Partition function of bosons vs fermions

I have two atoms, both of which are either bosons or fermions, with four allowed energy states: $E_1 = 0$, $E_2 = E$, $E_3 = 2E$, with degeneracies 1, 1, 2 respectively.

What's the difference between the partition functions of a pair of two bosons and that of a pair of two fermions?

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What have you tried so far? What does the partition function look like if you don't assume any particular statistics? How would you expect it to change when you add the statistics to the problem? – Colin McFaul Dec 7 '12 at 15:59

1 Answer

For the partition sum, you have so sum $e^{-E}$ ($T=1$) over all possible eigenstates of the system where $E$ is the energy of the corresponding state.

Two bosons can be in the states 10 $|kl\rangle$, with $1\leq k \leq l \leq 4$ where we accounted for the degeneracy by introducing an additional state with $E_4 =2E$. The corresponding partition sum reads (we assume the particles to be noninteracting) $$Z_B = \sum_{k\leq l} e^{-E_k- E_l} = 1+ e^{-E} + 3 e^{-2E} +2 e^{-3 E} +3 e^{-4E}.$$

Similarly, for fermions we have 6 states $|kl\rangle$, with $1\leq k < l \leq 4$ with the partition sum $$Z_F = e^{-E} + 2 e^{-2E} +2 e^{-3 E} + e^{-4E}.$$

So the difference of the partition functions of a pair of two bosons and that of a pair of two fermions is ;-) $$Z_B - Z_F = 1 + e^{-2E} +2 e^{-4E}.$$

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+1 for giving an answering with a technically correct interpretation of ‘difference’. – Claudius Dec 7 '12 at 21:42
I would just add a clarification that you consider states like $|12\rangle$ and $|21\rangle$ as indistinguishable (i.e., a proper boson\fermion state would be $|12\rangle \pm |21\rangle$) – Benji Remez Dec 8 '12 at 0:40
@BenjiRemez: thank you for the clarification! – Fabian Dec 8 '12 at 7:10
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