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Consider a system of two identical particles described by a wavefunction $\psi(x_1, x_2)$. There are two kinds of exchange operators one can define:

  • Physical exchange $P$, i.e. swap the positions of the particles by physically moving them around.
  • The formal coordinate exchange $F$, where $F\, \psi(x_1, x_2) = \psi(x_2, x_1)$.

Since $F^2 = 1$, the eigenvalues of $F$ are $\pm 1$. Some books incorrectly say this proves that only bosons or fermions can exist. This is wrong because the argument also works in 2D, where anyons exist.

The real argument is to consider the eigenvalues of $P$, which are $\pm 1$ only in three dimensions due to topology. In the 3D case, wavefunctions with $P$ eigenvalue $+1$ describe bosons, and those with $P$ eigenvalue $-1$ describe fermions.

However, all treatments of bosons and fermions say that bosons have $F$ eigenvalue $+1$, and fermions have $F$ eigenvalue $-1$. For example, you'll see the equation $$\psi(x_1, x_2) = -\psi(x_2, x_1)$$ for fermions. I'm not sure where this comes from; I accept the $P$ eigenvalue is $-1$, but as far as I can tell $F$ and $P$ are totally distinct. In particular, their eigenvalues must be different in two dimensions.

For identical particles in 3D, why are the $F$ and $P$ eigenvalues the same?

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That the behaviour under an actual, physical switch is equivalent to the behaviour under the formal exchange is precisely the content of the spin-statistics theorem. The spin (behaviour under physical rotations, in particular the representation of a "$2\pi$ rotation" on the space of states) determines the statistics (behaviour under formal exchange, or equivalently the (anti-)commutation behaviour of the creation operators).

This is a fundamentally relativistic quantum field theoretic result. In a non-relativistic quantum mechanical setting, there is no reason at all for the bosonic/fermionic behaviour to be related to the behaviour under physical exchange. Spin-0 fermions and spin-1/2 bosons are not inconsistent in the non-relativistic QM setting.

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  • $\begingroup$ I'm not entirely convinced, because the spin-statistics theorem seems to also work in 2D, where the $F$ and $P$ eigenvalues are not the same. How does that work out? $\endgroup$ – knzhou Mar 26 '16 at 21:15
  • $\begingroup$ @knzhou: What do you mean, "it seems to work in 2D"? It fails, there are anyons. And the reason it fails is not that anyons are non-relativistic (which would also be an allowed cop-out), the deeper reason is that $\mathrm{SO}(2)$ doesn't have the usual double cover as its universal cover like all the higher-dimensional rotation groups, but instead has $\mathbb{R}$ as its $\mathbb{Z}$-fold cover, allowed for different (anyonic) projective representations. If you examine e.g. the derivation in Weinberg, it relies on there being an even/odd spin label $j$, which doesn't exist in that way in 2D. $\endgroup$ – ACuriousMind Mar 26 '16 at 21:22

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