# Is there a wave function for anyons?

People talk about anyons a lot.

But i have never seen an anyon wave function.

I suspect that there is no such thing as a wave function for anyons. I mean, anyons are not generalizations of bosons or fermions. For bosons and fermions, one can have many-body hilbert spaces, but for anyons, there is no such thing.

I don't think that is the case. A useful reference is: http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.80.1083.

One way to approach a theory of anyons is to start by writing down a list of particle types along with their fusion rules. Once doing this one may obtain consistency equations from solving the hexagon and pentagon equations arising from modular tensor categories. If successful in solving these equations, you have a viable anyon theory. If multiple solutions exist you have multiple theories.

Now once doing this, we can label each state in the hilbert with a fusion tree. Hence our hilbert space is very well defined - albeit abstract. We can provide interactions on this hilbert space by creating projectors which favor nearby (in real space) anyons to fuse together in various channels. Whence doing this, in principle we could diagonalize a finite sized system and extract the wavefunctions.

A simple example is to consider Ising anyons. These appear in the Kitaev honeycomb model and manifest them selves as Majorana zero modes in the 1 dimensional p-wave wires. (see for example http://www.sciencedirect.com/science/article/pii/S0003491605002381, and). In the case of 1D p-wave wires we can certainly write down the wavefunctions of the Majorana zero modes as they happen to be solutions of the BdG equations.

• I do not understand what you mean. But, anyway, there is no such wave function in the usual sense for anyons. My point is that, many popular articles on anyons are very misleading! – Jiang-min Zhang Apr 22 '14 at 11:08
• @PKSer I could be more precise if you like - I was really just trying to point out some useful references. Could you be more specific on what you you mean by no wave function? We certainly have a Hilbert space, and can define Hamiltonian's over that Hilbert space. I do agree that popular articles can be misleading. – Stackexchange_user23 Apr 22 '14 at 22:20
• @ Stackexchange_user23. Consider two anyons. Suppose there is a wave function $f(x1, x2)$ for them. Then, by the popular articles, I personally have the impression that $f(x2, x1) = a f(x1, x2)$, where a is a phase |a|=1. Now exchanging them again, we have $f(x1, x2) = a^2 f(x1, x2)$. Therefore, $a= \pm 1$. That is, we can only have fermions or bosons. We cannot have anyons. The precise meaning of anyons is still unknown to me, but the popular articles are bullshit! – Jiang-min Zhang May 4 '14 at 18:42
• @Jiang-minZhang : Your definition of exchange is incorrect. – Name YYY May 13 '17 at 12:18

You should try to check that the $n$ particles lagrangian $$L = L_{0} - \alpha \sum_{i =1, j > i}^{n}\frac{d}{dt}\theta(\mathbf r_{i} - \mathbf r_{j}),$$ where $L_{0}$ is "standard" many particles lagrangian and $\theta (\mathbf r_{i} - \mathbf r_{j})$ is azimuthal angle between $i$th and $j$th particles position vectors, describes the statistics of anyons (the amplitude given of particles intercharging acquires additional phase proportional to $\alpha$).

Next, try to compute the Hamiltonian and ensure that anyons can be described by adding fictious non-propagating gauge field with some constrain.

Finally, the only thing which is remained to do is to solve corresponding Schroedinger equation...

P.S. About your comment ("...That is, we can only have fermions or bosons. We cannot have anyons..."): you don't take into account that the first homotopy group for 2-dimensional space is not the permutation group $Z_{2}$, as for spaces with dimension $> 3$. Therefore a loop encircled around the point in configuration space of particles living in 2D can't be shrinked to a point (unlike the 3D case). It's very simple to understand, if You imagine the closed loop with a point inside it in 2D and in 3D. This means, that the anyon's wave function need not be single valued under particles double interchange, and hence $\alpha$ from your comment need not be modulus 1.