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Is there a categorical difference between "discretized" statistical mechanics and quantum mechanics from a theory perspective?

For example, we can count possible configurations of a system using the binomial formula when position and time are discretized. I showed in this post that this state counting leads us to the familiar result for an $NVE$ ensemble in statistical mechanics when Stirling's approximation is applied.

Or, stated another way, is discretizing the methods in statistical mechanics satisfactory in reproducing the main results of QM?

Edit a year later

I don't like how I worded this question, nor how I created this post, which didn't even ask a question. I was mainly wondering if there was anything interesting about how I derived $\Omega_0 = \Omega_A \Omega_B$, for the case of discretized space.

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  • $\begingroup$ I don't see how your answer corresponded to discretised space, it looked more like the usual derivation with discretised energy levels. $\endgroup$
    – Emil
    Mar 3, 2023 at 6:17
  • $\begingroup$ I guess it is vague - I said there are $n$ particles distributed in $r$ discrete locations. And now that I'm thinking about it, not sure how the math would change if I had started with $n$ particles distributed among $r$ energy levels. $\endgroup$
    – michael b
    Mar 3, 2023 at 15:22

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Since neither space nor time are generically discrete in quantum theories, it's not clear to me to which results you are referring.

Furthermore, the fundamental distinction between classical and quantum theories is the presence of incompatible observables in the latter (represented in the standard formulation of QM by non-commuting linear operators). The discretization (or lack therof) of e.g. the spatial degrees of freedom for a system would not affect this distinction.

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  • $\begingroup$ I haven't studied QM nearly as extensively as stat mech, so I didn't know if there was a theoretical way of producing the wave function that was similar to how we produced the canonical ensemble (or something like that). Thanks for your answer! $\endgroup$
    – michael b
    Feb 17, 2022 at 12:18
  • $\begingroup$ In hindsight my original post and this question weren't worded very well. Any possibility of returning to this question a year later? I'm fairly confident I was able to show that $\Omega_0=\Omega_A \Omega_B$, where the number of microstates is simply based on all possible discrete locations a particle can be in (discretized) space. Is this a coincidence? Anything noteworthy in my linked post above? $\endgroup$
    – michael b
    Mar 3, 2023 at 5:34

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