Learn The Main Conceptual Ideas of Anyon Particles
Every quantum physicist knows that all particles are either bosons or fermions. And the standard textbook argues that this so does 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. However, my intention is not to present a summary of the standard literature. I want to explain it in my 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 the 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 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 exotic statistics (exotic exchange factor ##e^{i\alpha}##), while the “intrinsic” statistics of particles (i.e. statistics in the absence of exotic interaction) are either bosonic or fermionic.
The only non-trivial question is, does such interaction exist? 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 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 the standard literature.
Theoretical physicist from Croatia
Without going into much detail, it seems logical that an anyon would be a 2-d vector class, simply due to its similarities with dimensionality equations: the lower the dimension, the larger number of applicable equations, and the more parallels exist.
[QUOTE="Silviu, post: 5613347, member: 588158"]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.
[QUOTE="Silviu, post: 5613347, member: 588158"]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.
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!
Really no such thing as particles
[QUOTE="DrDu, post: 5610049, member: 210532"]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.
[QUOTE="Demystifier, post: 5605666, member: 61953"]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.
If I understand you correctly, you are claiming that the quasi particles making up anyons are distinguishable?
[QUOTE="rubi, post: 5608296, member: 395236"]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:[QUOTE="Currie, Jordan & Sudarshan"]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.
[QUOTE="stevendaryl, post: 5608309, member: 372855"]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.
[QUOTE="stevendaryl, post: 5608333, member: 372855"]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.)
[QUOTE="rubi, post: 5608322, member: 395236"]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.
[QUOTE="stevendaryl, post: 5608309, member: 372855"]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.[QUOTE="Demystifier, post: 5608312, member: 61953"]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".[/quote]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.
[QUOTE="rubi, post: 5608284, member: 395236"]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.
[QUOTE="rubi, post: 5608284, member: 395236"]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.
[QUOTE="A. Neumaier, post: 5608287, member: 293806"]but not in 1+1.”Why not?
[QUOTE="A. Neumaier, post: 5608291, member: 293806"]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.
[QUOTE="rubi, post: 5608284, member: 395236"]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.
[QUOTE="Demystifier, post: 5608131, member: 61953"]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.
[QUOTE="Demystifier, post: 5608220, member: 61953"]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.[QUOTE="stevendaryl, post: 5608271, member: 372855"]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.
[QUOTE="stevendaryl, post: 5608271, member: 372855"]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.
[QUOTE="Demystifier, post: 5608212, member: 61953"]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.
[QUOTE="rubi, post: 5608202, member: 395236"]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.
[QUOTE="stevendaryl, post: 5608170, member: 372855"]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.
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.
[QUOTE="stevendaryl, post: 5608156, member: 372855"]Or are you saying that Baez is just wrong about the loophole in 2+1 dimensions?” 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.
[QUOTE="stevendaryl, post: 5608155, member: 372855"]I'm still trying to reconcile your claim that the theorem is valid in 2+1 dimensions with John Baez' claim that it is only valid in 4 (and higher) dimensions. Are you saying that there is a general theorem that doesn't assume Lorentz invariance, which is valid in 4 or more dimensions, and then a specialization that does assume Lorentz invariance, which is valid in any number of dimensions?”Or are you saying that Baez is just wrong about the loophole in 2+1 dimensions?
[QUOTE="Demystifier, post: 5608133, member: 61953"]See my post above. So the spin-statistics theorem, with all the explicit assumptions of the theorem, is valid even in 2+1. In 2+1, anyons are OK as non-relativistic (first-quantized) QM, but not as relativistic local (second-quantized) QFT.”I'm still trying to reconcile your claim that the theorem is valid in 2+1 dimensions with John Baez' claim that it is only valid in 4 (and higher) dimensions. Are you saying that there is a general theorem that doesn't assume Lorentz invariance, which is valid in 4 or more dimensions, and then a specialization that does assume Lorentz invariance, which is valid in any number of dimensions?
[QUOTE="stevendaryl, post: 5605704, member: 372855"]Hmm. John Baez in an old article claims that the spin-statistics theorem only applies for 4 or more spacetime dimensions:http://math.ucr.edu/home/baez/braids/node2.html"Now for the catch: the spin-statistics theorem only holds for spacetimes of dimension 4 and up."”See my post above. So the spin-statistics theorem, with all the explicit assumptions of the theorem, is valid even in 2+1.
[QUOTE="A. Neumaier, post: 5605698, member: 293806"]But the Lorentz group is based on 4-dimensional Minkowski space if nothing is said.”Yes, but you can repeat their proof step by step in any number of dimensions, including 2+1. Indeed, as I already said, there is no consistent local Lorentz invariant QFT in 2+1 dimensions with intrinsic anyon statistics.
[QUOTE="A. Neumaier, post: 5606344, member: 293806"]Is this article about this year's Nobel price in physics (which was given for theoretical achievements in lower-dimensional physics) more understandable?”In fact, it was this year's Nobel prize that motivated me to read more about anyons. :smile:
Scan read it on phone, definitely be printing it off for a proper read. Thanks.
[QUOTE="houlahound, post: 5606340, member: 551046"]OK that caliber of paper is above my skill level.”Is this article about this year's Nobel price in physics (which was given for theoretical achievements in lower-dimensional physics) more understandable?
OK that caliber of paper is above my skill level.
[QUOTE="houlahound, post: 5606328, member: 551046"]What's the basis for you saying this?”The literature on the subject. Read some of the links given here: http://www.physicsoverflow.org/32114/[QUOTE="houlahound, post: 5606328, member: 551046"]Well then I'm free to make up any particle thats true in some dimension.”You are free to make up anything. The question is whether Nature will be described by your make-up.People had studied low-dimensional quantum physics and their peculiarities for theoretical reasons, long before it was found that there are many interesting systems in Nature described by them.
Well then I'm free to make up any particle thats true in some dimension. What's the basis for you saying this?
It depends on the detailed level of modeling. A point particle has zero dimensions, indeed. For quantum chemistry, nuclei are treated as point particles. If one models a wire in full detail, it becomes 3-dimensional. But in mechanics one usually treats it as a 1-dimensional object. The same holds for much of the physics of nanowires. Fact is that these materials behave like predicted by lower-dimensional quantum field theory.
Then what's an atom, zero dimensions?
Then what's an atom, zero dimensions?Obviously you wouldn't say that.
[QUOTE="houlahound, post: 5606154, member: 551046"]How is this possible when reality is clearly not 2 dimensional?”Because surfaces or thin films can be modeled in two space dimensions, and thin wires in one.
[QUOTE="Demystifier, post: 5605610, member: 61953"]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.”How is this possible when reality is clearly not 2 dimensional?
[QUOTE="Demystifier, post: 5605649, member: 61953"]Well, if you take e.g. the book by Streater and Wightman "PCT Spin Statistics and All That", which is a standard book with a rigorous derivation of spin-statistics theorem, they say nothing about the number of dimensions in the proof of the theorem.”Hmm. John Baez in an old article claims that the spin-statistics theorem only applies for 4 or more spacetime dimensions:http://math.ucr.edu/home/baez/braids/node2.html"Now for the catch: the spin-statistics theorem only holds for spacetimes of dimension 4 and up."
[QUOTE="vanhees71, post: 5605669, member: 260864"]Okay, then I need some education on this. Any nice review(s)?”http://www.physicsoverflow.org/32114/http://www.physicsoverflow.org/14488/http://www.physicsoverflow.org/16826/
[QUOTE="vanhees71, post: 5605669, member: 260864"]Okay, then I need some education on this. Any nice review(s)?”http://www.physicsoverflow.org/32114/http://www.physicsoverflow.org/14488/http://www.physicsoverflow.org/16826/
[QUOTE="Demystifier, post: 5605649, member: 61953"]they say nothing about the number of dimensions in the proof of the theorem. They assume Lorentz invariance,”But the Lorentz group is based on 4-dimensional Minkowski space if nothing is said.
[QUOTE="vanhees71, post: 5605669, member: 260864"]Okay, then I need some education on this. Any nice review(s)?”If the book above is not available to you (or is simply too big), try alsohttps://arxiv.org/abs/hep-th/9209066
[QUOTE="vanhees71, post: 5605662, member: 260864"]This I don't understand. Perhaps it's worth to write an Insight with the sufficient amount of math!”I don't think that I can explain it better that the standard literature, so I would rather refer you to some standard literature. For instance, the explanation in the bookhttps://www.amazon.com/Fractional-Statistics-Quantum-Theory-2nd/dp/9812561609is quite good.
Okay, then I need some education on this. Any nice review(s)?
[QUOTE="vanhees71, post: 5605661, member: 260864"]What about the famous paper by Laidlaw and C. de Witt:M. G. G. Laidlaw and C. M. DeWitt, Feynman Functional Integrals for Systems of Indistinguishable Particles, Phys. Rev. D, 3 (1970), p. 1375.http://link.aps.org/abstract/PRD/v3/i6/p1375“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.
[QUOTE="Demystifier, post: 5605656, member: 61953"]The anyon statistics is not based on the symmetric group, but on the braid group. It is an infinite group which has a finite symmetric group as a subgroup, even for 2 particles. In other words, anyons can be discussed even for 2 particles.”This I don't understand. Perhaps it's worth to write an Insight with the sufficient amount of math!
[QUOTE="Demystifier, post: 5605649, member: 61953"]Well, if you take e.g. the book by Streater and Wightman "PCT Spin Statistics and All That", which is a standard book with a rigorous derivation of spin-statistics theorem, they say nothing about the number of dimensions in the proof of the theorem. They assume Lorentz invariance, while quantum field theories with anyon statistics do not obey Lorentz invariance. (Attempts to construct 2+1 dimensional Lorentz invariant QFT's with intrinsic anyon statistics lead to problems.)If you mean arguments based on non-relativistic QM (not QFT), then, in most general QM textbooks, the principle that only two statistics are possible is justified heuristically (not derived rigorously) by arguments which do not depend on number of dimensions. Of course, these general QM textbooks don't mention anyons.”What about the famous paper by Laidlaw and C. de Witt:M. G. G. Laidlaw and C. M. DeWitt, Feynman Functional Integrals for Systems of Indistinguishable Particles, Phys. Rev. D, 3 (1970), p. 1375.http://link.aps.org/abstract/PRD/v3/i6/p1375
[QUOTE="vanhees71, post: 5605617, member: 260864"]Also with two indistinguishable particles, you have only the bosonic and fermionic representation of the symmetric group of two elements. You need at least 3 indistinguishable particles to discuss anyons.”The anyon statistics is not based on the symmetric group, but on the braid group. It is an infinite group which has a finite symmetric group as a subgroup, even for 2 particles. In other words, anyons can be discussed even for 2 particles.
[QUOTE="vanhees71, post: 5605617, member: 260864"]Hm, the only argument, why there are only bosons and fermions and no anyons I know about goes with the number of space dimensions”Well, if you take e.g. the book by Streater and Wightman "PCT Spin Statistics and All That", which is a standard book with a rigorous derivation of spin-statistics theorem, they say nothing about the number of dimensions. They assume Lorentz invariance, while quantum field theories with anyon statistics do not obey Lorentz invariance. (Attempts to construct 2+1 dimensional Lorentz invariant QFT's with intrinsic anyon statistics lead to problems.) If you mean arguments based on non-relativistic QM (not QFT), then, in most general QM textbooks, the principle that only two statistics are possible is justified heuristically (not derived rigorously) by arguments which do not depend on number of dimensions. Of course, these general QM textbooks don't mention anyons.
Congrats on your first Insight!
Hm, the only argument, why there are only bosons and fermions and no anyons I know about goes with the number of space dimensions, and indeed the anyons (i.e., anyonic quasiparticles in various condensed-matter contexts) live in 2 dimensions.Also, indeed, if there is a non-symmetric interaction potential for two particles, then the two particles are in fact distinguishable, and you have no restriction concerning the symmetry whatsoever of the wave functions/quantum states under exchange of the particles. Also with two indistinguishable particles, you have only the bosonic and fermionic representation of the symmetric group of two elements. You need at least 3 indistinguishable particles to discuss anyons. So even on a semi-popular level there is a bit to demystify in your demystification :-).