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Oct 8, 2018 at 19:55 comment added user26866 I guess you probably have to assume that the tensor is divergenceless and traceless for a group theory argument to work? Otherwise it's not an irrep and the tensor will have scalar bits
Oct 8, 2018 at 19:31 comment added user26866 There's no U(1) symmetry in scenario I have in mind. Just a vanilla scalar field coupled to an odd spin boson which can be massive or massless; doesn't matter. I think the rotation argument you're mentioning is the one I have in mind. Do you have a link?
Oct 7, 2018 at 18:58 comment added octonion tbt's argument has nothing to do with gauge symmetry. It's about global U(1) symmetry. Do you have charge conservation? Then you have U(1) symmetry. There is also the argument (Furry's theorem) in the comments due to charge conjugation. If now you are talking about an odd spin particle with two scalars, that vanishes due to a similar idea to the U(1) argument applied to rotation. The name of these kind of arguments in general is called the Wigner-Eckart theorem, although it's usually presented as something more complicated than it really is.
Oct 7, 2018 at 9:22 comment added user26866 None of these are the answer I'm looking for. There's a simple proof that the correlator between two identical scalars and an odd spin particle vanishes which only relies on spacetime symmetry arguments. The spinning particle can even be massive (I believe) in which case gauge invariance plays no role.
Oct 6, 2018 at 4:07 comment added octonion tbt's answer is correct and is the simple proof you are looking for. But note also that even correlation functions like $\langle \phi^\dagger \phi A\rangle$ don't make much sense since $A$ is not gauge invariant.
Oct 2, 2018 at 20:52 answer added tbt timeline score: 1
Sep 27, 2018 at 22:17 history asked user26866 CC BY-SA 4.0