I am taking an introductory course in particle physics and am quite confused about conservation of angular momentum at a vertex. This trouble arose specifically when considering the decay $\pi^0 \rightarrow \gamma \gamma$. I am asked to draw the Feynman diagram. I thought that it would most likely happen via a a) $\gamma$, b) $Z$ boson or c) gluon intermediate. I have now seen a diagram with a quark loop (I will upload an image when possible; for some reason all of my attempts are 'failing to upload' at the moment)

But I wanted to figure out why (if) my suggestions of an S diagram with a single propagator intermediate were wrong.

I discarded a) because I think it is not possible for a photon to decay into two photons. But I was also wondering if one can argue that angular momentum is not conserved at the first vertex. This would require that a photon cannot have angular momentum.

I am not at all sure about b) and c). I am not sure if it is permissible for Z bosons and gluons to couple to the photon. With the Z boson, on the one hand within the framework of electroweak unification, it is a linear combo of the same gauge bosons as the photon, so I would think it can interact. On the other hand, it is uncharged. The gluon is also uncharged, so maybe it cannot couple to the photon.

In any case, if it is impossible for a single paticle intermediate to have zero angular momentum, then the pseudoscalar mesons can never decay in this way!


In the standard model


Here are the lowest order $π^0$ decay diagrams:


The quark-antiquark pair are bound by the strong force, for a very short time until they annihilate :when particle meets antiparticle there is annihilation. It cannot decay into two gluons because there are no free gluons, and the pions are the lowest mass hadrons into which gluon jets can appear .

The electromagnetic interaction of the charged quarks takes over and the annihilation goes to two photons.

There cannot be an annihilation to one photon, because the photon has mass zero but the $π^0$ has a mass, and the photon no center of mass. There would be violation of energy and momentum conservation. There are no three photon vertices in the standard model, i.e a photon does not interact with a photon.

The off mass shell of a pi0 decay through a Z into electron positron or neutrino antineutrino or other energy momentum and quantum number conserving pairs, are very much suppressed, due to the weak interaction.

Here are the possible decays of the $π^0$, and a list of the reasons other are forbidden.

As for the title:

Orbital angular momentum for single particle in particle physics

Elementary particles have an intrinsic angular momentum called spin. The pi0 has zero spin, this means that the photons, each with a spin vector 1, have to have spin projection opposite to add to the zero spin of the pi0.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.