For gas of identical particles, when we cool it down to extremely low temperature we can see one of two types of behaviour depending on the symmetry of wavefunction with respect to argument interchange, fermion and boson. This happens because particles become indistinguishable at quantum level and their wavefunction will be either symmetric or anti-symmetric.

But what if we consider a gas of non-identical particles? Let us say their masses are different. If we cool down this gas, because of different masses particles will still be distinguishable.

What will happen then, will the quantum effects (indistinguishability, wave-particle duality) appear then? Because of distinguishability of masses, could we use the Maxwell distribution at very low temperature too?


I would start by saying that Bose-Einstein and Fermi-Dirac distributions are all single-particle equilibrium distributions.

That is, for any general many-body (i.e. interacting, not single-particle) state, the thermal density matrix is $$e^{-\beta \hat{H}} = \sum_ie^{-\beta E_i}\,|\Psi_i\rangle\langle \Psi_i|.$$
So in a sense, a generic many-body state is always "Boltzmann" distributed.

If you can write your $|\Psi_i\rangle$ a single particle state $|n\rangle$, then you can show that from the above you can get the Bose-Eintein (by letting the sum go from $n=0$ to $\infty$ and taking the limit of the geometric series) or the Fermi-Dirac (by restricting $n$ to $0,1$) distributions.

Also, The irreducible representations of the Poincaré group are labelled by mass $m$ and the spin $s$. So as soon you have particles that have different mass, they are intrinsically different and obey their own Bose/Fermi/... statistical distribution.


So, if you have non-interacting distinct particles and reduce the temperature (/density), eventually they will independently display quantum effects. I.e. if you have two bosonic species, they will both Bose condense, not caring about what the other does.

If, however, the two species are interacting, then it is a completely different story and it is not trivial to answer. You would have to do numerics to know the answer.


At some point, confinement and quantization will become important, and this will make that the Maxwell distribution does not apply.

For example He-4 which has six distinguishable particles: two electrons, two protons, two neutrons (two different spins of each), all in the ground state of a potential well.

  • $\begingroup$ That is true, but He-4 constituents do not form a "gas" in the usual sense of the word. I think the question is rather about a gas made of electrically neutral, weakly interacting molecules, each with unique mass. $\endgroup$ – Ján Lalinský Jan 14 at 12:44
  • $\begingroup$ @JánLalinský But one would want to do the (thought) experiment in a "box". This implies confinement, and quantization of energy levels. All those molecules can be in the ground state of the box. It is like the nucleons of a helium nucleus. $\endgroup$ – Pieter Jan 14 at 12:48
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    $\begingroup$ I just thought the helium constituents is a strange example, it is very microscopic and never behaves according to the Maxwell-Boltzmann distribution. But otherwise I agree with you, there should be quantization of energy levels of molecules due to confinement, even without symmetric psi function (so the statistics will be complicated, depending on the mass histogram). $\endgroup$ – Ján Lalinský Jan 14 at 12:56

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