This is related with Equation 8.58 in Kerson Huang's 2nd edition of Statistical Mechanics.

The partition functions for the ideal gases are given as $ Q_N (V,T) =\sum_{\{ n_p \}} g\{n_p \}e^{-\beta E\{n_p \}} $ where $E\{n_p \} =\sum_p \epsilon_p n_p$ and the occupation numbers are subject to the $\sum_p n_p =N$

The textbook says, for a bose gas and a boltzmann gas $n_p=0,1,2, \cdots $ and for the fermi gas $n_p=0,1$. I understand this is due to the pauli-exclusion principles. But what i couldn't understand is eq 8.58 which is following

$g\{n_p\} = 1$ for both bose and fermi gas

$g\{n_p\} = \frac{1}{N!} \left( \frac{N!}{\prod_p n_p !} \right)$ for boltzaman gas

Please explain why we have this results.


Bose particles cannot be identified as different in a given state, whereas boltzmann particles can (even though both types can occupy a given energy state with more than one particle). Thus boltzmann statistics need to take into account the permutations ($n!$) of the $n$ particles into a given state, in contrast to the bose particles (which are not identified as separate).

For example for Bose-Einstein statistics the configurations:

(ike, mike) and (mike, ike) are considered the same.

Whereas for Maxwell-Boltzmann statistics the configurations:

(ike, mike) and (mike, ike) are considered different (each particle "ike" or "mike" can be identified as "ike" or "mike").

wikipedia Identical particles

  • $\begingroup$ How about fermi gas? still the $g\{n_p\}$ are given as same in bose gas. Also in bose gas you mean that $\left( \frac{N!}{\prod_p n_p !}\right) =1$? $\endgroup$
    – phy_math
    Oct 23 '14 at 14:30
  • $\begingroup$ for fermi-dirac particles, you already stated the answer, in each given state there can be either $1$ or $0$ due to pauli-exclusion principle. It all breaks down to particle identification in quantum mechanics (see the link) (plus it is not clear what exactly the factors mean in your question, but in any case this is the core of the answer) $\endgroup$
    – Nikos M.
    Oct 23 '14 at 14:33
  • $\begingroup$ Here i am considering the degeneracy $g\{n_p \}$ of given particles. I understand your points, but i still got confused why both boson and fermion have degeneracy 1. $\endgroup$
    – phy_math
    Oct 23 '14 at 14:41
  • $\begingroup$ @phy_math, great made a small update to elucidate further $\endgroup$
    – Nikos M.
    Oct 23 '14 at 14:50

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