Feynman amplitude $\mathcal{M}$ for non-physical processes in QED

Question

I know $e^-e^+\rightarrow\gamma$ is an example of a non physical process in QED because it violates four-momentum conservation. For that reason, the Feynman amplitude $\mathcal{M}$ is never written or calculated explicitly in the bibliography, so I don't really know if it is assumed that it is zero or if it doesn't make sense calculating $\mathcal{M}$ for such a process.

I've tried to perform such a calculation for $e^-e^+\rightarrow\gamma$ following the Feynman rules and got $$i \mathcal{M} = (-ie)u^s(p)\gamma^{\mu} \bar{v}^{s'}(p')\epsilon_{\mu}^{\lambda^*}(k)$$ where $u^s(p)$ represents the incoming electron, $\bar{v}^{s'}(p')$ the incoming positron and $\epsilon_{\mu}^{\lambda^*}(k)$ the outgoing photon.

I doesn't look like $\mathcal{M} = 0$ here. Should it not vanish? It seems to me that it should be zero for a process that can never occur.

I've used an example to clarify my question, but I would also like to know if the amplitudes for all other non-physical processes should be zero or not and why.

Answer 1

Your initial approach is correct for finding the QED current of the vertex. However, the amplitude of a Feynman diagram (and therefore a particular interaction) is found by integrating over the momenta of the virtual bosons in the diagram, which couple to currents to make amplitudes. Neglecting the fact that there aren't any virtual bosons in the non-physical process you describe, as Vladimir mentioned, you would also need to include a term $2\pi\delta(p+p'-k)$ in the integral which maintains four-momentum conservation. Since four-momentum isn't conserved in the process described, this delta function would vanish.