Why do we assume that the particles in the gas are indistinguishable? In QM, a set of N particles are indistinguishable only if their combined wave function is either symmetric (bosons) or antisymmetric (fermions) under interchange of two particles. Why do we make this assumption for the combine...
I am not sure how this explains why the particles in the gas are indistinguishable. If indistinguishability defines the notions of bosons and fermions, then why not just assign the particles in the gas some other name/properties? In the classical model, the particles in the gas are distinguishable. You can distinguish them by their respective trajectories. To my understanding, particles are indistinguishable in QM only if their "trajectories" have mixed, such that the probability that particle 1 is at x_1 and 2 is at x_2 is equal to the probability that particle 1 is at x_2 and 2 is at x_1.
And surely two fermions don't have to be indistinguishable. If you have an unentangled system of a pair of electrons, then you can distinguish the particles by their single-particle wavefunctions (because of the Pauli principle, these wavefunctions must be different, and hence you can distinguish the particles).
Not quite, particles are indistinguishable in QM if swapping them shouldn't change the physics, i.e. $|x\rangle_1 \otimes |y\rangle_2$ should be the same as $|y\rangle_1\otimes |x\rangle_2$. The only thing that matters is "there is a particle at position $x$ and a particle at position $y$", not which of the two particles is in which position, because the particles being identical in every other way except position, their position is the only way to distinguish them. "there is a particle at position $x$ and a particle at position $y$" is translated mathematically in Fock space.
From this comes the fact that a proper wave function will always look something like $|x\rangle|y\rangle+|y\rangle|x\rangle$ or similar, because these are the only superpositions allowed. Classical particles are also indistinguishable if they are only characterized by their trajectory, but this as far as I know doesn't have very profound implications.
Example with classical particles: you have two balls: ball $1$ and ball $2$. They are perfectly identical except that one is in my right hand and the other one is in my left hand. Now I ask you to close your eyes, swap them a few times, and ask you: which one is ball 1 and which one is ball 2? You won't be able to tell, because there wasn't a ball 1 or a ball 2 in the first place, just one in the right hand and one in the left hand. The balls are just distinguished by the point they occupy in phase space, i.e. just their position.
If the balls were electrons and you say "their state is $|x\rangle |y\rangle$", you must be doing something wrong, because in that case there would be some experiment you could make that tells you whether the particles have been swapped or not, which means the particles weren't identical after all.
Yes, but I am of course talking about the quantum indistinguishability, not the classical one you referred to in your last comment. As you said, the combined wave function will be given as a superposition of all the permutations of the particles in the system, when we talk about quantum indistinguishability. My question is why we assume the particles in the gas are indistinguishable in this manner, when there are other possible states for the combined system.
I'm not sure what you mean by "when there are other possible states for the system". We assume particles in a gas are indistinguishable because it makes intuitive sense and yields correct experimental predictions. How else are you going to label the particles in a gas if not by their position and momentum? The particles in the gas are like the balls in the classical examples, they are only characterized by their state, unless you explicitely label them. Particles in a gas are assumed to be identical except for their position, momentum, spin or whatever other properties you're modelling.
Now, if you had a mixed gas of say, oxygen and nitrogen, then those are distinguishable. You could actually say "the oxygen atom is here and the nitrogen atom is there", and if you swap them you now have "the nitrogen atom is here and the oxygen atom is there", but this is because nitrogen and oxygen are distinguishable by other properties that are not position. In other words, you would notice if they were swapped.
"they are only characterized by their state, unless you explicitly label them". Exactly. If they are not completely indistinguishable, then it's possible to distinguish them in principle using their quantum states (or their position and momentum, in the classical case). But complete indistinguishability is established if there is no way to distinguish in this case, since the probability that particle 1 and 2 is in state 1 and 2 respectively is the same as the probability that particle 1 and 2 is in state 2 and 1 respectively.
but by using the states to distinguish them you are distinguishing the states, not the particles. Again, with the classical balls, you can distinguish that there is a ball in the right hand and a ball in the left hand, but not which one is ball number $1$ and which one is ball number $2$. That we force those two probabilities to be the same is a result of the fact that we consider them as indistinguishable, not the definition of indistinguishable.
@user2723984 Maybe I'm just terrible at explaining what I mean, but if you go to this article and read the section ""Distinguishing between particles", then maybe you get what I mean. In particular, the second and third paragraph gets at the heart of what I mean.
In the end, the whole notion of labelling the two Hilbert spaces is artificial, because you are only labelling them to eliminate the labels by second quantization. The non artificial notion is "there is one particle in state 1 and one particle in state 2", doesn't matter which one is which.
My point is that the second kind of indistinguishability mensioned in the article requires some special properties of the combined wavefunction (as discussed in the next section on symmetry and antisymmetry). Why do we assume that the combined wave function of the gas has these properties, ie. why do we assume the particles are completely indistinguishable?
Ok, first, one point, if the gas mentioned in your question is a classical gas and you're referring to the Gibbs' paradox, then I misunderstood, and you should refer to J.Murray's answer.
@user2723984 I actually didn't know there were two kinds of ideal gas. So I actually can't be sure which one it is I am referring to, but yes, I think it's related to the Gibbs paradox (though I haven't really got that far yet, but it's the next topic in my stat mech book).
But regarding this last comment of yours, again, I think you have it backwards, we are not assuming that the wave function of the gas has a particular property, we are assuming that the particles are indistinguishable, and concluding that the wave function must have those properties. That the particles are indistinguishable is a physical assumption, and we make it because, as explained in the paragraphs you pointed me to, the only way we can imagine distinguish them isn't compatible with QM.
It's not that there are two kinds of ideal gas, but that if you have the Hamiltonian for N free particles, you can treat it quantumly or classically. If you treat it classically, there will be no wave function at all, and the only "indistinguishability" consequence is this $1/N!$ factor in the partition function. If you treat it quantumly (i.e. you treat x and p as operators) and assume indistinguishability, you'll have to decide whether it's a boson or fermion gas.
@user2723984 But syrely two particles (fermions, say) don't HAVE to be indistinguishable in QM? For instance, if the combined wave function is unentangled and separable into the product of various single-particle wavefunctios, then it makes sense to talk about the identity of a single particle (hence, distinguishability).
as soon as you assumed your particles are fermion, you already assumed they are indistinguishable, and hence cannot have a separable unentangled wavefunction that would allow you to distinguish them through measurement.
indistinguishability is an assumption
If you assume the particles are distinguishable, then sure, you can have your $|x\rangle_1 |y\rangle_2$ wave function, but then if you measure a particle at position $x$, you have to have a way to reliably tell whether it's particle $1$ or particle $2$, for example, one of the two particles could be charged.
But then this isn't a system of two fermions.
this is done very often in physics as I mentioned, in lattice spin systems, where you assume you have a particle with spin at each point in some grid (like the Ising model). The particles there are distinguishable, hence they are neither bosons nor fermions.
@user2723984 So you're saying that the particles in all ideal gases we might ever come over is automatically indistinguishable because we assume they're fermions/bosons?
Not quite, being fermionic or bosonic is something you can say only after you assumed indistinguishability. Whether you assume indistinguishability or not depends on the physical situation, it makes sense to assume it for the quantum ideal gas.
So you know the Hamiltonian for N free quantum particles, and intuition tells you they should be indistinguishable, and consequently you can model two different kind of gases: free bosons or free fermions. Experiment tells us that these two kind of gases really exist (bosonic and fermionic gases behave differently!) which in turn justify your initial indistinguishability assumption.
If you never assumed indistinguishability, you would never have ran into the concept of fermions or bosons.
@user2723984 So it's more of an experimental fact that the gases we find are bosonic/fermionic, rather than a purely theoretical fact? I mean, surely it would have been possible to imagine a gas where the wave function is separable into single-particle wave functions, in which case the particles are distinguishable.
@user2723984 But then the particles won't be fermions or bosons, I guess?
It is surely possible to imagine! But this already leads to a non extensive entropy in the classical case (Gibbs' paradox), which is why people thought about indistinguishability in the first place.
In short you can imagine anything, but it gives unphysical results. There are no purely theoretical facts in physics :)
and no, they wouldn't be fermions or bosons, because this distinction only comes from the indistinguishabillity assumption.
@user2723984 Alright, that was a bit better. I thought there were purely theoretical reasons for believing in indistinguishability (aside from the extensivity property of entropy, as you mentioned). Meaning that indistinguishability for an ideal gas could be "derived" from simpler principles, instead of being based on a more experimental basis.
but to reiterate: if your book is still talking about a purely classical treatment of the ideal gas, then there is no wavefunction at all, hence no bosons or fermions or anything. There the indistinguishability was more or less a "trick" that then gave the right result. The reason why you have to use this trick is of course more or less that classical physics is not the whole story.
The Gibbs' paradox really puzzled physicists at the end of the 20th century.
@user2723984 Yeah, sure. So just to be clear (to maybe confirm what I thought I knew till now): the correction to the partition function (and entropy) which fixed extensivity is a purely quantum mechanical effect (derived from the assumption of indistinguishability), and cannot be explained through classical physics?
Yes, more or less. The resolution of Gibbs' paradox requires assuming the particles are indistinguishable. This is puzzling if you're a physicist in the 1800s. The advent of quantum mechanics gave a good justification as to why particles should be treated as indistinguishable.
@user2723984 Alright, thank you for all the help and time you've provided, and I think I understand it better now. I don't think I have the right to waste any more of your time ;)
Discussion between user2723984 and Fe…
Discussion between user2723984 and Fe…