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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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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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