The system in grand canonical emsemble together with the surrounding reservoir is isolated, thus have conserved particle number $N$. However, the system itself only has fixed average $\langle N\rangle$, and the quantum mechanical operator $\hat{N}$ necessarily does not commute with the Hamiltonian. Why can we assign both energy and particle number to each state?
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1You don't assign energy and particle number to each state, right? You just evaluate $\mathrm{Tr}[\exp [-\beta(H - \mu N)]]$ in a basis which diagonalizes at most one of them. – Connor Behan Nov 08 '21 at 02:54
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@ConnorBehan Yes, but when evaluating it, it seems that we always simply write out $\sum_n \exp[-\beta(En-\mu N_n)]$ as if we are doing it in a basis which diagonalizes both of them? – RicknJerry Nov 08 '21 at 04:49
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I've only seen that done in classical stat-mech systems. – Connor Behan Nov 08 '21 at 05:25
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1Possibly related: How can μ be nonzero if grand canonical density operator commutes with $\hat H$ ? and Does non-conservation of number of particles imply zero chemical potential?. – Tobias Fünke Nov 08 '21 at 09:09