Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is there some condition in the N=1 SUSY algebra telling that the spin of the superpartners of gauge bosons (either for colour or for electroweak) must be less than the spin of the gauge boson? I am particularly puzzled because sometimes a supermultiplet is got from sugra that contains one spin 2 particle, four spin 3/2, and then some spin 1, 1/2 and 0. If this supermultiplet is to be broken to N=1 it seems clear that the graviton will pair with the gravitino and the rest of spin 3/2 should pair with spin 1, so in this case it seems that a superpair (3/2, 1) is feasible. Why not in gauge supermultiplets?

share|improve this question
    
If my (very superficial) memory serves there is no problem with these multiplets from the algebraic point of view but they cause problems for other reasons, like breaking some gauge symmetric stuff, causing anomalies, etc. Good question, I hope someone more knowledgeable will answer it. –  Marek Aug 2 '11 at 13:01
    
my understanding was that the theory with a 3/2 gaugino superpartner would not be renormalizable cf the SUSY primer arxiv.org/abs/hep-ph/9709356 p6 –  luksen Aug 2 '11 at 14:53
    
"The next-simplest possibility for a supermultiplet contains a spin-1 vector boson. If the theory is to be renormalizable, this must be a gauge boson that is massless, at least before the gauge symmetry is spontaneously broken. A massless spin-1 boson has two helicity states, so the number of bosonic degrees of freedom is nB = 2. Its superpartner is therefore a massless spin-1/2 Weyl fermion, again with two helicity states, so nF = 2. (If one tried to use a massless spin-3/2 fermion instead, the theory would not be renormalizable.)" –  luksen Aug 2 '11 at 14:53
    
@luksen that could be the germ of an answer; is it just power counting and superficial divergence? The OP was for gluinos, but, what happens if we look for a massive supermultiplet for Z or W, can it have some combo of 3/2 and 1/2? And, is there some argument beyond renormalizability? Of course we do not ask renormalizability of sugra. –  arivero Aug 2 '11 at 19:01
    
Related answer in quora: quora.com/… –  arivero Aug 3 '11 at 10:24

1 Answer 1

up vote 5 down vote accepted

Fermionic spin 3/2 fields, much like bosonic fields with spin 1 and higher, contain negative-norm polarizations. Roughly speaking, a spin 3/2 field is $R_{\mu a}$ where $\mu$ is a vector index and $a$ is a spinor index. If $\mu$ is chosen to be 0, the timelike direction, one gets components of the spintensor $R$ that creates negative-norm excitations.

This is not allowed to be a part of the physical spectrum because probabilities can't be negative. It follows that there must be a gauge symmetry that removes the $R_{0a}$ components - a spinor of them. The generator of this symmetry clearly has to transform as a spinor, too. There must be a spinor worth of gauge symmetry generators. The generators are fermionic because the original field $R$ is also fermionic, by the spin-statistics relation.

It follows that the conserved spinor generators are local supersymmetry generators and their anticommutator inevitably includes a vector-like bosonic symmetry which has to be the energy-momentum density. This completes the proof that in any consistent theory, spin 3/2 fields have to be gravitinos. The number of "minimal spinors" - the size of the gravitinos - has to be equal to the number of supercharges spinors which counts how much the local supersymmetry algebra is extended. In particular, it can't be linked to another quantity such as the dimension of a Yang-Mills group.

So while both 3/2 and 1/2 differ by 1/2 from $j=1$, more detailed physical considerations show that it is inevitable for the superpartner of a gauge boson, a gaugino, to have spin equal to 1/2 and not 3/2. Similarly, one can show that the superpartner of the graviton can't have spin 5/2 because that would require too many conserved spin-3/2 fermionic generators which would make the S-matrix essentially trivial, in analogy with the Coleman-Mandula theorem. Gravitinos can only have spin 3/2, not 5/2.

share|improve this answer
    
But, what does it happen when we break susy down to, say, N=1? The number of spin 3/2 fields is still the same, but only one generator survives. –  arivero Aug 3 '11 at 11:05
    
Dear Alejandro, I am afraid that you have totally misunderstood or ignored my answer. The spin 3/2 fields are gravitino fields, so their number is the $N$ measuring the extended supersymmetry. Spontaneous breaking of SUSY has no impact on $N$; instead, SUSY breaking may make the gravitinos massive (much like electroweak SUSY breaking makes W+Z gauge bosons massive) but they're still spin 3/2 fields and they're still gravitinos. It's wrong to say that SUSY or other symmetry generators "don't survive" spont. SUSY breaking. They do: they just don't annihilate the vacuum. –  Luboš Motl Aug 4 '11 at 6:23
    
In fact I am almost fully happy with your answer; I was just waiting for some alternative attempt to it, before to mark it as answered. As for susy breaking, perhaps it is my misunderstanding, but I thought that the same generator that links the spin 2 graviton to a given 3/2 gravitino, will link the other 3/2 gravitinos to spin 1 particles, and that this will still be true after the breaking. It can be a trivial thing, but I think it was relevant to the answer, so I was pushing about it. –  arivero Aug 5 '11 at 14:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.