Formulation of Statistical Problem

Consider a gas consisting of $N$ identical non-interacting particles occupying volume $V$ , and in thermal equilibrium at absolute 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 $r$ , et cetera, is simply

$$E_R = \sum_r n_r \epsilon_r,$$ (8.16)

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

$$N = \sum_r n_r.$$ (8.17)

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,

$$Z =\sum_R {\rm e}^{-\beta E_R} = \sum_R {\rm e}^{-\beta (n_1 \epsilon_1+n_2 \epsilon_2+\cdots)}.$$ (8.18)

Here, the sum is over all possible states, $R$ , of the whole gas. That is, 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, et cetera. Thus, the mean number of particles in quantum state $s$ can be written

$$\bar{n}_s = \frac{\sum_R n_s \exp[-\beta (n_1 \epsilon_1+n_2 ... ...on_2+\cdots)]} {\sum_R \exp[-\beta (n_1 \epsilon_1+n_2 \epsilon_2+\cdots)]}.$$ (8.19)

A comparison of Equations (8.18) and (8.19) yields the result

$$\bar{n}_s = -\frac{1}{\beta}\frac{\partial \ln Z}{\partial\epsilon_s}.$$ (8.20)

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