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Hitoshi Murayama writes in his 221A Lecture Notes on Spin

How do we choose spin when you introduce a field, then? A consistent ( i.e. , renormalizable) quantum field theory can include only spin 0, 1/2, and 1. Renormalizable interactions are only those interactions that can appear without extra suppressions of $G_N E^2 / ̄\hbar c^5 5 = ( E/ 10^{19} \mathrm{\ GeV^2}) \ll 1$. Therefore, there is a reason why we see only particles of spin 1/2 and 1. Well, what about spin 0? We have not seen any fundamental particle of spin 0 yet. We are looking for one: the Higgs boson, which is expected to permeate our entire Universe, dragging the foot of all quarks, leptons, W and Z bosons, making it impossible for them to reach the speed of light. It is expected to be found this decade thanks to higher-energy accelerators being built. Once it is found, it exhausts all theoretical possibilities of spin 0, 1/2, and 1.

Is this really true? And if yes, why?

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  • $\begingroup$ I am sure that Hitoshi will tell you that the Higgs has been found meanwhile, and probably can be convinced to reword things in such a way that they do not suggest that consistent implies renormalizable. These notes must be very old, aren't they? $\endgroup$ – user178876 Jul 7 at 11:23
  • $\begingroup$ I think they were created around 2003 (web.archive.org/web/20030515000000*/hitoshi.berkeley.edu/221A/spin.pdf) $\endgroup$ – jak Jul 7 at 11:29
  • $\begingroup$ Yes, the Higgs discovery was announced 2012. As for your main question, it is believed that it is impossible to write down a renormalizable QFT in 4D with fundamental states with spin greater than 1, but AFAIK this has not been fully proven by now, yet no one has found a counter example either. Notice that you can have composite states with larger spin, some of our nuclei have larger spins. $\endgroup$ – user178876 Jul 7 at 11:33
  • $\begingroup$ @AccidentalFourierTransform While your answer there contains information that would also answer this question, I don't see how the two questions are supposed to be duplicates at all. $\endgroup$ – ACuriousMind Jul 7 at 11:44

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