# Follow-up on "Conservation of electric charge in Feynman diagram"

I took the following Feynman diagrams from page $$5$$ of this document:

For the right diagram, I drew the circled $$1$$ and $$2$$ myself. What I unfortunately don't understand yet is:

(i) This question is very similar to this one, though the answers there don't really answer my question, I'm afraid: Why is the left diagram allowed? The propagator would probably be an electron $$e^{-}$$ or positron $$e^{+}$$, but wouldn't charge conservation be violated at one vertex? For example: If we choose as the propagator an $$e^{-}$$, then for the upper vertex, we have one incoming $$e^{-}$$ and an outgoing $$e^{+}$$.

(ii) Up to now, I haven't really understood $$u$$-channels for Feynman diagrams. Where I drew the circled $$1$$ and $$2$$, are these vertices? If so, then I have the same question as in (i): Why isn't charge conservation violated?

• I'm no expert on Feynman diagrams, but I don't think that the diagram on the right is correct, either. I think it's just a "shorthand" version. See this answer by anna v about photon-photon interaction. Commented Dec 4, 2020 at 11:21
• @PM2Ring I'm not sure how the linked answer corresponds to this question Commented Dec 4, 2020 at 12:13
• @NiharKarve Both questions are about 2 photon interactions. True, the time direction of the linked question is opposite to this one, but that's not really relevant in Feynman diagrams. But I'm certainly not proposing the linked question as a duplicate. I linked it because of its Feynman diagram. Commented Dec 4, 2020 at 12:18
• Nothing wrong with either diagram Commented Dec 4, 2020 at 15:01

So the first diagram can be read as follows: $$e^-$$ emits $$\gamma$$ (lower vertex), and then annihilates with $$e^+$$ (upper vertex) producing another $$\gamma$$. Or another (but equivalent) reading: $$e^+$$ emits $$\gamma$$ (upper vertex), and then annihilates with $$e^-$$ (lower vertex) producing another $$\gamma$$.
• Okay, I think I see my mistake now: I assumed that for the first diagram, the positron $e^{+}$ would be an outgoing particle, not an incoming one ... :-)