I'm an amateur on a quest to understand QM. In various places (such as early in chapter III.4 of the Feynman Lectures) I have seen an argument that looks like it's trying to convince me that any class of indistinguishable particles must necessarily behave either like bosons or like fermions -- i.e., that this is not only an empirical fact, but a deductive consequence of the basic structure of QM.
I have trouble understanding these arguments. They seem to pull an assumption out of nowhere, that interchanging two of the identical particles must necessarily correspond to multiplying the wavefunction by a constant. Once this is given, I can see that this constant must be either $1$ or $-1$, but where does that assumption come from? If I write down a wavefunction without knowing in advance that the particles are identical, there are oodles of ways to write one that doesn't react to an interchange merely by a constant factor. Is there something that a priori excludes all those states from consideration?
I've been able to reconstruct an argument that works for a world containing only $2$ instances of the indistinguishable particles, to wit:
What it formally means for the particles to be indistinguishable must be that the operator $P$ which interchanges the two particle commutes with every observable, in particular with the Hamiltonian.
If we have any random state of the system, we can write it as a sum of eigenstates of $P$, and because $P^2=1$, the eigenvalues must be $\pm 1$. Now whenever time passes (we apply $H$) or we make an observation (and apply some other observable operator), the operator we use commutes with $P$, and therefore we can diagonalize $P$ and the operator simultaneously and see that the $+1$ eigenstates never contribute any amplitude to the $-1$ eigenstates and vice versa.
Thus if at some point in history we have made an experiment to determine out whether our two particles are bosons or fermions, right after the experiment the state of the world is going to be a pure $+1$ or $-1$ eigenstate -- and because of the above reasoning it is going to stay in that eigenspace for all eternity henceforth.
So in this case it's of no use to consider mixed states.
The problem with this argument is that it doesn't seem to scale beyond two particles. With $n$ particles there are a lot of different permutations of them one could consider, and these permutations don't commute internally, and so I cannot be sure that they will all diagonalize simultaneously with the operator I'm observing.
This is probably not just a problem with my proof skills, as dimension-counting shows. Let's consider a discrete system where each of $3$ particles can be in one of $3$ states. The state space has dimension $27$ over $\mathbb C$ -- but the "fermion" subspace (consisting of those states that go to their negative under an interchange of any pair of particles) has dimension $1$, and the "boson" subspace (those states that stay unchanged under any transposition) has dimension $10$ if I count it correctly. There are $16$ dimensions left unaccounted for! Is there a principled reason why all of those states are necessarily unphysical or uninteresting?
In Zee's Quantum Field Theory in a Nutshell (which I bought by mistake but don't have anywhere near the prerequisites to understand) there's some mumbling about "anyons", which as far as I can decipher it seems to claim that a strict Bose-Fermi dichotomy is only necessary if one has a continuous set of base states of dimension at least $3$ -- by some topological considerations that mostly went over my head.
Am I misunderstanding it when I think Feynman and others claim that the dichotomy is a derived result?