Simple reason why the correlator of a vector with two identical scalars vanishes?

Question

In scalar QED, the photon interacts with a charged scalar and the three point function of a vector, scalar and scalar bar is nonzero.

I remember an argument that very simply proved that if you try to write down a similar correlator with a vector and two identical scalars, the answer is zero. Same is true if the vector is replaced by any odd spin particle.

I'm trying to recall the proof of this fact. I think the argument only assumed Lorentz invariance, though maybe parity also.

Can anyone provide a simple proof of the type I'm thinking of?

Answer 1

Correct me if I've misunderstood your question...

$$\langle \phi_1 \phi_2 \rangle = \int \mathcal D \phi \mathcal D \phi^* \,\, \phi_1 \phi_2 e^{-iS[\phi,\phi^*, \ldots]} = \int \mathcal D \phi' \mathcal D \phi^{*'} \,\, \phi_1' e^{i\lambda}\phi_2 ' e^{i\lambda} e^{-iS[\phi',\phi^{*'} , \ldots]} =e^{2i\lambda} \langle \phi_1 \phi_2 \rangle $$ This implies $\langle \phi \phi \rangle$ to be zero. In the second equality I've used the global U(1) symmetry of the action.

Edit: in case it wasn't clear, the same argument applies to the correlator $\langle \phi \phi A \rangle $, because the gauge field remains inert under the global $U(1)$ symmetry.

Moreover, you can convince yourself (exercise; use the same technique as above, or, if you prefer using the operator formalism, remember that the vacuum state is invariant) that any correlator of fields must be an invariant tensor under trasformations that belong to an internal symmetry of theory.