Why is the scalar QED vertex not inconsistent with angular momentum conservation? In scalar QED, there is a vertex where two spin $0$ particles come in and a spin $1$ photon comes out. Naively this can't possibly be consistent with angular momentum conservation, because two spin $0$ things can't add up to spin $1$. 
It is claimed here that this is okay because the photon is "off-shell", so it has spin $0$. I don't believe this argument. While it is true that the usual formulation of QED has longitudinal photons, the whole point is that they decouple. And it's perfectly possible to formulate QED with only physical states from the start. Being off-shell is weird, but not so weird that it can reach outside the Hilbert space and produce a new scalar state out of nowhere.
A more sensible possibility, in my opinion, is that the two spin zero particles must have orbital angular momentum $\ell = 1$. But that seems rather complicated, and I don't know how to show that.
What is the resolution to this puzzle?
 A: Indeed, the two spin $0$ particles must have angular momentum $\ell = 1$. The crucial feature is that, unlike the case of regular QED, the scalar QED interaction $\mathcal{H}_{\text{int}} \sim A_\mu \phi \partial^\mu \phi$ contains a derivative, so it is proportional to the momentum $p^\mu$.
Then a first-order transition amplitude from scalar and photon states $|i \rangle, |i' \rangle$ to $|f \rangle, |f' \rangle$ is
$$\langle f, f' | \mathcal{H}_{\text{int}} | i, i' \rangle \sim \langle f | p^\mu \phi_{\mathbf{p}} \phi_{\mathbf{0}} | i \rangle \, \langle f'| A_\mu | i' \rangle.$$
Focusing on the first term here, by the Wigner-Eckart theorem, the total angular momentum of the final state is the sum of the angular momentum of the initial state and that of the transition operator. Since the $\phi$ fields and $\phi$ particles are scalars, and the vector $p^\mu$ carries spin $1$, that means one unit of angular momentum is transferred to the orbital angular momentum of the $\phi$ particles, for each scalar QED vertex.
