A common way to show that anyons exhibit fractional statistics in 2D is by arguing that the paths of two anyons winding round each other cannot be continuously deformed to zero. This seems to assume that the particles cannot pass through each other. Why is this assumption valid?

  • $\begingroup$ The configurations in which the particles overlap is singular (e.g. because the interaction energies etc. blow up) so a priori, one isn't allowed to assume that things are smooth around this point. Indeed, for the existence of nontrivial anyons, the point is singular. The burden of proof is on the opposite side than you suggest. If you wanted to prove that there are no anyons because the paths may be deformed - through the coincident positions - you would have to show that it's OK to get through this singular point. In trying to prove so, you would fail because it's not legitimate. $\endgroup$ – Luboš Motl Apr 28 '12 at 9:22
  • $\begingroup$ So, does that mean that particles with no mutual interaction cannot exhibit fractional statistics? $\endgroup$ – leongz Apr 28 '12 at 10:04
  • 1
    $\begingroup$ Statistics has nothing to do with interactions, just with dimensions. What Lubos is saying [please correct me if I am mistaken] is that if you try to make a path contractible through the point where the particle at time $t$ is, you have to show that contracting through that singular point is a an allowed operation, but it is not. $\endgroup$ – DaniH Apr 28 '12 at 10:36
  • $\begingroup$ Why is contracting through that singular point not allowed, other than the fact that the interaction energy blows up? $\endgroup$ – leongz Apr 29 '12 at 7:39
  • $\begingroup$ It is not just that the interaction energy blows up. The point where the particle is encodes a [phase] singularity and the configuration space is no longer smooth. $\endgroup$ – DaniH Apr 29 '12 at 8:34

We do not need to make the assumption that "the paths of two anyons winding round each other cannot be continuously deformed to zero".

To define fractional statistics, we only require that the phase of exchanging two particles do not depend on the smooth deformation of the exchange path, as long as two particles are always well separated during the exchange.

Two anyons can coincide with a finite energy cost.

| cite | improve this answer | |
  • $\begingroup$ Would you please elaborate a bit more about your last sentence "Two anyons can coincide with a finite energy cost." If I understand correctly, it would cost infinite energy to put two fermions to coincide (a poor-man Pauli principle, say), zero energy for bosons, and some energy for anyons. Is it dependent on the model, or only on the anionic statistics for instance ? Thanks in advance. $\endgroup$ – FraSchelle May 23 '13 at 21:29

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.