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In a proof of the equivalence of the canonical and grand canonical ensembles (in the thermodynamic limit) in Jochen Rau's Statistical Physics and Thermodynamics (highly recommended!), the author evaluates the grand canonical partition function as $$Z_G = \textrm{Tr}(\exp(-\beta \hat{H} + \alpha \hat{N})).$$ In evaluating this trace, it seems to me that the author uses a joint eigenbasis of $\hat{H}$ and $\hat{N}$ and, thus, the fact that they commute.

My question is, what sorts of restrictions exist on $\hat{H}$ (i.e. on the type of system to which this implies) for $[\hat{H},\hat{N}] = \hat{0}$?

EE18
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    Glad you like the book ;) You can also try out his QM book... Anyway, this was asked before, let me search the relevant PSE posts. But what kind of restrictions do you mean? The only answer I can imagine is: The Hamiltonian must preserve the particle number (and thus commute with the number operator). – Tobias Fünke Apr 20 '23 at 15:15
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    See e.g. this and this. – Tobias Fünke Apr 20 '23 at 15:16
  • It is superb, thank you again for the suggestion! Is his QM book worthwhile/at the level of Ballentine too? Also, I'm not sure I completely follow the answers in your links (none were accepted by their respective OPs); is the conclusion I am to derive that Rau is simply assuming a system for which $[\hat{H},\hat{N}] = \hat{0}$ is true? @TobiasFünke – EE18 Apr 20 '23 at 15:56
  • I really like the operational approach of the author, also in the context of QM. But IMHO it is, for a physicist at least, supplementary, because he does not discuss many things usual QM (intro) courses discuss (but instead discusses many things not found in "standard" books). Anyway: Yes, I mean if you have a system where the Hamiltonian does not conserve the particle number (say through creation or annihilation of particles), then its does not commute with the respective number operator. So yes, you basically have to assume this... – Tobias Fünke Apr 20 '23 at 16:02
  • ...as the most upvoted answer in the second link states: If you want to describe equilibrium conditions, such that the density operator is constant in time, you are forced to conclude that $H$ preserves the particle number (in the context of the grand canonical ensemble) as long as $\mu \neq 0$. – Tobias Fünke Apr 20 '23 at 16:03
  • @TobiasFünke Sorry, are you referring to the answer by Roger Vadim or ReasonMeThis? – EE18 Apr 20 '23 at 16:06
  • I've edited my previous comment. What is important to note that $H$ here refers to the Hamiltonian of the system (as also the answer of the user ReasonMeThis states in the first link). So you are right that depending on the physical situation at hand, it might not be true that the Hamiltonian conserves the particle number, which @Roger Vadim in the same post also points out. Does that clarify some things? – Tobias Fünke Apr 20 '23 at 16:12
  • @TobiasFünke Yes, I believe so, thank you! – EE18 Apr 20 '23 at 16:18
  • I guess there are some experts here which can elaborate much more and better than me. – Tobias Fünke Apr 20 '23 at 16:18

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