Chapter 6
Free Electron Fermi Gas
(FEFG)
Phys 175A
Dr. Ray Kwok
SJSU
Classification of Solids
Many ways to classify solids
(a) lattice structure and
(b) crystal bonding
(c) measurable properties
One of the easiest and earliest properties
observed is the dc electrical conductivity
Electrical Properties
Some solids conduct current at all temperatures and,
generally, the resistivity of such solids increases when
temperature increases. We call them METALS
Other solids stop conducting at low temperatures and their
resistivity decreases with increasing temperature. We call
them INSULATORS
SEMI-CONDUCTORS are basically insulators with higher
conductivity (due to smaller band gap).
Another way to understand this is to examine the band
structure of the compound (next chapter).
The Free Electron Gas Model
(FEG)
U(x) for a 1-D
crystal lattice:
Simple and
crude finitesquare-well
model:
U
U=0
Can we justify this model?
How can one replace the entire lattice by a constant (zero) potential?
Conduction Band
R
0
3p
-5
V(r)˜ -1/r
Atom 2
Atom 1
R
Energy (eV)
Discrete Electronic
Energy States
-10
-25
Split Energy
Levels
3s
Free electron band
Na: 1s22s22p63s1
monovalent alkali
-30
Atom 1
-35
0
Atom 2
2p
3.67 5
10
Interatomic Distance (Å)
Justifications of the FEG Model
1. Metals have high electrical conductivity and no apparent activation
energy, so at least some of their electrons are “free” and not bound to atoms.
Measured V=IR.
2. Coulomb potential energy of positive ions U ∝ 1/r is screened by bound
electrons and is weaker at large distances from nucleus. It’s further
screened by mobile electrons around to minimize U ~ 0.
3. Electrons would have lowest U (highest KE) near nuclei, so they
spend less time near nuclei and more time far from nuclei where U is
not changing rapidly. Pauli Exclusion Principle also forbid mobile
electrons to be right too close to the core.
4. Therefore model the behavior of the free
electrons with U = 0 inside the volume of the
metal and a finite potential step at the
surface. Each atom has n0 free electrons,
where n0 = chemical valence. Resistance
comes from electrons interacting with lattice
through occasional collisions:
Particle in a box (QM - wave)
Time-independent
Schrödinger Equation:
With U = 0:
h2 2
−
∇ ψ = Eψ
2m
1. Wave functions:
2. Energies:
h2 2
−
∇ ψ + U ψ = Eψ
2m
ψ = Ae
h2k 2
E=
2m
vv
i ( k ⋅r −ωt )
∇ 2ψ = −
2mE
ψ
2
h
∇ 2ψ = −k 2ψ
Traveling waves (plane waves)
E
electronic dispersion
Parabolic energy “bands”
kx
Particle in a 3D box
2
 πx 
ψ n (x) =
sin  n 
L  L 
1-D Potential Well
Here, n is the quantum number
π2h 2 2
En =
n
2
2mL
n>0
3-D “Cubic” Potential Well (with sides length L)
2
ψ n1n 2n3 (x) =  
L
E n1n 2n3
3/ 2
 πx   πy   πz 
sin n1  sin n 2  sin n 3 
L  L  L 
π2 h 2 2
2
2
2
n
n
n
E
n
=
+
+
=
1
2
3
0
2mL2
(
)
Degenerate states with n1, n2, n3. (ni > 0)
Example: 34 Electrons in a box
Quantum numbers (n1, n2, n3), E = n2Eo
The lowest energy for this system is 3E0, which corresponds
to n1 = n2 = n3 = 1
Thus only 2 (two) electrons can have this energy:
one with spin ↑ and one with spin ↓
Next energy level (6E0), for which one of n’s is 2 (3
combinations)
Thus total of 6 (six) electrons can have this energy
Next energy level (9E0) can also accommodate 6 electrons
What are the combinations of n’s for this energy level?
Next energy level (11E0) also accommodates 6 electrons
What are the n’s?
Next energy level (12E0) is two-fold degenerate
34 Electrons in a box (cont.)
So far we have placed 22 electrons, so we need
to add 12 more electrons
What is the next energy level? 14E0
What are n1, n2, and n3?
What is degeneracy?
This is the energy carried by each of the 12
electrons (what kind of energy is this?)
We placed all 34 electrons!!
Reading Energy Diagram
Energy is plotted in terms
of Eo
18
17
(3 ,2 ,2 )
16
In this example EF = 14 Eo
Configuration shown is at
T = 0.
Probability to find
electrons with energy
greater than 14Eo is 0,
and below 14 Eo is 1
(at T = 0)
0
Energy (in units of E )
Values of n’s are shown
on the right.
15
14
(3 ,2 ,1 )
13
12
(2 ,2 ,2 )
11
(3 ,1 ,1 )
10
9
(2 ,2 ,1 )
8
7
6
(2 ,1 ,1 )
5
4
3
2
1
0
(1 ,1 ,1 )
Density of States: FEG
By using periodic boundary conditions for a cubic solid with edge L and
volume V = L3, we define the set of allowed wave vectors:
2πn x
kx =
L
such that:
ky =
2πn y
L
kz =
2πn z
L
n x , n y , n z = ±1, ± 2, ± 3, ...
(No upper limit)
r
r
r
r
ψ( r + L x ) = ψ( r ) = ψ( r + L y ) = ψ( r + L z )
 2π 
so the volume in k-space per state is:  
 L
and the density of states in k-space is:
3
1
V
N(k ) =
= 3
3
(2π / L ) 8π
Just as in the case of phonons!
Fermi Surface
Since the FEG is isotropic, the surface of
constant E in k-space is a sphere. Thus for
a metal with N electrons we can calculate
the maximum k value (kF) and the maximum
energy (EF).
 # states 
 V 4

(volume _ k ) → N = 2  3  πk F3 
# e − = 2

 8π  3
 volume _ k 
1/ 3
 2 N
k
=
 3π

Fermi wave vector
F
V


Fermi energy
(
2
= 3π n
h 2 k F2 h 2
EF =
=
3π 2 n
2m
2m
(
2 spin states per k
N = # of mobile electrons
ky
)
1/ 3
kF
kx
)
2/3
kz
Fermi sphere
Fermi Parameters
Fermi wavevector
Fermi Energy
(
k F = 3π 2ηe
EF =
h 2 k F2
)
2
=
1
Fermi Energy, EF
(
h
3π 2ηe
2m
2m
hk
h
3π 2ηe
Fermi Velocity: vF = F =
m
m
E
Fermi Temp.
TF = F
kB
(
Element Electron
Density, ηe
28 -3
[10 m ]
Na
2.65
Cu
8.47
Ag
5.86
Au
5.90
Fe
17.0
Al
18.1
Sn
14.8
Φ: Work Function
3
Fermi
Energy
EF [eV]
3.24
7.00
5.49
5.53
11.1
11.7
10.2
)
1
)
2
Vacuum
Level
Energy
3
3
Band Edge
Fermi
Temperature
4
TF [10 K]
3.77
8.16
6.38
6.42
13.0
13.6
11.8
Fermi
Wavelength
λF [Å]
6.85
4.65
5.22
5.22
2.67
3.59
3.83
Fermi
Velocity
6
vF [10 m/s]
1.07
1.57
1.39
1.40
1.98
2.03
1.9
Work
Function
Φ [eV]
2.35
4.44
4.3
4.3
4.31
4.25
4.38
Density of States N(E)
The differential number of
electron states in a range of
energy dE or wavevector dk is:
Free electron gas
dN = N ( E )dE = 2 N (k )d 3 k
# of e per E
density of state
h2k 2
E=
2m
1/ 2
 2mE 
k = 2 
 h 
2
N ( E ) = 2 N (k )
N (E) =
Or
2
4πk dk
V
k
mV  2mE 
 V  4πk
= 2 3 
= 2 2
= 2 2 2 
dE
 8π  dE / dk π h k / m h π  h 
(
)
1/ 2
V
3
2
m
E
π 2 h3
 V 4
 V  2mE 
N = 2  3  πk 3  = 2  2 
 3π  h 
 8π  3
1/ 2
N (E) =
1/ 2
2
N(E)
3/ 2
E
1/ 2
dN
V  2mE   2m 
V
3
=
 2  = 2 3 2m E
2 
2 
dE 2π  h   h  π h
(
)
EF
Fermi-Dirac Distribution
The probability distribution function, f(E), which describes
the probability that a state with energy E is occupied for
electrons is the Fermi-Dirac Distribution Function
at T = 0
0, for E > E F
f (E) = 
1, for E ≤ E F
Fermi-Dirac Distribution T > 0
Occupation Probability, f
f (E) =
1
e ( E −µ ) / k BT + 1
µ is the Chemical Potential
which is equal to EF at T=0.
Often, µ and EF are mixed up.
kBT
1
1/2
T=0K
Vacuum
Energy
T →∞
IncreasingT
0
Electron Energy,E
EF
Work Function,Φ
Heat Capacity of the FEG
19th century puzzle: each monatomic gas molecule in sample at
temperature32 kT
T has energy, so if the N free electrons in a metal make up
a classical “gas” they should behave similarly.
Eel = N ( 32 kT )
or expressed per
mole (n):
So the electronic contribution to the
molar heat capacity would be expected
to be
Eel 3 N
= 2 kT = 32 N A kT = 32 RT
n
n
d
Cel =
dT
 Eel  3

= 2R
n


This is temperature independent!!! Only half of the 3R we found for
the lattice heat capacity at high T. But experiments show that the
total C for metals is only slightly higher than for insulators, which
conflicts with the classical theory!
Heat Capacity of the FEFG
FEG + Pauli + Fermi-Dirac (QM)
So at temperature T
the total energy is:
And the electronic
heat capacity is:
∞
∞
EN(E )
E el = ∫ Ef (E ) N(E ) dE = ∫ (E −µ )/ kT
dE
e
1
−
0
0
dE el
1 EN (E)(E − E F )e (E − E F )/ kT
C el =
= 2∫
dE
2
(
)
E
−
µ
/
kT
dT kT 0
e
−1
∞
(
)
This integral is complicated because µ is also a function of T.
Treatment of this approximation is given in advance texts.
The result is:
C el =
π2
3
k 2 N ( E F )T
Cel ∝ T
Understand linear T dependence
# electrons that can
absorb “extra”
thermal energy
≅ (2kT ) N ( EF ) f ( EF )
≅ kT N ( EF )
Only electrons near the Fermi surface
can gain energy !!
total thermal energy E (T ) ≅
of electrons at T
≈2kT
N(E)
N(E)f(E)
E0 + ( 32 kT )kT N ( EF )
E (T ) ≅ E0 + 32 k 2T 2 N ( E F )
FEG heat
capacity at T
Remarkably
close to the
exact result!
Cel =
dEel
≅ 3k 2 N ( EF )T
dT
2
Cel = π3 k 2 N ( EF )T = γ T
E
EF
Total C
But the linear dependence is impossible to measure directly, since the
heat capacity of a metal has two contributions. Now for a metal at low
temperatures we can write the total heat capacity:
C(T ) = Cel + Clattice = γ T + αT 3
Heat Capacity of Metals (expt)
Results for simple metals (in
units mJ/mol K) show that
the FEG values are in
reasonable agreement with
experiment, but are always
higher:
The discrepancy is
“accounted for” by defining
an effective electron mass
m* that is due to the
neglected electron-ion
interactions.
i.e. phonons drag electrons
& make them move slower.
γexpt/ γFEG =
Metal
γexpt
γFEG
Li
1.63
0.749
2.18
Na
1.38
1.094
1.26
K
2.08
1.668
1.25
Cu
0.695
0.505
1.38
Ag
0.646
0.645
1.00
Au
0.729
0.642
1.14
Al
1.35
0.912
1.48
m*/m
Ohm’s Law
V E⋅L
=
R= =
i J⋅A
L 1
R=ρ =
A σ
E L
⋅ 
1 E
J A
⇒ ρ = =
L 
σ J
A 
1
J =σ ⋅E = E
ρ
Examples of Resistivity (ρ)
Ag (Silver): 1.59×10-8 Ω·m
Cu (Copper): 1.68×10-8 Ω·m
Graphite (C): (3 to 60)×10-5 Ω·m
Diamond (C): ~1014 Ω·m
Glass: ~1010 - 1014 Ω·m
Pure Germanium: ~ 0.5 Ω·m
Pure Silicon: ~ 2300 Ω·m
At Room Temp
Microscopic Model
r
r
r
vavg = vd ( E ) + vRMS (T )
v RMS =
3 k BT
m
≈ 10 6
at RT
me
s
1
3
m e v 2 = k BT
2
2
J
J = e nv d ⇒ vd =
en
n ~ 10 23 cm − 3 = 10 29 m − 3 ; e = 1.6 × 10 −19 C
2
v2 =
-6
2
vRMS>>vd
vavg ≈ vRMS (speed)
i = 1.6 A and A = 1 mm (10 m )
i
1.6
−4
depends on T only, not E
vd =
=
=
10
m/s
!??
enA (1.6 × 10 −19 )(10 29 )(10 -6 )
0.1 mm/s? Takes forever to turn on the light??
Drude Model - conductivity
The FEG model was developed by Paul Drude
(1900) in order to describe the electrical and
thermal conductivity of metals. This work greatly
influenced the course of “solid-state physics” and it
introduces basic concepts we still use today.
F = qE = m∆v/∆t ≈mvd/τ
vd = qEτ/m
J = nqvd = σE
σ = nq2τ/m
τ is the relaxation time = l /vavg
l is the mean free path
Temperature dependence of σ is just that of τ
Conductivity in Reciprocal Space
The Fermi sphere contains all occupied
electron states in the FEG. In the
absence of an electric field, there are
the same number of electrons moving
in the ±x, ±y, and ±z directions, so the
net current is zero.
But when a field E is applied along the
x-direction, the Fermi sphere is shifted
by an amount related to the net change
in momentum of the FEG:
ky
Fermi surface
kx
kF
kz
h∆k x = ∆p x = mvx = −eE xτ
eE τ
∆k x = ∆p x = mvx = − x
h
ky
kx
E
The shift in Fermi sphere creates a net current
flow since more electrons move in the –x direction
than the +x direction. But the excess current
carriers are only those very near the Fermi
surface. So the current carriers have velocity vF.
Analysis of Mean Free Path
Since the velocity of current-carrying electrons is essentially independent
of T, we need to examine the behavior of the mean-free-path.
Electrons collide with atoms dominates… Coulomb.
e-
The probability of a collision in a distance ∆x is:
vF
P=
∆x
collision cross-section = π<r2>
cross-sectional area of slab = A
atomic density = na
total cross − sectional area for collision
cross − sectional area of slab
P=
na A∆x π r 2
A
= na ∆x π r 2
Now in a distance ∆x =l , P = 1 is
true, so we can solve for Λ:
1
l=
na π r2
ρ(T)
Now if we assume that the collision cross-section is due to vibrations of
atoms about their equilibrium positions, then we can write:
r 2 = x2 + y2
And the thermal average potential energies can be written as:
1
2
Therefore
Cx 2 =
r2 ∝ T
1
2
Cy 2 = 12 kT
and
l∝
1
1
∝
< r2 > T
σ = nq2τ/m ≈ nq2 l /mvF
Finally
1
σ∝
T
or
ρ∝T
At low T, l is limited by size or impurities, so ρ becomes constant
Example : ρ(T) of K
The Hall Effect
This phenomenon, discovered in 1879 by American physics
graduate student (!) Edwin Hall, is important because it allows us
to measure the free-electron concentration n for metals (and
semiconductors!) and compare to predictions of the FEG model.
It also confirmed the electron-hole description of charge carriers.
Hall probe
FB = q v x B
I
Hall Coef.
A hypothetical charge carrier of
charge q experiences a Lorentz
force in the lateral direction:
FB = qvB
I
t
w
As more and more carriers are deflected, the
accumulation of charge produces a “Hall field” EH
that imparts a force opposite to the Lorentz force:
Equilibrium is reached when these two opposing
forces are equal in magnitude, which allows us to
determine the drift speed:
From this we can write the current density:
FE = qE H
qvB = qE H
J = nqv =
And it is customary to define the Hall coefficient in terms of
the measured quantities: (EH is transverse to J)
n is # of charge carriers per unit volume
v=
nqE H
B
RH =
EH
1
=
JB nq
EH
B
Hall Effect Results!
In the lab we actually measure the Hall voltage VH and the current I, which
gives us a more useful way to write RH:
EH
VH / w
VH t 1
I
=
JA
=
Jwt
=
=
=
=
R
VH = EH w
H
JB (I / wt )B IB nq
If we calculate RH from our measurements
and assume |q| = e (which Hall did not
know!) we can find n. Also, the sign of VH
and thus RH tells us the sign of q! (carriers)
The discrepancies between the FEG
predictions and expt. nearly vanish when
liquid metals are compared. This reveals
clearly that the source of these
discrepancies lies in the electron-lattice
interaction. But the results for Be and Zn
are puzzling. How can we have q > 0 ???
Look at the diagram for +/- charges again.
RH (10-11 m3/As)
Meta
l
n0
solid
liquid
FEG value
Na
1
-25
-25.5
-25.5
Cu
1
-5.5
-8.25
-8.25
Ag
1
-9.0
-12.0
-12.0
Au
1
-7.2
-11.8
-11.8
Be*
2
+24.4
-2.6
-2.53
Zn*
2
+3.3
-5
-5.1
Al
3
-3.5
-3.9
-3.9
Thermal Conductivity of Metals
In metals at all but the lowest temperatures, the electronic
contribution to κ far outweighs the contribution of the lattice.
So we can write:
The electron mean-free path can be
rewritten in terms of the collision time:
l = vτ
Also, the electrons that can absorb extra thermal energy and
therefore contribute to the heat capacity have energies very
near EF, so they essentially all have velocity vF. This gives:
From our earlier discussion the electronic heat capacity is:
It is easy to show that
N(EF) can be expressed:
N(E F ) =
3N
2E F
1
κ ≅ κ el = C el vl
3
1
κ = C el v 2 τ
3
1
κ = C el v 2F τ
3
C el =
π2
3
or per unit volume N ′( E F ) ≡
Which gives the heat capacity per unit volume: Cel′ ≡
Cel π 2 2 T
= 2 k n
V
EF
k 2 N ( E F )T
N ( EF )
3n
=
V
2 EF
Wiedemann-Franz Law
Now the thermal conductivity
per unit volume is:
or…
κ′ =
π 2 k 2 nτ
3m
1 π2 2 T 2
1 π2 2
T
κ′ = 2 k n
vFτ = 2 k n 1 2 vF2τ
3
EF
3
2 mv F
T
τ ~ 1/T, so κ’ ~ independent of temperature!!
Now long before Drude’s time, Gustav Wiedemann and
Rudolf Franz published a paper in 1853 claiming that the ratio
of thermal and electrical conductivities of all metals has nearly
the same value at a given T:
κ
σ
= constant
Not long after (1872) Ludwig Lorenz (not
Lorentz!) realized that this ratio scaled
κ
linearly with temperature, and thus a
Lorenz number L can be defined:
σT
Gustav Wiedemann
≡L
very nearly constant for all
metals (at room T and above)
Lorenz Number
We can readily compare the prediction of the
FEG model to the results of experiment:
π 2 k 2 nτ
LFEG
T
2 2
κ
k
π
=
= 3m
=
2
ne τ
σT
3e 2
m
LFEG = 2.45 × 10 −8
T
(CKJ )2
This is remarkable…it is
independent of n, m, and even τ !!
L = κ/σT 10-8 (J/CK)2
Metal
0 °C
100 °C
Cu
2.23
2.33
Ag
2.31
2.37
Au
2.35
2.40
Zn
2.31
2.33
Cd
2.42
2.43
Mo
2.61
2.79
Pb
2.47
2.56
Agreement with experiment is
quite good, although the value of L
is about a factor of 10 less at
temperatures near 10 K…Can you
speculate about the reason?
An Historical Footnote
Drude of course used classical values for the electron velocity v and heat
capacity Cel. By a tremendous coincidence, the error in each term was about two
orders of magnitude in the opposite direction! So the classical Drude model
gives the prediction:
2
−8
LDrude = 1.12 × 10
(CKJ )
[ C ~ 3/2 Nk, a factor T/TF missing,
½ mv2 = 3/2 kT instead of EF, v2 = (2kT/m) (TF/T), the factor TF/T is missing
κ = 1/3 Cv2τ becomes right order!! ]
But in Drude’s original paper, he inserted an erroneous factor of two, due to a
mistake in the calculation of the electrical conductivity. So he originally
reported:
J 2
L = 2.24 × 10 −8 CK
!!!
( )
So although Drude’s predicted electronic heat capacity was far too high, this
prediction of L made the FEG model seem even more impressive than it really
was, and led to general acceptance of the model.