Thermal Properties of Crystal Lattices

dTvCQ

Remember the concept of capacity heat

Amount of heat Temperature increase

Heat capacity at constant volume

pdvQdU

Internal energy

constant volumedV=0

0

vT

UvC

How to calculate the vibrational energy of a crystal ?

Classical approach

2

2

1xDEpot

x

222

1

2

2rm

m

pE

•In the qm descriptionapproach of independentoscillators with singlefrequency is calledEinstein model

•In classical approachdetails of Epot irrelevant

dTvCdUQ

Click for details about differentials

Let us calculate the thermal average of the vibrational energy NEU

Classical: )r,p(EE

defines state (point in phase space)

Energy can change continuously to arbitrary value

rdpd)r,p(E

e

rdpd)r,p(E

e)r,p(E

E

33

33

Average thermal energyof one oscillator

Boltzman factor, whereTBk

:1

rdpd

rmm

p

e

rdpd

rmm

p

ermm

p

E

3321

2

3321

22221

2

2

22

2

22

2

pdm

p

e

pdm

p

em

p

32

322

2

2

2

rdrm

e

rdrm

erm

321

321

2221

22

22

TBk2

3 TBk

2

3 TBk3

zdpydpxdpm

ppp

epdm

p

e

zyx

232

2222

3

2

2

dpm

p

e

32 2

dxxem

pm

x2

3

2

m

zdpydpm

pp

exdpmp

emxp

pdm

p

em

pzyx

222

2332

2

2222

2

2

223

22

dpmp

edpmp

e

mm 22

3

mm 232

3

pdm

p

e

pdm

p

em

p

32

322

2

2

2

mm 232

3

32

m

343

m

m

TBk2

31

2

3

The same applies for the second integral

rdrm

e

rdrm

erm

32

1

32

122

2

1

22

22

TBk2

3

TBkE 3 TBkNU 3

Classical value of the thermal average of the vibrational energy TBkNU 3

Understanding in the framework of: Theorem of equipartition of energy

every degree of freedom TBk2

1

Example: diatomic molecule

TBkTBkTBkU

2

1

2

1

2

3

Only rotations relevant where moment of inertia 0J

Vibration involveskinetic+pot. energy

TBk

vT

UvC

with nRBkNvC 33

# of moles

Gas constant R=kB NA = 8.3145J/(mol K)

= n 24.94J/(mol K)

Classical limit

Solid: N atoms 3N vibrational modes TBkNU 3

Requires quantum mechanics

1D Quantum mechanical harmonic oscillator

)x(E)x(xmdx

)x(d

m

222

12

2

2

2Schrödinger equation:

Solution: Quantized energyn

1E (n )

2

E

n nE (x)

x

2

n nE (x)

x

0

1E

2

1

1E (1 )

2

5

1E (5 )

2

CORRESPONDENCE PRINCIPLE

Large quantum numbers: correspondence between qm an classical system

2

20(x)

x Classical point of reversial

Classical probability density

Einstein model: N independent 3D harmonic oscillators

x y zn n nE(3D) E E E 3 E

n

n

En

n 0

E

n 0

E eE

e

Energies labeled by discrete quantum number n

Boltzman factor weighting everyenergy value TBk

:1

n

n

E

nE

n 0

ep

e

Probability to find oscillator in state n

nn 0

p 1

Quantum mechanical thermal average of the vibrational energy N)D3(EU

1(n )

2

n 0n 1

(n )2

n 0

1(n ) e

2E

e

Let us introduce

1(n )

2

n 0

Z e

partition function

n

1 ZE

Z

therefore calculate Z

n/ 2

n 0

e e

1

(n )2

n 0

Z e

/ 2 1

e1 e

/ 2 1Z e

1 e

/ 2 / 2Z 1 1e e

1 e 1 e

/ 2 2/ 2e

e 1 e e2 1 e

1 ZE

Z

e

2 1 e

1 1E

2 e 1

1E n

2

where1

ne 1

Bose-Einsteindistribution

With NEU 3 and 1E n

2

1n

e 1

In the Einstein model where E for all oscillators

2

3

1

3 ETBk/

E N

e

NU

zero point energy

vT

UvC

Heat capacity:

2

2

13

TBk/E

TBk/EBE

Bve

eTk/kNC

Note: typing error inEq.(2-57) in J.S. Blakemore, p123

Classical limit

2

2

13

TBk/E

TBk/EBE

Bve

eTk/kNC

1 for

EBTk TBk/E

B

E eTk

2

for

EBTk

•good news: Einstein model explains decrease of Cv for T->0

•bad news: Experiments show

3TCv for T->0

0 1 2 30,0

0,5

1,0

CV

/3N

k B

T/TE

Assumption that all modes have the same frequency E unrealistic

refinement

Debye Model

We know already: )k(

wave vector k labels particular phonon mode

1)

2)

3) total # of modes = # of translational degrees of freedom

3Nmodes in 3 dimensions N modes in 1 dimension

Let us remind to dispersion relation of monatomic linear chain

N atoms N phonon modes

labeled by equidistant k values within the1st Brillouin zone of width

a

2

distance between adjacent k-values

LaNdk

221

Ldk

2

.constE

A more detailed look to the origin of k-quantization

D a

Quantization is always the result of the boundary conditions

Let’s consider periodic boundary conditions

Atom position n characterized by

After N lattice constants a we end up again at atom n

( ( ) )i k n N a tn Nu A e

( )i kna tnu A e

( )i kna tnu A e

1ikNae 2kNa n

2k n

Na

2 2k n and k

L L

In 3D we have: ,L

dkx

x

2

yy L

dk

2

andz

z Ldk

2

One phonon mode occupies k-space volume

VLLL

dkdkdkkdzyx

zyx

33 2222

Volume of the crystal

How to calculate the # of modes in a given frequency interval , d ?

11 d

22 d

k

D( ) (k) Density of states

3

3k

Vd k

2

Blakemore callsit g(), I prefer D()

)k( 1 )k( 2

))k(( 1

11 d))k((

k

vL

vT,1=vT,2=vT

Let us consider dispersion of elastic isotropic medium

d)(Dmax

0

total # of phonon modes In a 3D crystal Nd)(Dmax

30

Particular branch i: kv i

kd)()(

V)(D k

332

dkkkd 23 4here

kv)k()k( i

22

i

k

vk

ki

dv

dk 1

k

ii

kk d

vv)(

)(

V)(D

14

2

2

3 3

2

22 iv

V

Taking into account all 3 acoustic branches

332

2

21

2 TL vv

V)(D

What is the density of states D(ω) good for ?Calculate the internal energy U

00

Ud)T,(n)(DUmax

# of modes in , d

Energy of a mode # of excited phonons )T,(n

temperature independentzero point energy

= phonon energy

D(ω)

00

Ud)T,(n)(DUmax

00

2

332 1

21

2Ud

evv

VU

max

TL

How to determine the cut off frequency max ?also called Debye frequency DDensity of states of Cu

determined from neutron scattering

2)(D

Nd)(DD

30

choose D such that both curves enclose the same area

Nd)(DD

30

332

2

21

2 TL vv

V)(D

3332

921

2 DTL

N

vv

V

00

2

3 1

9Ud

e

NU

max

D

withvT

UvC

de

e

Tk

NC

max

TBk/

TBk/

BDv

02

2

231

9

de

e

Tk

NC

max

TBk/

TBk/

BDv

02

2

231

9

D

energy

BD k/

temperature

D:

Tkx

B

Substitution:

dxTk

d B

Debye temperature

dxe

exTNkC

T/D

x

x

DBv

0

2

43

19

0T dxe

exdx

e

exx

xT/D

x

x

02

4

02

4

11

34

5

12

DBv

TNkC

Click for a table of Debye temperatures