3Quantum gases
II Statistical Physics
3.6 Bose–Einstein condensation
We can write our previous
N
V
equation as
g
3/2
(z) =
N
V
λ
3
. (†)
Recall that
λ ∝ T
−1/2
. As we decrease
T
with
N/V
fixed, the right hand side
grows. So the
g
3/2
(
z
) must grow as well. Looking at the series expansion,
g
3/2
is evidently an increasing function of
z
. But
z
cannot grow without bound! It is
bounded by z ≤ 1.
How does g
3/2
grow as z → 1? By definition, we have
g
n
(1) = ζ(n),
the Riemann
ζ
-function. It is a standard result that this is finite at
n
=
3
2
. In
fact,
ζ
3
2
≈ 2.612.
We have secretly met this
ζ
function before. Recall that when we did the black
body radiation, we had an integral to evaluate, which was
Z
∞
0
dx x
3
e
x
− 1
= Γ(4)g
4
(1) = Γ(4)ζ(4) = 3! ·
π
4
90
=
π
4
15
.
Of course, getting the final answer requires knowledge of the value of
ζ
(4), which
fortunately the number theorists have figured out for us in their monumental
special values of L-functions programme.
Going back to physics, we have found that when
T
hits some finite value
T
=
T
c
, the critical temperature, we have
z
= 1. We can simply find
T
c
by
inverting the relation (†), and obtain
T
c
=
2π~
2
km
1
ζ(3/2)
N
V
2/3
.
Then we can express
T
c
T
=
λ
3
ζ(3/2)
N
V
2/3
.
So
T
c
is the temperature at which
λ
3
becomes comparable to the number density.
But physically, there shouldn’t be anything that stops us from going below
T
c
.
But we can’t. What does this mean? Maybe
N
should decrease, but particles
cannot just disappear. So something has gone wrong. What has gone wrong?
The problem happened when we replaced the sum of states with the integral
over the density of states
X
k
≈
V (2m)
3/2
4π
2
~
3
Z
∞
0
dE E
1/2
.
The right hand side ignores the ground state
E
= 0. Using the formula for
hn
r
i
,
we would expect the number of things in the ground state to be
n
0
=
1
z
−1
− 1
=
z
1 − z
.
For most values of
z ∈
[0
,
1), this is not a problem, as this has just a few particles.
However, for
z
very very very close to 1, this becomes large. This is very
unusual. Recall that when we described the validity of the classical approximation,
we used the fact the number of particles in each state is unlikely to be greater
than 1. Here the exact opposite happens — a lot of states try to lump into the
ground state.
Now the solution is to manually put back these ground states. This might
seem a bit dodgy — do we have to fix the first exciting state as well? And the
second? However, we shall not worry ourselves with these problems. Just fixing
the ground states is a decent approximation.
We now just put
N =
V
λ
3
g
3/2
(z) +
z
1 − z
.
Then there is no problem keeping
N
fixed as we take
T →
0. Then we find
z → 1 −
1
N
as T → 0.
For
T < T
c
, we can set
z
= 1 in the formula for
g
3/2
(
z
), and we can
approximate
N =
V
λ
3
g
3/2
(1) +
1
1 − z
,
and
n
0
=
1
1 − z
.
If we divide this expression through by N , we find
n
0
N
= 1 −
V
Nλ
3
ζ
3
2
= 1 −
T
c
T
3/2
.
This is the fraction of the particles that are in the ground state, and we see that
as T → 0, they all go to the ground state.
For
T < T
c
, there is a macroscopic number of particles that occupy the
ground state. This is called a Bose–Einstein condensate. This is a theoretical
prediction made in the early 20th century, and was confirmed experimentally
much later in 1995. It took so long to do this because the critical temperature is
∼ 10
−7
K, which is very low!
Let’s look at the equation of state in this condensate. We just have to take
our previous expression for the equation of state, but put back in the contribution
of the ground state. For example, we had
pV = −Φ = −
1
β
X
r
log(1 − e
−β(E
r
−µ)
).
We previously just replaced the sum with an integral, but this time we shall not
forget the ground state term. Then we have
pV =
2
3
E −
1
β
log(1 − z).
Recall that we had
E
V
=
3kT
2λ
3
g
5/2
(z). (‡)
We don’t need to add terms for the ground state since the ground state has zero
energy. Substitute this into the equation of state, and use the fact that
z ≈
1
for T < T
c
to obtain
pV = N kT
V
Nλ
3
ζ
5
2
− kT log(1 −z).
For
T < T
c
, the first term is
O
(
N
), and the second term is
O
(
log N
), which is
much much smaller than
N
. So the second term is negligible. Thus, we can
approximate
p =
kT
λ
3
ζ
5
2
∼ T
5/2
.
There is a remarkable thing about this expression — there is no
V
or
N
in here.
This is just a function of the energy, and very unlike an ideal gas.
We can also compute the heat capacity. We have
C
V
=
∂E
∂T
V,N
.
Using ‡, we find that
C
V
V
=
15k
4λ
3
g
5/2
(1) +
3kT
2λ
3
g
0
5/2
(1)
∂z
∂T
V,N
.
When T T
c
, we have z ≈ 1. So the second term is negligible. So
C
V
V
=
15k
4λ
3
ζ
5
2
∝ T
3/2
.
That wasn’t too exciting. Now consider the case where
T
is slightly greater than
T
c
. We want to find an expression for
z
in terms of
T
. Our previous expressions
were expansions around
z
= 0, which aren’t very helpful here. Instead, we have
to find an expression for z near 1.
Recall that z is determined by
g
3/2
(z) =
λ
3
N
V
.
It turns out the best way to understand the behaviour of
g
3/2
near
z
= 1 is to
consider its derivative. Using the series expansion, it is not hard to see that
g
0
n
(z) =
1
z
g
n−1
(z).
Taking n = 3/2, we find
g
0
3/2
(z) =
1
z
g
1/2
(z).
But
g
1/2
(
z
) is not a very nice function near
z
= 1. If we look at the series
expansion, it is clear that it diverges as
z →
1. What we want to know is how it
diverges. Using the integral formula, we have
g
1/2
(z) =
1
Γ(1/2)
Z
∞
0
dx
x
−1/2
z
−1
e
x
− 1
=
1
Γ(1/2)
Z
ε
0
dx x
−1/2
z
−1
(1 + x) − 1
+ bounded terms
=
z
Γ(1/2)
Z
ε
0
dx x
−1/2
1 − z + x
+ ···
=
2z
Γ(1/2)
√
1 − z
Z
√
ε/(1−z)
0
du
1 + u
2
+ ···
where we made the substitution
u
=
q
x
1−z
. As
z →
1, the integral tends to
π/
2
as it is arctan, and so we find that
g
0
3/2
(z) =
π
Γ(1/2)
√
1 − z
+ finite.
We can then integrate this back up and find
g
3/2
(z) = g
3/2
(1) −
2π
Γ(1/2)
√
1 − z + O(1 − z).
So from (†), we know that
2π
Γ(1/2)
√
1 − z ≈ ζ(3/2) −
λ
3
N
V
.
Therefore we find
z ≈ 1 −
Γ(1/2)ζ(3/2)
2π
2
1 −
λ
3
N
ζ(3/2)V
2
= 1 −
Γ(1/2)ζ(3/2)
2π
2
1 −
T
c
T
3/2
!
2
.
We can expand
1 −
T
c
T
3/2
≈
3
2
T −T
c
T
c
for T ≈ T
c
. So we find that
z ≈ 1 − B
T −T
c
T
c
2
for some positive number B, and this is valid for T → T
+
c
.
We can now use this to compute the quantity in the heat capacity:
∂z
∂T
V,N
≈ −
2B
T
2
c
(T −T
c
).
So we find that
C
V
V
≈
15k
4λ
3
g
5/2
(z) −
3Bk
λ
3
g
0
5/2
(1)(T −T
c
).
Finally, we can compare this with what we have got when
T < T
c
. The first
part is the same, but the second term is not. It is only present when T > T
c
.
Since the second term vanishes when
T
=
T
c
, we see that
C
V
is continuous
at T = T
c
, but the derivative is not.
T
C
V
T
c
3
2
Nk
This is an example of a phase transition — a discontinuity in a thermodynamic
quantity.
Note that this is possible only in the thermodynamic limit
N → ∞
. If we
work with finite
N
, then the partition function is just a finite sum of analytic
function, so all thermodynamic quantities will be analytic.
Superfluid Helium
There are two important isotopes of Helium. There is
4
He, which has two protons,
two neutrons and two electrons, hence a boson. There is also a second isotope,
namely
3
He, where we only have one neutron. So we only have 5 fermions, hence
a fermion. Both of these are liquids at low temperatures.
It turns out when temperature is very low, at
2.17 K
, helium-4 exhibits a
phase transition. Then it becomes what is known as a superfluid. This is a
strange state of matter where the liquid appears to have zero viscosity. It does
weird things. For example, if you put it in a cup, it just flows out of the cup.
But we don’t really care about that. What is relevant to us is that there is a
phase transition at this temperature. On the other hand, we see that
3
He doesn’t.
Thus, we figure that this difference is due to difference in quantum statistics!
One natural question to ask is — is this superfluid a Bose condensation?
One piece of evidence that suggests they might be related is that when we
look at heat capacity, we get a similar discontinuity in derivative. However,
when we look at this in detail, there are differences. First of all, we are talking
about a liquid, not a gas. Thus, liquid interactions would be important. The
details of the
C
V
graph are also different from Bose gas. For example, the low
energy part of the curve goes like
C
V
∼ T
3
. So in some sense, we found a fluid
analogue of Bose condensates.
It turns out
3
He becomes superfluid when the temperature is very very low,
at
T ∼ 10
−3
K
. Why is this? Apparently, there are very weak forces between
helium-3 atoms. At very low temperatures, this causes them to pair up, and
when they pair up, they become bosons, and these can condense.