Experimental confirmation of indistinguishability of identical particles?

[Xverse]

Does the experiment confirming Bose Einstein condensation prove that identical particles are indistinguishable?

How does that reconcile conceptually with Maxwell distribution to which Bose Einstein distribution converts? Is this a result of the strength of interaction among the identical particles?

 

[Ken G]

Yes, you would not get the condensate for distinguishable particles, because there are a lot more ways for those distinguishable particles to be in states that are not the ground state. But indistinguishable particles cannot be permuted around the excited states in nearly as many different ways, so there is much less of a premium on not being in the ground state. At low temperature, they will thus tend to aggregrate in the ground state. That this is seen experimentally demonstrates the power of the indistinguishability constraint.

The Maxwellian does not require any interactions between the particles, beyond those necessary to allow the system to have access to many different possible states. You get the Maxwellian when the temperature is high enough that it is unlikely that two particles would ever be in the same state anyway, regardless of whether they are distinguishable or fermions or what. When this is true, it doesn't matter if they are bosons or fermions or indistinguishable or distinguishable-- each particle has an effectively independent relationship with the thermal reservoir that is definining both the meaning of T and the probability that the particle will be in various states (which is then distributed like a Maxwellian).

 

[Xverse]

Thats good. Thank u. But can you explain why u said that it doesn't matter to the Maxwell distribution whether paticles are distinguishable or not? I thought, looks incorrectly, that Maxwell distribution is based on distinguishable particles statistics.

Thanks

 

[PatrickPowers]

Does the experiment confirming Bose Einstein condensation prove that identical particles are indistinguishable?

How does that reconcile conceptually with Maxwell distribution to which Bose Einstein distribution converts? Is this a result of the strength of interaction among the identical particles?

As a note I'd suggest considering the cores of neutron stars, which share the property of superfluidity with BECs. It's also worth noting that in said cores the neutrons are superfluid, the protons are superfluid, but the electrons are not. So superfluids of differing types of particles can share the same space, and share that space with ordinary fluids as well.

I have read that even in a BEC most of the atoms can be an ordinary fluid, so such sharing can occur there too. As the BEC cools then more and more atoms are in the superfluid, but I would think it is never 100%.

 

[Ken G]

Thats good. Thank u. But can you explain why u said that it doesn't matter to the Maxwell distribution whether paticles are distinguishable or not? I thought, looks incorrectly, that Maxwell distribution is based on distinguishable particles statistics.

The Maxwellian distribution doesn't care if particles are distinguishable or not, what it cares is that particles are independent of each other (so have a probability distribution that depends only on their contact with a reservoir at T). That can be achieved if they would be unlikely to be in the same state if they were distinguishable, and when that is true, it doesn't matter if they aren't distinguishable. Indistinguishable particles can be treated as if they were distinguishable when they don't overlap at all-- indeed that's just what we do when we say things like "a single electron was shot out of cathode and that same electron was detected at the anode." That statement pretends the electron has an identity, which it does not, but it doesn't matter-- we get the same answer imagining it does, and it sure simplifies our language.