In QED, when two photons collide, they can turn into an electron and positron pair. We know from $U(1)$ gauge symmetry that the total charge of the initial and final states must be conserved. On the other hand, I expect that the total spin must also be conserved. But I do not quite get the details of how this works.
In this post the total spin of two-photon-state is discussed. Based on the transversality argument, OP argues there are three distinct spin states associated with the two-photon system. Two of them correspond to the spin-0 representation and the remaining one corresponds to a spin-2 state.
Based on the above argument, if the total spin in pair production is to be conserved, I would assume that the incoming photons must be in the spin-0 state, excluding the spin-2 state because the spin-state of the created electron-positron pair does not have a spin-2 representation. As far as I know, this spin state can have one spin-0 rep. and three spin-1 rep.
Edit: Also, in Wikipedia page there is Landau–Yang theorem, stating that a massive particle with spin 1 cannot decay into two photons. I suspect this selection rule follows from the requirement of the conservation of the total spin. Because as suggested in the linked question two-photon state does not have a spin-1 rep.
Is this reasoning correct?
The second point is about symmetry. If the total spin is to be conserved, what is the associated symmetry? I am thinking it must the rotational invariance of pair production amplitude. But what do the generators of this rotational symmetry look like? and where do they act? These generators must not correspond to the ordinary rotations in space. Because this would correspond to the conservation of orbital angular momentum, not spin.