5
$\begingroup$

If we have a state $\psi(x_1,x_2)$ of two identical particles and an exchange operator $O$ which swaps the particles. Obviously the physics must be the same and hence $O$ can only introduce an overall phase.

Secondly it must square to the identity and hence the phase is $\pm1$.

However I have since heard that this argument is wrong and that you can get anyons (or at least you can get quasi particles which are anyons in 2+1 d).

What is wrong with the argument above?

$\endgroup$
  • $\begingroup$ It seems that it relates to parity anomaly in odd dimensions $\endgroup$ – Artem Alexandrov Feb 14 at 9:51
7
$\begingroup$

The classical argument is the following: exchanging twice the position of the two particles is similar to having particle $2$ "loop around" particle $1$. Because the system you end with is the same as the one you started with, the wavefunction describing it must be the same up to a phase. However, any simple loop circling around a point can be deformed continuously to a trivial loop (a point) in 3 spatial dimensions and onwards. This is because if your loop was to "touch" the point while you're deforming it, you could just avoid it by bending the loop slightly out of plane. Since the loop can be reduced to basically nothing, the only acceptable phase for exchanging twice the particles must be $1$, and the phase for exchanging them once must be $1$ or $-1$.

In two spatial dimensions however, if your loop circles around a point (the second particle), there is no way for you to reduce it to a trivial loop without "touching" the other particle. Thus the previous argument does not hold and you can have other statistics than bosons and fermions.

Note that the picture is even more extreme in one spatial dimension, because then the particles cannot pass each other at all. As a consequence, in $1\mathrm{D}$, fermions and bosons are completely equivalent as any collection of fermionic operators can be mapped onto bosonic operators (bosonization) or vice-versa (fermionization).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.