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It is a typical introductory problem in classical statistical physics to calculate the entropy of a two-level-system: say we have a N particle system in which particles can have energy E or 0. Assuming the particles are distinguishable we simply calculate the number of microstates using the microcanonical ensemble and relate it with entropy.

My question is: what if the particles (of each energy level) are indistinguishable? The number of microstates is simply one, and therefore the entropy nule. I guess something is wrong in the reasoning but can't see what.

(of course we don't take into account if particles are bosons or fermions)

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  • $\begingroup$ And can you more clearly articulate what you see here as a contradiction? $\endgroup$ – Aleksey Druggist May 11 at 21:44
  • $\begingroup$ It just seems weird to me that a system like this has nule entropy. $\endgroup$ – Andoni Royo May 13 at 6:15
  • $\begingroup$ If we put this system in a thermostat (canonical ensemble) the situation will not look so strange. By the way, the Gibbs distribution is just derived as the probability of having a certain energy to a single microstate of the system, and the presence or absence of microstates of a system with a given energy depends on the model of the system. $\endgroup$ – Aleksey Druggist May 13 at 22:11

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