1

Chapter Six Free Electron Fermi Gas

What determines if the crystal will be a metal, an insulator, or a semiconductor ?

Band structures of solids

filledstates

empty states

Conduction bandpartially filled

filledstates

empty states

Eg

Eg>>kBT

empty states

filledstates

Eg

Metal semiconductor Insulator

Conduction electrons are available

at high T or by doping

No conduction electrons

E

Valence band filled / Conduction band emptyEg<kBT

Conduction electrons are available

2

Basic idea : pushing atoms together to form a crystal

free atoms crystalsmolecules

discrete energy levels splitting of levels band of states

Low energy levels remain discrete and localized on atoms. Core states

High energy levels split to form bands of closely energy levels that can extend through the crystal

valence and conduction bands

3

Free electron model – treat conduction electrons as free particles

Continuum states – density of states

Fermi statistics – occupancy of states

Thermal properties – Thermal energy, heat capacity, …

Electrical and thermal transports – scatterings of conduction electron

Magnetic field effect

4

Free conduction electrons in the box

Not interacting electrons(except w/. walls of the box)

( )2mk

2mPK.E.ε

22 h===

In reality, interactions of electrons :

Ions – steady Coulomb interaction (electron binding)

but

Screening by core electrons weakens the attraction at large distancePauli exclusion principle requires that conduction electrons stay away

from core electrons localized at the atoms.

Electrons – strong Coulomb repulsionbut• Coulomb repulsion • Pauli exclusion principle

Electrons tend to stay apart

5

0 Lx

U→ ∞U(x)= 0 0 ≤ x ≤ L

∞ elsewhere

Ψ=Ψ+Ψ− εU(x)dxd

2m 2

22hSchrödinger equation

0

In one dimension

m

0(L)(0) nn ==ϕϕBoundary condition

( ) ( )2

222n

nnnn 2mL) nπ (

2mkε and

Lnπk w/.xkAsin hh

====ϕtherefore,

How to accommodate N electrons on the line ?Pauli exclusion principle + spin degeneracy (two spins↑↓ per level)

Start to fill the levels from the bottom (n=1) and continue to fill higher levels with electrons until all N electrons are accommodated.

1,…, nF, where nF is the value of n for the uppermost filled level.

6

In general cases, such as periodic chain

Density of states

One state every k-interval ∆k=2π/L

Boundary condition L)(x(x) nn +=ϕϕ Ln2π

n ±=k

2πL

/L)(2π1

∆k1D(k)

n

===

uniform

dk/ dε) L/2π(2

dε2D(k)dk) ε D(

)dεD(ε2D(k)dk

==

=

and ε=h2k2/2m

( )

ε1

π2Lm

k2m

π2Lm

k/m) 2(L/2π) ε D( 22

h

hhh

=

==

singly spin density of states in one dimension

k

ε

ε

D(ε)

∝ ε -1/2

7

In three dimensions,

Ψ=Ψ+Ψ

∂∂

+∂∂

+∂∂

− εz)y,U(x,zyx2m 2

2

2

2

2

22h

SchrÖdinger equation0

Boundary condition : Ψ is periodic in x, y, and z with period L

,...L4π ,

L2π ,0k; ,...

L4π ,

L2π ,0k; ,...

L4π ,

L2π ,0k zyx ±±=±±=±±=

One state every k-volume interval ∆kx ∆ky ∆kz=(2π/L)3

( )3

3

3zyx 2π

V2πL

/L)(2π1

∆k∆k∆k1D(k) =

===

( )

( )( )

dεε2m

4πVdε

km

2πVk 4π

dεdk / dε

1k 4π2πVdkk D(k)4π)dε ε D(

3

3

223

2

23

2

hh==

==

8

ε2m4πV) ε D(

2/3

22

=h

x 2 for spin degeneracy

Conduction electrons : free to move through the crystal

Density of conduction electrons n = N/V (Table 1)typically n ~ 1022 ~ 1023 cm-3

mostly “s” orbital electrons but also “p” and “d”

∝ ε 1/2D(ε)

ε

singly spin density of states in three dimensions

ε2m2πV) ε D(

2/3

22

=h

density of states in three dimensions

9

Difference between electrons and phonons

Electrons Phonons

Number N=nV fixed N ~ kBT varies w/. T

Degeneracy Fermions Bosons(Fermi-Dirac statistics) (Planck distribution)two per orbital state ↑↓ n per mode excited

Dispersion ε ∝ k2 ω ∝ k

Density of states D(ε) ∝ ε 1/2 D(ω) ∝ ω2

up to ωD DebyeGround states T=0, Fill energy level from bottom : 2 per level ↑↓

εn

ε3

ε4

ε1

ε2

εF highest level occupied w/. ε F

Fermi energy

Maximum energy : ε F = h2kF2/2m

10

kF

kzStates w/. k ≤ kF are occupied

Fermi sphere – volume in k-space occupiedby electrons in the ground states

ky

Fermi surface – kF states w/. ε = εFkx

electrons of # ) (2π

Vπk342N 3

3F ==

spin volume of Fermi sphere

D(k)

V

N3π2m

ε and V

N3πk3/222

F

3/12

F

=

=

h

typically, ~ 10-8 cm-1 ~ 1 – 10 eV

11

n εFTF

12

f(ε)

ε0

1

εF

εF

T=0

D(ε)

ε

∫

∫

=

=∞

Fε

0

0

)dε ε D(

)dε ε )f( ε D(N

f(ε) is the probability that a state ofenergy ε is occupied

f(ε)=1 , ε ≤ εF

0 , ε > εF

Fermi energy is important because electronic properties are dominatedby states near εF only

kBT << εF

13

Finite temperatures

Kinetic energy of electron increases due to the increase of thermal energy

occupy higher energy levels

What is the probability of occupancy of an electron state w/. energy ε at T ?

Boltzmann factor exp(- ε/kBT) ? For phonons (Bosons)

Electrons are Fermions : quantum effects such as Pauli exclusion principle

Standard problem in statistics (see appendix D)

Fermi-Dirac distribution( )[ ] 1T/kµεexp

1) ε f(B +−

=

where µ is the chemical potential to conserve electron number

At T=0 µ= εF, when ε= µ= εF, f(ε) changes discontinuouslyAt finite T, when ε =µ, f(ε)=1/2When (ε-µ) >> kBT, f(ε) Boltzmann distribution

14

(1)0 ≤ f(ε,T) ≤ 1

0 2 4 6 8 100.00.10.20.30.40.50.60.70.80.91.01.1

f(ε)

ε(A.U.)εF

T=0.01TF

T=0.02TF

T=0.05TF

T=0.5TF

T=1.0TF

(2) when T<0.1TF, µ ≈ εF, and f(ε, T)=1/2 when ε=EFwhen ε< µ, f(ε,T)>1/2when ε> µ, f(ε,T)<1/2

15

-df/dε

εεF0.8εF0.6εF 1.4εF1.2εF

T=0, δ-function

T=0.001TF

T=0.01TF

T=0.02TF

T=0.05TF

(3) Electrons excited from below εF to above εF as T is increased

∆ε ~ kBT

Spread energy region increases with increasing temperature.

16

(4) µ=µ(T) decreases as T increased

why ? D(ε) ∝ ε1/2 non-uniform

What does determine µ ? Total number of electrons is conserved

∫∞

=0

T) , f(ε ) ε D( dεN( )∫

∞

+=

0 B3

3

2 1Tµ)/k-(εexp1

ε(2m)4πV dεN

h

−=

2

F

B2

F εTk

12π1εµ(T)

0.00 0.01 0.02 0.03 0.04 0.050.9980

0.9985

0.9990

0.9995

1.0000

µ (ε

F)

kBT/εF

0.0 0.1 0.2 0.3 0.4 0.5

0.80

0.85

0.90

0.95

1.00

µ (ε

F)

kBT/εF

Hence,

17

(5) Useful expression for D(ε)

εcε(2m)

4πV

dεdnε) D( 3

3

2state ===

h

3F

ε

0

0

ε32cεc dε

ε) f( ε) D( dεN

F

==

=

∫

∫∞

T=0K

FF3

F3F

2ε3N) ε ( D and ,

ε2ε3N ) ε D( ,

ε23Nc ===

Total thermal energy and heat capacity of electrons at TClassical point of view, U = Ne (3kBT/2) and CV= Ne (3kB/2)

In reality, much smaller at room TNot every electrons gains energy 3kBT/2

18

ε T) , ε f( ) ε D( dεU0∫∞

=

At ground state, T=0

εFD(ε

)f(ε,

T)

ε

0.6εFF

5

3

ε

03

ε5

3N

ε52

ε2 3Nε ε

ε2 3N dεU

F

F

F

F

=

== ∫

Average energy of each electron <ε> = 0.6εF

At finite temperature (T≠0), electrons are excited to higher energy statesand U(T) increases.

εFεF

D(ε

)f(ε,

T)

ε 0.00.10.20.30.40.50.60.70.80.91.01.1

ε

f(ε,T

) T=0T≠0T=0

T≠0

19

U(T)U(0K)ε T) , ε f( ) ε D( dεU(T)0

∆+== ∫∞

( )) ε f(-1) ε D( ε)-(ε dε) ε f( ) ε D( )ε-(ε dεUF

F

ε

0F

εF ∫∫ +=∆

∞

∫∫ ∫

∫∫

=+

==

∞

∞

FF

F

F

ε

0F

ε

0F

ε

ε

00

) ε D(ε dε) ε f( ) ε D(ε dε

) ε D( dε) ε f( ) ε D( dεN

( )∫∞

∂∂

==0

Fe TT) , ε f( ) ε D(ε-ε dε

dTdUC

In general, T/TF<0.01, df/dT has non-zero value within couples of kBTD(ε) is about D(εF) in the energy regime εF± kBT

( )∫∞

∂∂

=0

FFe TT) , ε f(ε-ε dε ) ε D(C

( )( ) ( )

( )( )

( ) Tkεε xere wh

1ee

Tx

1T)k/()εε(expT)k/()εε(exp

Tkεε

1T)k/()εε(exp1

dTd

TT), f(ε

B

F2x

x

2BF

BF2

B

F

BFε

−=

+=

+−−−

=

+−

=∂

∂

( )∫∞

+=

T/kε-2x

x22

BFe

BF 1eedx x T)k ε D(C

( )∫∞

∞ +=

-2x

x22

BFe1e

edx x T)k ε D(C

20

( ) 3π

1eedx x

2

-2x

x2 =

+∫∞

∞

FB

2

2B

FB

222BFe

TTNkπ

21

TkT2k

3N3π

3π T)k ε D(C

=

==

∝ TFree electrons contribution to

heat capacity

Ce

T

∝T

T

U

∝T2

0.6NεF

In general, when T<<ΘD and T<<TF=εF/kB

C = γT + AT3 sum of electron and phonon contributions

21

F

B2

TNk

2π γ = ∝ TF

-1 ∝ m (mass of electron)

mth, obtained from measured γobserved, is different from me.

• Interaction between conduction electrons with periodic potential of thecrystal lattice. ------ Band effective mass

• Interaction between conduction electrons with phonons.moving electrons drag nearby ions along

• Interaction between conduction electrons with themselves. A moving electron causes an inertial reaction in the surrounding

electron gas.

For some materials, mth can be 1000me. Heavy Fermions

such as CeAl3, CeCu2Si2, … and other exotic superconductors.

Series in Modern Condensed Matter Physics – Vol 11 edited by H. Radousky 2000

“Magnetisms in Heavy Fermion systems”

22

Transport propertiesApplying

uJ ,J T ,Errr

∇driving field current density

( )TκJ

E σJ

U ∇−=

=rr

rr ( )TLT ∇−+r

E TLT

r+

Electric current density

Heat current density

σ : electrical conductivity

κ : thermal conductivity

LT: thermalelectric coefficient

Consider all physics about

carriers and scatterings

coupling both electric and thermal responses

coefficients

23

Electrical conductivity

Applying an electric field

Equation of motion Er

dtkd

dtPd

dtvdmEe)(F 2

2r

h

rrrv===−=

At a constant E, Electric field accelerates electronsk increases linearlyh

rrr tEe(0)k(t)k −

=−

kx

kyE=0

k<kF occupied

kF

E

kkx

ky

k’

δk

E shifts Fermi sphere in k-space

Each k increases by τEek δh

rr −

=

24

Current density

kδmen

kδn2me

nm

kδe2nmke2

nm

kδmke2

nmke2

nve2J

o

oo

o

k

ok

ok

k

ok

o

k

ok

o

k

kk

kkk

rh

rh

rh

rh

rh

rh

rh

rr

=

∑=

∑+∑=

+∑=

∑=

∑=

0

unshifted

What limits δk ?

scatteringsElectrons can scatter to states of lower energy and reduce current.

Assume collision time is τ

τEek δh

rr −

=

EmτneτEe

menJ

2 r

h

rhr

−=

−=

rr

mτne2

=σkokk gnn +=

Thermal equilibrium

Deviation from non-equilibrium

E σJ ≡And Ohmic devices

Electric conductivity Free electron model

25

EEmτne

meEτnenevJ

2

d σ≡===

Approaches to a “steady state” value

non-equilibrium

δk

timeτ

∆k

0@ In classical picture, all e−s carry charge –e at a constant velocity vd.

@ Only electrons near the Fermi surface contribute to current.

δk<<kF

kF

kx

ky

newly filled nf

newly emptied ne

( )( ) Fef

FeFf

kk

k

evnn venevn

gveJ

+=−−=

= ∑ rr

participating states

all at vF

Paul Drude1863-1906

266

F

d

F

10v

2v~k

k 2δnδn −=≈

( ) m101.7coulomb106.1m108.45kg101.9

nemσ τ

mτneσ

8-219328

31

2

2

Ω×××

×==

=

−−

−

@ Current is carried only by a fraction of electrons traveling at vF.

Both newly filled and newly emptied states contribute same current.nf electrons

ne holesCopper

V

Cross sectional area A=wtLength l

R=ρl/A

ρ(300K)= 1.7µΩcmn=8.45×1028 1/m3

vF= 1.57×106 m/sec

I

V

= 2.5×10-14 sec

l = vFτ = 4×10-8 m= 40 nm

For E = 1 volt/cm vd ~0.43 m/sec

Fraction of states participating

27

mτne2

=σElectron scattering processes

Conductivity σ is limited by scatterings (τ, l)for a perfect crystal, no scattering σ → ∞

Scattering mechanisms

ρ(T)

T0

IV III II

I

ρo

Regime ILarge e-ph scatterings ρ(T) ∝ T

Regime IISmall e-ph scatterings ρ(T) ∝ T5

Regime IIIe-e scatterings ρ(T) ∝ T2

Regime IVimpurity scatterings ρ(T) ∝ T0

Free electron model~ ρo

28

Regime I Large angle Inelastic scattering ( electron-phonon at high T )

Scattering rate ∝ # of phonon ( ∝ T )

τ ∝ T-1

Therefore, σ ∝ T-1

ρ ∝ T

kr q

r

'kr

This neglect umklapp process which gives a different result. (exponential as for k in the insulator)

Umklapp should dominates in an intermediate range of temperature.

Regime II Small angle Inelastic scattering ( electron-phonon at low T )

'kr

kr

qr

qkk ωEEqkkh

rrr

±=±=′

′

Scattering rate τ-1 ∝ # of phonon ( ∝ T 3) Debye

Effectiveness factor of collision∝ T q << kF 2

29

Regime IV Elastic scattering (impurities, boundaries, defects,…)

Impurity concentration etc .. determine τ, l

kr

'kr Energy is conserved

Ek=Ek’

τ = constant

l =vFτ = constant , independent of T

Therefore, σ(T) = σo , ρ(T) = ρo

constant

Regime III Electron-electron scattering'2kr

'2

'121

'2

'121

EEEE

kkkk

+=+

+=+rrrr

2kr

1kr τ-1 ∝ T2 possible states

σ(T) ∝ T-2 , ρ(T) ∝ T2'1kr

30

Two additional rules :

(1) Multiple scattering mechanisms

...ρρρρ

...σ

1σ

1σ

1σ1

...τ

1τ1

τ1

τ1

impurityeeephe

impurityeee

impurityeeephe

phe

+++=

+++=

+++=

−−−

−−

−−−

− Matthiessen’s rule

not exact but pretty good

(2) Residual resistance ratio

oρ ρ(300K)

R(0K)R(300K)RRR =≡

Phonon dominates

Impurity dominates

RRR → ∞, perfect crystal

In general, RRR ~ 102 to 104 (pure metal)

31

Experimental evidences for Matthiesen’s Rule

Three different samples w/. different defect concentrations.

McDonald and Mendelssohn (1950).J. Linde, Ann. Phys. 5, 15 (1932).

32

Motion in magnetic fields

Electric field change magnitude of

Magnetic field change direction of dtkdBkqBvqFB h

rrhrrr

EqFE h==

=×=×=m

dtkd

r

rrr

zBB=r

kr

Lorentz force ⊥ motion directionExample :

0dt

dk

kmqB

dtkd

kmqB

dtkd

z

y

2

2y

2

x

2

2x

2

=

−=

−=

0dt

dk

kmqB

dtdk

kmqB

dtdk

z

xy

yx

=

−=

= solutions

C(t)k

t)sin(ωA (t)k t)cos(ωA (t)k

z

cy

cx

=

==

Helical circular motion ⊥ Bωc=qB/m “cyclotron frequency”

33

Circular motion in both real and k- spaces in free electron model

εk ∝r

ky

kz

kx

B

= constant

Electron at εF moves in orbits along the Fermi surface sphere.

True for all Fermi surfaces, not only for free electrons.

For transport properties, important factor is ωcτ , phase change of electron between two successive scattering events.

34

Hall Effect

xnevxjJ

zBB

d==

=r

r

- - - - - -

+ + + + + +E

FB

e-, vdFE

Magnetic field

current density

zy

x

e-, vdx

y

z FBB

In general,

ne1

BjE

Rx

yH −=≡

Hall coefficient

Hall effect revealsdensity and sign of charge carriers.

yjBRynejBE

)y(njBEeBvqEqF

H≡=

−+=×+=

r

rrrvr

( )thickness×==I

VjE

ρ y

x

yH

Hall resistivity [ Ωm ]

Transverse

35

0.91Cs

1.11Rb

1.01K

1.01Na

0.81Li

valenceMetal expH

theorH /RR

Alkali metals : OK

1.51Au

1.21Ag

1.41Cu Noble metals :numerically incorrect

-0.33Al

-0.83Zn

-1.22Cd

-0.22Be Higher-valent metals : wrong sign

one hole

36

Thermal conductivity

TH TL∇T

jU

dxdTκjU −= the energy transmitted

across unit area per unit timeThe flux of the thermal energy

0E

U

Tjκ

=∇

−≡ r κ : thermal conductivity coefficient

( )( )TκE TLJ

TLE σJ

TU

T

∇−+=

∇−+=rrr

rrr thermal electric current densityElectric current density

Heat current density

In a open-circuit heat measurement,

TκTL

TκTLTLJ

TLE 0J

2T

TTu

T

∇

−=

∇−

∇=

∇=→=

r

rr

rrr

σ

σ

σ

σ

2T

0J

U*

TLκ

Tjκ

−=

∇−≡

=

r

In fact, the 2nd term, LT, is very small in most metals and semiconductors.

Hence, κ* = κ

37

Heat current from phonon – previous chapter

τCv31Cv

31κ 2

gg == l

2FF

F

BB

2e

mv21ε

εTknkπ

21C

=

=Tτ

mnkπ

31κ

2B2

e =

In pure metal, the electronic contribution is dominant at all Ts.

In impure metals or disordered materials,τ is reduced by collisions with impurities, and the phonon contribution may be comparable with the electronic contribution.

Apply to free electrons

Ratio of Thermal to Electrical Conductivity2

B2

ek

3πL

=LTT

ek

3π

τ/mne3m / Tτnkπκ 2

B2

2

2B

2e ≡

==

σLorenz number

Wiedemann-Franz law Lth = 2.45 × 10-8 Watt-Ω/K2

38

A temperature-independent Lorenz number depends on the relaxation processes for electrical and thermal conductivity being the same – which is not true at all temperatures.