solid state theory physics 545 - bilkent universitygulseren/phys545/pdf/phonons-cv.pdf · solid...

Solid State TheorySolid State TheoryPhysics 545Physics 545

The lattice specific heatThe lattice specific heat

Statistical thermodynamics of Solids:

Kinetic energyKinetic energyIntroduction of structured solidsLaw of Dulong and Petit (Heat capacity) 1819w o u o g d e ( e c p c y)Einstein Model of Crystals 1907Born and von Karman approach 1912Debye Model of Crystals 1912

Electronic energyElectronic energyFermi level 1926Fermi-Dirac distribution

Law of Dulong and PetitThe crystal stores energy as:

- Kinetic energy of the atoms under the form of vibrations.According to the equipartition of energy, the kinetic internal energ is f ½ k T here f is the degree ofinternal energy is f . ½ . k. T where f is the degree of freedom. Each atom or ion has 3 degrees of freedom

EK = 3/2 N k TEK 3/2 N k T

- Elastic potential energy. Since the kinetic energy convert to potential energy and vice versa the average values areto potential energy and vice versa, the average values are equal Epot = 3 N (½ K x2) = 3 N x (½ k T)

h d l i hThe stored molar energy is then:

E = E + E = 3 N k T = 3 R T C = dE/dT = 3 RE = EK + Epot = 3 NA k T = 3 R T C = dE/dT = 3 R

Law of Dulong and PetitWithin this law, the specific heat is independent of:

- temperature- temperature- chemical element- crystal structure

At low temperatures, all materials exhibit a decrease of their specific heatexhibit a decrease of their specific heat

Classical harmonic oscillator Quantum + Statistical mechanicsQ

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.p

• 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.

PhononPhonon Energy

Distance

•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

hωω⎟⎠⎞

⎜⎝⎛ +=

21nu

• 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

1⎞⎛

=n

Bose-Einstein statistics:

Equilibrium distribution1exp −⎟⎟

⎠

⎞⎜⎜⎝

⎛TkB

ω

Total Energy of Lattice VibrationTotal Energy of Lattice Vibration

( ) 1 ⎤⎡ ( ) pKpKp

l nE ,, 21 ωω∑∑ ⎥⎦

⎤⎢⎣⎡ +=

K

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

Experimental observations of lattice specific heatpreceded inelastic neutron scattering.

Model crystal: p atoms per unit cell )Model crystal: p atoms per unit cell )N unit cells ) 3pN harmonic oscillators

Thermal energy (quanta) excites crystal and any number of gy (q ) y yexcitations into quantized energy levels for oscillators.

BoseBose--Einstein StatisticsEinstein Statistics

1ns ω⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠

=kk

Number of excitationsi i l d kexp 1kB

sT

ω ⎛ ⎞⎝ ⎠⎜ ⎟⎝ ⎠ −k in particular mode, k, s

In harmonic approximation, total energy density

1 112

sU sV Vω

ω⎛ ⎞⎜ ⎟⎛ ⎞ ⎝ ⎠

⎜ ⎟ ⎛ ⎞⎝ ⎠= +∑ ∑

kk

k2,s ,sexp 1sV V s

k TB

ω ⎛ ⎞⎝ ⎠⎜ ⎟⎝ ⎠

∑ ∑−

kk k

s: polarization(LA,TA, LO, TO)

B

k: wave vector

Specific Heat (at Constant Volume)Specific Heat (at Constant Volume)

( )⎞⎛ k( )( )

1TV1

TUC

s

s

sVV ω

ω∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= ∑ kk

k( )

1Tk

expB

ss,V −⎠⎝ k

s: polarization(LA,TA, LO, TO)k: wave vectork: wave vector

8 3πVolume of k-space per allowed k value isV

8π=Δk

V( ) ( ) kkkkk

Δ⋅π

= ∑∑ F8

VF3

For Δk → 0 (i.e., V → ∞)

( ) ( )kkk F8

dFV1

Lim 3∫∑π

=k 8VV π∞→

( )d ω∂∫

kk ( )( )3

s8exp 1

k

sV

s

dCT

T

ωωπ

∂=

∂−

∑∫kkk

Bk T

Density of Phonon States in 1D

A li h i f N 10 t

a

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

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

Only 1 K state lies within a ΔK interval of 2π/L# of states per unit range of K is: L/2π # of states per unit range of K is: L/2π DOS ≡ # of K-vibrational modes between ω and ω+dω :

L 1dKd

LD/1

2)(

ωπω =

Density of States in 3D

LN

LLKKK zyx

πππ±±±= ;...;4;2;0,,

LLLy

N3: # of atoms

Ky

Kz

Kx 2π/L2

32

2 ;

/1

2)( LV

dKdVKD ==

ωπω

Density of States

Define D(ω) such that D(ω) dω is total no. of modes withfrequencies in range ω to ω + dω per unit volume of crystal.

( )3 3 3

1 1 1( ) )8 8 8n

d dSD d d dk dS ωω ω δ ω ωπ π π ω

⋅⋅ = = =

∇∫ ∫∫ ∫∫sk

k ( -k( )

( )( )3

18

dSD ωπ ω

=∇∫∫

k k( )k

and for any function Q(ωs(k))dk ( )( ) ( ) ( )38 s

s

d Q d D Qω ω ω ωπ

= ⋅ ⋅∑ ∫k k

Hence( )( ) ( )

exp 1VC D d

Tω

ω ωω

∂= ⋅ ⋅

∂−

∫kk

B

exp 1k T

Lattice Specific HeatLattice Specific Heat

( ) pKpKp

l nE ,, 21 ωω∑∑ ⎥⎦

⎤⎢⎣⎡ +=

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

( )( )

( )( )∫∑∫∑ ⎥⎦

⎤⎢⎣⎡ +=⎥⎦

⎤⎢⎣⎡ += 3

2

,,3,,2

421

221

LdKKn

LdnE pKpKpKpKl

πωωωω K( ) ( )∫∫ ⎥⎦⎢⎣⎥⎦⎢⎣ 2222 LL pp ππ

( )Dispersion Relation: ( )ωgK =

Energy Density: ( ) ( ) ωωωω dDn∫∑ ⎥⎤

⎢⎡ +=∈

1Energy Density: ( ) ( ) ωωωω dDn

pl ∫∑ ⎥⎦⎢⎣

+=∈2

dKdVKD

/1

2)( 2

2

ωπω =

Density of States (Number of K-vibrational modes between ω and ω+dω)

ndd ∈

dKd /2 ωπ

Lattice Specific Heat: ( ) ωωω dDdT

nddT

dCp

ll ∫∑=

∈=

High-Temperature “Classical” Limit:g p

x=ω

« 1xTk B

= « 1

( ) ( ) 3V B B BC k T D d k D d pNkT

ω ω ω ω∂= ⋅ ⋅ = ⋅ =

∂ ∫ ∫

hi h i h h l i l l ( l d i l 3

T∂

which is the same as the classical result (Dulong and Petit law: 3R J/mole/K for a monatomic solid). The reason for this is because at

this level of approximation the energy associated with a quantum ofthis level of approximation the energy associated with a quantum of lattice vibration, ω, exactly cancels out and therefore it doesn't

matter how big that quantum is (including zero).g q ( g )

Low Temperature Limit:pOnly low-frequency acoustic modes excited.

f h b h h i i i i lsωfor each branch, where is initial slope of the particular phonon dispersion

curve (Note that cs is related to the elastic

scsω= ks sc kω =

curve. (Note that cs is related to the elastic constant for the mode, e.g., for [100]L elastic

waves

[ ] == 11100s

cvc [ ] ρL100s vc

h i l i dwhere c11 is an elastic constant and ρthe density

Low Temperature Limit:p

( ) [ ]2 2 2 21 sin 1 1 1cosk k kD d

ππθ ωω θ θ∑ ∑ ∑ ∑∫( )

( ) ( )[ ]2 2 2 2 30

0

cos2 22 2s s s ss s s s

D dc c c c

ω θ θπ ππ π

= = − = =∑ ∑ ∑ ∑∫

32

2

3T

3T

3L

2

2

c2

3

c

1

c

1

c

1

221

π

ω=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡++

π

ω=

21 ⎥⎦⎢⎣

ω⋅ω

⋅ω∂

= ∫ω

d3C2

V

max

Tkx ω

= ω⋅= dTk

dxωπ−

ω∂ ∫ dc21

TkexpT

C32

0B

V

maxω

Tk B Tk B

( )( )

3B

B2Tk

0x

3

23

4B

V cTk

k5

2dx1e

x

2

3

c

TkT

CB

⎟⎠

⎞⎜⎝

⎛π=⋅

−π∂∂

= ∫

at low temperatures the specific heat is proportional to T3.0

Einstein ModelEinstein Model

Each molecule in the crystal lattice is supposed to vibrateEach molecule in the crystal lattice is supposed to vibrate isotropically about the equilibrium point in a cell delimited by the

first neighbors, which are considered frozen.g ,

System of N molecules

the system can be treated as 3Nthe system can be treated as 3N independent one- dimensional harmonic oscillator

Motions in the x, y and z axis areIndependent and equivalent

Einstein ModelSystem of 1-Dim Harmonic Oscillator

Quantized expression of the energy:εv = hυ (v+1/2) v = 0, 1, 2, ...v ( ) , , ,

Partition function (without attributing 0 to the ground state)q = Σ e-(hυ(v+1/2)/kT) = e-(hυ/2kT) Σ e-(hυ/kT) v

Considering the vibrational temperature θ = θΕ = hυ/k

q = e-θ/2T

θ/

The molecular internal energyU l l = - d[Ln(q)] / dβ ]N V = k T2 d[Ln(q)] / dT ]N V

q 1- e-θ/T

Umolecular d[Ln(q)] / dβ ]N, V k T d[Ln(q)] / dT ]N, V

Umolecular = k θ (1/2 + )1

θ/molecular (1/2 )eθ/T - 1

Einstein Model energy of the system

System of N 3-Dim Harmonic Oscillators

gy y

Q = q3N U = 3N Umolecular

18000

20000

gy

U = 3 N k θ (1/2 + )1eθ/T - 1

12000

14000

16000

3/2 Nhυ + 3 NkT

nal e

nerg

6000

8000

10000 3/2 Nhυ

Inte

rn

0 0 0 5 1 0 1 5 2 0 2 52000

4000

6000

0.0 0.5 1.0 1.5 2.0 2.5

kT/hυ

Einstein Model the heat capacity of the system

System of N 3-Dim Harmonic Oscillators

The heat capacity of the crystal is thenC = dU / dTC dU / dT

1.03Nk

10.6

0.8

paci

ty

U = 3 N k θ (1/2 + )1eθ/T - 10.4

Hea

t ca

C = 3 Nk (θ/T)2 ––––––eθ/T

(e 1)2θ/T0 0 0 5 1 0 1 5 2 0 2 5

0.0

0.2

(e – 1)20.0 0.5 1.0 1.5 2.0 2.5

T/θ

Einstein ModelAssumed model for crystal to be 3n harmonic oscillators, each of frequency, Eω ( )EEBk ω=θ

( ) ( )3 ED nω δ ω ω= − ( )( ) ( )VC D d

ωω ω∂

= ⋅ ⋅∫k

Substituting this into Equation ( ) ( )

B

exp 1k

VC D dT

T

ω ωω∂

−∫ k

2⎞⎛⎞⎛

2

E2

E2

B

E2

B

E

BVT

exp

TR3

Tkexp

Tknk3C

⎞⎛ ⎞⎛ θ

⎟⎠

⎞⎜⎝

⎛ θ

⎟⎠

⎞⎜⎝

⎛ θ=

⎞⎛ ⎞⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ ω⎟⎟⎠

⎞⎜⎜⎝

⎛ ω

= ⎟⎠

⎞⎜⎝

⎛ θ⋅=

TFR3 E

EE

B

E 1T

expT

1Tk

exp ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠

⎞⎜⎝

⎛ θ⎠⎝⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛ ω ⎠⎝ T

T » θE, ⎟⎠

⎞⎜⎝

⎛ θT

F EE → 1, so CV → 3R (the classical high-temperature limit).

T « θE, ⎟⎠

⎞⎜⎝

⎛ θ−⎟

⎠

⎞⎜⎝

⎛ θ≈⎟

⎠

⎞⎜⎝

⎛ θT

expTT

F E2

EEE

CV dominated by the exponential term, which is not foundexperimentally at low temperatures.

Einstein Model comparison with experiment

Th l f θ 1325 KThe value of θΕ = 1325 K was given to produce an agreement with the experiment at 331 1 Kthe experiment at 331,1 K.θΕ or ωΕ = kθΕ/∋ is the parameter that distinguishes different substances:

ωΕ ≈ Α (a Ε/m)1/2

whereE i Y ’ d l

Comparison of the observed molar

E is Young’s modulusm is atomic massand a is the lattice parameterComparison of the observed molar

heat capacity of diamond (+) with Einstein’s model. (After Einstein’s

and a is the lattice parameterEinstein model gives also a qualitatively quite good agreement (

original paper-1907) on term of θΕ calculated from the elastic properties

Einstein Model results and limitation

The Einstein Model of crystals takes into account the alteration of the heat capacity by:of the heat capacity by:

- temperature- chemical elementchemical element- crystal structure

This model explained the decrease of the heat capacity at low temperature.

However:However:This decrease is too fast! The experimental results evolve as T3

Reason is that the Einstein model does not consider the collective motion and only consider one vibrational frequency.

Born and von Kármán approach

System of N atoms possess 3N degrees of freedom, all expressing vibrational motion.Thus, the whole crystal has 3N normal modes of vibration characterized by their frequencies υi =ωi/2π

THE LATTICE VIBRATIONS OF THE CRYSTAL ARE EQUIVELENT TO 3N INDEPENDENT OSCILLATORSEQUIVELENT TO 3N INDEPENDENT OSCILLATORS

E = Σ hυi (1/2 + (e(hυi /kt)-1)-1)3N

E Σ hυi (1/2 (e 1) )

Propagation of sound wave in solids notion

This propagation could be solved using the classical concepts since the atomic structure (dimensions) can be ignored in comparison to the wavelength of a sound wave.

The 3-D wave equation ∇2 φ(r) + k2 φ(r) = 0where: k is the magnitude of the wave vector k = 2π/λwhere: k is the magnitude of the wave vector k 2π/λ

Wave phase velocity v = λ υ =λ ω/2π = ω/k

Propagation of sound wave in solids standing waves in a boxg

The 3-D wave equation of motion solved in a cubic box with the side LΦn1 n2 n3 (r) = A sin(n1π x / L) sin(n2π y / L) sin(n3π z / L)

The wave vector in the Cartesian coordinates is k(πn1/L, πn2/L, πn3/L)

In the k space, formed by the allowed values of k(ni = 1, 2, ...), is composed of cubic point lattice with the separation of π/L and the

3volume of Vu= (π/L)3.

Propagation of sound wave in solids Density of statesy

Defining the density of states come to the determination of the number of normal modes of standing waves with the lying magnitude between kof normal modes of standing waves with the lying magnitude between k and k+dk.f(k) dk = (1/8) (4πk2) dk /(π/L)3 = Vk2 dk/(2π2)f(k) dk (1/8) (4πk ) dk /(π/L) Vk dk/(2π )

In term of circular frequency:In term of circular frequency:f(k) dk = f(ω) dω

= (Vk2/2π2) (dk/dω) dω= (Vk2/2π2) (dk/dω) dω= V ω2 dω /(2 v2 vg π2)

Where vg= dω/dk i h l iis the group velocity

Propagation of sound wave in solids Density of statesy

In a non dispersing medium v = vIn a non dispersing medium vg = v

f(k) dk = f(ω) dω = V ω2 dω /(2 v3 π2)f(k) dk f(ω) dω V ω dω /(2 v π )

The wave vector has three independent modes:The wave vector has three independent modes: 1 longitudinal and 2 transversal modes

f(ω) dω = V ω2 dω /(2 π2) (1/vL3 + 2/vT

3 )I i t i M diIn an isotropic Medium vL = vT =vm

f(ω) dω = 3V ω2 dω /(2 v 3 π2)f(ω) dω = 3V ω dω /(2 vm π )

Debye ModelLattice vibrations are regarded as standing waves of the atomic

planes´ displacement

It is assumed that all normal mode frequencies satisfy the equation f th d it f t tof the density of states

An upper limit for frequencies is, however, set such asAn upper limit for frequencies is, however, set such as

ƒωD f(ω) dω = 3Nf( ) d 9N 2 d / 3f(ω) dω = 9Nω2 dω/ωD

3

Now the sum can be replaced with an integral

Σ3N ƒ f( ) dΣ3N..... = ƒωD .....f(ω) dω

Debye ModelDebye Model

uenc

y,ω Kvs=ωKvs=ωDebye Approximation:

( ) ( ) 22 dgg ωωDebye Density of

Freq

u( ) ( )322 22 svd

dggDπω

ωπωω ==Debye Density of

States

33

4 KDπ

0 /a

Number of Atoms: ( )33

2L

KN D

π

π=

( ) 312 Wave vector, K0 π/a( ) 326 ηπ=DKDebye cut-off Wave Vector

Deb e C t off Freq Kv=ωDebye Temperature [K]

Debye Cut-off Freq. DsD Kv=ω

( )sDD

v 3126 ηπωθ ==Debye Temperature

C(dimnd) 1860 Ga 240Si 625 NaF 492Ge 360 NaCl 321B 1250 NaBr 224

BBD kk

θDebye Temperature B 1250 NaBr 224Al 394 NaI 164

Debye Model The energy of the crystal

U Σ ε3N

U = Σ εi1

= Σ [(1/2)ħω + ]3N ħωi= Σ [(1/2)ħωi + ⎯⎯⎯ ]1 ħωi

kTe -1ħ= ƒ [(1/2)ħω + ⎯⎯⎯ ] D(ω) dωħω

e 1ħω kT0

ωD

e -1kT

9 9 3xD

= ⎯ NkθD + ⎯ NkT ƒ ⎯⎯ dx98

9xD

3 e -1x3

x0

xD

Where θD =ħωD/k xD = ħωD/kT x = ħω/kT

D

Debye Model The heat capacity of the crystal

dxe

xNkTx

NkEDx

xD ∫ −+=

3

3 19

89 θ

exD 0 18

T 0T ∝X 0T 0

X ∝

x 3/3

X 0

π4/15xD

3/3E = 9Nkω /8 + 3NkT

E = 9NħωD/8 Vibrational zero-point energy

E = 9NkωD/8 + 3NkT

Cv = 3Nkp gy

Cv = dE/dT]v = 0

v

Debye Model The heat capacity at low temperature

Cv =(dU/dT) v T enters this expression only inCv (dU/dT) v T enters this expression only in the exponential term (β)

ƒ3 4xD

Cv= 3Nk {⎯ ƒ ⎯⎯⎯ dx }3xD

3x4 ex

(ex -1)20D ( )

When T<<θD ƒ ⎯⎯⎯ dx = 4π4/15x4 ex

(ex 1)20

xDƒ (ex -1)20

C = π4 N k12 T 3CV = ⎯ π4 N k ⎯

5 θD

Debye Model-Experiment

The Debye Model gives good fits to the experiment; however, it is only an interpolation formula between two correct limits (T = 0 and infinite)an interpolation formula between two correct limits (T = 0 and infinite)

Lattice Specific Heat Lattice Specific Heat ⎤⎡ θ

( ) 4

0

3

30

32

3

19

221 T

edxxkd

vn

T

xD

B

spl

DD

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥⎦

⎤⎢⎣⎡ +=∈ ∫∫∑

θω

θηω

πωωEnergy Density Tk

xB

ω=

⎦⎣

Specific Heat( ) ⎥

⎥

⎦

⎤

⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛= ∫

TD

x

x

DBl

e

dxxeTkC

θ

θη

02

43

19

( ) ⎦⎣⎠⎝ e0 1

10 6

10 7

C = 3ηkB = 4.7 ×106 Jm3 −K

When T << θD,

34 TCT

10 5

(J/m

-K

)3 Diamond

34, TCT ll ∝∝∈

10 3

10 4

Spec

ific

Hea

t, C

C ∝ T 3

1

10 2S

θD =1860 KQuantumRegime

ClassicalRegime

10 410 310 210 110 1

Temperature, T (K)

Regime

Einstein-Debye Modelsste ebye ode sLattice structure of AlCubic Closest PackingCubic Closest Packing

Θ / Θ 0 79ΘE / Θelst = 0,79

ΘD / Θelst = 0,95

The lattice parameter a = 0,25 nmThe density ρ=2,7 g/cm3

The wave velocity v =3,4 km/sΘelst

Lattice parameter

Cubic close packed, (a) Hexagonal close packed (a, c)

Body centered cubic (a)

Cu (3.6147) Be (2.2856, 3.5832) Fe (2.8664)Ag (4.0857) Mg (3.2094, 5.2105) Cr (2.8846)Au (4.0783) Zn (2.6649, 4.9468) Mo (3.1469)Al (4 0495) Cd (2 9788 5 6167) W (3 1650)Al (4.0495) Cd (2.9788, 5.6167) W (3.1650)Ni (3.5240) Ti (2.506, 4.6788) Ta (3.3026)Pd (3.8907) Zr (3.312, 5.1477) Ba (5.019)Pt (3 9239) Ru (2 7058 4 2816)Pt (3.9239) Ru (2.7058, 4.2816)Pb (4.9502) Os (2.7353, 4.3191)

Re (2.760, 4.458)

Debye Temperature

Ε = ⎯ NkθD + ⎯ NkT ƒ ⎯⎯ dx98

93 e 1

x3x

xD

D ƒ

Where θ =ħω /k x = ħω /kT x = ħω/kT

8 xD3 e -1x

0

C = π4 N k12 T 3

Where θD =ħωD/k xD = ħωD/kT x = ħω/kT

CV = ⎯ π4 N k ⎯125

TθD

The limit of the Debye Modely

The electronic contribution to the heat capacity was not consideredp y

Electronic contribution Fermi level

At absolute zero temperature, electrons pack into the lowest available energy, respecting the Pauli exclusion principle

“ h t t t h b t l ti l ““each quantum state can have one but only one particle“

Electrons build up a Fermi sea and the surface of this sea is theElectrons build up a Fermi sea, and the surface of this sea is the Fermi Level. Surface fluctuations (ripples) of this sea are induced by the electric and the thermal effects.

So, the Fermi level, is the highest energetic occupied level at zero b l tabsolute

41

Electronic contribution Fermi functionec o c co bu o

The Fermi function f(E), drown from the Fermi-Dirac statistics,express the probability that a given electronic state will be occupied at a given temperatureoccupied at a given temperature.

1.0

E E 0E-EF < 0

)0.5f(E

0 200 400 600 800 10000.0

E-EF > 0

TemperatureTemperature

Electronic contribution to the internal energinternal energy

Orbitals are filled starting from the lowest levels, and the last filled or orbital will be characterized by the Fermi wave vector KForbital will be characterized by the Fermi wave vector KFThe total number of electron in this outer orbital is:

∫∫FK

dkkLdkkf 23

)(Because electrons can

∫∫ ==T dkkLdkkfN0

22

32)(2

πBecause electrons can Adopt 2 spin orientations

0

V Nk FT

VN 323π

=3 23 3π

VNk T

F =3π

Electronic contribution to theElectronic contribution to the internal energyinternal energy

The wavefunction of free electron is:

).(),( tkxiAetx ω+=ΨIts substitution in the Schrödinger equation:

)(

222 ),()()( txtxtxE Ψ∂

−=Ψ∇=Ψ 22),(),(

xmtxtxE

∂Ψ∇Ψ

22

kE2

km

E =

Electronic contribution to theElectronic contribution to the internal energyinternal energy

2Fermi Energy

[ ]3 22

2

3

22

2

322

π⎥⎦⎤

⎢⎣⎡==

NTFF kE [ ]

22 ⎥⎦⎢⎣ Vmm FF kEFermi Temperature

EFT kEF

FT =

Temperature effect on electronse pe u e e ec o e ec o sMetal K Na Li Au Ag CuMetal K Na Li Au Ag Cu

NT /V (10^22 cm^3) 1.34 2.5 4.6 5.9 5.8 8.5KF (1/A°) 0.73 0.9 1.1 1.2 1.2 1.35EF (eV) 2 1 3 1 4 7 5 5 5 5 7EF (eV) 2.1 3.1 4.7 5.5 5.5 7TF (K) 24400 36400 54500 64000 64000 81600

Only electrons near from F i l l ff d b hFermi level are affected by the temperature.

Electronic contribution in the heat capacity of a metal

[ ]ET

C VNee

v ,∂∂

= [ ]T VNv ,

⎤⎡∂

∂

∑∞

EfET

CVNi

iiev

,1

)(2 ⎥⎦

⎤⎢⎣

⎡∂∂

= ∑=

dNEfECe

,

)(2 ⎥⎤

⎢⎡∂

∫∞

dNEfET

CVN

iiev

,0

)(2 ⎥⎦

⎢⎣∂

= ∫

TkEmVC Fev

22/12/3

222

⎟⎠⎞

⎜⎝⎛= kC Fv 222

⎟⎠

⎜⎝π

SummaryThe nearest model describing the thermodynamic properties of crystals at low temperatures is the one where the energy is y p gycalculated considering the contribution of the lattice vibrations in the Debye approach and the contribution of the electronic motion ( hi i f i h l di d)(this is of importance when metals are studied).

3 TTCv .. 3 γα +=v γ

Summary

Terms that replaced the partition function are:Density of state (collective motion)Fermi function (electronic contribution)( )