# Deriving Boltzmann statistics from the maximum entropy principle

In some lecture notes I have, the author derives the expectation value of the occupation numbers for a discrete system of fermions as follows:

Consider all states that have a certain energy $$\varepsilon_s$$. There shall be $$a_s$$ such states, and $$n_s$$ particles occupying these states. Then, the number of configurations in this energy level is $$W_s = \frac{a_s!}{n_s!\left(a_s-n_s\right)!}$$

If one now considers all energy levels, the total number of possible states for the whole system is $$W = \prod_s W_s$$ where $$s$$ enumerates the energy levels. (I think countably infinitely many levels should not be a problem, would they?)

Now, one can try and maximize the entropy $$S=k\ln\left(W\right)$$ under the constraints that the particle number is fixed, $$N=\sum_s n_s$$, as is the total energy, $$E=\sum_s n_s\varepsilon_s$$. Introducing the Lagrangian multipliers $$\alpha$$ and $$\beta$$ in $$\Lambda := \frac{S}{k} - \alpha \left(\sum_s n_s - N\right) - \beta \left(\sum_s n_s \varepsilon_s - E\right)$$ one indeed finds an extremum for $$\Lambda$$ for $$n_i = \frac{a_i}{1+\exp\left(\alpha+\beta\varepsilon_i\right)}$$ which is, up to a factor, the Fermi-Dirac statistic once the Lagrangian multipliers are identified to be $$\alpha = - \frac{\mu}{k_B T}$$ and $$\beta=\frac{1}{k_B T}$$.

Now, in a side remark, the lecture notes claim that, if one had assumed a classical system where $$W = a_s^{n_s}$$, one would have obtained Boltzmann statistics: $$n_s = \exp\left(-\alpha-\beta \epsilon_s\right)$$.

I assume that instead of $$W$$, the author meant to write $$W_s$$. From the form of $$W_s$$, I conclude that the system under consideration has discrete energy levels, and that the particles do not obey the Pauli principle and are distinguishable from one another. In these circumstances, $$a_s^{n_s}$$ seems to give the right number of configurations within the energy level $$\epsilon_s$$, the total number of configurations again being $$W=\prod_s W_s$$.

However, $$S$$ now is linear in the $$n_s$$, so that differentiation of the new $$\Lambda$$ with regard to some $$n_i$$ gives an expression independent of any of the $$n_k$$.

What went wrong? Why did the procedure seem to work in the first case, but not in this? Or did I make a (conceptual?) mistake somewhere along the line?

Any input would be greatly appreciated!

• I think you are missing a $\frac{1}{n_s!}$ from the product (i.e. $W = \prod_s \frac{a^{n_s}}{n_s!}$). See e.g. this. – alarge Feb 4 '15 at 22:46
• Thank you very much, @alarge! Deviating from what is said in the lecture notes and including a factor $\frac{1}{n_s!}$, i.e. assuming the particles are indistinguishable, I get what the author claimed, except that I get an additional factor of $a_i$, which intuitively seems to be correct: $n_i = a_i \exp\left(-\alpha-\beta\varepsilon_i\right)$. So I guess this is what the author originally intended. Can I somehow mark your comment as the accepted answer? – RQM Feb 4 '15 at 23:32

You are missing the term $\frac{1}{n_s!}$ from the product, i.e. $$W = \prod_s \frac{a^{n_s}}{n_s!}$$ from which the wanted result follows.