Chapter Six Free Electron Fermi Gas
What determines if the crystal will be a metal, an insulator, or a semiconductor ?
Band structures of solids
E
empty
states
filled
states
Conduction band
partially filled
Metal
empty
states
empty
states
Eg
Eg
filled
states
filled
states
Valence band filled / Conduction band empty
Eg<kBT
Eg>>kBT
semiconductor
Insulator
Conduction electrons Conduction electrons No conduction electrons
are available
are available
at high T or by doping
1
Basic idea : pushing atoms together to form a crystal
free atoms
discrete energy levels
molecules
splitting of levels
crystals
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
2
Free electron model – treat conduction electrons as free particles
9 Continuum states – density of states
9 Fermi statistics – occupancy of states
9 Thermal properties – Thermal energy, heat capacity, …
9 Electrical and thermal transports – scatterings of conduction electron
9 Magnetic field effect
3
Free conduction electrons in the box
Not interacting electrons
(except w/. walls of the box)
P 2 (hk )
ε = K.E. =
=
2m
2m
2
In reality, interactions of electrons :
Ions – steady Coulomb interaction (electron binding)
but
z Screening by core electrons weakens the attraction at large distance
z Pauli exclusion principle requires that conduction electrons stay away
from core electrons localized at the atoms.
Electrons – strong Coulomb repulsion
but
• Coulomb repulsion • Pauli exclusion principle
Electrons tend to stay apart
4
In one dimension
U→ ∞
U(x)= 0
∞
m
0≤x≤L
elsewhere
Schrödinger equation
0
L
Boundary condition
therefore,
ϕn = Asin(k n x )
x
0
h2 d2
−
Ψ + U(x)Ψ = εΨ
2
2m dx
ϕn (0) = ϕn (L) = 0
(
nπ
hk n ) h 2 ( nπ ) 2
w/. k n =
and ε n =
=
L
2m
2mL2
2
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.
5
In general cases, such as periodic chain
Boundary condition
ϕn (x) = ϕn (x + L)
n2π
kn = ±
L
Density of states
D(k) =
One state every k-interval ∆k=2π/L
2D(k)dk = D(ε )dε
2D(k)dk 2(L/2π )
D( ε ) =
=
dε
dε / dk
1
1
L
=
=
∆kn (2π /L) 2π
ε
ε=h2k2/2m
and
D( ε ) =
2(L/2π )
mL
=
h 2 k/m
2πh
2m
(hk )2
mL 1
=
2πh ε
singly spin density of states
in one dimension
uniform
k
D(ε)
∝ ε -1/2
ε
6
In three dimensions,
0
SchrÖdinger equation
h2  ∂2
∂2
∂2 
 2 + 2 + 2 Ψ + U(x, y, z)Ψ = εΨ
−
2m  ∂x
∂y ∂z 
Boundary condition : Ψ is periodic in x, y, and z with period L
2π 4π
2π 4π
2π 4π
k x = 0, ± , ± ,... ; k y = 0, ± , ± ,... ; k z = 0, ± , ± ,...
L
L
L
L
L
L
One state every k-volume interval ∆kx ∆ky ∆kz=(2π/L)3
3
1
1
V
L
=
=  =
D(k) =
3
∆k x ∆k y∆kz (2π /L)  2π  (2π )3
V
1
2
dε
D( ε )dε = D(k)4π k dk =
4π k
3
dε / dk
(2π)
2
V
4π k 2 V m
dε
=
=
4π 2
(2π)3 h 2k
(2m)3 ε
h
3
dε
7
D( ε ) =
V  2m 
2  2 
4π  h 
3/ 2
singly spin density of states
in three dimensions
ε
x 2 for spin degeneracy
D(ε)
∝ ε 1/2
ε
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”
V  2m 
D( ε ) = 2  2 
2π  h 
3/ 2
ε
density of states in three dimensions
8
Difference between electrons and phonons
Electrons
Number
Degeneracy
Dispersion
Density of states
Phonons
N ~ kBT varies w/. T
N=nV fixed
Fermions
(Fermi-Dirac statistics)
two per orbital state ↑↓
Bosons
(Planck distribution)
n per mode excited
ε ∝ k2
ω∝k
D(ε) ∝ ε 1/2
D(ω) ∝ ω2
up to ωD
Debye
Ground states T=0, Fill energy level from bottom : 2 per level ↑↓
εn
εF
ε4
ε3
ε2
ε1
highest level occupied w/. ε F
Fermi energy
Maximum energy : ε F = h2kF2/2m
9
kz
States w/. k ≤ kF are occupied
Fermi sphere – volume in k-space occupied
by electrons in the ground states
kF
ky
Fermi surface – kF states w/. ε = εF
kx
4 3 V
N = 2 πkF
= # of electrons
3
3
(2π )
spin volume of
Fermi sphere
1/ 3
 3π N 

k F = 
 V 
2
typically,
D(k)
~ 10-8 cm-1
h  3π N 


εF =
2m  V 
2
and
2
2/3
~ 1 – 10 eV
10
n
εF
TF
11
T=0
∞
N = ∫ D( ε )f( ε )dε
D(ε)
0
εF
εF
= ∫ D( ε )dε
0
ε
f(ε)
1
f(ε) is the probability that a state of
energy ε is occupied
1
, ε ≤ εF
f(ε)=
0
, ε > εF
{
0
ε
εF
Fermi energy is important because electronic properties are dominated
by states near εF only
12
kBT << εF
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
1
f( ε ) =
exp[(ε − µ )/k BT ] + 1
where µ is the chemical potential to conserve electron number
z At T=0 µ= εF, when ε= µ= εF, f(ε) changes discontinuously
z At finite T, when ε =µ, f(ε)=1/2
Boltzmann distribution
z When (ε-µ) >> kBT, f(ε)
13
(1) 0 ≤ f(ε,T) ≤ 1
f(ε)
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
T=0.01TF
T=0.02TF
T=0.05TF
T=0.5TF
T=1.0TF
0
2
4
εF
ε(A.U.)
6
8
10
(2) when T<0.1TF, µ ≈ εF, and f(ε, T)=1/2 when ε=EF
when ε< µ, f(ε,T)>1/2
when ε> µ, f(ε,T)<1/2
14
(3) Electrons excited from below εF to above εF as T is increased
∆ε ~ kBT
-df/dε
T=0, δ-function
T=0.001TF
T=0.01TF
T=0.02TF
T=0.05TF
0.6εF
0.8εF
εF
1.2εF
ε
1.4εF
Spread energy region increases with increasing temperature.
15
(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
∞
∞
V (2m)3 ε
1
N = ∫ dε 2
3
4π
h
exp((ε - µ)/kBT) + 1
0
N = ∫ dε D(ε ) f(ε , T)
0
 π2  k T 2 
B
 
Hence, µ(T)= εF 1− 
 12  εF  


1.00
1.0000
0.95
µ (εF)
0.9995
µ (εF)
0.90
0.9990
0.85
0.9985
0.80
0.9980
0.0
0.1
0.2
0.3
kBT/εF
0.4
0.5
0.00
0.01
0.02
0.03
k BT/ε F
0.04
16
0.05
(5) Useful expression for D(ε)
dn
V
D( ε) = state = 2
dε
4π
(2m)3
h3
ε =c ε
∞
N = ∫ dε D( ε) f( ε)
T=0K
0
εF
= ∫ dε c ε =
0
3N
2c 3
εF
3
3N ε
3N
c=
, D( ε ) =
, and D ( ε F ) =
3
3
2ε F
2 εF
2 εF
Total thermal energy and heat capacity of electrons at T
Classical point of view, U = Ne (3kBT/2) and CV= Ne (3kB/2)
In reality, much smaller at room T
Not every electrons gains energy 3kBT/2
17
∞
U = ∫ dε D( ε ) f( ε , T) ε
0
D(ε)f(ε,T)
At ground state, T=0
εF
U = ∫ dε
0
0.6εF
=
εF
3N
2 ε 3F
εε=
2 5
εF
3 5
2 εF
3N
3N
εF
5
Average energy of each electron
<ε> = 0.6εF
ε
f(ε,T)
D(ε)f(ε,T)
At finite temperature (T≠0), electrons are excited to higher energy states
and U(T) increases.
T=0
T≠0
εF
ε
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
T=0
T≠0
εF
ε
18
∞
U(T) = ∫ dε D( ε ) f( ε , T) ε = U(0K) + ∆U(T)
0
∞
εF
0
0
N = ∫ dε D( ε ) f( ε ) = ∫ dε D( ε )
∞
εF
εF
∫ + ∫ dε ε D( ε ) f( ε ) = ∫ dε ε D( ε )
F
∞
εF
εF
0
0
εF
F
0
∆U = ∫ dε (ε - ε F ) D( ε ) f( ε ) + ∫ dε (ε F - ε) D( ε )(1 - f( ε ) )
∞
∂f( ε , T)
dU
= ∫ dε ( ε - ε F )D( ε )
Ce =
∂T
dT 0
In general, T/TF<0.01, df/dT has non-zero value within couples of kBT
D(ε) is about D(εF) in the energy regime εF± kBT
∞
∂f( ε , T)

d 
1
∂f(ε , T)
∂T
=


0
∂T ε dT  exp((ε − ε F ) /(k BT) ) + 1
∞
x
e
C e = D( ε F )k 2BT ∫ dx x 2
2
x
(
ε − ε F )  exp((ε − ε F ) /( k BT) ) 
e
+
1
-ε F /k B T
=


k BT 2  (exp((ε − ε F ) /(k BT) ) + 1)2 
C e = D( ε F ) ∫ dε ( ε - ε F )
(
∞
C e = D( ε F )k T ∫ dx x
2
B
-∞
2
(e
ex
x
)
)
+1
2
x ex
=
T ex + 1 2
(
)
where x =
ε − εF
19
k BT
∞
π2
∫-∞ dx x e x + 1 2 = 3
2
(
ex
)
2
2
π
π
3N 2
C e = D( ε F )k 2BT
=
k BT
3
3 2k BTF
1 2
T
= π Nk B
2
TF
Free electrons contribution to
heat capacity
∝T
U
Ce
∝T
∝T2
0.6NεF
T
T
In general, when T<<ΘD and T<<TF=εF/kB
C = γT + AT3 sum of electron and phonon contributions
20
π 2 Nk B
γ =
2 TF
∝ TF-1 ∝ m (mass of electron)
mth, obtained from measured γobserved, is different from me.
• Interaction between conduction electrons with periodic potential of the
crystal 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”
21
Transport properties
Applying
r r
E, ∇T
r
J, Ju
current density
driving field
Electric current density
Heat current density
(
)
(
σ : electrical conductivity
coefficients
)
r
r
r
J = σ E + L T − ∇T
r
r
r
J U = κ − ∇T + TL T E
κ : thermal conductivity
LT: thermalelectric coefficient
Consider all physics
about
carriers and scatterings
coupling both electric and thermal responses
22
Electrical conductivity
Applying an electric field
r
E
r
r
r
2
r
Equation of motion v
d v dP
dk
F = (−e)E = m 2 =
=h
dt
dt
rdt
r
r
− eEt
At a constant E, k(t) − k(0) =
Electric field accelerates electrons
h
k increases linearly
ky
ky
E
E=0
k<kF occupied
kx
δk
k
k’
kx
kF
E shifts Fermi sphere in k-space
r
r − eE
Each k increases by δ k =
τ
23
h
Current density
r
r
J = 2∑ ev k n k
k
r
hk
= 2∑ e n k
k
m
r
r unshifted
 hk o hδk  o
n k
= 2∑ e
+

ko
m
m


r
r
0
hk o o
hδk o
n k + 2∑ e
nk
= 2∑ e
ko
ko
m
m
r
eh 
o
=  2∑ n k δk
m  ko 
eh r
= n δk
m
Deviation from non-equilibrium
n k = n ok + g k
Thermal equilibrium
What limits δk ?
scatterings
Electrons can scatter to states of
lower energy and reduce current.
Assume collision
r time is τ
r − eE
δk=
τ
h
r
r eh  − eE 
ne2τ r
τ = −
J = n 
E

m h 
m
r
r
And
J ≡ σ E Ohmic devices
ne2τ
σ=
m
Electric conductivity
Free electron model
24
δk
Approaches to a “steady state” value
∆k
0
non-equilibrium
time
τ
@ In classical picture, all e−s carry charge –e at a constant velocity vd.
eEτ ne 2 τ
=
E ≡ σE
J = nev d = ne
m
m
@ Only electrons near the Fermi surface contribute to current. Paul Drude
δk<<kF
ky
kF
newly filled
nf
kx
newly
emptied ne
r
r
J = ∑ ev k g k
1863-1906
k
= n f ev F − n e e(− v F )
= (n f + n e )ev F
participating
states
all at vF
25
@ Current is carried only by a fraction of electrons traveling at vF.
Both newly filled and newly emptied states contribute same current.
electrons
nf
ne
holes
Cross sectional area A=wt
Length l
Copper
R=ρl/A
I
ρ(300K)= 1.7µΩcm
2
V
n=8.45×1028 1/m3
vF= 1.57×106 m/sec
ne τ
m
V
mσ
9 .1 × 10 − 31 kg
τ= 2 =
2
28
−3
−19
ne
8.45 × 10 m 1 .6 × 10 coulomb 1.7 × 10 -8 Ω m
σ=
(
= 2.5×10-14 sec
Fraction of states participating
l = vFτ = 4×10-8 m= 40 nm
For E = 1 volt/cm
)
vd ~0.43 m/sec
δn 2δ k 2v d
−6
≈
~
= 10
26
n
kF
vF
Electron scattering processes
ne 2 τ
σ=
m
Conductivity σ is limited by scatterings (τ, l)
for a perfect crystal, no scattering
σ→∞
Scattering mechanisms
Regime I
Large e-ph scatterings ρ(T) ∝ T
ρ(T)
I
IV
III
II
ρo
Regime II
Small e-ph scatterings ρ(T) ∝ T5
Regime III
e-e scatterings ρ(T) ∝ T2
0
Free electron model
T Regime IV
impurity scatterings ρ(T) ∝ T0
~ ρ27
o
Regime I Large angle Inelastic scattering ( electron-phonon at high T )
Scattering rate ∝ # of phonon ( ∝ T )
r
q
r
k
r
k'
τ ∝ T-1
Therefore, σ ∝ T-1
ρ∝T
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 )
r
k
r
q
r
k'
r r r
k′ = k ± q
E k′ = E k ± hωq
q << kF
Scattering rate τ-1 ∝ # of phonon
( ∝ T 3) Debye
Effectiveness factor of collision
28
∝T2
Regime IV Elastic scattering (impurities, boundaries, defects,…)
Impurity concentration etc .. determine τ, l
r
k
r'
k
Energy is conserved
Ek=Ek’
τ = constant
constant
l =vFτ = constant , independent of T
Therefore, σ(T) = σo , ρ(T) = ρo
Regime III Electron-electron scattering
r'
k2
r
k2
r
k1
r r
r' r'
k1 + k 2 = k1 + k 2
E1 + E 2 = E1' + E '2
r'
k1
τ-1 ∝ T2
possible states
σ(T) ∝ T-2 , ρ(T) ∝ T2
29
Two additional rules :
(1) Multiple scattering mechanisms
1
1
1
1
=
+
+
+ ...
τ τ e − ph τ e −e τ e −impurity
1
1
1
1
=
+
+
+ ...
σ σ e −ph σ e −e σ e −impurity
Matthiessen’s rule
ρ = ρ e − ph + ρ e −e + ρ e −impurity + ...
not exact but pretty good
(2) Residual resistance ratio
R(300K) ρ(300K)
RRR≡
=
R(0K)
ρo
Phonon dominates
Impurity dominates
RRR → ∞, perfect crystal
In general, RRR ~ 102 to 104 (pure metal)
30
Experimental evidences for Matthiesen’s Rule
Three different samples w/. different
defect concentrations.
McDonald and Mendelssohn (1950).
J. Linde, Ann. Phys. 5, 15 (1932).
31
Motion in magnetic fields
Electric field
Magnetic field
Example :
r
r
r
dk
FE = qE = h
change magnitude of
dt
r
r
r
r
r r qh
dk
FB = qv × B =
k×B = h
change direction of
dt
m
r
B= B ẑ
dk x qB
=
ky
dt
m
dk y
qB
=−
kx
dt
m
dk z
=0
dt
r
k
Lorentz force ⊥ motion direction
2
d 2k x
 qB 
= −
 kx
2
dt
m
2
d ky
2
2
 qB 
= −
 ky
m
dt
dk z
=0
dt
solutions
k x (t) = A cos(ωc t)
k y (t) = A sin(ωc t)
k z (t) = C
Helical circular motion ⊥ B
ωc=qB/m “cyclotron frequency”
32
Circular motion in both real and k- spaces in free electron model
r
k ∝ε
= constant
kz
Electron at εF moves in orbits
along the Fermi surface sphere.
B
ky
True for all Fermi surfaces, not
only for free electrons.
kx
For transport properties, important factor is ωcτ , phase change of
electron between two successive scattering events.
33
Hall Effect
Magnetic field
r
B = Bẑ
r
J = jx̂ = nev d x̂
z
y
current
density
y
z
In general,
-
e , vd
x
FB
B
+ + + + + +
E
FE
x
RH ≡
Ey
jx B
=−
1
ne
Hall coefficient
-
e , vd
- - - - - FB
r
r jB
v
r r
F = qE + qv × B = eE + (− ŷ)
n
Transverse r
jB
ŷ ≡ jBR H ŷ
E=
ne
Hall effect reveals
density and sign of charge carriers.
ρH =
Ey
jx
=
Vy
I
× (thickness )
Hall resistivity [ Ωm ]
34
Metal
valence RHtheor /RHexp
Li
1
0.8
Na
1
1.0
K
1
1.0
Rb
1
1.1
Cs
1
0.9
Cu
1
1.4
Ag
1
1.2
Au
1
1.5
Be
2
-0.2
Cd
2
-1.2
Zn
3
-0.8
Al
3
-0.3
Alkali metals : OK
Noble metals :
numerically incorrect
Higher-valent metals :
wrong sign
35
one hole
Thermal conductivity
TH
∇T
jU
TL
dT
The flux of the thermal energy jU = − κ
dx
j
κ ≡ − rU
∇T E = 0
the energy transmitted
across unit area per unit time
κ : thermal conductivity coefficient
Electric current density
Heat current density
(
)
r
r
r
J = σ E + L T − ∇T
r
r
r
J U = TL T E + κ − ∇T
(
)
thermal electric
current density
In a open-circuit heat measurement,
r
r LT r
∇T
J=0 → E=
σ
r
 LT r 
∇T  − κ∇T
J u = TL T 
σ

 TL2T
r
= 
− κ ∇T
 σ

j
κ * ≡ − rU
∇T J = 0
=κ−
TL2T
σ
In fact, the 2nd term,
LT, is very small in
most metals and
semiconductors.
36
Hence, κ* = κ
Heat current from phonon – previous chapter
1
1
κ = Cv g l = Cv g2 τ
3
3
Apply to free electrons
1 2 nk 2B
κe = π
Tτ
3
m
1 2
k BT
C e = π nk B
2
εF
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.
Ratio of Thermal to Electrical Conductivity
1
ε F = mv 2F
2
2
2
B
2
2
2
2
π  kB 
 
3  e 
π nk Tτ / 3m π  k B 
=
=   T ≡ LT
σ
ne τ/m
3  e 
Lorenz number
Wiedemann-Franz law
Lth = 2.45 × 10-8 Watt-Ω/K2
κe
L=
2
37
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.
38