Physics 413: Statistical Mechanics - Homework Solutions 8

Problem 1: Quantum corrections to classical ideal gas

The classical (Boltzmann) limit corresponds to

⟨n⟩ = 1

eβ(ϵ−µ) + δ≪ 1

Expanding the denominator gives

⟨n⟩ = e−β(ϵ−µ)(1− δe−β(ϵ−µ)) +O[(e−β(ϵ−µ))3]

Average particle number and internal energy read

N =

∫dϵ a(ϵ) e−β(ϵ−µ)(1− δe−β(ϵ−µ))

U =

∫dϵ ϵ a(ϵ) e−β(ϵ−µ)(1− δe−β(ϵ−µ))

with

a(ϵ) =V

2π2

(m

h2

)3/2√2ϵ

The energy per particle is

U

N=

∫dϵ ϵ3/2 e−βϵ(1− δe−β(ϵ−µ))∫dϵ ϵ1/2 e−βϵ(1− δe−β(ϵ−µ))

= kBT(3/4)

√π (1− δeβµ/25/2)

(1/2)√π (1− δeβµ/23/2)

=3

2kBT (1 + δeβµ

√2/8)

The factor eβµ in the correction term can be calculated in lowest order, i.e. for the classical idealgas.

eβµ =N

V

(2πh2

mkBT

)3/2

=N

Vλ3

where λ is the thermal wave length. Therefore

U

N=

3

2kBT (1 + δNλ3

√2/(8V ))

The corrections to the pressure most easily follow from the general relation p = (2U)/(3V ) whichcan be proven as follows:

p = −(∂Ω

∂V

)T,µ

= −Ω

V

=kBT

VlnZµ =

kBT

V

1

δ

∫dϵ a(ϵ) ln[1 + δe−β(ϵ−µ)]

partial integration

=2

3

kBT

V

1

δ

∫dϵ a(ϵ)ϵ

βδe−β(ϵ−µ)

1 + δe−β(ϵ−µ)=

2

3

U

V

Problem 2: Bose-Einstein temperature of Helium