Per the Wikipedia page on Maxwell-Boltzmann statistics, the mean occupation number describes the average number of particles in the i-th single-particle state for distinguishable particles.
To be clear about my understanding, it describes distinguishable particles and it describes the mean number of particles in the i-th state of the single particle.

Please note that I won’t be including the degeneracies, $g_i$, here as I think occupation numbers are more meaningful.
I understand how the second equality was proved:
$$\langle N_i \rangle = \frac{N}{Z} e^{-\varepsilon_i/k_BT}.$$ This effectively states that the average number of particles in their single-particle state i is the total number of particles times the probability of being in that state ($p_i=e^{-\varepsilon_i/k_BT}$).

However, I don’t understand how the other equality was derived: $$\langle N_i \rangle = e^{(\varepsilon_i-\mu)/k_BT}.$$ This absolutely follows from the classical limit of Fermi-Dirac or Bose-Einstein statistics: $$\langle N_i \rangle = \frac{1}{e^{(\varepsilon_i-\mu)/k_BT}\pm1}.$$ In this limit, the fugacity, $e^{\mu/k_BT}$, tends to infinity, quantum correlations can be ignored, the energy levels are high enough that the exponential is far greater than 1.
Therefore the $\pm1$ can be ignored.

However, my puzzle concerns the distinguishability component.
If one attempts to derive the Maxwell-Boltzmann statistics second equation from the grand canonical ensemble, one has to include the Gibbs factor, $1/N!$, to recover the correct equation.
That factor asserts that the particles are not distinguishable.

So are they distinguishable or not in Maxwell-Boltzmann statistics?

Thanks for your help.

  • $\begingroup$ You are missing a minus sign in the part you said you did not understand how the fugacity comes in. You could just take it that the fugacity is defined so that the two expressions agree, in the trivial sense. The Maxwell-Boltzmann distribution is the unique coinciding limit of both Bose-Einstein and Fermi-Dirac statistics. In a sense, it is not that it is distinguishable, but rather that it is irrelevant whether it is distinguishable or indistinguishable. As for the Gibbs factor, your assumed initial thing is indistinguishable, so you need to insert that factor to end up with distinguishable. $\endgroup$ Commented Apr 18 at 8:06
  • $\begingroup$ @naturallyInconsistent, yes, my apologies, I missed the minus sign. The fugacity argument is actually the one I understand. Apologies if that was unclear. I also understand why the Gibbs factor has to come into play when the particles are indistinguishable (in the classical limit). The argument I don’t agree with is the mathematical derivation from the grand canonical ensemble. However, your point offers a fairly satisfying resolution: the Maxwell-Boltzmann statistics are, by definition, the unique ones that coincide with both the Fermi-Dirac and Bose-Einstein ones in the classical limit. $\endgroup$
    – TheorVHP
    Commented Apr 20 at 7:19

1 Answer 1


For the record, the three regimes agree in the dilute limit, i.e. when $N_i\to0$ when fugacity $z = e^{\beta\mu}$ vanishes (not when $z\to+\infty$ as you claimed, this is perhaps due to your sign error spotted by naturallyInconsistent). Check out a previous answer of mine for more.

You can answer the question by studying the formulas for the canonical and grandcanonical ensembles. The canonical partition function is: $$ Z_N = \frac1{N!}\sum_{\sum N_i = N}\frac{N!}{\prod N_i!}e^{-\beta\sum n_i\epsilon_i} $$ which factories to: $$ Z_N = \frac1{N!}\left(\sum e^{-\beta\epsilon_i}\right)^N $$ The corresponding grand canonical ensemble is: $$ \mathcal Z = \sum_{N=0}^\infty Z_Nz^N $$ which factorises to: $$ \mathcal Z = \prod \exp\left(ze^{-\beta\epsilon_i}\right) $$ and gives the MB statistics.

It may look paradoxical because the particles are treated as distinguishable due to the multinomial coefficient. However, they are also treated as indistinguishable due to the overall division by the factorial. Actually, the second step is not a translation of indistinguishability, but rather an ambiguity in the labelling of the particles. In short, what matters is the distinguishability within an orbital, as expressed by the formula: $$ Z_N = \sum_{\sum N_i = N}\frac1{\prod N_i!}e^{-\beta\sum n_i\epsilon_i} $$ which is different from the BE statistics: $$ Z_N = \sum_{\sum N_i = N}e^{-\beta\sum n_i\epsilon_i} $$ (for FD statistics, since $N_i=0,1$ so $N_i!=1$ and there is technically no combinatorial distinction)

Hope this helps.

  • $\begingroup$ Yes, I missed the minus sign. I’m familiar with the derivations you posted via the grand canonical and canonical ensembles. What bothered me was the Gibbs factor ($1/N!$) if we’re claiming that they are distinguishable. However, your point that that factor actually describes a labelling matter rather than a distinguishability one is very sound, and effectively resolves my conundrum. Thanks. $\endgroup$
    – TheorVHP
    Commented Apr 20 at 7:24

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