Is Photon Decay Allowed in QED? Investigating Conservation Laws and Amplitudes

[Viky553]

I'm starting to read up a bit on QFT, starting with Griffith's intro to elementary particles book. I've gone through the background stuff and I'm now into the QED chapter. I was trying to get a feel for how the number of loops introduces free momentum variables, and I ended up drawing a diagram that got me wondering about photon decay (time flows from left to right for the sake of this post):

[PLAIN]http://img832.imageshack.us/img832/613/12photondecay.png

I couldn't figure out any reason for it to be kinematically forbidden (provided the decayed photons were collinear in the same direction as the original photon), yet I also couldn't find any literature on photon decay. After MUCH searching I found https://www.physicsforums.com/showthread.php?t=512811 which suggested that that diagram was forbidden by conservation of C-parity (since photons have -1 C-parity).

-So I guess my first question is, is this reasoning correct and does it prove that QED does not allow for a single photon to decay into two photons?

-Next, do the Feynman rules for calculating amplitudes explicitly encode conservation of angular momentum and C-parity? I can see the explicit conservation of energy-momentum via the delta function for each vertex. I can't quite see angular momentum conservation right away, although I wouldn't be surprised to see it follow somehow from some coordinate transform on the 4-momentum or something. The C-parity conservation I have no clue.

The thread that I linked made it sound like all photon decays are forbidden. However, the cited paper was a little over my head. So can anyone explain why the following diagram is not allowed?

[PLAIN]http://img853.imageshack.us/img853/9843/13photondecay.png

Three photons can combine to form spin 1, and the C-parity checks out fine.

-In light of the previous questions, if you actually try to do the integrals for these two diagrams do the amplitudes actually go to 0? i.e. do the Feynman rules explicitly forbid these interactions? This would surprise me just a tad (in the second case) considering that the following diagram is allowed:
[PLAIN]http://img26.imageshack.us/img26/9198/photonscattering.png
although clearly conservation of angular momentum checks out in that case.

-Now, even with all the above considered, can an even number of photons decay into a larger even number of photons? And likewise can an odd number of photons (greater than 1) decay into a larger odd number of photons? Or vice versa, can even/odd numbers of photons "combine" into less even/odd number of photons? Neither of these would cause an issue with the C-parity nor angular momentum. An example would be:

[PLAIN]http://img828.imageshack.us/img828/2918/24photondecay.png

(Sorry for the large images...I didn't realize it until previewing, and I don't feel like redrawing them all and reuploading them. I'll resize them next time.)

 

[bcrowell]

-So I guess my first question is, is this reasoning correct and does it prove that QED does not allow for a single photon to decay into two photons?

You don't have to depend on those of us who participated in that thread to have gotten that right. The Fiore paper http://arxiv.org/abs/hep-th/9508018 confirms it.

The thread that I linked made it sound like all photon decays are forbidden. However, the cited paper was a little over my head. So can anyone explain why the following diagram is not allowed?

This is the case that we were all frustrated with. Fiore offers technical arguments to the effect that it's impossible, but all of us felt that it should be possible to rule it out on more straightforward grounds. If those fermions in your diagram are massive, then this post https://www.physicsforums.com/showthread.php?t=512811&page=4 by PAllen offers an argument against the decay that I personally find persuasive.

You may also be interested in this: http://physics.stackexchange.com/questions/12488/radioactive-decay-of-massless-particles

Re your question about 2\gamma\rightarrow4\gamma, I don't see any reason why it couldn't happen. There isn't any requirement for the decay products to be collinear as there is in \gamma\rightarrow3\gamma, so the phase-space volume doesn't vanish. But it has six vertices, so I would expect it to be very low in cross-section compared to 2\gamma\rightarrow2\gamma.

 

[Viky553]

As far as I'm concerned, if QED is a complete theory of...well...QED then it should take care of any possible conservations laws already (can anyone confirm this for me? such as angular momentum and C-parity which I raised earlier). If this is the case, which it better be if I'm going to buy QED at all, then the questions should be able to be answered by just doing the calculation.

Now, as a total newb to the field this is way over my head to do. I haven't even worked through the simple tree diagram examples yet except for a toy QFT that Griffith's introduces. However, I'm also aware that these integrals very quickly become next to impossible for even QED gurus. That said, how hard would it be for someone familiar with QED to just do the calculation? I mean they're all one loop diagrams. You'd think a similar calculation had been done for the photon-photon scattering box diagram already (albeit with a much different phase space working volume). The sixth order diagram would probably be ridiculous, but what about the fourth order one (lowest order \gamma\rightarrow3\gamma)?

I just find it surprising that after all these years that this isn't a more discussed and studied issue. You'd think it'd be the first question someone would ask after hearing why a single photon can't decay into a massive pair.

I don't know if those fermions are massive, they're virtual particles so they can be whatever they feel like (right?) and there would probably be massless versions to integrate over in addition to the massive options.EDIT: In fact, I'd even be happy to hear/see that someone did the \gamma\rightarrow2\gamma single loop third order diagram and found it to have 0 amplitude (and thus agreeing with the C-parity argument), which you'd think would be feasible within current mathematical abilities.