The photon doesn't have a well-defined quantity such as spin. Instead, it is characterized by helicity $h$.
Let's assume the state of two photons in CM frame (with $\mathbf k$ being the momentum of one of the photons), each of which has definite helicity $h_{i}$: $$ |\Psi \rangle = |h_{1},\mathbf k;h_{2},-\mathbf k\rangle $$ How to calculate the total spin (NOT total helicity) of such a system? Or, if it is possible, how to make a conclusion about the spin of an arbitrary two-photon state by having its helicity?
For example, suppose the following basis of two-photon states (here I omit momenta labels): $$ \tag 1 |L,R\rangle , \quad |R,L\rangle, \quad |L,L\rangle \pm |R,R\rangle $$ The first two states have total helicity $\pm 2$, so it seems that they correspond to a spin 2 state (naively, there can't be a higher spin for two photons from spin group representation point of view). The last two states have zero total helicity, so it seems that they correspond to spin zero. But I'm not sure about this statement. I can then suppose the last two states have zero helicity, but their spin is non-zero in general (in the sense that they belong to a non-zero spin representation of the Lorentz group).
Also, there is the problem with treating a two-photon system as a system with a well-defined spin. The total spin of a two-photon system must be calculated just as the sum of spins of each photon. The spin of a single photon as angular momentum at rest, however, isn't defined, so, from this point of view, there is no quantity of total spin for a two-photon system. Also, in some sense, the spin can be introduced as the quantity, which determines the non-coordinate transformation properties of the quantity under rotation transformation. So, for such a tensor, which is irrep of the spin group (say, the Lorentz group), the number $N$ of its independent components defines the spin $S$ by the relation $N = 2S+1$. Unfortunately, this interpretation quickly fails when we take into account gauge symmetry (or just use the true massless representation, which is $F_{\mu\nu}$, not $A_{\mu}$). Instead of spin, such a quantity is rather helicity.
P.S. This topic is relevant if we want to determine selection rules for orbital angular momentum of a two-photon state for a given total angular momentum $J$ and (suppose that it is defined) total spin $S$. Allowed values of $L$ are $$ \tag 2 |J-S|\leqslant L\leqslant J+S $$ But if we know only the helicity $h$ of the state, then we can't use $(2)$ for getting selection rules for $L$. In any case, independently of the fact of the definition of spin, there are selection rules for basis $(1)$, and I'm interested in them.