3Quantum gases
II Statistical Physics
3.4 Quantum ideal gas
Finally, we go back to talk about actual gases. Previously, for an ideal gas, we
wrote
Z = Z
N
1
,
and we found there was a problem with entropy, and then we argued that we
should have a factor of
1
N!
there, because we over-counted identical states. How-
ever, it turns out this is still not quite true. This is really just an approximation
that is valid at certain circumstances.
We let single particle states be
|ii
, with energies
E
i
. For simplicity, we assume
they are bosons. Let’s consider the simplest non-trivial example, which is when
N
= 2. Then since we have two bosons, we know the states of the particles must
be symmetric. So the possible states are of the form
|ii|ii,
1
√
2
(|ii|ji + |ji|ii),
with
i 6
=
j
. Let’s now calculate the partition function by summing over all these
states. We have
Z =
X
i
e
−β2E
i
+
1
2
X
i6=j
e
−β(E
i
+E
j
)
,
where we had to divide by 2 to avoid double counting. We compare this with
1
2!
Z
2
1
=
1
2
X
i
e
−βE
i
!
X
j
e
−βE
j
=
1
2
X
i
e
−β2E
i
+
1
2
X
i6=j
e
−β(E
i
+E
j
)
.
We see that the second parts are the same, but the first terms differ by a factor
of
1
2
. Thus, for the approximation
Z ≈
Z
2
1
2!
to be valid, we need that the probability of two particles being in the same
one-particle state is negligible. Similarly, for N particles, the approximation
Z =
Z
N
1
N!
(∗)
is valid if the probability of 2 or more particles being in the same state is
negligible. This would be true if
hn
i
i
1 for all
i
, where
n
i
is the number of
particles in
|ii
. Under these circumstances, we can indeed use (
∗
). This is also
true for Fermions, but we will not do those computations.
When does this assumption hold? We will somewhat circularly assume that
our model is valid, and then derive some relation between
hn
i
i
and other known
quantities.
Last time, we fond
hn
i
i = −
1
β
∂ log Z
∂E
i
≈ −
N
β
∂ log Z
1
∂E
i
=
N
Z
1
e
−βE
i
for all i, where we used the approximation (∗) in deriving it.
For a monoatomic gas, we had
Z
1
=
X
i
e
−βE
i
≈
Z
∞
0
dE g(e)e
−βE
=
V
λ
3
,
where
λ =
r
2π~
2
mkT
.
Substituting this into our expression for hn
i
i, we need
1
Nλ
3
V
e
−βE
i
for all
i
. So we need
λ
V
E
1/3
. So we see that this is valid when our gas is
not too dense. And by our definition of λ, this is valid at high temperatures.
Recall that when we did diatomic gases classically, we had
H = H
trans
+ H
rot
+ H
vib
Z
1
= Z
trans
+ H
rot
+ H
vib
.
We re-examine this in the quantum case.
We still have
H
trans
=
p
2
2m
.
So we find that
Z
trans
=
Z
∞
0
dE g(E)e
−βE
=
V
λ
3
,
as before. So this part is unchanged.
We now look at the rotational part. We have
H
rot
=
J
2
2I
.
We know what the eigenvalues of J
2
are. The energy levels are
~
2
j(j + 1)
2I
,
where j = 0, 1, 2, ···, and the degeneracy of each level is 2j + 1. Using this, we
can write down the partition function
Z
rot
=
∞
X
j=0
(2j + 1)e
−β~
3
j(j+1)/2I
.
Let’s look at the high and low energy limits. If
kT
~
2
2I
,
then the exponents up there are small. So we can approximate the sum by an
integral,
Z
rot
≈
Z
∞
0
dx (2x + 1)e
−β~
3
x(x+1)/2I
.
This fortunately is an integral we can evaluate, because
d
dx
x(x + 1) = 2x + 1.
So we find that
Z
rot
=
2I
β~
2
.
This is exactly what we saw classically.
But if
kT
~
2
2I
, all terms are exponentially suppressed except for
j
= 0. So
we find
Z
rot
≈ 1.
This is what we meant when we said the rotational modes are frozen out for low
energy modes.
We can do the same thing for the vibrational modes of the degree of freedom.
Recall we described them by harmonic oscillators. Recall that the energy levels
are given by
E
n
= kω
n +
1
2
.
So we can write down the partition function
Z
vib
=
∞
X
n=0
e
−βkω
(
n+
1
2
)
= e
−β~ω/2
1
1 − e
−β~ω
=
1
2 sinh(~βω/2)
.
We can do the same thing as we did for rotational motion. Now the relevant
energy scale is ~ω/2.
At high temperatures
kT ~ω/
2, we can replace
sinh
by the first term in
its Taylor expansion, and we have
Z
vib
≈
1
β~ω
.
On the other hand, at low temperatures, we have a sum of exponentially small
terms, and so
Z
vib
≈ e
−β~ω/2
.
So we find that
E
vib
= −
∂
∂β
log Z
vib
=
~ω
2
.
So for N molecules, we have
E
vib
≈
N~ω
2
.
But this is not measurable, as it is independent of
T
. So this doesn’t affect the
heat capacity. So the vibrational modes freeze out.
We can now understand the phenomenon we saw previously:
T
C
V
/Nk
~
2
/2I
~ω/2
1.5
2.5
3.5
Note that in theory it is certainly possible that the vibrational mode comes first,
instead of what we drew. However, in practice, this rarely happens.