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I am in undergraduate quantum mechanics, and the TA made an off-hand comment that currently no one knows how to describe fundamental particles with spin > 1 without supersymmetry. I was curious and tried to look up info on this, and wikipedia does make some comments about troubles with spin 3/2 http://en.wikipedia.org/wiki/Rarita%E2%80%93Schwinger_equation

So my questions are

  1. Is there an easy to understand reason why photons aren't a problem, but a hypothetical particle with spin 3/2 doesn't work?
  2. Does this also mean there are troubles explaining 3/2 composite particles in a low energy regime where we can treat the composite as strongly bound / 'fundamental'?
  3. How does supersymmetry help here?
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In his Lectures on Gravitation (available as a book), Feynman explains why the graviton field must be integer (0, 1, 2, 3,...), then explains why 0 and 1 are out of the question. He then proceeds to attempt to construct a spin-2 theory, because it's the simplest that could work. In the end, Feynman abondoned his program for this quantization of gravitation using QED as an analogy. –  mtrencseni Sep 30 '11 at 10:38
    
Related: physics.stackexchange.com/q/12647/2451 –  Qmechanic Oct 7 '11 at 21:33

1 Answer 1

Most of this is covered in this answer: Why do we not have spin greater than 2?

For question 2, this is indeed the case--- you always resolve the composite structure of spin-3/2 particles (which are not gravitinos) at a scale comparable to the mass of the particles. The phenomenon is that the spin-3/2 particle must come in a family of other bound-states, called a Regge trajectory, which unitarizes the scattering by exchange of this particle. Without other degrees of freedom which are at a similar energy, you can't take a pointlike limit. This allows you to predict either new particles related to the original, or a breakup into substructure of some sort.

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I've always wondered, why is this concern with "taking a pointlike limit" or requesting that the particle is "fundamental"? Is it just because we love/need/can work with renormalizable theories? At low enough energies, a single Hydrogen atom can be in total angular momentum state you like and act as an "elementary" particle"... –  Slaviks Oct 8 '11 at 7:18
    
@Slaviks: One reason is that no-poinlike-limit allows you to predict new particles--- something must pop up at the scale where unitarity would be violated. This is how you can give mass bounds on the Higgs. Regarding H-atom, one of the most damning limitations of modern physics is that there is no good pointlike relativistic description of bound-states like nuclei or H-atoms. I have one which I try to finish every once in a while, but I never got it to work fully. The nonrelativistic version works, so that LePage and others used nonrelativistic H + v^2/c^2 corrections. But this is annoying. –  Ron Maimon Oct 8 '11 at 20:32

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