4. Heat Capacity of a Free Electron Fermi Gas
Prediction from classical statistical mechanics (for N free electrons):
Cv,free electrons ~ (3/2)NkB
However, the observed electronic contribution to the heat capacity
at room temperature is usually less than 1% of this predicted value.
The reason for this discrepancy is that the free electrons in a
metal must obey the Pauli exclusion principle.
When a metal specimen is heated
from absolute zero, not every
conduction electron gains an energy
~kBT,
T as expected classically
classically.
kB T
kz
F
..........
...............
..................
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...........................
.............................
F
...............................
................................
kF
.................................
…...............................
…...............................
…...............................
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................................
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…...............................
…...............................
…...............................
…...............................
kBT
1/2
F
ky
kx
For a simple, qualitative argument:
If there are N free electrons in the metal, only a
fraction (in the order of kBT/F ) can be excited
thermally, the energy of each electron is increases by
~ kBT upon thermal excitation.
kB T
F
The total thermal energy of the electrons
k T
U  N  B
 F
The electronic heat capacity:
C ele 

T 
 k BT  N   k BT

 TF 
T 
U
 Nk B  
T
 TF 
Celel is proportional to T,
T
agreeing with experimental
measurements.
At room temperature Cele is smaller than the classical value NkB/2) by
a factor ~0.01. Clearly, electrons cannot be treated as classical particles.
1
The quantitative expression for Cele can be derived for kBT<<F
When the temperature is increased from 0 K
to T, the energy of the system is increased by:

F
0
0
U  U (T )  U (0)   d D ( ) f ( )   d D ( )
D()
1/2
Df
T=0
For a system
y
with N free electrons:

F
0
0
N   d D ( ) f ( )   d D ( )
T>0
which can be rewritten as:

d F
  F   d  D( ) f ( )   F d  D( )
F
F
0
 0  F 
The expression of U can then be written as:
F

F
F

F
U       d D ( ) f ( )   d D ( )       d  F D ( ) f ( )   d  F D ( )
F 
0
F 
0
 0
 0

F
F
0
  d    F D ( ) f ( )   d  F    [1  f ( )]D ( )
Since the only temperature dependent term in U is f(), we have
Cele 


df ( )
dU
df ( )  F
df ( )
 d    F D ( )
  d  F   
D ( )   d    F D ( )
0
0
dT
dT  F
dT
dT
F
The electronic heat capacity

C ele   d    F D ( )
0
df ( )
dT
For T << TF, the term df()/dT is only significant
around  = F, so

Cele  D ( F )  d    F 
0
f ( ) 
1
e (   ) / k BT  1
df ( )
dT
df ( )    F
e (   F ) / k BT

dT
k BT 2 e (   F ) / k B T  1 2
At low temperatures T<< TF,
≈ F, independent of T,

We have:

   F e (   ) / k T
Cele  D ( F )  d    F 
 k B2TD ( F )
0
k BT 2 e (  ) / k T  12

F
F
Since kBT << F, and
finally


dx x 2
e

Cele  k B2TD ( F ) 


ex
x

1
2

dx x 2


B

B
F
/ k BT
d
dx
e
x 2e x
x
 1
2
2
3
ex
1
  2 k B2TD ( F )
3
e 1

x

2
2
Cele  k B2TD ( F ) 


dx x 2
e
ex
x

1
2
1
  2 k B2TD ( F )
3
D()
Using the relationship: D ( )  3 N ( )
2
We get:

1
 2 Nk B2T  2 Nk BT

C ele   2 D ( F ) k B2T 
3
2 F
2TF
F

Cel is proportional to D(F), the density of states at the Fermi level, this is
because only these electrons can be excited.
Experimental heat capacity of metals
At low temperatures (T <<  and T << TF) the heat capacity of metals can expressed
as the sum of the electronic contribution and the phonon (lattice vibration)
contribution: C =C
C +C = T + AT3
A l
A = slope
ele
l
pho
h
 is called the Sommerfeld constant.

 2 Nk B2  2 Nk B

2 F
2TF
To obtain the values of
A and , plot C/T as a
function of T2:
intercept
C/T=  + AT2
C = T + AT3

 2 Nk B2  2 Nk B

2 F
2TF
measured predicted mth/m
Experimentally measured
results on simple metals show
that the Free Electron Model
can reasonably
bl predict
di the
h
electric contribution of to the
heat capacity.
3
The discrepancy between the measured
 values and predicted values using the
free electron model is caused by the
approximations made in the model. The
concept of a “thermal effective mass”
is introduced to interpret experimental
data.
measured predicted mth/m
2/3

3 N
The conduction electrons in metals are


F 
2 mel  V 
not completely free, interactions (with ion
cores, phonons and other electrons) make
the electrons appearing more “massive”,
with an effective mass, mth.
2/3
 2 Nk B2  2 Nk B2  V   2mel 

 
 2   2   mel
2 F
2  3 N    
2

2

The thermal effective mass mth is defined as
2/3
 measured 
 2 Nk B2  V   2 mth 

 

2  3 2 N    2 
 measured mth

 free
mel
Heave Fermions (Steglich et al., 1976)
For most metals mth~ 1-2 mel, meaning that the electrons are basically
free to move throughout the lattice.
However, some materials (metallic compounds) have mth~ 100 -1000
mel. These are called heavy fermion materials. For example
measured
- These are known as highly correlated electron systems, and they arise from
complicated f-electron physics.
- These materials often show bizarre behaviors, such as strange magnetic
order and strange superconductivity.
4
The Principal Quantum(Energy level): n = 1, 2, 3, … …
Angular Momentum Quantum Number (Orbital shape): l = 0, 1, 2, … (n-1)
l:
0
1
2
3
4
5
Name: s
p
d
f
g
h
The Magnetic Quantum Number (# sub-orbitals): ml = -l… 0…+l
The Spin Quantum Number: ms = -1/2 or +1/2
Depending the chemical environment, felectrons can be localized or delocalized.
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5. Electrical Conductivity and Ohm’s Law
The conductivity σ is defined by
j = σ E,
where j = current density
How to derive an expression for σ using the Free Electron Model?
When in an electromagnetic field, an
electron experiences a Lorentz force:

 1 
F  e( E  v  B )
c


p  mv  k
The momentum of a free electron is:






In the case B = 0, the
eE
dv
dk
or
dk  
dt

 eE
F m
equation of motion is:

dt
dt


k (t ) 
If the electric field is applied at t = 0, then at
eE t
dk  
dt

time t, the k vector will change according:
k (0)
0

Equilibrium
q
- no field
ky
v
kx
-v


With applied
pp
field
ky


eE
k   t

E
k
kx
Every electron feels a
shift in its k-value by k.
Net flow of electrons
No net flow of electrons


eE
k   t

With applied field
ky
This simple theory predicts that the velocity of the
electrons would keep increasing under a constant
electric field, which would suggest that, when an
electric field is applied to a metal wire, the electric
current would grow with time,
time with no limit
limit.
E
k
kx
Net flow of electrons
Question: What stops the electrons from moving faster and faster in the electric field?
The velocity of the electrons is limited by collisions with
- Impurities in the lattice.
- Lattice imperfections.
- Phonons.
Phonons
If the average time between two collisions is τ, then
the average electron velocity will be

  
eE
v  k   
m
m
Strained region by impurity exerts a
scattering force F = - d(PE) /dx

I


6

  
eE
v  k   
m
m

v

E
The current density is


 eE  ne 2 


j  nqv  n(e)    
E
m
 m 


j  E
The Ohm’s law
where n is the density of
conduction electrons
The electrical conductivity

The electrical resistivity
ne 2
e
 ( ne) 
m
m
Collision time
Charge density
determines the acceleration 
1


m
ne 2
vy
Electrical conduction
Classical picture: the current is carried
equally by all electrons, each moving with a
very small drift velocity vd.
Quantum-mechanical picture : the current
is carried only by a very small fraction of
electrons, all moving with the Fermi velocity
vF. , since k is very small.
E
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
vF
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
……………………………...
vx
vF ~ 106 m/s
Experimental electrical resistivity of metals

ne 2

m

1


m 1
ne 2 
Strained region by impurity exerts a
scattering force F = - d(PE) /dx
At T ~ 300 K,  is dominated by collisions between
conduction electrons with phonons
At T ~ 4 K,  is determined by collisions between
conduction electrons with impurities.
i
I
L

The collisions with phonons and impurities are often independent of one another,
therefore the collision time can be expresses as:
1


1
L

1
i
   L  i (Matthiessen’s rule)
Phonon contribution
T-dependent
impurity contribution
T-independent
As T => 0, ρL=> 0
therefore ρ= ρi, called the residual resistivity
Sample purity can be estimated by the
Resistivity ratio = ρ(RT)/ρ(0)
Large for pure specimens.
Small for impure specimens.
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