Can a lone photon disintegrate? While studying the very basics of QFT, I became aware of scattering interactions. After some research, it seems that a massive particle can give two new particles.
However, I cannot find anything about something I would call "photon disintegration". The fact that a lone photon disintegrates into two photons (or two massless particles).
Can a lone photon disintegrate into two (or more) massless particles?
Maybe I'm mixing things up since I still understand barely anything about QFT.
 A: tl;dr: In principle, yes at the amplitude level, but no at the rate level. In the Standard Model, no, not at all.
I assume you are talking about something like "photon decay in vacuum" because you say "lone photon." So I won't consider photon conversion due to interactions with some other particle, for which there are more possibilities. There are two issues here. The first is kinematics. The second is interactions that would mediate such a conversion.
Massless particles have the special property that the sum of the 4-momentum vectors of two massless particles with momenta $p_1$ and $p_2$ pointing in the same direction gives another massless 4-momentum vector with momentum $p_1+p_2$. So as far as energy-momentum conservation is concerned, a massless particle can be split into two as long as those two are both exactly in the direction of the original. You can check that that is the only possibility. If the two produced particles had different directions, the sum of their 4-momenta would be massive, and that violates energy conservation, because the original photon is massless.
Now let's assume you have some interaction that could couple a photon to two other massless particles. Then in general there could be a diagram $\gamma\to A+B$ and the energy-momentum conserving delta functions in that diagram would be satisfied for the collinear kinematics described above. However, to turn that diagram into a decay rate, one needs to integrate over phase space. However, as we just discussed, the decay is only allowed at one point in phase space (where the outgoing momenta are collinear with the original momentum). So the phase space volume is zero, and the rate is zero.
There is the additional question of what kind of interaction could allow this. In order to preserve Lorentz invariance, the operator would need to have all indices contracted. In order to preserve gauge invariance, it needs to be made of $F_{\mu\nu}$. You can quickly check that there is no set of massless particles in the Standard Model that you can put together to get something like this. But you could imagine adding new fields. Let's add a new massless vector and call the field strength $F'_{\mu\nu}$, and a new massless scalar $X$. We can write an operator $$\Lambda^{-1}F_{\mu\nu} F'^{\mu\nu} X$$ where $\Lambda$ is a parameter with mass dimension 1. This gives a diagram for $\gamma\to\gamma'X$. The rate for this is still zero as discussed above though. For more than two particles there would be other possibilities that you can play with. And of course I'm not saying such extensions to the Standard Model are allowed under experimental constraints. This is just an exercise.
