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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – alarge
    Feb 4, 2015 at 22:46
  • $\begingroup$ 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? $\endgroup$
    – RQM
    Feb 4, 2015 at 23:32

1 Answer 1


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.

Obviously in a quantum sense classical particles are distinguishable, so the term does not immediately arise from indistinguishability per se. Rather, the reason the "extra" term appears is because we ought to be looking at the macroscopic realizations of the system, and it does not matter for the macroscopic state if we switch two particles, and it is in this sense that they are indeed indistinguishable.


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