= Solution
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 $L^2(2\pi k\,dk)/(2\pi)^2$ states. With $\epsilon=\hbar^2k^2/(2m)$, $k\,dk=m\,d\epsilon/\hbar^2$, giving
$$
\boxed{g(\epsilon)=\frac{L^2m}{2\pi\hbar^2}\quad(\epsilon>0).}
$$
There is no spin degeneracy. At temperature $T>0$, set $z=e^{\mu/(k_BT)}<1$. The Bose occupation integral gives
$$
\frac N{L^2}=\frac{m}{2\pi\hbar^2}\int_0^\infty\frac{d\epsilon}{z^{-1}e^{\epsilon/(k_BT)}-1}
=\boxed{-\frac{mk_BT}{2\pi\hbar^2}\log(1-z).}
$$
The integral follows by expanding the denominator as $\sum_{\ell\geq1}z^\ell e^{-\ell\epsilon/(k_BT)}$ and integrating each positive term.
As $\mu\uparrow0$, 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 \b[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.
Back to article page
