3Quantum gases

II Statistical Physics

3.5 Bosons

In the previous derivation, we needed to assume that

λ =

r

2π~

2

mkT

V

N

1/2

.

At very low temperatures, this is no longer true. Then quantum effects become

important. If there are no interactions between particles, then only one effect is

important — quantum statistics.

–

Bosons have integer spin, and the states have to be are symmetric with

respect to interchange of 2 particles. For example, photons are bosons.

–

Fermions have spin

1

2

, and the states have to be antisymmetric, e.g.

e, p, n

.

Since spins add up, atoms made from an even (odd resp.) number of electrons,

protons and neutrons is a boson (fermion resp.). In an atom, since the charge is

neutral, we know we have the same number of electrons and protons. So we are

really counting the number of neutrons.

For example, hydrogen has no neutrons, so it is a boson. On the other hand,

deuterium has 1 neutron, and so it is a fermion.

We suppose we have bosons, and the single particles states are

|ri

with energy

E

r

. We let

n

r

be the number of particles in

|ri

. We assume the particles are

indistinguishable, so to specify the state of the

n

-particle system, we just need

to specify how many particles are in each 1-particle state by

{n

1

, n

2

, n

3

, ···}.

Once we specify these numbers, then the energy is just

X

r

n

r

E

r

.

So we can write down the partition function in the canonical ensemble by

summing over all possible states:

Z =

X

{n

r

}

e

−β

P

s

n

s

E

s

.

This is exactly the same as we had for photons. But there is one very important

difference. For photons, the photon numbers are not conserved. But here the

atoms cannot be destroyed. So we assume that the particle number is conserved.

Thus, we are only summing over {n

r

} such that

X

n

r

= N.

This makes it rather difficult to evaluate the sum. The trick is to use the grand

canonical ensemble instead of the canonical ensemble. We have

Z =

X

{n

r

}

e

−β

P

s

n

s

(E

s

−µ)

,

and now there is no restriction on the allowed values of

{n

i

}

. So we can factorize

this sum and write it as

Z =

Y

s

Z

s

,

where

Z

s

=

∞

X

n

s

=0

e

−β(E

s

−µ)n

s

=

1

1 − e

−β(E

s

−µ)

.

For this sum to actually converge, we need

µ < E

s

, and this has to be true for

any

s

. So if the ground state energy is

E

0

, which we can always choose to be 0,

then we need µ < 0.

Taking the logarithm, we can write

log Z = −

X

r

log(1 − e

−β(E

r

−µ)

).

We can now compute the expectation values hn

r

i. This is

hn

r

i = −

1

β

∂

∂E

r

log Z =

1

e

β(E

r

−µ)

− 1

.

This is called the Bose–Einstein distribution. This is the expected number of

variables in the 1-particle state r.

We can also ask for the expected total number of particles,

hNi =

1

β

∂

∂µ

log Z =

X

r

1

e

β(E

r

−µ)

− 1

=

X

r

hn

r

i.

Similarly, we have

hEi =

X

r

E

r

hn

r

i,

as expected. As before, we will stop writing the brackets to denote expectation

values.

It is convenient to introduce the fugacity

z = e

βµ

.

Since µ < 0, we know

0 < z < 1.

We now replace the sum with an integral using the density of states. Then we

obtain

log Z = −

Z

∞

0

dE g(E) log(1 − ze

−βE

).

The total number of particles is similarly

N =

Z

∞

0

dE

g(E)

z

−1

e

βE

− 1

= N(T, V, µ).

Now we are using the grand canonical ensemble just because it is convenient,

not because we like it. So we hope that we can invert this relation to write

µ = µ(T, V, N). (∗)

So we can write

E =

Z

∞

0

dE

Eg(E)

z

−1

e

βE

− 1

= E(T, V, µ) = E(T, V, N ),

using (∗). As before, we have the grand canonical potential

pV = −Φ =

1

β

log Z = −

1

β

Z

∞

0

dE g(E) log(1 − ze

−βE

).

We now focus on the case of monoatmoic, non-relativistic particles. Then the

density of states is given by

g(E) =

V

4π

2

2m

~

2

3/2

E

1/2

.

There is a nice trick we can do. We integrate by parts, and by integrating

g

(

E

)

and differentiating the log, we can obtain

pV =

2

3

Z

∞

0

dE Eg(E)

z

−1

e

βE

− 1

=

2

3

E.

So if we can express

E

as a function of

T, V, N

, then we can use this to give us

an equation of state!

So what we want to do now is to actually evaluate those integrals, and

then invert the relation we found. Often, the integrals cannot be done exactly.

However, we can express them in terms of some rather more standardized

function. We first introduce the Gamma functions:

Γ(s) =

Z

∞

0

t

s−1

e

−t

dt

This function is pretty well-known in, say, number-theoretic circles. If

n

is a

positive integer, then we have Γ(n) = (n −1)!, and it also happens that

Γ

3

2

=

√

π

2

, Γ

5

2

=

3

√

π

4

.

We shall also introduce the seemingly-arbitrary functions

g

n

(z) =

1

Γ(n)

Z

∞

0

dx x

n−1

z

−1

e

x

− 1

,

where

n

need not be an integer. It turns these functions appear every time we

try to compute something. For example, we have

N

V

=

1

4π

2

2m

~

2

3/2

Z

∞

0

dE E

1/2

z

−1

e

βE

− 1

=

1

4π

2

2m

~

2

3/2

1

β

3/2

Z

∞

0

dx x

1/2

z

−1

e

x

− 1

=

1

λ

3

g

3/2

(z).

Similarly, the energy is given by

E

V

=

3

2λ

3

β

g

5/2

(z).

It is helpful to obtain a series expansion for the

g

n

(

z

) when 0

≤ z ≤

1. Then we

have

g

n

(z) =

1

Γ(n)

Z

∞

0

dx

zx

n−1

e

−x

1 − ze

−x

=

z

Γ(n)

Z

∞

0

dx zx

n−1

e

−x

∞

X

m=0

z

m

e

−mx

=

1

Γ(n)

m

X

m=1

z

m

Z

m

0

dx x

n−1

e

−mx

=

1

Γ(n)

∞

X

m=1

z

m

m

n

Z

∞

0

du u

n−1

e

−u

=

∞

X

m=1

z

m

m

n

.

Note that it is legal to exchange the integral and the summation because

everything is non-negative.

In particular, this gives series expansions

N

V

=

z

λ

3

1 +

z

2

√

2

+ O(z

2

)

E

V

=

3z

2λ

3

β

1 +

z

4

√

2

+ O(z

2

)

.

Let’s now consider different possible scenario. In the limit

z

1, this gives

z ≈

λ

3

N

V

.

By the assumption on z, this implies that

λ

N

V

1/3

,

which is our good old classical high-temperature limit. So

z

1 corresponds to

high temperature.

This might seem a bit counter-intuitive, since

z

=

e

βµ

. So if

T → ∞

, then we

should have

β →

0, and hence

z →

1. But this is not a valid argument, because

we are fixing N, and hence as we change T , the value of µ also varies.

In this high temperature limit, we can invert the

N

V

equation to obtain

z =

λ

3

N

V

1 −

1

2

√

2

λ

3

N

V

+ O

λ

3

N

V

2

!!

.

Plugging this into our expression for E/V , we find that

E =

3

2

N

β

1 −

1

2

√

2

λ

3

N

V

+ ···

+

1

4

√

2

λ

3

N

V

+ ···

.

So we find that

pV =

2

3

E = NkT

1 −

1

4

√

2

λ

3

N

V

+ O

λ

3

N

V

2

!!

.

We see that at leading order, we recover the ideal gas law. The first correction

term is the second Virial coefficient. These corrections are not coming from

interactions, but quantum statistics. This Bose statistics reduces pressure.

This is good, but it is still in the classical limit. What we really want to

understand is when things become truly quantum, and

z

is large. This leads to

the phenomenon of Bose–Einstein condensation.