[Physics] Electron-positron annihilation: Feynman diagram

feynman-diagramsquantum-electrodynamicsquantum-field-theory

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:
enter image description here

Second one:
enter image description here

Best Answer

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.