Paraparticles and Parastatistics

[yossell]

Empirically, all known fundamental particles are either bosons or fermions. However, I understand that it is theoretically possible that there be other systems of indistinguishable particles -- and that these are known as paraparticles. Classical QM represents bosons by symmetric kets: kets that are unchanged under a permutation; and bosons by anti-symmetric kets: kets that are either mapped onto themselves or onto their negative by a permutation, depending on the parity of the permutation.

Does anybody roughly know the behaviour of kets for paraparticles under a permutation? For instance, for such a ket, would a permutation map a ket k to e^{i \theta} k? Or would permutations still always map a ket onto itself or its negative -- but which somehow depends on something more subtle than the parity of the permutation.

I've looked at http://en.wikipedia.org/wiki/Parastatistics, but it was a bit above me.

Thanks.

 

[Bill_K]

I believe it's true that a particle obeying parastatistics is equivalent to a normal particle with an unobserved internal quantum number such as color. See the Wikipedia remark:

QCD can be reformulated using parastatistics with the quarks being parafermions of order 3

 

[yossell]

Thanks for the reply Bill_K, I appreciate your help. I'm afraid that I don't know anything about QCD and that the wiki remark is above me. Even my Quantum Field Theory isn't too strong.

As far as I can tell -- though if I'm wrong on this it would be helpful to know -- the behaviour of paraparticles and parastatistics can be discussed within the framework of classical, non-relativistic Quantum Physics. This, I take it, would mean a weakening of the postulate that pure states representing indistinguishable particles are all totally anti-symmetric or totally symmetric.

Now, I note that one might continue to impose the requirement that all admissible states are such that the permutation of any two particles always results in the same state or the negative of that state *without* imposing that such states be totally anti-symmetric or totally symmetric. That is, we could impose this constraint by allowing wavefunctions \psi(x y z w) such that a swap of the first two places leaves the function unchanged, while a swap of the second two places results in the negative of the original wavefunction. Is this essentially how the representation of paraparticles behaves in QM? Is this an oversimplification? Or am I missing the point.

Again, thanks in advance for any help with this question.

 

[Bill_K]

It doesn't have anything to do with QCD or QFT. Isn't it just a matter of counting how many particles you can put in the same state?

For a parafermion of order 3, you can put 3 of them in the same state. But in the case of quarks the "explanation" is that they are really normal fermions but with a hidden property - color - which can be R, G or B. When it appears you're putting three identical particles in the same state, it's because they have three different colors, they just look identical.

 

[DrDu]

Bill_K is right. E.g. in the non-relativistic limit spin is not observable (considering light as a relativistic phenomenon) as magnetic fields are small by a factor v/c which vanishes for v to 0. Then electrons behave as parafermions which are sometimes called "freeons" in non-relativistic quantum chemistry.

 

[yossell]

Bill_K is right.

And I've never doubted it. I just don't see how it answers my question about the behaviour under permutations of kets representing paraparticles in classical quantum mechanics.

 

[Bill_K]

I believe the answer to your question is... neither. For normal bosons and fermions, if you interchange two of them you always get back +/- the same wavefunction. This is not the case for paraparticles.

Beyond that, I have to defer to a survey paper on the subject, which treats it from several angles.