Why must a consistent theory with a rarita schwinger field (i.e. massless gravitino in the spectrum) be supersymmetric? I was reviewing the GSO projection, Spin Structure, etc. & wasn’t able to make out the argument. Is this an actual theorem (like Weinberg-Witten for composite gravitons for instance) or is it just a belief?
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Not sure what you mean by "belief". Composite spin 3/2 fields are permeating our world, viz the Δ(1232) baryon.
Fundamental Rarita-Schwinger (R-S) fields propagate acausally, as proven in a historic paper by Velo & Zwanzinger, Phys Rev 188 2218 (1969) available online.
One of the earliest triumphs of supergravity was Das & Freedman, Nucl Phys B114 (1976) 271-296 demonstrating that local supersymmetry invariance saves the day by ensuring causality.
The propagator grows like $s\sqrt s$ at high energies, by Mandelstam, but unphysical scattering exchanges thereof are prevented by its coupling to the conserved supercurrent, of which it is the gauge field. The same supercurrent is not simultaneously Poincare & local susy invariant, evading the W-W theorem analogously to vector gauge fields. But I assume you are not focussed on loopholes to the exclusions of the W-W theorem; physics theorems normally rely on facts on the ground to ensure they are carefully conditioned to prevent deductions of the nonexistence of fish.
answered Apr 17, 2020 at 14:15
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1$\begingroup$ requiring unitarity of GSO projection (sum over spin structures) on string worldsheet gives one a SUSY spectrum to work with where as unitarity of a different projection will remove the gravitino (so long as it is massless) from non-SUSY spectrum. I could look for an exact reference with these results in the literature on modular invariance, spin structures, etc. but I reckon this is a fairly well known result. The preceding point is a reason one might expect a consistent theory with a massless gravitino must be a SUSY theory (there may be loopholes or a more exact statement I am unaware of). $\endgroup$ Commented Apr 19, 2020 at 0:21
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1$\begingroup$ I studiously avoided stringery in favor of "ancient", standard, dull QFT. $\endgroup$ Commented Apr 19, 2020 at 0:26
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$\begingroup$ @AlexanderNordal thank you, that is what I had in mind (not sure how to give this the check mark so I just upvoted instead). $\endgroup$ Commented Apr 20, 2020 at 0:52