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I'm currently working on my master thesis, and I need to know a bit of supersymmetry. I have been looking the theory and I have a basic knowledge about it.

I have a problem with understanding the spin of the superpartners. When the multiplets are build, we have the operators which rise and lower the spin of the state by 1/2. Therefore the superpartners will have a spin which differs by 1/2 respect to the usual particle. But, why the squarks have spin 0 instead of 1? Quarks have spin 1/2, thus I understand that the superpartner should have either spin 0 or 1. But why 0 and not 1? This question applies of course to other particles, such as the photino(why not 3/2?), slepton(why not 1?), etc.

Finally, we know that the gauge fields (graviton, photon, etc) are the connections for forces. Do the superpartners of the gauge fields play a role in the forces as well? I know that the gravitino can't, since there are not interacting theories of massless particles with spin greater than 2. What about the other gauge fields?

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You're right in that a spin-1/2 particle could hav superparteners of spin 0 or 1. This is encoded in the statement that in supersymmetry, there are two supermultiplets:

  • the chiral multiplet contains a complex scalar and a spin-1/2 fermion
  • the vector multiplet contains a vector (spin-1) boson and a spin-1/2 fermion

(Note that this is true for what is called $\mathcal{N}=1$ SUSY in four dimensions, which is the standard setting for standard model extensions such as the MSSM. There are many more possibilities -- a nice introduction can be found at https://iktp.tu-dresden.de/Lehre/SS2009/SUSY/literatur/sohnius_article.pdf.)

Now what about the standard model fermions? The vector multiplet, conating a vector field, comes with a gauge symmetry and is in the adjoint representation. In particular, the fermoin cannot be chiral, as the standard model fermions are. So the quarks and leptons cannot be part of a vector multiplet, and their superpartners can only be scalars.

For the superpartners of the gauge bosons, you have to recall two things:

  • For renormalisable theories (i.e. guage theories, such as the standard model, but excluding gravity), you cannot have particles of spin larger than one (under reasonable assumptions, such as nontrivial scattering).
  • If you allow higher spins in a nonrenoamlisable theory, you can go up to spin two. However, that possibility is quite restircted: There basically is one spin-two field (the graviton) which can have one superpartner of spin 3/2 (the gravitino), and both have more or less fixed interactions (e.g. they are not charged unde gauge groups). This basically comes about because supergravity is possible only if supersymmetry is gauged, and the gravitino is the gauge field of supersymmetry (i.e. $\delta \psi_\mu\sim\partial_\mu \epsilon+\dotsm$)

(Again, this is a bit schematic, and I refer you to the introduction by Sohnius (of course, there are other textbooks as well)).

Together, these statements imply that the superpartners of the gauge bosons can only be of spin 1/2 (i.e. ghauge boson and gaugino form a vector multiplet), and graviton and gravitino come in an additional multiplet by themselves. (The gauginos themselves transform homogeneously under the gauge groups, in the adjoint.)

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  • $\begingroup$ Hey Toffomat, I understand your answer, but still I don't see the point. I understand that both cases are present, but in the literature the gravitino is always the spin 3/2 particle, and not the other one. $\endgroup$ – Jordi Dec 5 '18 at 12:59
  • $\begingroup$ @Jordi I have extended the answer, hope that answers your question $\endgroup$ – Toffomat Dec 5 '18 at 13:19

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