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Which is the correct partition function for an ideal (bosonic) gas at high $T$:

1) Sum over the number of particles in each momentum state:

$$ z_{\vec{p}} = 1 + e^{- \varepsilon_{\vec{p}}/T} + ... = \frac{1}{1-e^{- \varepsilon_{\vec{p}}/T}}$$

and then take the product over all states:

$$Z = \prod_{\vec{p}} z_{\vec{p}}$$

and

$$F = -T \log Z = \sum_{\vec{p}} \log z_{\vec{p}} = -VT\int\!\frac{d^3p}{(2\pi\hbar)^3}\log z_{\vec{p}}\simeq VT \int\!\frac{d^3p}{(2\pi\hbar)^3} e^{-\varepsilon_{\vec{p}}/T}$$

or

2) Sum over the states of each particle:

$$z_1 = V\int\frac{d^3p}{(2\pi\hbar)^3}e^{-\varepsilon_{\vec{p}}/T}$$

and then take the product over all particles:

$$Z = \frac{z_1^N}{N!}$$

with

$$F\propto N\log \frac{V}{N}?$$

Why is the right one right, and more importantly, why is the wrong one wrong?

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  • $\begingroup$ Where 1) and 2) are coming from? Additional details could help to answer. Although the final result of 2) looks very suspicious... $\endgroup$ – GiorgioP Mar 24 at 17:25
  • $\begingroup$ How are you getting the expression for $\log{z_p}$? $\endgroup$ – user3518839 Mar 24 at 17:38
  • $\begingroup$ Why $F\propto \log V$? Rather $F\propto N \log (V/N)$ $\endgroup$ – Aleksey Druggist Mar 24 at 19:45
  • $\begingroup$ @GiorgioP and others I am just trying to calculate $Tr(e^{-H/T})$ and am confused. I added a few lines of explanation. $\endgroup$ – Eric David Kramer Mar 25 at 10:17
  • $\begingroup$ @Aleksey I think what they are doing over there is basically what I did in my (2). But this contradicts the volume scaling of number (1)! $\endgroup$ – Eric David Kramer Mar 25 at 13:08
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These expressions are both correct but describe different situations. In 2) you have taken a fixed set of $N$ particles (you are working in th cannonical ensamble) and summed over there possible states, using Maxwell-Boltzmann statitistics, which are the correct high $T$ limit for a bosonic gas.

In 1) on the other hand, notice that the total number of particles never enters the calculation. Indeed you have summed over all possible particle numbers (for each state individually) in your first line. So in this case you have a variable number of particles and are working in the grand cannonical ensamble. Since you don't have an explicit chemical potential, either it has been set to $0$ (which would be the case for, for example, a photon gas) or it has been absorbed into the energies $\epsilon_p$.

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  • $\begingroup$ Thanks! But then in principle the two free energies should be related by a simple Legendre transform. But $$< N > = V\int\!dp'\,({\rm something}),$$ and so $$F \sim V\int\!dp'\,({\rm something}) + \mu V\int\!dp'\,({\rm something}) = V\int\!dp'\,({\rm something}),$$ while in (2) we have $$F\sim \log (V/N).$$ How can they scale differently with $V$? $\endgroup$ – Eric David Kramer Mar 25 at 12:57

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