

• finite temperature gives them kinetic energy

• even at zero temperature quantum mechanics doesn’t allow simultaneous localisation of position and momentum

• they are excited e.g. by an incident neutron beam

Some consequences of the motion of ions:

• a lattice can absorb significant amounts of heat (the electronic contribution to heat capacity is very small compared to that of the lattice)

• inelastic scattering of particles from the lattice due to some of the incident particle’s energy being absorbed by the lattice

• electrical resistance

A lattice can sustain coherent, elastic vibrations

Ionic MotionA lattice can sustain coherent, elastic vibrations

To find out what the vibrations look like, need to find and solve the equations of motion for the ions.

ui is the displacement of ith ion from its equilibrium position.

U−∇== amFTo find the force on the ith ion, with mass mi , Eq. (1) becomes

(1)

( )n21i

ii u,....,u,uu

aF Um i ∂∂−==

i - 4 i - 3 i - 2 i - 1 i i + 1 i + 2 i + 3

(2)

ui

Potential Experienced by the IonsFinding the exact potential U(u1, u2,…., un) experienced by the ions is not possible in a system larger than a couple of atoms.

To start with assume that the ions experience an average effective potential due to the electrons.

Justification:

The ions are not moving very far from equilibrium positions

i.e. of order 0.2Å compared to lattice spacing of order 2-3Å

Secondly assume that the potential experienced by the ions can be written as a sum over pairs of ions of a function which only depends upon how far the ions are apart.

( )∑∑==

−Φ=n

j

n

iU

1ji

1n21 RR

21)R,....,R,R(

Only justifiable because the ions are not moving very far from equilibrium positions.

Potential Experienced by the Ions

Because ions don’t move very far from equilibrium positions, introduce the harmonic approximation, i.e replace Φ(|Ri-Rj|) by a parabola centred on the equilibrium separation of the ions.

-10

-5

0

5

10

15

20

25

30

35

0 1 2 3 4 5 6

Angstroms

eV

Ionic potential for KCl as a function of the separation of the ions

Crystal Vibration

s-1 s s+1

Mass (M)

Spring constant (C)

x

Transverse wave:

Energy

Distancero

Parabolic Potential of Harmonic Oscillator

Eb

Interatomic Bonding

The Harmonic ApproximationMathematically, we are expanding Φ(|Ri-Rj|) about the equilibrium separation of ions i and j and only keeping terms up to second order in the displacement of ions i and j from their equilibrium positions.

Write Φ(|Ri-Rj|) in terms of the equilibrium positions and the displacement from the equilibrium position:

( ) ( )jieqj

eqiji uuRRRR −+−Φ=−Φ

To expand Φ(|Ri-Rj|) about the equilibrium positions of the ions, we need to perform a Taylor series expansion.

To find an approximate expression for the function f(r) at the point a in terms of a polynomial expression we have

However, we want to expand a function of a vector, f(r), about place in space where r equals the vector a, so we need the more general form:

( ) ( ) ( ) ( ) ( ) ( ) ( ) .....f) -(!3

1 f) -(2!1 f) -(ff 32 +∇•+∇•+∇•+= aaraaraarar

( ) ( ) ( ) ( ) .....r

fa -r !3

1r

fa -r 2!1

rf a) -r(afrf 3

33

2

22 +

∂∂+

∂∂+

∂∂+=

=== ararar

The Harmonic Approximation

e q e q e q e qR R u u R R

i j i j i j

e q e qu u R R

i j i j2

e q e q1 u u R R2 i j i j

Φ − + − ≈ Φ −

+ − • ∇ Φ −

+ − • ∇ Φ −

Expanding Φ(|Ri-Rj|) to second order about the equilibrium positions of the ions gives:

( ) ( ) ( ) ( ) ( )

( ) ( )

1 2f f ( - ) f ( - ) f2 !

1 3 ( - ) f . . . . .3 !

= + • ∇ + • ∇

+ • ∇ +

r a r a a r a a

r a a

The Harmonic Approximation cont.

( ) ( ) ( )

( )( ) ( )∑∑

∑∑∑∑

==

====

−Φ∇•−+

−Φ∇•−+−Φ=

n

j

n

i

n

j

n

i

n

j

n

iU

1

eqj

eqi

2ji

1

1

eqj

eqiji

11

eqj

eqi

1

RRuu41

RRuu21RR

21

The potential energy to second order in the displacement of the ions is

• the first term is simply a constant Ueq

• the second term sums over all the forces, Grad Φ, on the ions at their equilibrium positions, where by definition the net force on the ions is zero

•the third term can be written more compactly as

( ) ( ) ( )

( ) ( )ηηη

µηµµ

µ

ηηη ηµ

µµµ

j,i,,1

j,i,,1

j,i,,1 RR

2

j,i,,1

uuuu41

uuuu41

eqj

eqi

−−=

−∂∂

Φ∂−

∑∑

∑∑

==

= −=

n

jij

n

i

n

j

n

i

K

rrr

where the sums over µ and η are over the coordinates x,y,z

The Harmonic Approximation cont.

( ) ( ) ( )

( ) ( )ηηη

µηµµ

µ

ηηη ηµ

µµµ

j,i,,1

j,i,,1

j,i,,1 RR

2

j,i,,1

uuuu41

uuuu41

eqj

eqi

−−=

−∂∂

Φ∂−

∑∑

∑∑

==

= −=

n

jij

n

i

n

j

n

i

K

rrr

In one dimension the potential energy has the following simple form

( )∑∑==

−+=n

jij

n

i

eq KUU1

2ji

1uu

21

•The lattice is behaving as if all the ions are coupled by springs.

•We can now find the equations of motion for any lattice.

where the sums over µ and η are over the coordinates x,y,z

Crystal Vibration of a Monoatomic Linear Chain

a

Spring constant, g Mass, m

xn xn+1xn-1

Equilibrium Position

Deformed Position

Longitudinal wave of a 1-D Array of Spring Mass System

us: displacement of the sth atom from its equilibrium position

us-1 us us+1

M

Equation of motion for a 1D chainConsider a 1D chain of identical ions, with equilibrium spacing a.

i - 4 i - 3 i - 2 i - 1 i i + 1 i + 2 i + 3a

Let us consider interactions between nearest neighbours only, and let the strength of that interaction be K.

The potential energy is now

( ) ( )∑=

−+ −+−+=n

i

eq KKUU1

21ii

21ii uuuu

21

Hence the force on ion i is

( ) ( )( )11

2

2

i

2221

F

−+ −+−−=

∂∂−=

=

iiii

i

ii

uuKuuK

Uu

dtudm

( )112

22 −+ −−−= iii

i

i uuumK

dtud

Leading to the equation of motion

Vibrations of small chainsConsider transverse vibrations of two ions in isolation:

Ions can vibrate coherently πout of phase.

λ = 2a, k = 2 π/λ = π/a

Now take three ions Possible coherent phase difference,

ϕ = 2π/3, with λ = 3a, k = 2 π/3a

Now take four ions Possible coherent phase difference,

ϕ = π/2, with λ = 4a, k = π/2a

2a

3a

4a

ui(t)

Revisiting reciprocal space

• each value of k describes a particular way of combining the atoms in a chain, or to put it another way, a mode of vibration

• maximum value of k = π/a

• as number of ions increases, so does number of possibilities for k

• for an infinite chain, k can vary continuously between 0 and π /a

• values of k greater than π /a give no new information, i.e. they do not describe any additional possible vibrational modes

We are now (re)discovering the power of reciprocal space.

Mathematically easier to deal with quantities that vary continously than those that vary discretely. In a lattice, the wavelength associated with each possible vibration is a discrete multiple of the lattice constant. However, for an infinite lattice, the reciprocal quantity, k, is a continuous variable.

The real space lattice can be completely described in terms of combinations of its basis vectors {a}.

In exactly the same way, a reciprocal space lattice can be constructed from reciprocal space basis vectors {b}.

λ=2a/3, k = 3π/a λ=2a, k = π/a

2a2a

Solutions of the equations of motion for 1D chain

( )112

22 −+ −−−= sss

s

s uuumK

dtud

The following functions are solutions:)()()()( e ,e ,e ,e ksatiksatiksatiksati −−+−−+ ωωωω

• first two functions describe a wave propagating backward in time, so reject them.

• second two are both valid, so general solution is a linear combination of the two

)()( ee)( ksatiksatis BAtu −−+− += ωω

What determines A and B?

The boundary conditions, i.e. the physics of the situation, whether the ends are held fixed or free, or whether the ends of the chain are joined (Born-von Kármán boundary conditions)

We will now examine three type of one dimensional chain

• a monatomic 1D chain of finite length Na

• a monatomic1D chain of infinite length

• a diatomic 1D chain of infinite length

Monoatomic 1D chain of length Na

Let us examine two possible boundary conditions

1. Fixed ends, i.e. displacement of 0th and Nth ions is zero.

a

( ) BABABAtu tititi −=⇒+=+== −− eee0)( -0

ωωω

Putting A = -B means solutions are standing waves

)()( ee)( ksatiksatis BAtu −−+− += ωω

( ))- eee)( iksaiksatis Atu −−= ω

( )1,0 ; sine2

eee0)(

−=±=⇒=

−==

−

−−

NnNankkNaAi

Atu

ti

ikNaikNatiN

πω

ωHaving the Nth ion also fixed imposes conditions on k

Physically: for n=0 and n=N all ions are at rest. Hence there are N - 1 non-trivial solutions for us(t), consistent with the N-1 degrees of freedom of the chain.

• the N-1 us(t) are the N-1 modes of vibration of the chain

• the set of us(t) are mutually orthogonal set

• all possible linear combinations of the ions in the chain can be described as sums the over the us(t)

• the set of us(t) constitute a basis known as the normal modes.

Finite monatomic chain continuedSubstituting us(t) back into the equation of motion we find

( )

( ) ( ) ( )( )( )

2sin4)cos22(ee2

ee2eee2eee

-2

22

)2

1-s12

2

kaKkaKKm

Km

uuuKdt

udm

ikaika

ikaikaiksaikaikaiksaiksaiksa

sss

=−=−−=

−−−−−−=−−

−−=

−

−−−−

+

ω

ω

us(t) satisfies the equation of motion if for each value of k there is a frequency

2sin2 ka

mK=ω

k

ω

π/a−π/a

N=4

Dispersion relation

Note that grey points simply replicate green points, they are an equivalent set of k

Finite monatomic chain continued2. Join the ends of the chain (Born-von Karman boundary conditions)

( ) ( ) ( )ikNaikNatiN

ti BAtuBAtu −+==+= eee)(e --0

ωω

Set B = 0, then boundary conditions are satisfied if

2 1 e ; , 0 , 1 o r , e q u i v a l e n t l y , 1 , 2 , . . .2

i k N a n Nk n N nN aπ= = = − = ± ± ±

with N modes of vibration between k = −π/a and π/a, the 1st Brillouin zone. Note the spacing of k values is 2π/Na = 2π/L

Setting B=0 corresponds to the travelling wave solution

( ) iksatis Atu ee - ω=

k

ω

π/a−π/a

N=4

Any value of A and B satisfying the Born-von Kármán boundary conditions arrives at a set of N modes of vibration for the lattice.

Dispersion relation

Allowed Wavevectors (K)

Solution: us ~uK(0)exp(-iωt)sin(Kx), x =saB.C.: us=0 = us=N=10

K=±2nπ/(Na), n = 1, 2, …,NNa = L

a

A linear chain of N=10 atoms with two ends jointed x

Only N wavevectors (K) are allowed (one per mobile atom):

K= -8π/L -6π/L -4π/L -2π/L 0 2π/L 4π/L 6π/L 8π/L π/a=Nπ/L

Kx

Ky

Kz

2π/L

Allowed Wave Vectors in 3D

LN

LLKKK zyx

πππ ±±±= ;...;4;2;0,,

N3: # of atoms

One dimensional infinite monatomic latticeAs N approaches infinity, spacing between the allowed values of k becomes infinitesimal.

kπ/a−π/a

Dispersion relation

2sin

2sin2 max

kakamK ωω ==

Important features:

• ω> ωmax does not propagate in the lattice

•both longitudinal and transverse polarizations are valid solutions

• a single mode wave has a defined phase velocity, vp= ω/k

• for a wave packet consisting of a number of modes, group velocity is more meaningful

vg= dω/dk

3π/a2π/a 4π/a−2π/a−3π/a−4π/a

ω

At the first Brillouin zone boundary: i.e wave solution is a standing wave or a superposition of two travelling waves reflected at lattice points.

02

cos2

v maxg === kaa

dkd ωω

In long wavelength limit dispersion is light-like pgmax vv,

2 and ,0 =→→→

kkak ωωω

Solution of Lattice Dynamics

Identity:

Time dep.:

cancel

Trig:

s-1 s s+1

Same MWave solution:u(x,t) ~ uexp(-iωt+iKx)

= uexp(-iωt)exp(isKa)exp(±iKa)

ω: frequency K: wavelength

ω-K Relation: Dispersion Relation

K = 2π/λλmin = 2aKmax = π/a-π/a<K< π/a

2aλ: wavelength

Polarization and VelocityPolarization and Velocity

( ) ( )[ ] ( )

( ) 21

cos12

cos12expexp22

KaMC

KaCiKaiKaCM

−=

−=−−−=

ω

ω

Freq

uenc

y,ω

Wave vector, K0 π/a

Longitudinal Acoustic (LA) M

ode

Transverse

Acoustic (T

A) Mode

Group Velocity:

dKdvgω=

Speed of Sound:

dKdv

Ks

ω0

lim→

=

1D infinite diatomic lattice

a

Two common variants of this problem

1. Ions have alternating masses, but same spring constant between nearest neighbours. This is the case first we will be considering.

2. Ions have equal masses but the spring constant alternates between neighbouring pairs, a chain of “molecules”

i-2b i-1w i-1b iw ib i+1w i+1b i+2w

jF ˆUu

Us

s ∂∂−=−∇=

Expect solutions to be linear combinations of the solutions for two infinite monatomic chain with spacing a.

The equations of motion for ion s are obtained from

iksatibs

iksatiws

Btu

Atu

ee)(

ee)(ω

ω

−

−

=

=

( )∑∑==

−+=n

jij

n

i

eq KUU1

2ji

1uu

21

The lattice potential energy, U, in the harmonic approximation is

( )

( )wi

wi

bi

bi

b

bi

bi

wi

wi

w

uuuKdt

udm

uuuKdt

udm

12

2

12

2

2

2

+

−

−−−=

−−−=

Hence, with only nearest neighbours interacting,

for a transverse vibration

Substituting in usw(t) and us

b(t)

( )( ) ( )( )( ) ( ) 02e1

e2

0e12

e2

2

2

2

2

=+−++−⇒

−−−=−

=+−+−⇒

−−−=−

+

−

−

KmBA

AABKBm

BKmA

BBAKAm

bika

ikab

ikaw

ikaw

ωω

ωω

which has a solution when the determinant of the following matrix is zero

( )( ) 2 e1-

e1 22

2

Km

-Km

bika

ikaw

+−+

++− −

ωω

i.e. when( )

( ) 0cos12

22

24

=−+

+−

kaK

mmKmm wbwb ωω

Note values of A and B depend on boundary conditions

Solutions for ω2 simply those for quadratic equation( ) ( ) ( )

bw

bwbwbw

mmkaKmmmmKmmK

2cos1842 222

2 −−+±+=ω

ω

k2π/aπ/a−2π/a −π/a

transverse optic transverse acoustic

optic

acoustic

( )wb

wb

mmmmK +2

bmK2

wmK2bw mm >

Two types of vibration (both of which may be transversely or longitudinally polarised):

Dispersion relation:

bandgap

Lattice Constant, a

xn ynyn-1 xn+1

( )

( )nnnn

nnnn

yxxfdt

ydM

xyyfdt

xdM

2

2

12

2

2

12

2

1

−+=

−+=

+

−

Two Atoms Per Unit CellTwo Atoms Per Unit Cell

Solution:

Ka

M2 M1

f: spring constant

1/µ = 1/M1 + 1/M2

What is the group velocity of the optical branch? What if M1 = M2 ?

Acoustic and Optical Branches

K

Ka

Lattice Constant, a

xn ynyn-1 xn+1

PolarizationPolarization

Freq

uenc

y,ω

Wave vector, K0 π/a

LA TA

LO

TO

OpticalVibrationalModes

LA & LO

TA & TO

Total 6 polarizations

• System now has 2N degrees of freedom, hence there are 2N normal modes.

• 1st Brillouin zone extends from -π/a to π/a

• there is a gap in which vibrations of a certain frequency cannot be sustained by the lattice

• a 3D crystal composed of a total of N unit cells, with p ions per unit cell will have 3 acoustic and 3(p – 1)N optic modes of vibration

• if ω2 is negative, ω is imaginary and the mode is unstable

• an optic mode with ω close to 0 is called a soft mode and is often involved in phase transitions

0→kaIn the long wavelength limit,

( ) ( )( )2

2

kaOmm

mmK

kamm

K

bw

bwo

bwa

++→

+→

ω

ω

so that( ) 2

21cos kaka −≈

leading to the following expressions for the frequencies of acoustic and optic modes:

Oscillations of diatomic lattice. A B

K K’

unun-1 vn-1vn un+1 vn+1

State 1

State 2

)v(uK')uK(vvM

)u(uK')uK(vuM

n1nnnB

1-nnnnA

−+−−=

−−−=

+

••

••

n

n

Equations of motion are:

Substituting:

na na+b (n+1)a+b

Xa=na Xb=na+bCoordinates

t))-Vexp(i(nkavt))-Uexp(i(nkau

n

n

ω=ω=

ω=

∆VU

MVU 2The equations can be written as

+η−−

η−−+=∆

''*''

KKKKKKKK )exp(ika=η

=

B

A

MM

M0

0

222, DCBATO ++±=ω

The solution of the equation exists for all values of ω except for the ones that are in between of the special frequencies, given by

++=

BA MMKKA 11)'(

21

−+=

BA MMKKB 11)'(

21

))cos('(1 kaKKMM

CBA

+−=BAMM

kaKC )sin('−=

Oscillations of diatomic lattice. Analysis. There are two types of the excitations

(phonons) in diatomic crystals: a) acoustic (bottom curve), when A

and B atoms oscillate in- phase .b) optical (top curve), when A and

B atoms oscillate out of phase.

-There are excitations ωA(k) <ω< ωO (k) that cannot propagate in the lattice in the lattice at all. - The stronger the bonds between the atoms (larger K), the higher the excitation frequencies and the larger the gap between ωA and ωO.- The heavier the atoms, the smaller the frequencies and the smaller the gap between ωA and ωO .

sTT CkMKakthenkaifkaMK ==ω<<=ω /12

;2

sin/2

Note ωO≠0 if k=0

Monatomic lattices have only acoustic (bottom) branch with the frequencies:

sT CkMKakthenkaif ==ω<< /12

;2

sin/2 kaMKT =ω

Cs is is the speed of sound. For crystals Cs=103-104 m/s.Cs increases as the stiffness of the lattices increases and decreases if the atoms substituted by the heavier ones. In general Cs≈(E/ρ)½ , where E is the Young’s modulus and ρ is the density of the crystal.

Acoustic phonons are responsible for heat conductivity of the crystal lattice.

What crystal has the highest thermal conductivity?

Chapter 4, figure 8 in Kittel (7th edition) illustrates measured dispersion relations for Ge and KBr along the [111] direction in the reciprocal lattice.

Ge and KBr are cubic lattices.

Note that the transverse modes are doubly degenerate because of symmetry direction.

Different scales reflect different strength of ion-ion interaction in Ge and KBr.

Dispersion in Si

0 0.2 0.4 0.6 0.8 1.00.20.40

2

4

6

8

(111) Direction (100) DirectionΓ XL Ka/π

LA

TATA

LA

LO

TO

LO

TO

Freq

uenc

y (1

0 H

z)12

Dispersion in GaAs (3D)Dispersion in GaAs (3D)

PhononPhonon

hω

Energy

Distance

Equilibrium distribution1exp

1

−

=

Tk

n

B

ω

• where ħω can be thought as the energy of a particle called phonon, as an analogue to photon

• n can be thought as the total number of phonons with a frequency ω, and follows the Bose-Einstein statistics:

ω

+=

21nu

•The linear atom chain can only have N discrete K ω is also discrete

• The energy of a lattice vibration mode atfrequency ω was found to be

Normal modes and phonons• Description of lattice vibrations has so far been purely classical because we solved classical equations of motion to find the vibrational modes and dispersion relation of the lattice.

• In the case of a harmonic potential, the classical approach gives the same modes and dispersion relation as the quantum approach.

• Each mode is the mode of vibration of a quantum harmonic oscillator with wave vector k and polarisation s and quantised energy:

( ) ( ) 1e1 ,

21

/,,, −=

+= Tkkskssksk Bs

nknE ωω

where n is the number of phonons in the mode k,s.

A phonon is a bosonic particle with wave vector k and polarization s

• The more phonons in the mode, the greater the amplitude of vibration.

Total Energy of Lattice VibrationTotal Energy of Lattice Vibration

( ) pKpKp

l nE ,, 21 ωω∑∑

+=

K

p: polarization(LA,TA, LO, TO)K: wave vector