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This question is a follow-up to David Bar Moshe's answer to my earlier question on the Aharonov-Bohm effect and flux-quantization. What I forgot was that it is not the wavefunction that must be single-valued, but rather, the probability density (wavefunction-squared).

But if that is so, how do I justify the quantization condition $m=\{\ldots,-1,0,1\ldots\}$ based on the boundary condition on the azimuthal part of spherically symmetric wavefunctions $\psi(x,y,z)\sim R(r)\Theta(\theta)\Phi(\phi)$:

$$\Phi(\phi)\sim e^{im\phi}$$

whose square (probability density) is unity, independent of $m$?

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Hopefully David Bar Moshe can give a more rigorous explanation in terms of cohomology, but I have the following intuitive understanding of the difference between the two situations.

In the Aharonov-Bohm effect, the particle is constrained to move around an (effectively) infinite solenoid. It then suffices to consider a problem on a plane, but with a hole in the middle that is threaded by a magnetic flux (the solenoid). A path on which the charge loops $n$ times around the solenoid cannot be continuously deformed into a path with $m\neq n$ loops. In topological language, we say that the two paths are not homotopic to each other, because the winding numbers $n$ and $m$ are different. In physical terms, paths with different winding numbers correspond to quite distinct physical situations, so it is reasonable that the two paths could have different phases even if the actual lengths of the paths are the same.

In Dirac's monopole, the singularity is now just a point in 3D space rather than a cylinder. The configuration space of the problem is essentially the surface of a sphere, with the monopole at the centre. Any path that takes the particle on a closed path around the monopole is homotopic to the trivial path, in which the particle just stays where it is. Put another way, it is not possible to reasonably define which paths constitute 'going around' the monopole and which do not, since any path which goes exactly around the equator of the sphere is just an $\epsilon$ away from a path that goes 'above' or 'below' the monopole.

If a non-trivial phase was picked up on spanning a loop around the monopole, you could envisage the following confusing situation. Assume that the charge sits very close to the monopole. Paths which are infinitesimally close to the trivial path now pick up a large phase relative to the amplitude for just staying still. Put in terms of the path integral formulation of quantum mechanics, where $\mathrm{phase} \sim \mathrm{action}$, this would basically imply a discontinuity in the action functional. See Zee's book Quantum Field Theory in a Nutshell for a nice explanation of this in path integral language (I don't have the book to hand but will post an exact chapter reference when I can).

Incidentally, the same topological reasoning explains why anyons can exist in two dimensions but not in three or more. A path where two identical particles exchange positions twice is topologically equivalent to a path where one spans a loop around the other. Since in three dimensions, this path is homotopically trivial, this restricts the phase acquired on a double exchange to be $+1$. Therefore the phase on a single exchange can only be $\pm 1$, leaving fermionic and bosonic statistics as the only possibilities. In two dimensions, however, double exchanges are homotopically nontrivial (so long as the particles cannot pass right through each other!), meaning that in principle any phase can be acquired on exchange of identical particles (hence the name anyons).

See the beautiful paper by Leinaas and Myrheim for more on this geometrical viewpoint on particle statistics.

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Your extra info on the existence of 'anyons' is very much welcomed.! Thanks! –  QuantumDot Oct 28 '12 at 23:05
    
+1 on the anyons. I never really got that. –  Emilio Pisanty Nov 28 '12 at 1:31
    
@Emilio thanks. if you like that, you should check out the last chapter of Preskill's notes as well for a really nice topological justification for the spin/statistics connection in 2D that can be checked for yourself easily with, say, a belt. –  Mark Mitchison Nov 28 '12 at 23:04

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