Simplest "physical" photon generating Feynman diagram

Question

I'm half way through the excellent "Student Friendly Quantum Field Theory" and I read that single vertex Feynman diagrams in QED are "not physical" because their corresponding amplitudes are zero. For example the diagram $$ e_{\mathbf{p_1}}^- + e_{\mathbf{p_2}}^+ \to \gamma_{\mathbf{k_1}} $$ has a probability amplitude that (when you calculate it) includes a factor of $\delta^{(4)}(k_1 - p_1 - p_2)$, where the boldface p's and k are 3-momenta and the normal typeface are 4-momenta. The argument goes that since the photon is massless we must have $k_{1\mu}k_1^{\mu} = 0$, but if you work out $(p_1+p_2)_{\mu}(p_1 + p_2)^{\mu}$ it turns out non-zero, therefore we can't find a real photon momentum that makes the dirac delta, and consequently the amplitude $\langle\gamma_{\mathbf{k_1}}\lvert e_{\mathbf{p_1}}^- e_{\mathbf{p_2}}^+\rangle$ non zero. A similar reasoning shows that every other single vertex Feynman diagram (e.g. $e_{\mathbf{p_1}}^-\to \gamma_{\mathbf{k_1}} + e_{\mathbf{p_2}}^-$) are also "non physical".

So my questions are:

1. If these diagrams are non-physical what's the simplest diagram that generates a photon that is physical. Or, "where do all the photons come from?"
2. Are there any interpretations (for the fact the amplitude for single vertex diagrams are zero) other than that they are "non-physical"? For example, perhaps photons with $k^2 \ne 0$ are possible, but live too short a time to be observed.

Please be gentle, I'm not actually a student, just an enthusiast and this is my lockdown reading!

Answer 1

The reason you state for why the amplitude $e^+e^-\to\gamma$ vanishes is correct. But I would like to simplify it a bit. Mainly because it is not a consequence of QFT but of Special Relativity.

Suppose you are in the center-of-mass frame of the electron-positron pair. Momentum conservation tells you that in this frame the resulting particle will be produced at rest and will have a mass of $M^2 = (p_1+p_2)^2 = (E_1+E_2)^2>4m_e^2$. The photon is never at rest and its mass squared is zero. So there is no way to conserve momentum.

Note that there simply cannot exist a photon with $k^2\neq 0$. Such a photon would not travel at the speed of light, so it's inherently a contradiction. Also note that, by the same argument as above done for the opposite process, we can show that photons cannot decay. And the same conclusions hold for all particles with zero mass.

The simplest process that gives photons is $e^+e^-\to2\gamma$. The presence of two photons makes it kinematically viable because we can have a nonzero invariant mass in the center-of-mass frame of the two.