Insights What is the explanation for the existence of anyons in 2 dimensions?

[Demystifier]

The argument I heard, which I didn't really understand, is that if you consider paths in configuration space, the path where two identical particles exchange places is continuously deformable to a path were one of the particles rotates through 360 degrees and nothing else happens. So somehow there must be a relationship between particle exchange and rotations. But the deformation requires at least 3 dimensions, so there is a 2-D loophole. Something like that.

There are two ways to define quantum statistics. One is in terms of particle exchange. The other is in terms of algebra of field operators. The former is a definition for QM, where the fundamental degrees are particles. The later is a definition for QFT, where the fundamental degrees are fields. The two definitions are closely related, but not fully equivalent.

The argument above is a valid argument in QM. But it is not a valid argument in QFT. QM and QFT are different theories. Even if they are not so different in the case of bosons and fermions, they are very different in the case of anyons. The standard spin-statistics theorem is a theorem about relativistic QFT, not a theorem about QM. We have a consistent QM theory of anyons in 2+1, but not a consistent local relativistic QFT of anyons in 2+1. The standard spin-statistics theorem and the anyon theory cannot be directly compared because they talk about different objects; one is talking about fields and the other about particles.

It seems that Baez (and apparently many others) failed to realize that two different ways of defining statistics cannot be directly compared. The fact that anyons as particles can live in 2+1 QM does not contradict the other fact that anyons as fields can not live in 2+1 local relativistic QFT.

 

[Demystifier]

The proof of spin-statistics in Streater & Wightman depends heavily on 4 dimensions through the theory of complexified Lorentz transformations and the analytic continuation of the Wightman functions to the extended tube. It's just that Streater & Wightman prove these facts in earlier chapters and the proof of spin-statistics just silently uses them.

You may be right about that, but the original Pauli version of spin-statistics theorem does not depend on it.

For various derivations of spin-statistic relation see the book
https://www.amazon.com/dp/9810231148/?tag=pfamazon01-20

 

[stevendaryl]

There are two ways to define quantum statistics. One is in terms of particle exchange. The other is in terms of algebra of field operators. The former is a definition for QM, where the fundamental degrees are particles. The later is a definition for QFT, where the fundamental degrees are fields. The two definitions are closely related, but not fully equivalent.

This is a topic for another thread, but I am intrigued by the possibility of a pure-particle formulation of relativistic quantum mechanics that is equivalent to QFT. It would at first seem impossible, because of particle creation, but there are at least two hand-wavy approaches to getting around that: (1) the Dirac sea idea, and (2) viewing particle creation in terms of a particle that can travel back and forth in time. I don't know whether there has been any serious effort to make either of these work rigorously.

 

[Demystifier]

This is a topic for another thread, but I am intrigued by the possibility of a pure-particle formulation of relativistic quantum mechanics that is equivalent to QFT. It would at first seem impossible, because of particle creation, but there are at least two hand-wavy approaches to getting around that: (1) the Dirac sea idea, and (2) viewing particle creation in terms of a particle that can travel back and forth in time. I don't know whether there has been any serious effort to make either of these work rigorously.

I guess I could say something about that too, but not in this thread.

 

[rubi]

You may be right about that, but the original Pauli version of spin-statistics theorem does not depend on it.

Well, Pauli's proof also requires 4 dimensions, since it also makes use of the $SU(2)$ angular momentum theory.

This is a topic for another thread, but I am intrigued by the possibility of a pure-particle formulation of relativistic quantum mechanics that is equivalent to QFT. It would at first seem impossible, because of particle creation, but there are at least two hand-wavy approaches to getting around that: (1) the Dirac sea idea, and (2) viewing particle creation in terms of a particle that can travel back and forth in time. I don't know whether there has been any serious effort to make either of these work rigorously.

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.

 

[A. Neumaier]

but you can repeat their proof step by step in any number of dimensions, including 2+1.

but not in 1+1. And it applies only in the relativistic case. The observable anyons are nonrelativistic.

 

[A. Neumaier]

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.

C, J & S only handle the classical case.

 

[rubi]

C, J & S only handle the classical case.

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.

 

[Demystifier]

but not in 1+1.

Why not?

 

[stevendaryl]

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.

 

[Demystifier]

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.

 

[rubi]

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.

 

[stevendaryl]

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.

 

[rubi]

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.)

 

[A. Neumaier]

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.

 

[A. Neumaier]

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.

 

[DrDu]

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

 

[DrDu]

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.

 

[Demystifier]

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.

 

[Silviu]

Hi! I am new to this topic and a bit confused. In the normal boson and fermions statistics, as you said, we have identical particles. So, depending whether the wave function is symmetric or not, we can argue that that particle is a boson or fermion. But in the example you gave, if instead of $\pm 1$ we have another factor, we have an anyon. But you also said that this is the case only if the 2 particles are not identical, so we can have an asymmetric potential. So what exactly is called an anyon, as we have 2 different particles, that individually are either bosons or fermions? Is the wave function itself that is called anyon? Thank you!

 

[A. Neumaier]

we have 2 different particles, that individually are either bosons or fermions?

No. A single particle alone in the world is neither boson nor bermion nor anyon. The latter are properties of the corresponding many-body system.

 

[Demystifier]

Hi! I am new to this topic and a bit confused. In the normal boson and fermions statistics, as you said, we have identical particles. So, depending whether the wave function is symmetric or not, we can argue that that particle is a boson or fermion. But in the example you gave, if instead of $\pm 1$ we have another factor, we have an anyon. But you also said that this is the case only if the 2 particles are not identical, so we can have an asymmetric potential. So what exactly is called an anyon, as we have 2 different particles, that individually are either bosons or fermions? Is the wave function itself that is called anyon? Thank you!

The two particles, which in the absence of this specific interaction would behave as bosons or fermions, behave as two anyons due to interaction. Since they behave as two anyons, it is common to say that they "are" two anyons.