# Electron-positron annihilation: Feynman diagram

Maybe my question is just a blunder. Consider Compton interaction: $e^{-}+\gamma \rightarrow e^{-}+\gamma$. There are two Feynman diagrams related to this process to the lowest order of $\alpha$.

Now consider $e^{-}+e^{+} \rightarrow \mu^{-}+\mu ^{+}$. I think here there are two different diagrams as well. First, connect electron and positron to a virtual photon (propagator) with two free points that can be connected to a fermion called $A$ and $B$. I guess one can connect $A$ to muon or to anti-muon so there are two different Feynman diagrams for this process to the lowest order of the electron charge. However, I checked this interaction on QFT reference books such as Peskin (Chapter 5) and they have just calculated one of this diagrams. What is the problem?

Note: There are other diagrams too.

First one:

Second one:

• Can you draw the diagrams you're talking about? Jul 15 '16 at 17:39
• @knzhou I edited my question Jul 15 '16 at 17:41
• Put the arrows on the fermion lines and you should see the problem with your first diagram ;) Jul 15 '16 at 17:57
• @annav done! seems there is only 1 diagram in the lowest order but I don't know why the other (maybe others) isn't allowed Jul 15 '16 at 18:15

The diagrams you are drawing are not allowed. In the last one you have an electron going into an muon, by the emission of a photon. Try to isolate that part. If it works one way, it should also work the other way - a muon should be able to decay into an electron by the emission of a photon. This cannot happen. Muons decay to electrons in the weak interaction, not QED.

In the first one, you have the same problem, plus an additional one: you don't obey charge conservation in your vertices. An electron (charge -1) cannot become an anti-muon (charge 1), no matter how many photons it emits.

If you're only considering QED, there is only one vertex for this interaction -- electron and positron annhilate to photon, photon pair-produces muon and antimuon. Any electron, photon, muon vertex will have flavor conservation issues.

If you're not JUST considering QED, there is a second diagram, identical to the first, with the photon replaced with a $Z^{0}$. For low energies, the contribution of this one is small, compared to the first, because it's suppressed with factors of $m_{Z}$

Basically, your proposed vertex of $e^{-} \rightarrow \gamma + \mu^{-}$ violates symmetries that you haven't yet encountered in Peskin and Schroder.

• I edited my question Jul 15 '16 at 17:41
• @KNP: yes, in your edit, you include an $e^{-}$ hitting a vertex, and becoming a $\mu^{-}$ and a $\gamma$. This is forbidden by flavor conservation. Jul 15 '16 at 20:02

The two diagrams you draw are all wrong, for the vortex is determined by Lagrangian.

• I suggest to add more details. Jul 15 '16 at 19:05
• Yes. None of them are true. I want to know why they are not allowed? (or why they are zero?) Jul 15 '16 at 19:15