Hi, I have a question in Srednicki's QFT textbook. In p.460 section 75(about Chiral gauge theory), it says "In spinor electrodynamics, the fact that the vector potential is odd under charge conjugation implies that the sum of these diagrams(exact 3photon vertex at one-loop) must vanish." That's good, because it's just the Furry's theorem for odd number of external photons. But I find the subsequent statement confusing. "For the present case of a single Weyl field(coupled to U(1) field), there is no charge conjugation symmetry, and so we must evaluate these diagrams." But shouldn't the transformation rule of [itex]A^\mu (x)[/itex] under C,P or T be universal regardless of specific theory?
In spinor electrodynamics, the Lagrangian is *invariant* under the C operation, which among other things takes ##A \to -A##. One consequence of this is that there should be an equality between the following three-point functions: ##\langle A(x) A(y) A(z) \rangle = \langle(-A(x))(-A(y))(-A(z))\rangle = -\langle A(x) A(y) A(z) \rangle## The only way this equality can be satisfied is if ##\langle A(x) A(y) A(z) \rangle## vanishes. This is Furry's theorem. This result relies crucially on the fact that the C operation, under which ##A \to -A## and also ##\psi## transforms nontrivially, is a *symmetry of the Lagrangian*. That's why we have the equality above. In the chiral case, you can still define a transformation under which ##A \to -A## but it is no longer a symmetry of the Lagrangian, no matter what transformation rule you choose for ##\psi##. Therefore you cannot make the same argument and Furry's theorem does not go through.
But doesn't Furry's theorem still hold even if the langrangian has no C-symmetry? I thought it is valid as long as the vacuum is invariant under C. What am I missing here?
Well... Maybe because the vacuum contains no particle? Are the invariance of the vacuum and invariance of lagrangian related to each other?
Right. If your theory has a symmetry and you have a unique vacuum state, then the vacuum is invariant under that symmetry. If you have some operation that is not a symmetry of your theory, there's no reason to expect the vacuum to be invariant under that symmetry.
It's very clear now! I always assumed that the vacuum is invariant under any operation. Thank you, it was very helpful :)