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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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  • $\begingroup$ Did you ever come to clarity on this issue? $\endgroup$
    – Craig
    Dec 22, 2019 at 20:38
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    $\begingroup$ @Craig Yup, on reflection I realized that the answer I got below is totally right -- and that, depressingly, almost all textbooks are wrong. $\endgroup$
    – knzhou
    Dec 22, 2019 at 20:44
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    $\begingroup$ @Craig For a correct account you can see, e.g. chapter 3 of this draft book. $\endgroup$
    – knzhou
    Dec 22, 2019 at 20:45
  • $\begingroup$ Thank you immensely! $\endgroup$
    – Craig
    Dec 22, 2019 at 20:56
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    $\begingroup$ @knzhou you’re severe in condemning most textbooks; technically the argument $F^2=1$ thus eigenvalues are $\pm 1$ is correct, even in 2d. The issue occurs when one particle goes around another - which is equivalent from $F^2$ - in the punctured plane. You’re right textbooks are not precise and often overlook this but it doesn’t make the argument wrong: just incorrect in some cases of the punctured 2d plane. The link you posted looks quite nice and I’m eager to download and read it. $\endgroup$
    – ZeroTheHero
    Dec 23, 2019 at 14:02

1 Answer 1

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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, 2016 at 21:15
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    $\begingroup$ @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, 2016 at 21:22
  • $\begingroup$ I would be incredibly grateful if you did a Q&A on this topic (at the level if this last comment) or if you could recommend books which treat this topic with proper care. $\endgroup$
    – Craig
    Dec 23, 2019 at 4:58
  • $\begingroup$ @Craig I tried to write something up but realized I don't really know what you want to know about. Maybe you could ask a question yourself? ;) $\endgroup$
    – ACuriousMind ♦
    Jan 20, 2020 at 18:28

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