Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 34D (Statistical Physics)

Periodic boundary conditions give wavevectors $\mathbf{k}=(2\pi /L)(n_1,n_2)$. Each state occupies area $(2\pi /L{)}^2$ in wavevector space. A ring of radius $k$ contains $L^2(2\pi kdk)/(2\pi {)}^2$ states. With $ϵ={\hslash }^2{k}^2/(2m)$, $kdk=mdϵ/{\hslash }^2$, givingThere is no spin degeneracy. At temperature $T>0$, set $z=e^{\mu /(k_{B}T)}<1$. The Bose occupation integral givesThe integral follows by expanding the denominator as ${\sum }_{\ell \ge 1}{z}^{\ell }{e}^{-\ell ϵ/(k_{B}T)}$ and integrating each positive term.

As $\mu \uparrow 0$, this density diverges. Thus the excited states never have a finite maximum capacity at a positive temperature: every finite prescribed density can be accommodated with a strictly negative chemical potential. The ground-state occupation then stays finite as area tends to infinity and its fraction vanishes. Hence there is no Bose-Einstein condensation at any positive temperature in this homogeneous ideal two-dimensional gas. At exactly zero temperature the ground state is occupied; the printed phrase about any temperature is understood as absence of a nonzero critical temperature, not exclusion of that zero-temperature limit.