Why can many distinguishable particles be in the same quantum state?

Question

In Boltzmann distribution, it says we have no restrictions on how many particles there are in a same quantum state. BUT the contradiction is: Distinguishable particles mean we think their wave functions have no overlapping. Being on the same quantum state means their wave function is 100% the same, i.e., 100% overlapping. How to solve this paradox?

Answer 1

So, you are thinking of matter in a 1600s way, as entities which are defined by the fact that they occupy space. We just don't think of matter that way any more.

Even before we got into quantum field theory, we had a notion that matter is spread out over space only in a wavy probabilistic way that simultaneously expresses a deep fundamental uncertainty about both where it is and how much momentum it has: because it turns out that momentum is tied closely to the wavelength and accurately specifying that requires many wavelengths, hence a very spread-out spatial state.

There is nothing wrong with two wavefunctions occupying the same position; it means that if you measure whether the particles are in a box centered at that position, they have some nonzero chance to both be found inside that box. Of course the smaller the box the higher the resulting uncertainty about the momentum, as you will have a very poorly defined wavelength afterwards.

Answer 2

You are confusing quantum state with the (approximate) classical momentum state used in classical statistical mechanics.

Indeed, the Maxwell–Boltzmann distribution gives the probability of finding a particle with momentum p, and implies many distinguishable particles will be in the same momentum state, or more precisely, a particle will on average have momentum with kinetic energy proportional to k T.

At high temperatures a gas of indistinguishable quantum particles with fixed volume and particle number can be effectively described as a gas of distinguishable particles (with correct Boltzmann counting) because the thermal de Broglie wavelength becomes much smaller than the average intermolecular separation: the particles become spatially localised at different points.

Imagine filling the volume with boxes of side lengths equal to the average intermolecular separation distance. In each box we can say whether or not there is a particle and if there is we can say what is it’s momentum to good approximation.

Effectively we know the momentum and position of the particle and this is why we can say the particle is distinguishable. But this is precisely the limit where the quantum effects of the particle are not relevant. The uncertainty in the position ( intermolecular separation) multiplied by the uncertainty in momentum ( some factor times by \sqrt{2mkT}) is much greater than hbar/2.

Said another way, many particles will have approximately the same energy/momentum (kT), but they will not be in the same state because their positions will be different.

Answer 3

I am not sure whether i understood your question or not but i'll give it a try.

In you example if particles of ideal gas are indeed distinguishable, then there has to be at least one parameter that make their wave functions different (e.g their mass, position, etc) in a way or another. If not, how they are distinguishable in the first place? See this link What are distinguishable and indistinguishable particles in statistical mechanics?

Of course two distinguishable particles can have same energy, spin, etc, but that doesn't necessarily make their wave functions same, because there is always at least one parameter that make theirs different. In ideal gas, i think the only way to consider particles distinguishable would be difference in mass of particles, or knowing position of them beforehand, (either way, they would have different wave functions). If there's another way and that way doesn't affect wave functions, please enlighten me.

Edit: this might help too: Distinguishable particles in Maxwell-Boltzmann distribution