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Me and a few friends were wondering about the origin of indistinguishability in QM and thought of the following thought experiment. Consider a toy model of a classical universe which has only one kind of particle(lets call them unons) of some mass m and radius a. Also assume that unons obey newtons laws, and the only interactions between them are elastic collisions. Keeping in mind that there are nothing else in this universe except for unons, I have the following statement and questions

1) If you have a collision between two unons there is no way to observe where the individual unons went.

2) You cannot follow the trajectory of individual unons without disturbing them.

3) 1 and 2 if true would imply that unons have to be indistinguishable to an observer external to the system but still part of the universe ( i.e they too have access to only unons as particles)

4) Would that mean that the equilibrium statistics of unons should not be Boltzmann statistics and should be something like Fermi statistics?

5) Now however if can add another class of point particles to observe unos, then suddenly unons become distinguishable (like in our world) and their statistics become Boltzmann statistics.

6) Thermodynamic properties however are derivable from statistical mechanics and we would expect the same thermodynamics for both 4 and 5, since they should not depend on the point particles. This would be paradoxical if they have different statistics

My gut feeling tells me that 4 is wrong and the systems of pure unons should exhibit Boltzmann statistics as well. Statics of a system should not be Bayesian

However would that mean that indistinguishability is a purely quantum phenomena? Why should that be? After all in my first universe it is impossible to track the trajectories of particles with arbitrary high precision.

Is that because quantum mechanics doesn't exhibit realism? What about (non-local) hidden variable interpretations then? Why would they have indistinguishable particles?

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I think you are misinterpreting something in the concept of indistinguishability.

Which statistics your particles follow does not depend on indistinguishability.

If they have integer spin they follow Bose-Einstein.

If they have non integer spin they follow Fermi-Dirac.

If they are big enough to neglect quantum effects they follow Boltzmann (this may be suggested to be right answer due to the fact that you mention that they follow Newton's law...).

The problem of indistinguishability arises in classical mechanics too (see mixing paradox and correct Boltzmann counting). Huang's book explains it quite effectively: if you do not account for identical particles being not distinguishable, which (see later) is due to quantum mechanics, you get paradoxes in the classical domain.

Unons, and particles in general, are always distinguishable from a classical mechanics point of view. This, as I said, leads to some paradoxes because quantum-ly they are really not distinguishable. But this is not because our universe only contains unons. Even in a universe where unons are detectable and there are other particles, you could not discriminate between two different unons (provided they are small enough to follow QM).

Indistinguishability is not merely a problem of detection: it is intrinsic to quantum mechanics and its wave-like description of matter.

In classical mechanics you can say "particle 1 is in position r and with momentum p". Even if you do not detect it, you can in principle say so and discriminate, or label, particles according to this.

In QM however you cannot, as every particle is everywhere. If you had to particles, A and B, even if you could detect a particle with high precision, there would still be a slight chance that you by trying to measure A you are actually measuring B, as the probability of finding B where you are looking for A is not vanishing. Thus whatever the type (unons, point like, electrons...), particles of the same type are not distinguishable and thus you must account for it using the (corrected) Boltzmann statistics or using FE/BE statistics.

Indistinguishability is not only in the fact that we can not distinguish particles in an experiment because we do not have the means. It is because we cannot describe them not even in principle.

Your unons are quite easy to simulate on a computer and they would behave as a hard spheres's gas. So the observer does not need to measure unons to know how they behave: since you said they follow classical mechanics, he can predict their behavior albeit only in principle. So they can be treated both as distinguishable and not distinguishable particles depending on your point of view.

Answering to your question, in general, I would say: whatever distribution unons follow (and that is due to their size, spin, difference in energy level...) that does not depend on the presence of the point like particles. If (for instance) you thought they were fermions (then FD) because you could not detect them, then you detect them and find out they're boson, they have not changed in nature, it is you that weren't able to detect it earlier. They were (and still are) following BE from the beginning.

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  • $\begingroup$ FD e BE are related to the properties of the wave function under particles exchamge. Such properties are a consequence of indistinguishability. $\endgroup$ – JalfredP Nov 6 '16 at 2:32
  • $\begingroup$ Sorry I deleted my previous comment by mistake when trying to edit it. $\endgroup$ – elenasto Nov 6 '16 at 2:34
  • $\begingroup$ (follows from before..) You could see FD and BE as classical distributions in the grand canonical ensemble - but with a 'strange' way of counting states due to QM. It's not a limiting case of Boltzmann, that is for sure. Anyways, distinguishability and statistics are linked, also for Boltzmann distribution, despite distinguishability is a purely quantum concept. $\endgroup$ – JalfredP Nov 6 '16 at 2:39
  • $\begingroup$ FD and BE are a consequence of indistinguishability and some thing else, i.e commuting or anti commuting nature of their fields. (btw In my original question, I wasn't suggesting that unons will obey FD). Also I'm not sure what you mean by FD and BE being boltzmann, isn't it known that the latter is a limiting case of the former and not vice versa? Moreover the wiki article you have linked to on mixing paradox claims that entropy is subjective which seems to support my question $\endgroup$ – elenasto Nov 6 '16 at 2:41
  • $\begingroup$ I was assuming you know which kind of particles you are considering. If you only have classical unons they are distinguishable. Quantum unons are not (and.you have to account for that using correct Boltzmann counting). $\endgroup$ – JalfredP Nov 6 '16 at 2:51

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