Formulation of the statistical problem

Consider a gas consisting of $N$ identical non-interacting particles occupying volume $V$ and in thermal equilibrium at temperature $T$. Let us label the possible quantum states of a single particle by $r$ (or $s$). Let the energy of a particle in state $r$ be denoted $\epsilon_r$. Let the number of particles in state $r$ be written $n_r$. Finally, let us label the possible quantum states of the whole gas by $R$.

The particles are assumed to be non-interacting, so the total energy of the gas in state $R$, where there are $n_r$ particles in quantum state $r$, etc., is simply

$$E_R = \sum_r\,n_r\,\epsilon_r,$$ (579)

where the sum extends over all possible quantum states $r$. Furthermore, since the total number of particles in the gas is known to be $N$, we must have

$$N = \sum_r n_r.$$ (580)

In order to calculate the thermodynamic properties of the gas (i.e., its internal energy or its entropy), it is necessary to calculate its partition function,

$$Z =\sum_R {\rm e}^{-\beta\,E_R} = \sum_R {\rm e}^{-\beta\,(n_1\,\epsilon_1+n_2\, \epsilon_2+\cdots)}.$$ (581)

Here, the sum is over all possible states $R$ of the whole gas: i.e., over all the various possible values of the numbers $n_1, n_2,\cdots$.

Now, $\exp[-\beta\,(n_1\,\epsilon_1+n_2\,\epsilon_2+\cdots)]$ is the relative probability of finding the gas in a particular state in which there are $n_1$ particles in state 1, $n_2$ particles in state 2, etc. Thus, the mean number of particles in quantum state $s$ can be written

$$\bar{n}_s = \frac{\sum_R n_s\,\exp[-\beta\,(n_1\,\epsilon_1+... ...sum_R \exp[-\beta\,(n_1\,\epsilon_1+n_2\,\epsilon_2+\cdots)]}.$$ (582)

A comparison of Eqs. (581) and (582) yields the result

$$\bar{n}_s = -\frac{1}{\beta}\frac{\partial \ln Z}{\partial\epsilon_s}.$$ (583)

Here, $\beta\equiv 1/k\,T$.