Say we have a system with 2 particles with energy levels $E_1=0,E_2=E,E_3=2E,E_4=2E$. I want to figure out the partition function of the system depending on whether

i) the particles are distinguishable and at most one can occupy any state

ii)the particles are indistinguishable and at most one can occupy any state

iii) the particles are distinguishable and any number of them can occupy a state

iv) the particles are indistinguishable and any number of them can occupy a state

Then decide which is relevant for Fermi-Dirac , Bose-Einstein and Maxwell-Boltzmann distributions

Note : there is a very similar question posted here Partition function of bosons vs fermions , and indeed much of my reasoning is based of it, but the distinction between distinguishable and indistinguishable is not made clear , and as that is the main focus of my question I don't think it counts as a duplicate.

Here's what I think :

Firstly using the method in the link we have states $|kl>$ ,$1 \leq k \leq l \leq 4$

$Z=\sum_{k \leq l} e^{-E_k-E_l}$

i) there are only six possibilities of $|kl>$ if they can't occupy the same state , but they are aslo distinguishable so we must count all of these twice giving $Z=2e^{-E}+4e^{-2E}+4e^{-3E}+2e^{-4E}$

ii)For this we count the states once so $Z=e^{-E}+2e^{-2E}+2e^{-3E}+e^{-4E}$

iii)now we count the 10 states of $kl> $ twice so we have $Z= 2+ 2e^{-E} + 6 e^{-2E} +4 e^{-3 E} +6 e^{-4E}.$

iii) and here we count the two states once so $Z== 1+ e^{-E} + 3 e^{-2E} +2 e^{-3 E} +3 e^{-4E}.$

Now the fermi dirac distribution is for indistinguishable particles obeying the pauli exclusion principle so ii satisfies this

Bose Einstein is for indistinguishable particles of which any number can occupy a state so this applies to iv

Maxwell Boltzmann is for distinguishable particles so this applies to i and iii.

Could anyone please tell me if I have all this right ? ( I'm studying for an exam by the way not doing homework)

  • $\begingroup$ It is strange that you do not explicitly write out the temperature (or the inverse temperature $\beta$), equating $\beta$ to zero you could check the correctness using simple combinatorics $\endgroup$ – Aleksey Druggist Jan 21 at 11:40

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