# How are Anyons possible?

1. Apr 5, 2012

### lugita15

If |ψ> is the state of a system of two indistinguishable particles, then we have an exchange operator P which switches the states of the two particles. Since the two particles are indistinguishable, the physical state cannot change under the action of the exchange operator, so we must have P|ψ>=λ|ψ> where |λ|=1. Obviously switching the states of the two particles, and then switching them back, leaves the particles with their original states, so (P^2)|ψ>=(λ^2)|ψ>=|ψ>, so λ=±1, and thus the state of the system must be either symmetric or anti-symmetric with respect to exchange.

Now I've heard that this reasoning does not hold for two dimensions, leading to the possibility of "anyons", for which you can have λ be something other than 1 or -1. How in the world is that possible? Where is the flaw or oversight in the above reasoning, that makes it exclude the 2D case?

Any help would be greatly appreciated.

2. Apr 5, 2012

### Bill_K

See for example http://arxiv.org/abs/0707.1889v2, from which I quote:

"A process in which two particles are adiabatically interchanged twice is equivalent to a process in which one of the particles is adiabatically taken around the other. Since, in three dimensions, wrapping one particle all the way around another is topologically equivalent to a process in which none of the particles move at all, the wave function should be left unchanged by two such interchanges of particles... Two-dimensional systems are qualitatively different from three (and higher dimensions) in this respect. A particle loop that encircles another particle in two dimensions cannot be deformed to a point without cutting through the other particle... Then, when two particles are interchanged twice in a clockwise manner, their trajectory involves a non-trivial winding, and the system does not necessarily come back to the same state."

"Suppose that we have two identical particles in two dimensions. Then when one particle is exchanged in a counterclockwise manner with the other, the wavefunction can change by an arbitrary phase, ψ (r1, r2) → eψ (r1, r2). The phase need not be merely a ± sign because a second counter-clockwise exchange need not lead back to the initial state but can result in a non-trivial phase: ψ (r1, r2) → e2iθψ (r1, r2)."

3. Apr 5, 2012

### lugita15

But Bill K, what is the flaw in my reasoning above? Your quote says "The phase need not be merely a ± sign because a second counter-clockwise exchange need not lead back to the initial state", but I would like to know why that is.

4. Apr 5, 2012

### Bill_K

In two dimensions there are two distinct ways of "switching": clockwise switching and counterclockwise switching, and they may produce different phase factors. Only if you switch clockwise and then counterclockwise do you go back to the original wavefunction. Suggest you take a look at the paper I cited.