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If we're asked to find the partition function for a two state system with energies $ E_1 $ and $ E_2 $ for $ N $ indistinguishable, independent classical particles, then it makes sense to me that the total partition function $ Z $ is a product of individual partition functions $ Z_1 $ accounting for multiplicity.

$$ Z = \frac{1}{N!} Z_1^N = \frac{1}{N!} \bigg( e^{-\beta E_1} + e^{-\beta E_2} \bigg)^N $$

I am a bit confused however if the number of particles are known in each state. Say the $ E_1 $ state has $ N_1 $ particles and the $ E_2 $ state has $ N_2 $ particles. Could this system now be thought of as the sum of two 1-state systems so that the overall partition function is:

$$ Z = \frac{1}{N_1!} e^{-\beta E_1 N_1} + \frac{1}{N_2!} e^{-\beta E_2 N_2} $$

Or would the partition function not change because we know that an arbitrary particle could be in one of two possible states before we observe the state that it's in? Any help would be greatly appreciated!

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  • $\begingroup$ If you know $N_{1}$ and $N_{2}$, you have fixed the energy. That means you are no longer in the canonical ensemble at all, since the canonical ensemble fixes $T$ instead of $E$. Fixing $E$ makes the problem microcanonical instead. $\endgroup$
    – Buzz
    Jul 27, 2022 at 6:32

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I am a bit confused however if the number of particles are known in each state.

No, the number of particles in each state is not known. What is known is the total number of particles. The number of particles in each state fluctuates as particles transition between states. A good question to ask is what is the mean number of particles in each state (but since you did not ask it, I will not answer it :)

Your second equation is not correct: we never add partition functions. If we are lucky, we can multiply them, as in your Eq 1.

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