# Anyon Demystified

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Every quantum physicist knows that all particles are either bosons or fermions. And the standard textbook arguments that this is so do not depend on the number of dimensions.

On the other hand, you may have heard that in 2 dimensions particles can be anyons, which can have any statistics interpolating between bosons and fermions. And not only in theory, but even in reality. But how that can be compatible with the fact that all particles are either bosons or fermions? Where is the catch?

This, of course, is discussed in many papers (and a few books) devoted to anyons. But my intention is not to present a summary of the standard literature. I want to explain it in my own way which, I believe, demystifies anyons in a way that cannot be found explicitly in the existing literature. I will do it in a conceptual non-technical way with a minimal number of explicitly written equations. Nevertheless, the things I will say can be viewed as a reinterpretation of more elaborated equations that can easily be found in the standard literature. In this sense, my explanation is meant to be complementary to the existing literature.

Consider a 2-particle wave function ##\psi({\bf x}_1,{\bf x}_2)##. The claim that it is either bosonic or fermionic means that it is either symmetric or antisymmetric, i.e.
$$\psi({\bf x}_1,{\bf x}_2)=\pm \psi({\bf x}_2,{\bf x}_1) ….. (1)$$
But suppose that the wave function satisfies a Schrodinger equation with a potential ##V({\bf x}_1,{\bf x}_2)##, which has a property of being asymmetric
$$V({\bf x}_1,{\bf x}_2) \neq V({\bf x}_2,{\bf x}_1) .$$
In general, with an asymmetric potential, the solutions of the Schrodinger equation will not satisfy (1). And yet, no physical principle forbids such asymmetric potentials. It looks as if it is very easy to violate the principle that wave function must be either bosonic or fermionic.

But that is not really so. The principle that wave function must be symmetric or antisymmetric refers only to identical particles, i.e. particles that cannot be distinguished. On the other hand, if the potential between the particles is not symmetric, then the particles are not identical, i.e. they can be distinguished. In that case, (1) does not apply.

Now assume that the asymmetric potential takes a very special form, so that the wave function of two non-identical bosons or fermions takes the form
$$\psi({\bf x}_1,{\bf x}_2)=e^{i\alpha} \psi({\bf x}_2,{\bf x}_1)$$
where ##\alpha## is an arbitrary real number. This is the anyon. And there is nothing strange about it, it is simply a consequence of the special interaction between two non-identical particles. The effect of interaction is to simulate an exotic statistics (exotic exchange factor ##e^{i\alpha}##), while the “intrinsic” statistics of particles (i.e. statistics in the absence of exotic interaction) is either bosonic or fermionic.

The only non-trivial question is, does such interaction exists? It turns out (the details of which can be found in standard literature) that mathematically such an interaction exists, provided that the particles live in 2 dimensions and that the potential is not really a scalar potential ##V({\bf x}_1,{\bf x}_2)## but a vector potential ##{\bf A}({\bf x}_1,{\bf x}_2)##. And physically, that is in the real world, such interaction does not exist for elementary particles such as electrons, but only for certain quasi-particles in condensed matter physics. These are the main conceptual ideas of anyons, while the rest are technical details that can be found in standard literature.

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51 replies
1. stevendaryl says:

There is a nice theorem by Currie, Jordan & Sudarshan that says that (Hamiltonian) relativistic particle theories must be non-interacting, both classically and quantum mechanically.

Hmm. Well the classical version of that theorem would seem to be contradicted by Feynman's absorber theory, which I thought was (under certain assumptions) equivalent to the usual classical electrodynamics.

2. Demystifier says:

There is a nice theorem by Currie, Jordan & Sudarshan that says that (Hamiltonian) relativistic particle theories must be non-interacting, both classically and quantum mechanically.

If you replace particles by objects extended in one dimension, then you get strings. Remarkably, first quantization of free strings automatically produces interacting strings, including the string creation and destruction. That's one of the most beautiful properties of string theory which make some people believe that it could have something to do with the theory of "everything".

And related to the spin-statistics relation, first quantized strings in D dimensions can be viewed as a QFT in 1+1 dimensions, and this 1+1 QFT also obeys the standard spin-statistics relation.

3. rubi says:

Hmm. Well the classical version of that theorem would seem to be contradicted by Feynman's absorber theory, which I thought was (under certain assumptions) equivalent to the usual classical electrodynamics.

The theorem is about Hamiltonian theories only. There are interacting relativistic particle theories that can't be formulated in the Hamiltonian framework and the Wheeler-Feynman theory is one example. IIRC, the general form of these theories is discussed for example in the book "Special relativity in general frames" by Eric Gourgolhon. However, for a quantum theory, you need a Hamiltonian formulation, so the theorem still applies.

If you replace particles by objects extended in one dimension, then you get strings. Remarkably, first quantization of free strings automatically produces interacting strings, including the string creation and destruction. That's one of the most beautiful properties of string theory which make some people believe that it could have something to do with the theory of "everything".

Well, the theorem doesn't apply to strings, since they don't obey the particle algebra relations. Of course different objects like strings or fields can escape the assumptions of the theorem.

4. stevendaryl says:

The theorem is about Hamiltonian theories only. There are interacting relativistic particle theories that can't be formulated in the Hamiltonian framework and the Wheeler-Feynman theory is one example. IIRC, the general form of these theories is discussed for example in the book "Special relativity in general frames" by Eric Gourgolhon. However, for a quantum theory, you need a Hamiltonian formulation, so the theorem still applies.

Are you sure you need a hamiltonian? Certainly the usual canonical quantization requires a Hamiltonian, but I'm not sure that a path integral formulation necessarily does.

5. rubi says:

Are you sure you need a hamiltonian? Certainly the usual canonical quantization requires a Hamiltonian, but I'm not sure that a path integral formulation necessarily does.

The path integral formulation is just a different way to state an ordinary Hamiltonian quantum theory. If you have a well-defined path integral, you have reflection positivity, which allows you to reconstruct the Hilbert space. Invariance under time translations (which you need in a relativistic theory) then allows you to derive the Hamiltonian. (Similarly, you can also derive the other generators.)

6. A. Neumaier says:

Hmm. Well the classical version of that theorem would seem to be contradicted by Feynman's absorber theory, which I thought was (under certain assumptions) equivalent to the usual classical electrodynamics.

Feynman (before he worked on QED), created a classical relativistic but nonlocal many-particle theory, With the right (''absorbing'') boundary conditions you get more or less classical electrodynamics. But (like any classical multiparticle picture I have seen) its interpretation defies any rational sense of physics. In the present case, the dynamics of any particle depends on the past and future paths of all other particles!

Eric Gourgoulhon who discusses the theory in his book ''Special Relativity in General Frames'', mentions on p. 375 the disadvantage that it leads to integro-differential equations that have no well-defined Cauchy problem. This leads to interpretational difficulties.

7. A. Neumaier says:

That's not true. They only make use of the bracket relations, which are the same in the classical and the quantum formalism. They even emphasize this explicitely in their paper.

On p.363 of their paper in Reviews of Modern Physics 35.2 (1963), 350, they say explicitly:

The situation in quantum mechanics is much less clear because there a particle does not have a definite trajectory; that is, an exact value for its position at each time.

and then go on making plausibility arguments only (and say so almost explicitly on p.364 just before Section IV). The bulk of their paper is fully classical reasoning based on trajectories, and the summary on p.370 explicitly restricts the main conclusion to the classical case.

Indeed, in the quantum case there are counterexamples. See the topic ''Is there a multiparticle relativistic quantum mechanics?'' from Chapter B1 of my theoretical physics FAQ.

8. DrDu says:

If I understand you correctly, you are claiming that the quasi particles making up anyons are distinguishable?

9. DrDu says:

As I see from the Abstract, they rule out parastatistics, not anyons. These two exotic statistics should not be confused. Parastatistics is based on the symmetric group, anyons are based on the braid group.

This paper is one famous example of a paper where the abstract contains a claim (in this case a wrong one) which is not found in the paper itself.

10. Demystifier says:

If I understand you correctly, you are claiming that the quasi particles making up anyons are distinguishable?

Not exactly. In principle, if you could turn off the interaction that makes them behave as anyons, then they could be distinguishable. But in the case where anyons are actually realized, and the only such currently known case is the fractional Hall effect, I'm not sure that you can turn off the interaction without destroying the quasi-particles themselves.