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After reading these posts: Why is the partition function divided by $(h^{3N} N!)$?, What is the resolution to Gibb's paradox?, and these articles: The Gibbs Paradox and the Distinguishability of Identical Particles , The Gibbs Paradox, I understand that the division of the partition function by $N!$ has no relationship with the fact of identical particles.

Thus, I ask: If the partition function is given as a normalization factor for the probability of a system, that is: $$p_r = {e^{- \beta \epsilon_r} \over Z}$$ where $Z=\sum_r e^{- \beta \epsilon_r} $, I have read that the total partition function of the system is $$Z_\text{tot} = Z^N$$ where $N$ the total number of particles, so we can say that the total average energy of the system is $$\langle E\rangle=N \epsilon_1$$ with $$\epsilon_1 =-kT {\partial (\ln Z) \over \partial \beta} $$

Why don't we divide the total partition function $Z_\text{tot}$ by $N!$, if this division has no relationship with the fact that in quantum mechanics the particles are identical? Then we should have that $$Z_\text{tot}={Z^N \over N!}$$ So why don't we divide by $N!$?

Note: The references at the top of the posts are about the division by the factorial. As I have understood, they argue that the division by $N!$ is not because the particles are identical but because we want the thermodynamic entropy to be the same as the statistical entropy (a matter of convenience). So why isn't this division necessary here?

Note: After a first answer I am making a note here: If someone argues that the division happens because of the Gibbs paradox, they should at least present a counter-answer to the references I am making at the top of the post — arguing at least why those posts and articles contain errors or are entirely mistaken and, if possible, provide more sources for reading and study.

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  • $\begingroup$ Why would you divide by $N!$? This would only change the definition of $Z_{tot}$, but has no other consequence by itself. $\endgroup$ – Ján Lalinský Jun 14 '15 at 15:12
  • $\begingroup$ @JánLalinský The references at the top of the posts are about the division with the factorial. As I have understood, they argue that the division with the N! is not because the particles are identical but because we want the thermodynamic entropy to be the same with the statistical entropy(a matter of convenience ). So why this division isn't necessary here? Thanks. $\endgroup$ – Constantine Black Jun 14 '15 at 16:37
  • $\begingroup$ Notice that the paper by Jaynes already has the answer: spin systems have long range forces, which as mentioned in the answer by Kyle below, means that the partition function doesn't factorise, and indeed the entropy is not extensive anyway, so you don't want such factors in it. $\endgroup$ – genneth Aug 21 '15 at 18:45
  • $\begingroup$ @genneth Thanks for the note, but what if we make a theoretical assumption that the particles are non-interacting(very weak interactions). Mandle's book, when making an introduction to quantum ideal gas makes such a hypothesis. Your comment indeed answer in the case where we have interactions. Thanks. $\endgroup$ – Constantine Black Aug 22 '15 at 8:49
  • $\begingroup$ @ConstantineBlack If you subscribe to Jaynes's point of view, then you can approach in two ways. One is to identify what macroscopic variables you can control, and make sure that controlling them leads to reproducible consequences. Then, you can just apply the obvious formulae and whatever should happen will happen. The other approach is phenomenological, which requires an additional hypothesis about how the entropy should change with system size -- the Clausius relationship only tells you how it changes with energy and temperature. $\endgroup$ – genneth Aug 23 '15 at 11:09
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In the example you gave above and in calculating certain thermodynamic variables such as pressure: $$P=-\frac{dF}{dV}=\frac{d}{dV}k_bT\ln(Z)$$ the $N!$ cancels: $$\frac{d}{dx}\ln(Z'/N!)=\frac{d}{dx}\left(\ln(Z')-\ln(N!)\right)=\frac{d}{dx}\ln(Z')$$

you still need to included the $N!$ when you do calculations for chemical potential or entropy because the $N!$ doesn't cancel. What you read above is just short hand for the calculation they were doing.

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