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.