Bose-Einstein statistics

Consider a single particle energy level $E\,$ with degeneracy $g(E)\,$ , where each of the $g(E)\,$ states can be occupied by zero or more indistinguishable bosons. Let us consider one of those states, labeled by $s\,$ , as our system and the remaining states to form a reservoir, so that the combined number of particles stays fixed. The grand canonical partition function of the system is:

$\mathcal{Z} = \sum_{N=0}^\infty e^{-\frac{(E_{s(N)} - N \mu)}{k_B T}}\,$ .

The state $s(N)\,$ with $N\,$ particles has energy $N E_s\,$ . Thus

$\mathcal{Z}\,$	$= \sum_{N=0}^\infty e^{-N \frac{(E_{s} - \mu)}{k_B T}}\,$ ,
	$= \sum_{N=0}^\infty \left( e^{-\frac{(E_{s} - \mu)}{k_B T}}\right)^N\,$ ,
	$= \frac{1}{1 - e^{-\frac{(E_{s} - \mu)}{k_B T}} }\,$ ,

The expected number of particles in the state $s\,$ is given by

$\left\langle n\right\rangle\,$	$=\frac{1}{\beta} \left(\frac{\partial \ln \mathcal{Z}}{\partial \mu}\right)_{\beta, V}\,$ ,
	$= \frac{1}{e^\frac{E_s-\mu}{k_B T} -1}\,$ .

Since all $g(E)\,$ states are equally probable, the expected number of indistinguishable bosons with energy $E_i\,\,$ is

$\left\langle n(E_i)\right\rangle = \frac{g(E_i)}{ e^\tfrac{E_i-\mu}{k_B T}-1}\,$ .