3Quantum gases

II Statistical Physics 3.7 Fermions
We now move on to study fermions. One major application of this is the study
of electrons in a metal, which we can model as gases. But for the moment, we
can work in complete generality, and suppose we again have a non-interacting
fermion gas. Since they are fermions, they obey the Pauli exclusion principle.
Two fermions cannot be in the same state. Mathematically, this means the
occupation numbers
hn
r
i
cannot be arbitrary they are either 0 or 1. As in
the case of bosons, lets study these fermions using the grand canonical ensemble.
The partition function is
Z =
Y
r
Z
r
,
where
Z
r
=
1
X
n
r
=0
e
βn
r
(E
r
µ)
= 1 + e
β(E
r
µ)
.
This is just like the bosonic case, but we only have two terms in the sum. We
then find that
log Z =
X
r
log(1 + e
β(E
r
µ)
).
Then we have
hn
r
i =
1
β
E
r
log Z =
1
e
β(E
r
µ)
+ 1
.
This is called the Fermi-Dirac distribution. The difference between this and the
bosonic case is that we have a difference in sign!
We can get the total number of particles in the usual way,
hNi =
1
β
β
log Z =
X
R
1
e
β(E
r
µ)
+ 1
=
X
r
hri.
As before, we will stop drawing the h·i.
There is another difference between this and the bosonic case. In the bosonic
case, we required
µ <
0 for the sum
Z
r
to converge. However, here the sum
always converges, are there are just two terms. So there is no restriction on µ.
We assume that our particles are non-relativistic, so the energy is
E =
~
2
k
2
2m
.
Another thing to notice is that our particles have a spin
s Z +
1
2
.
This gives a degeneracy of
g
2
= 2
s
+ 1. This means the density of states has to
be multiplied by g
s
. We have
g(E) =
g
s
V
4π
2
2m
~
2
3/2
E
1/2
.
We can now write down the total number of particles in terms of the density of
states. We have
N =
Z
0
dE g(E)
z
1
e
βE
+ 1
,
where again
z = e
βµ
.
Similarly, the energy is given by
E =
Z
0
dE Eg(E)
z
1
e
βE
+ 1
.
Again, we have the density of states
pV = kT log Z = kT
Z
0
g(E) log(1 + ze
βE
).
Just as for bosons, we can integrate this by parts, and we obtain
pV =
2
3
E.
We now look at the high temperature limit. In the case of boson, we identified
this with a small
z
, and we can do the same here. This is done on example sheet
3, and we find that to leading order, we have
z λ
3
N
V
,
exactly as for bosons. This is exactly the same as in bosons, and again as in the
example sheet, we find
pV = N kT
1 +
λ
3
N
4
2g
s
V
+ ···
.
We see the sign there is positive. So Fermi statistics increases pressure. This is
the opposite of what we saw for bosons, which makes sense.
We now go to the opposite extreme, with zero temperature. Then
β
=
.
So
1
e
β(Eµ)
+ 1
=
(
1 E < µ
0 E > µ
Note that here the
µ
is evaluated at 0, as
µ
depends on
T
. Clearly the value of
µ at T = 0 is important.
Definition (Fermi energy). The Fermi energy is
E
f
= µ(T = 0) = lim
T 0
µ(T, V, N ).
Here we see that at
T
= 0, all states with
E E
f
are occupied, and states
with higher energy are not occupied. Now it is easy to work out what this Fermi
energy is, because we know the total number of particles is N. We have
N =
Z
E
f
0
dE g(E) =
g
s
V
6π
2
2m
~
2
3/2
E
3/2
f
.
Inverting this, we find that
E
f
=
~
2
2m
6π
2
g
s
N
V
2/3
.
We note that this is very much like the critical temperature of the Bose–Einstein
distribution. We can define the characteristic temperature scale
T
f
=
E
f
k
.
Now whenever we talk about whether the temperature is high or low, we mean
relative to this scale.
How high is this temperature? We saw that for bosons, this temperature is
very low.
Example. For
3
He, the Fermi temperature is
T
f
4.3 K.
While this is very low compared to everyday temperatures, it is pretty high
compared to our previous T
c
.
For electrons in a metal, we have
T
f
10
4
K
. This is very high! Thus, when
studying metals, we cannot take the classical limit.
For electrons in a white dwarf stars, we have
T
f
10
7
K
! One reason this is
big is that
m
is the electron mass, which is very tiny. So it follows that
E
f
is
very large.
From the Fermi energy, we can define a momentum scale
~k
f
= (2mE
f
)
1/2
.
While we will not be using this picture, one way to understand this is that in
momentum space, at
T
= 0, all states with
|k| k
f
are occupied, and these
states are known as a Fermi sea. States at the boundary of this region are known
as Fermi surface.
Let’s continue working at zero temperature and work out the equations of
state. We have
E =
Z
E
f
0
dE Eg(E) =
3
5
NE
f
.
This is the total energy at zero temperature. We saw earlier that
pV =
2
3
E =
2
5
NE
f
.
This shows the pressure is non-zero, even at zero temperature! This is called the
degeneracy pressure. This is pressure coming from the Fermi statistics, which is
completely unlike the bosons.
Now let’s see what happens when we are at low, but non-zero temperature,
say
T T
f
. While we call this low temperature, since we saw that
T
f
can be
pretty high, this “low temperature” might still be reasonably high by everyday
standards.
In this case, we have
µ
kT
E
f
kT
=
T
f
T
1.
Therefore we find that z 1.
Our goal now is to compute the heat capacity. We begin by trying to plot
the Fermi-Dirac distribution at different temperature scales based on heuristic
arguments. At T = 0, it is simply given by
E
n(E)
E
f
When we increase the temperature a bit, we would expect something that looks
like
E
n(E)
E
f
kT
Since
kT
is small compared to
E
f
, that transition part is very small. Only states
within kT of E
f
are affected by T .
We can now give a heuristic argument for what the capacity should look like.
The number of states in that range is
g
(
E
f
)
kT
. Relative to where they are in
T
= 0, each of these is contributing an excess energy of order
kT
. Therefore the
total energy is
E E|
T =0
+ g(E
f
)(kT )
2
.
So we expect the heat capacity to be
C
V
k
2
g(E
f
)T kN
T
T
f
.
This rather crude argument suggests that this heat capacity is linear in tempera-
ture. We shall now show that this is indeed correct by direct calculation. It is
the same kind of computations as we did for bosons, except it is slightly more
complicated. We have
N
V
=
g
s
4π
2
2m
~
2
2/3
Z
0
dE
E
1/2
z
1
e
βE
+ 1
.
Similarly, we find
E
V
=
g
s
4π
2
2m
~
2
3/2
Z
0
dE E
3/2
z
1
e
βE
+ 1
.
We again do a substitution x = βE, and then we find
N
V
=
g
s
λ
3
f
3/2
(z)
E
V
=
3
2
g
s
λ
3
kT f
5/2
(z),
where we similarly have
f
n
(z) =
1
Γ(n)
Z
0
dx
x
n1
z
1
e
x
+ 1
.
Recall that we have
Γ
3
2
=
π
2
, Γ
5
2
=
3
π
4
.
We know that
z
1, and we want to expand
f
n
in
z
. This is called the
Sommerfeld expansion. We have
Γ(n)f
n
(z) =
Z
βµ
0
dx
x
n1
z
1
e
x
+ 1
+
Z
βµ
x
n1
z
1
e
x
+ 1
| {z }
(2)
.
We can rewrite the first term as
Z
βµ
0
dx x
n1
1
1
1 + ze
x
=
(log z)
n
n
Z
βµ
0
x
n1
1 + ze
x
| {z }
(1)
,
using the fact that z = e
βµ
.
We now see that the two remaining integrals (1) and (2) look very similar.
We are going to make a change of variables to make them look more-or-less the
same. We let
η
1
= βµ x
in (1), and
η
2
= x βµ
in (2). Then we find that
Γ(n)f
n
(z) =
(log z)
n
n
Z
βµ
0
dη
1
(βµ η
1
)
n1
1 + e
η
1
+
Z
0
dη
2
(βµ + η
2
)
n1
e
η
2
+ 1
.
Since
βµ
1, we may approximate the first integral in this expression by
R
0
.
The error is e
βµ
1
z
1, so it is fine.
Calling η
1
= η
2
= η, we get
Γ(n)f
n
(z) =
(log z)
n
n
+
Z
0
dη
(βµ + η)
n1
(βµ η)
n1
1 + e
η
.
Let’s now look at the numerator in this expression here. We have
(βµ + η)
n1
(βµ η)
n1
= (βµ)
n1
"
1 +
η
βµ
n1
1
η
βµ
n1
#
= (βµ)
n1

1 +
(n 1)η
βµ
+ ···
+
1
(n 1)η
βµ
+ ···

= 2(n 1)(βµ)
n2
η
1 + O
η
βµ

.
We just plug this into our integral, and we find
Γ(n)f
n
(z) =
(log z)
n
n
+ 2(n 1)(log z)
n2
Z
0
dη
η
e
η+1
+ ··· .
We have to be slightly careful. Our expansion was in
η
βµ
, and while
βµ
is large,
µ
can get arbitrarily large in the integral. Fortunately, for large
µ
, the integrand
is exponentially suppressed.
We can compute
Z
0
dη
η
e
η
+ 1
=
Z
0
dη
ηe
η
1 + e
η
=
Z
0
dη η
X
m=1
e
(1)
m+1
=
X
m=1
(1)
m+1
m
2
=
π
2
12
.
Note that it was valid to swap the sum and integral because the sum is absolutely
convergent.
So we find that
f
n
(z) =
(log z)
n
Γ(n + 1)
1 +
π
2
6
n(n 1)
(log z)
2
+ ···
.
Hence we find that
N
V
=
g
s
6π
2
~
3
(2)
3/2
1 +
π
2
8
kT
µ
2
+ ···
!
. ()
Recall that the reason we did this is that we didn’t like
µ
. We want to express
everything in terms of
N
instead. It is an exercise on the example sheet to invert
this, and find
µ = E
f
1
π
2
12
T
T
f
2
+ ···
!
. ()
We can plug this into the epxression of
E
V
and obtain
E
V
=
g
s
10π
2
~
3
(2m)
3/2
µ
5/2
1 +
5π
2
8
kT
µ
2
+ ···
!
.
Dividing by () and using (), we find
E
N
=
3E
f
5
1 +
5π
2
12
T
T
f
2
+ ···
!
.
We can then differentiate this with respect to
T
, and find the heat capacity to
be
C
V
E
T
V,N
=
π
2
2
Nk
T
T
f
+ ··· ,
and this is valid for T T
f
.
Heat capacity of metals
In a metal, we have a lattice of positive charges, and electrons that are free to
move throughout the lattice. We can model them as a Fermi ideal gas. As we
mentioned previously, we have
T
f
10
4
K.
So we usually have
T T
f
. In particular, this means any classical treatment of
the electrons would be totally rubbish! We have the lattice as well, and we have
to think about the phonons. The Debye temperature is
T
D
10
2
K.
Now if
T
D
T T
f
,
then, assuming an equal number of electrons and atoms, we have
C
V
π
2
2
Nk
T
T
f
| {z }
electrons
+3Nk 3N k.
This agrees with the Dulong–Petit result.
But if
T T
D
, T
f
, then we need to use the low energy formula for the
phonons, and we get
C
V
π
2
2
Nk
T
T
f
+
12π
4
5
Nk
T
T
0
3
.
In this case, we cannot just throw the terms away. We can ask ourselves when
the contributions of the terms are comparable. When are their contributions
comparable. This is not hard to figure out. We just have to equate them and
solve for T , and we find that this happens at
T
2
=
5T
3
D
24π
3
T
f
.
If we plug in actual values, we find that this is just a few Kelvins.
While this agrees with experiments very well, there are some flaws. Most
notably, we neglected interactions with the lattice. We need to correct for a
periodic potential, and is not too difficult. This is addressed in the AQM course.
Even worse, we are ignoring the interaction between electrons! It is a puzzle
why our model works despite ignoring these. This is explained by the Landau–
Fermi liquid theory.
White dwarf stars
Another important application of this is to astrophysics. A star is a big ball
of hot gas that is not collapsing under gravity. Why does it not collapse under
gravity?
In our sun, there are nuclear reactions happening in the core of the star,
and these produce pressure that resists gravitational collapse. But this process
eventually ends. As the star runs out of fuel, the star will shrink. As this
happens, the density of the star increases, and thus the Fermi energy
E
f
for
electrons increase. Eventually this becomes so big that it exceeds the energy
required to ionize the atoms in the star. This happens at around
T
f
10
7
K
.
At this stage, then, the electrons are no longer bound to the atoms, and all
electrons can move through the star. We can therefore attempt to model these
electrons as a low temperature Fermi gas (because T
f
is now very high).
One important feature of Fermi gases is that they have pressure even at zero
temperature. In a white dwarf star, the degeneracy pressure of this gas supports
the star against gravitational collapse, even if the star cools down to
T
= 0! A
particular important example of a star we think will end this way is our own
sun.
Let’s try and understand some properties of these stars. We construct a very
crude model of a white dwarf star by treating it as a ball of constant density. In
addition to the kinds of energies we’ve talked about, there is also gravitational
energy. This is given by
E
grav
=
3GM
2
5R
.
This is just a statement about any star of constant density. To see this, we ask
ourselves what energy it takes to destroy the star by moving shells of mass
δM
off to infinity. The work done is just
δW = PE =
GM
R
δM.
So we have
dW
dM
=
GM
R(M)
=
GM
(M/(
4
3
πρ))
1/3
.
where
ρ
is the fixed density. Solving this, and using the expression of
R
in terms
of M gives
E
grav
= W =
3GM
2
5R
.
Let’s see what stuff we have got in our star. We have
N
electrons, hence
N
protons. This gives
αN
neutrons for some
α
1. In the third example sheet,
we argue that we can ignore the effect of the protons and neutrons, for reasons
related to their mass.
Thus, we are left with electrons, and the total energy is
E
total
= E
grav
+ E
kinetic
,
We can then ask ourselves what radius minimizes the energy. We find that
R M
1/3
N
1/3
.
In other words, the more massive the star is, the smaller it is! This is pretty
unusual, because we usually expect more massive things to be bigger.
Now we can ask ourselves how massive can the star be? As the mass goes
up, the radius increases, and the density goes up. So
E
f
goes up. For more and
more massive stars, the Fermi energy is larger and larger. Eventually, this
E
f
becomes comparable to the rest mass of the electron, and relativity becomes
important.
Consider E
f
mc
2
. Then we have E mc
2
, and we can expand
g(E)
V
π
2
~
3
c
3
E
2
m
2
c
4
2
+ ···
,
where we plugged in g
s
= 2. We can work out the number of electrons to be
N
Z
E
f
0
dE g(E) =
V
π
2
~
3
c
2
1
3
E
3
f
m
2
c
4
2
.
Here we are assuming that we are at “low” energies, because the Fermi tempera-
ture is very high.
Recall this is the equation that gives us what the Fermi energy is. We can
invert this to get
E
f
= ~c
3π
2
N
V
1/3
+ O
N
V
1/3
!
.
Similarly, we find
E
kinetic
Z
E
f
0
dE Eg(E)
=
V
π
2
~
3
c
3
1
4
E
4
f
m
2
c
2
4
E
2
f
+ ···
=
~c
4π
(3π
2
N)
4/3
V
1/3
+ O
V
N
V
2/3
!
.
We now note that
V
=
4
3
πR
3
, and
m m
p
m
n
, where
m, m
p
, m
n
are the
masses of an electron, proton and neutron respectively. Then we find that
E
total
=
3~c
4
9πM
4
4(α + 1)
4
m
4
p
1/3
3GM
2
5
!
1
R
+ O(R).
This is the total energy once the electrons are very energetic. We can plot what
this looks like. If the coefficient of
1
R
is positive, then we get something that
looks like this:
R
E
There is a stable minimum in there, and this looks good. However, if the
coefficient is negative, then we go into trouble. It looks like this:
R
E
This is bad! This collapses to small
R
. In other words, the degeneracy of the
electrons is not enough to withstand the gravitational collapse.
For stability, we need the coefficient of
1
R
to be positive. This gives us a
maximum possible mass
M
c
~c
G
3/2
1
m
2
p
.
This is called the Chandrasekhar limit.
Should we believe this? The above calculation neglects the pressure. We
assumed the density is constant, but if this were true, the pressure is also
constant, but this is nonsense, because a constant pressure means there is no
force!
A more careful treatment shows that
M
c
1.4M
sun
.
What if we are bigger than this? In the case of a very very very dense star, we
get a neutron star, where it is the neutrons that support the star. In this case,
we have protons combining with electrons to get neutrons. This would involve
actually doing some general relativity, which we will not do this.
This neutron star also has a maximum mass, which is now
M
max
25M
sun
.
If we are bigger than this, then we have to either explode and lose a lot of mass,
or become a black hole.