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In the assumption of Maxwell-Boltzmann statistics, particles are treated as distinguishable. So for identical particles with discrete energy levels, there are degeneracies for any specific occupation state: $|n_1,n_2,...\rangle$

$$ \frac{N!}{n_1!n_2!n_3!...} $$

The partition function in the Maxwell-Boltzmann statistics is:

$$ \begin{split} Z=tr(\exp{-(\beta H)})&=\frac{1}{N!}\sum^{N}_{n_1,n_2,...=0}'\frac{N!}{n_1!n_2!n_3!...}\langle n_1,n_2,...|\exp{-(\beta H)}|n_1,n_2,...\rangle\\ &=\frac{1}{N!}\sum^{N}_{n_1,n_2,...=0}'\frac{N!}{n_1!n_2!n_3!...}\exp{(-\beta\sum^{\infty}_{k=1}n_kE_k)}\\ &=\frac{1}{N!}\Big(\sum^{\infty}_{k=1}\exp{(-\beta E_k)}\Big)^N\\ &=\frac{1}{N!}(Z(T,V,1))^N, \end{split} $$ thereof prime on the summation means we calculate the partition function when the total number of particles is fixed, that is $n_1+n_2+...=N.$ $\frac{1}{N!}$ is Gibbs' factor. My question is why do we put a Gibb's factor here? In the precedent, we suppose the particles are distinguishable—the presence of Gibbs' factor here contradicts the assumption of distinguishable particles.

Note: Here I use canonical ensemble.

Sources:

From Thermodynamics and Statistical Mechanics by Walter Greiner, Ludwig Neise, Horst Stöcker, D. Rischke chapter 12, Grand canonical description of ideal quantum system

and

Statistical Mechanics, 4th edition (R.K. Pathria, Paul D. Beale) chapter 6.1-6.2

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  • $\begingroup$ Sources and references? $\endgroup$ Commented Aug 28, 2023 at 13:01
  • $\begingroup$ @TobiasFünke oh I forgot it. I found it from Thermodynamics and Statistical Mechanics by Walter Greiner, Ludwig Neise, Horst Stöcker, D. Rischke chapter 12, Grand canonical description of ideal quantum system. Also, I found the similar arguement from Statistical Mechanics, 4th edition (R.K. Pathria, Paul D. Beale) chapter 6.1-6.2 $\endgroup$
    – Hsu Bill
    Commented Aug 28, 2023 at 16:12
  • $\begingroup$ This paper (as well as earlier ones) by Swendsen are rather interesting, in my opinion. He argues quite convincingly that the usual derivation of the $N!$ from quantum indistinguishability is, in spite of its popularity, not the correct one. The basis of his argument is that the usual definition of the entropy as $k_B$ times the volume in phase space is incorrect and does not correspond to the definition Boltzmann actually proposed (which would be $k_B$ times the probability of the macrostate). $\endgroup$ Commented Aug 28, 2023 at 17:07
  • $\begingroup$ @YvanVelenik Thank you for the reply $\endgroup$
    – Hsu Bill
    Commented Aug 29, 2023 at 13:41

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I know of two arguments:

  1. This is necessary for the result to match the quantum result. Period. (This is my less favorite).

  2. It does not matter that the particles are distinguishable in principle. It matters whether we actually care about them being distinguishable. You are not following each particle in the gas, so if two of them swapped places, you wouldn't really mind. Hence, you are not really counting microstates, but rather some sort of "mesostate", which is what really matters to you. Since I don't care about where each individual particle is, but rather only about where there are particles, I treat them as if they were indistinguishable. Because in practice, that is what I actually want to do.

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  • $\begingroup$ Thank you for the reply $\endgroup$
    – Hsu Bill
    Commented Aug 29, 2023 at 13:41
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    $\begingroup$ @HsuBill You're welcome! If the answer solves your problem, please consider accepting it. $\endgroup$ Commented Aug 29, 2023 at 16:47
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    $\begingroup$ @HsuBill and the same holds for some other questions of you. Consider either to accept the answers there or clarify why the answers did not answer your questions or what is unclear. $\endgroup$ Commented Aug 31, 2023 at 13:53

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