# How can we derive the Feynman rule for the ordinary QED 3-vertex?

I have checked some Quantum Field Theory texts that include basic QED and they all include the Feynman rule that each vertex bring with it a factor of $$\pm i e \gamma^\mu$$ but I have yet to find a derivation of this rule. How can we start from the interaction term in the Lagrangian, $-e \bar{\psi}\gamma^\mu A_\mu \psi,$ and derive this rule?

• In the time-dependent perturbation theory, we get $i$'s because we expand $e^{-iV_{I}t}$ in a Taylor series. $i$'s appearing in Feynman rules have exactly the same origin. Dec 14, 2013 at 10:58
• @higgsss I am interested in a derivation of the entire contribution Dec 14, 2013 at 11:00
• It is really the $i$ from $\exp(-iV_{int}t)$, or in the language of the path integral, the $i$ from $\exp(i\int d^{4}\mathcal{L_{int}})$. Any QFT textbook should have the derivation in where it introduces the interaction picture, perturbation theory, and eventually Feynman rules. Dec 14, 2013 at 11:11
• @higgsss As I mentioned, I consulted two textbooks before asking and while both had the rule, neither has a real derivation. They just pluck the rule out of the air. Of course these rules come from the action in the path integral and thus also from the interaction Lagrangian, but how, exactly? Dec 14, 2013 at 11:14
• What books did you consult? Most QFT books do have a derivation. Could it be that you went directly to the parts about QED, while the reasoning behind diagrammatic perturbation theory where explained earlier in simpler settings (such as scalar $phi^3$ or $phi^4$ theory)? Are you ok with the derivation in those simpler situations? If so, you should explain what the problem with QED is, if not, check out those first.
– fqq
Dec 15, 2013 at 15:53