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This questions might have been asked several times, but I haven't seen a mathematical point of view, so here it is.

Based on Wigner classfication: A particle is a representation, because any theory that describes a particle in a space must teach us how to describe the change of state as we change coordinates, e.g. gradually rotating the resting frame. Therefore, a massive particle is at least a (projective) representation of $SO(3)$, and a massless particle is at least a (projective) representation of $SO(2)$. In this question, I focus on the later.

A projective representation of $SO(2)$ can be described in terms of a rational number $\frac{r}{s} \in \mathbb{Q}$, so it is natural to consider massless particles of $1/3, 1/4$ .. etc. My question is, why not?

A typical answer I got from my physics friends and profs is that

Yes, you can consider it, but they only exist in $2+1$ space-time. This is because in $3+1$ or above, exchanging two particles draws you a tangle in a $4$-space, which is trivial!

I understand you can un-tangle any tangles in $4$-spaces. What I fail to see is the relation between this reason and my question. I was never considering two particles! Why would everyone tell me the picture with 2 particles winding around with each other (even wikipedia:anyon does that)?

After all, what $1/3$ really means mathematically is: if you focus on that single particle, and slowly change coordinates with that particle fixed at the origin, you will find the the state got changed by a scalar multiplication by $\exp(2\pi i/3)$ after a full turn. This, to me, seems to work in any dimension. What's the fundamental difference for $2+1$, without invoking that un-tangling business? Or do I miss something?

Qmechanic
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Student
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    related (possible dup?) https://physics.stackexchange.com/q/221881/84967 – AccidentalFourierTransform Mar 12 '20 at 21:37
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    A projective rep of $SO(2)$ is described by any real number, not necessarily rational. This is because the universal cover of $SO(2)$ is $\mathbb R$. For $n>2$, the fundamental group of $SO(n)$ is $\mathbb Z_2$, which means there are two congruence classes of irreps: bosonic/fermionic (aka linear/projective, or tensor/spinor, etc.). For $n=2$, one has $\pi_1SO(2)=\mathbb Z$, thus an infinite number of congruence classes. – AccidentalFourierTransform Mar 12 '20 at 21:41
  • I see. I had an impression from my friend that even in dimension $3+1$, massless particles only have $SO(2)$-symmetries because there's no coordinate change for the particle to be at rest.. is this true? – Student Mar 13 '20 at 00:43
  • Massless particles (as explained in the linked post) are classified by representations of $SO(d-2)$. So, indeed, in $d=4$ this becomes $SO(2)$. – AccidentalFourierTransform Mar 13 '20 at 00:59
  • Yeah, so my question is still there.. in dimension $3+1$, can a massless particle have arbitrary real-valued spin? – Student Mar 13 '20 at 11:47
  • No? You want a rep of $SO(3)$, which is induced from a rep of $SO(2)$, but it is still a rep of $SO(3)$. Thus, a $4\pi$ rotation must be trivial, and so the spin (rather, helicity) must be integral or half-integral. It is true that (projective) reps of $SO(2)$ are labelled by $\mathbb R$, but those of $SO(3)$ (which you construct using a rep of $SO(2)$, but it is still a rep of $SO(3)$) are classified by $\mathbb Z_2$. – AccidentalFourierTransform Mar 13 '20 at 13:38
  • I'm getting close! Why do we need a representation of $SO(3)$? Why doesn't a representation of $SO(2)$ suffice? – Student Mar 13 '20 at 14:48
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    Sorry, I misspoke! You do not want a rep of SO(3). You want a rep of ISO(3,1), i.e., of Poincare! (Let's just step back for a minute, :: sips coffee :: what we want is a unitary rep of the symmetry group of your theory, to wit, the Poincare group in four dimensions). How do you construct a rep of ISO(3,1)? You induce it from a rep of the little group of some reference momentum. This little group is SO(2). So you take a rep of SO(2), and use it to build a rep of ISO(3,1). Now the universal cover of Poincare is a two-cover, so $4\pi$ has to be trivial, and the rest follows. Is this clear? – AccidentalFourierTransform Mar 13 '20 at 14:58
  • Almost clear! So the critical difference lies in the fact that the universal covering space of $ISO(2,1)$ is different than that of $ISO(d>2,1)$. Why would everyone tried to sell me the tangling picture? They seem two different stories to me: (a) one is about the topology of the symmetry group (b) another one is about the fact that in $3+1$-spaces you have rooms for tangles to untangle. – Student Mar 13 '20 at 16:33
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    In $d=3+1$ the massless little group is indeed SO(2), and its irreps are indeed classified by a continuous real number. So the mathematics does not forbid massless particles with continuous spin, but usually such representations are thrown out the window since we have not observed such particles in nature. At least that is always how I've seen the argument go (pretty sure Weinberg argues this way in QFT vol 1). In fact, there are many modern research groups who are looking closer into the continuous reps of massless particles (e.g. https://arxiv.org/abs/1805.09706 and refs therein). – NormalsNotFar Mar 15 '20 at 05:40
  • Or perhaps I misunderstood your question? – NormalsNotFar Mar 15 '20 at 05:40
  • @Normals I think you understand my question well. – Student Mar 15 '20 at 14:07
  • @NormalsNotFar I think you understand my question well. What you said seems to contradict to AccidentalFourierTransform's comments above. So how is spin defined? Should I build an $SO(3)$ representation from an $SO(2)$ representation? – Student Mar 15 '20 at 14:09
  • @AccidentalFourierTransform This discussion about the fundamental group of the rotation group ($\mathbb{Z}_2$ vs. $\mathbb{Z}$) deserves to be put into an answer rather than be buried in this comment thread. – MannyC Mar 15 '20 at 14:59
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    @Student The "tangling picture" is how you prove that $\pi_1SO(2)=\mathbb Z$ and $\pi_1 SO(n)=\mathbb Z_2$ for $n>2$, or at least the intuitive idea behind the formal proof. The "two ideas" are basically the same: the "room for tangles to untangle" is nothing but a fact about the topology of the rotation group! – AccidentalFourierTransform Mar 15 '20 at 15:00
  • @NormalsNotFar Not quite. The infinite spin reps of ISO(3,1) are induced from a rep of the little group of massless particles, to wit, ISO(2,1). These are not found in nature, and so we declare that the rep has to trivialise the normal subgroup of ISO(2,1), which lands us on SO(2). And this one has continuous spin as well, but we throw away those due to the topology of ISO(3,1), not due to lack of experimental confirmation. In short, there are two sources of continuous spin, one due to ISO(2,1) and the other due to SO(2). And only the former has to do with experiments. – AccidentalFourierTransform Mar 15 '20 at 15:03

1 Answers1

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The question and the comments discuss the mathematical models devised to describe particles. This is the experimental physics answer:

Physics theoretical models are modeling measurements and predicting new situations.

The answer to why are particles either bosons or fermions is that that is what we have observed in our measurements of particle interactions.

In order for angular momentum conservation to hold as a true law at the level of particle physics ( laws are physics axioms) when interacting and decaying, an angular momentum as specific as mass and charge had to be assigned to each particle. This has resulted in the symmetries seen in the quark model and the ability of the models to be predictive. They are validated continuously with any new experiments, up to now.

Thus the mathematical models you are discussing are necessary in order to fit data and observations and be predictive of new states.

If future experiments discover new particles where the assignment of spin in order to obey angular momentum conservation in its interactions, need a different spin , a new quantum mechanical wave equation has to be found other than Dirac,Klein Gordon and quantized Maxwell to model its wavefunctions and keep angular momentum conservation as a law.

anna v
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  • Good answer! Indeed, I should mention which theoretical framework I was considering. Otherwise, the ultimate answers in physics are always experimental. – Student Mar 15 '20 at 14:03
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    The question is clearly about any group theoretical reasons that particles are either bosons or fermions in d>2 (if the OP was asking about the QFT describing our universe the OP would not ask about d>2). The rest of the question hints at the same. So most of your answer about the specific QFT that might describe our universe is irrelevant. Are you just saying: There are no group theoretical reasons and that other representations (anyons) are allowed (but happen to be unobserved). That seems to just be outright wrong (see https://en.wikipedia.org/wiki/Anyon). – Marten Jul 22 '21 at 08:10
  • @Kvothe "This is the experimental physics answer:" Did you notice this ? – anna v Jul 22 '21 at 08:17
  • That is like saying this is the dermatological answer. The answer is completely irrelevant to this group theory question! (The only way an experimental answer could be useful was if it gave an example of the existence of a representation that was in question. On the other hand the absence of a field transforming in a certain representation in experiments says absolutely nothing about whether such an irrep could exist.) – Marten Jul 22 '21 at 08:21
  • And an important part of the predictiveness of QFT comes exactly from no-go theorems that will tell you that certain representations simply cannot occur in relativistic d=4 theories. There is an important difference between situations where a field/particle could exist but is simply not observed (yet) and situations where a field/particle is forbidden by the axioms of relativistic quantum field theory. You are misrepresenting which situation applies here. – Marten Jul 22 '21 at 08:27
  • @Kvothe I am stating clearly that the choice of the model QFT has to have representations that describe the data,. Particles means data. If there is no data to be fitted it is a mathematics, not a physics question. I just want to ground people to the notion that physics means observations and data , and the mathematics with extra axioms is used to model nature. Why don't you write an answer? – anna v Jul 22 '21 at 12:31
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    Well this is clearly what you would call a "mathematics" question then. It does not ask what representations have been seen in our universe. It asks for d>2 (and so definitely more general than our universe) whether certain representations can exist. The answer is: No they cannot for group theory reasons. And yet you answer: Yes it is possible but it hasn't been detected. – Marten Jul 22 '21 at 12:36
  • If he had asked whether we could have say operators transforming under a global SU(5) such an answer would have been reasonable since mathematically it would be allowed, it just hasn't been observed. For the case at hand such an answer is just wrong. – Marten Jul 22 '21 at 12:36
  • So it is not a choice not to have anyons representations (as least not when one has already chosen to use relativistic QFT). It is dictated by the mathematics. Compare how the physically measurable constant of $\pi$ is fixed by mathematics while the gravitational constant $G$ is not. It is important to distinguish these situations. Your answer implies this question is in the category of $G$ and not $\pi$. I believe the opposite is true. – Marten Jul 22 '21 at 12:40
  • @Kvothe I know nothing and said nothing about anyons and katyons in fanciful mathematical models. I know that the standard model QFT is continually validated and that experiments are trying hard to falsify certain regions in order to discover new physics. New physics may require new modeling is all I am saying. – anna v Jul 22 '21 at 12:51
  • And I am saying that in this case there is a better approach then throwing our hands up in the air and saying new data might mean we should consider new representations. There is a lot more predictiveness in QFT. After all we don't just want to make models that tell us about our history. We want to make future predictions. One of those predictions already answers this question (and also that was what was asked for). So there is no need to just shrug our shoulders and say, well we will see. We can just answer the question here and now. (I'll do it when I have time to do a good job at it.) – Marten Jul 22 '21 at 12:54