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The classical partition function for an open system is given as $$ Z_{\text{max}} = \sum_{N=0}^{\infty} \dfrac{h^{-N}}{N! } \prod_{j=1}^{N} \left( \sum_{i=0}^{\infty} e^{-\beta (E_{ij}-\mu)} g_{i} \right) $$ where $E_{ij}$ is $j$th particle with in the $j$th energy state, $g_i$ is the degenecery of the $i$th energy state, $\mu$ is the chemical potential, and $N!$ is removes over-counting of indistinguishable particles.

The Bose-Einstein partition function is given as $$ Z_{\text{bos}} = \prod_{j=1}^{\infty} \left( \sum_{i=0}^{\infty} e^{-\beta (E_{j}-\mu)i} \right)^{g_{j}} $$ where $j$ is evaluates all possible energies and $g_j$ is the degeneracies of the energies.

Why is it fundamentally impossible to reduce $Z_{\text{max}}$ into $Z_{\text{bos}}$, even though both systems describe particle number as not conserved? The average number of particles of both systems is given as $$ \langle N \rangle = \dfrac{1}{\beta} \dfrac{\partial \ln(Z)}{\partial \mu } $$

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  • $\begingroup$ crosspost to movies.stackexchange.com? $\endgroup$
    – arivero
    Aug 12, 2015 at 21:40

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Fundamentally it is that the $1/N!$ for the classical system only correctly compensates for overcounting of indistinguishable states if the particles are always in different states. For a system of Bosons at low temperature, where it is quite likely that many particles are in the same state, this breaks down. For a very understandable introduction to this I highly recommend the chapter on quantum statistics in the excellent thermal physics textbook by Schroeder.

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