3Quantum gases

II Statistical Physics


3.8 Pauli paramagnetism
What happens when we put a gas of electrons (e.g. electrons in a metal) in a
magnetic field? There are two effects — the Lorentz force, which is
∼ VB
, and
the spin of the electron coupling to B. We will examine the second first.
For simplicity, we assume there is a constant
B
field in the
z
-direction, and
the electron can either be aligned or anti-aligned. The electron given by the spin
is
E
spn
= µ
B
Bs,
where
s =
(
1 spin up
−1 spin down
Here
µ
B
=
|e|~
2mc
.
The Bohr magnetization. We see that being aligned gives negative energy,
because the electron has negative charge.
Due to the asymmetry, the
↑
and
↓
states have different occupation numbers.
They are no longer degenerate. Again, we have
N
↑
V
=
1
4π
2
2m
~
2
3/2
Z
∞
0
dE
E
1/2
e
β(E+µ
B
B−µ)
+ 1
=
1
λ
3
f
3/2
(ze
−βµ
B
B
).
Similarly, for the down spin, we have
N
↓
V
=
1
λ
3
f
3/2
(ze
+βµ
B
B
).
Compared to what we have been doing before, we have an extra external
parameter
B
, which the experimenter can vary. Now in the microcanonical
ensemble, we can write
E = E(S, V, N, B).
As before, we can introduce a new quantity
Definition (Magnetization). The magnetization is
M = −
∂E
∂B
S,V,N
.
Using this definition, we can write out a more general form of the first law:
dE = T dS − p dV + µ dN − M dB.
In the canonical ensemble, we worked with the free energy
F = E − T S.
Then we have
dF = −S dT − p dV + µ dN − M dB.
Thus, we find that
M = −
∂F
∂B
T,V,N
.
The grand canonical potential is
Φ = E − T S −µ dN.
So we find
dΦ = −S dT − p dV − N dµ − M dB.
So we see that in the grand canonical ensemble, we can write
M = −
∂Φ
∂B
T,V,µ
.
This is what we want, because we are working in the grand canonical ensemble.
How do we compute the magnetization? Recall that we had an expression
for Φ in terms of Z. This gives us
M = kT
∂ log Z
∂B
T,V,µ
.
So we just have to compute the partition function. We have
log Z =
Z
∞
0
dE g(E)
h
log
1 + ze
−β(E+µ
B
B)
+ log
1 + ze
−β(E−µ
B
B)
i
.
To compute the magnetization, we simply have to differentiate this with respect
to B, and we find
M = −µ
B
(N
↑
− N
↓
) =
µ
B
V
λ
3
(f
3/2
(ze
βµ
B
B
) − f
3/2
(ze
−βµ
B
B
)).
So we see that the magnetization counts how many spins are pointing up or
down.
As before, this function
f
3/2
is not very friendly. So let’s try to approximate
it in different limits.
First suppose
ze
±βµ
B
B
1. Then for a small argument, then is essentially
what we had in the high temperature limit, and if we do that calculation, we
find
f
3/2
(ze
±βµ
B
B
) = ze
±βµ
B
B
.
Plugging this back in, we find
M ≈
2µ
B
V
λ
3
z sinh(βµ
B
B).
This expression still involves
z
, and
z
is something we don’t want to work with.
We want to get rid of it and work with particle number instead. We have
N = N
↑
+ N
↓
≈
2V z
λ
3
cosh(βµ
B
B).
Now we can ask what our assumption
ze
±βµ
B
B
1 means. It implies this
cosh
is small, and so this means
λ
3
N
V
1.
It also requires high temperature, or weak field, or else the quantity will be big
for one choice of the sign. In this case, we find
M ≈ µ
B
N tanh(βµ
B
B).
There is one useful quantity we can define from
M
, namely the magnetic
susceptibility.
Definition (Magnetic susceptibility). The magnetic susceptibility is
χ =
∂M
∂β
T,V,N
.
This tells us how easy it is to magnetize a substance.
Let’s evaluate this in the limit B = 0. It is just
χ|
B=0
=
Nµ
2
B
kT
.
The fact that χ is proportional to
1
T
is called Curie’s law.
Let’s now look at the opposite limit. Suppose
ze
±βµ
B
B
1. Recall that for
large z, we had the result
f
n
(z) ≈
(log z)
n
Γ(n + 1)
.
Then this gives
f
3/2
(ze
±βµ
B
B
) ≈
[β(µ ± µ
B
B)]
3/2
Γ(5/2)
=
(βµ)
3/2
Γ(5/2)
1 ±
3µ
B
B
2µ
+ ···
.
To get N, we need to sum these things up, and we find
N = N
↑
+ N
↓
≈
2V
λ
3
(βµ)
3/2
Γ(5/2)
.
So we find that
µ ≈ E
f
. Taking the
log
of our assumption
ze
±βµ
B
B
1, we
know
β(µ ± µ
B
B) 1,
which is equivalent to
T
f
±
µ
B
B
k
T.
Since this has to be true for both choices of sign, we must have
T
f
−
µ
B
|B|
k
T.
In particular, we need
T T
f
. So this is indeed a low temperature expansion.
We also need
µ
B
|B| < kT
f
= E
f
.
So the magnetic field cannot be too strong. It is usually the case that
µ
B
|B|
E
f
.
In this case, we find that
M ≈
µ
B
B
λ
3
(βµ)
3/2
Γ(5/2)
3µ
B
B
µ
≈
µ
2
B
V
2π
2
2m
~
2
3/2
E
1/2
f
B
= µ
2
B
g(E
f
)B.
This is the magnetization, and from this we get the susceptibility
χ ≈ µ
2
B
g(E
f
).
This is the low temperature result. We see that
χ
approaches a constant, and
doesn’t obey Curie’s law.
There is a fairly nice heuristic explanation for why this happens. Suppose we
start from a zero
B
field, and we switch it on. Now the spin down electrons have
lower energy than the spin up electrons. So the spin up electrons will want to
become spin down, but they can’t all do that, because of Fermi statistics. It is
only those at the Fermi surface who can do so. Hence the magnetic susceptibility
is proportional to how many things there are on the Fermi surface.
Does this match what we see? The first thing we see is that
χ
is always
non-negative in our expression. Substances with
χ >
0 are called paramagnetic.
These are not permanently magnetic, but whenever we turn on a magnetic
field, they become magnetized. They are weakly attracted by a magnetic field.
Examples of such substances include aluminium.
We can also have paramagnetism coming from other sources, e.g. from the
metal ions. Most paramagnetic substances actually have complicated contribu-
tions from all sorts of things.
Landau diamagnetism*
There is another effect we haven’t discussed yet. The magnetic field induces a
Lorentz force, and this tends to make electrons go around in circles. We will not
go into full details, as this is non-examinable. The Hamiltonian is
H =
1
2m
p +
e
c
A(x)
2
,
where p is the momentum, −e is the charge, and A is the magnetic potential,
B = ∇ × A.
To understand the effect of the
B
-field, we can look at what happens classically
with this B-field. We look at the 1-particle partition function
Z
1
=
1
(2π~)
3
Z
d
3
x d
3
p e
−βH(x,p)
.
By making a change of variables in this integral,
p
0
= p +
e
c
A(x),
then we can get rid of
A
, and the partition function
Z
1
is independent of
A
!
Classically, there is no magnetization.
However, when we go to a quantum treatment, this is no longer true. This
leads to Landau levels, which is discussed more thoroughly in IID AQM. When
we take these into account, and see what happens at
T T
f
, then we find that
M ≈ −
µ
2
B
3
g(E
f
)B.
This gives a negative susceptibility. This gives an opposite effect to that coming
from spins. But since we have the factor of
−
1
3
, we know the net effect is still
positive, and we still have M
total
, χ
total
> 0. The net effect is still positive.
However, there do exists substances that have negative
χ
. They are repelled
by magnetic fields. These are called diamagnetic. Bismuth is an example of a
substance with such effects.