Identical particles: why a symmetric or antisymmetric wave function? I've read in many text books that the indistinguishability of two identical particles at $x$ and $y$ implies:
$$|\psi(x,y)|=|\psi(y,x)|\quad (1)$$
This sounds rather natural.
Then they say there are two solutions: the symmetric and antisymmetric ones leading to bosons and fermions:


*

*Symmetric: $\psi(x,y)=\psi(y,x)$

*Antisymmetric: $\psi(x,y)=-\psi(y,x)$


I can't help to wonder why there is no other solution. Is it a theorem that can be proven from (1)? Or is it just that these two solutions are the only ones observed in nature?
 A: Yes.  If you study the representation theory of the permutation group, you will learn there are only two representations with the property that the $n$-particle states comes back to any multiple of itself.
For the symmetric representation, the state comes back to + itself, and for the antisymmetric to - itself.
The requirement that the state come back to $\pm$ itself is an essential consequence of the indistinguishability of particles.  For any other representations (for $n\ge 3$ particles there are states of mixed symmetry transforming as linear combination of other states) there are problems in the computation of average values if the state is not fully symmetric or antisymmetric.  You can find references to literature on this in this related answer.
Note that anyonic states can come back themselves up to some (in principle) generic phase, but anyons are quasi-particles rather than fundamental particles.  Moreover, they live in 2d rather than 3d.  There are multiple posts on this site on anyons (see for instance this one) in addition to those already cited above.
A: $$ \left| \phi(x,y) \right| = \left| \phi(y,x) \right| \implies 
\phi(y,x) = e^{i \alpha} \phi(x,y) $$
Making exchange twice, we get
$$\phi(x,y) = e^{2 i \alpha} \phi(x,y)$$
Hence,
$$e^{2 i \alpha} = 1$$
so $\alpha = n \pi$ with $n$ integer. If it's an odd integer then $\phi(x,y) = -\phi(y,x)$. If it's an even integer then $\phi(y,x) = \phi(x,y)$ (assuming $\alpha$ is a constant)
A: The physically measurable quantity in QM is the probability density. This gives you $\left|\psi\left(x,y\right)\right|^2 = \left|\psi\left(y,x\right)\right|^2$ (since when we say two particles are indistiguishable, what we actually mean is that we cannot make any measurements to distinguish them). This would imply that $\psi\left(x,y\right) = \pm\psi\left(y,x\right)$, which tells you that the wavefunction must be odd or even.
Edit
As pointed out by eranreches, the reasoning above is flawed, since the wavefunction with particles interchanged could be multiplied by a complex phase factor, in general.
A better way of looking at the above problem would be to define a particle exchange operator $P$ such that $P\psi\left(x,y\right) = \psi\left(y,x\right)$. If the system does not change under the action of this operator, it means that the Hamiltonian is invariant under the action of this operator, which implies $\left[H,P\right] = 0$. This implies that the energy eigenstates are also eigenstates of $P$. We can also see that by definition, $P^2 = I$, where $I$ is the identity operator. Let $p$ be the eigenvalue for $P$ of some energy eigenstate $\psi\left(x,y\right)$. We can then see that $P^2\psi\left(x,y\right) = p^2\psi\left(x,y\right) = I\psi\left(x,y\right) = \psi\left(x,y\right)$. This implies $p^2 = 1$, which implies $p = \pm 1$. Thus we get $\psi\left(y,x\right) = P\psi\left(x,y\right) = \pm\psi\left(x,y\right)$.
This is not the complete story, however. If you want to delve further into why particles with half-integer spins have antisymmetric wavefunctions, and why particles with integer spin have symmetric wavefunctions, you would have to look at the Spin statistics theorem, as pointed out by ACuriousMind.
