I read this Under what conditions is a vector-spinor gamma trace free. And also read many papers about higher spin, but no one explains why irreducible spinor is gamma traceless spinor? Can anyone explain this?
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1$\gamma^\mu \psi_\mu$ has no free Lorentz indices, whereas it has free spinor indices, the ahllmark of a spinor, so spin 1/2. What is the problem? This answer looks fine. – Cosmas Zachos Aug 13 '19 at 15:30
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I read that fermions transform as reducible ((n+1)/2,n/2)+(n/2,(n+1)/2) (which is equivalently as direct product of n (1/2,1/2) and Dirac (1/2,0)+(0+1/2) rep )representation because this representation has space inversion as Weinberg says in their QFT Vol 1, page 233. For example n=1. We have (1/2,1/2)x((1/2,0)+(0,1/2))=(1,1/2)+(1/2,0)+(1/2,1)+(0,1/2) it is REDUCIBLE. He also says (in footnote) that doubling of j=1/2 removes by imposing Dirac equation remaining divergenceless condition (not completely, but intuitive clear- removing ghosts). Question is how we remove one of two j=3/2? – fika97 Aug 13 '19 at 21:45
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Right question is how gamma tracelles removed one j=3/2 ? – fika97 Aug 13 '19 at 21:52
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I really have no idea what you are trying to say. Perhaps if you refashioned your question, without reference to Weinberg's book!, and asked just one question... Most texts on supergravity are vastly clearer on this. – Cosmas Zachos Aug 13 '19 at 22:11
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Same answer. – Cosmas Zachos Aug 13 '19 at 22:24