Is there any method to somewhat rigorously determine the relevant Feynman diagrammes (no loops, $e^2$ approximation)? I don't by doing the full calculation of course. For example, I have the following Lagrangian: $$ \mathcal{L}=-\frac{1}{4} F^{\mu \nu} F_{\mu \nu}+\left(D^{\mu} \chi\right)^{*} D_{\mu} \chi-m^{2} \chi^{*} \chi $$


  • $F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$
  • $D_{\mu} \chi=\partial_{\mu} \chi-i e A_{\mu} \chi$
  • $\chi$ is an $\tilde{e}^{-}$ scalar field

I can look online and check that for a process like $\tilde{e}^{+} e^{-} \rightarrow \tilde{e}^{+} e^{-}$ we get something like a Bhabha scattering meaning with an s channel, a t channel and a seagull channel.

For another process like $\tilde{e}^{+} \tilde{e}^{-} \rightarrow \gamma \gamma$ i find a t channel, a u channel and the seagull channel.

If i take these as given I can find the matrix element and the cross sections using Feynman rules. I am aware that the seagull channel comes from the EM field term in the covariant derivative.

However, I don't understand how we get there? How do I use the Lagrangian to draw the Feynman diagrams and how do I choose which ones I'll keep.

Any help?

  • 1
    $\begingroup$ Do you know have to do it for $\phi^4$ theory? $\endgroup$
    – Qmechanic
    Commented Oct 8, 2021 at 12:14
  • $\begingroup$ I know how to draw the diagrammes I don't know how to pick the "relevant" channels $\endgroup$
    – jmstf94
    Commented Oct 8, 2021 at 12:36
  • 1
    $\begingroup$ A first step would be to expand out the Lagrangian and seperate out the terms that look like $A_\mu \chi^* \chi$ and $A^\mu A_\mu \chi^* \chi$. From this you have the different 3/4-point interactions. $\endgroup$ Commented Oct 8, 2021 at 13:11
  • $\begingroup$ Could you explain better how you get the seagull diagram in the Bhabha scattering? I don't that's possible. $\endgroup$
    – Einj
    Commented Oct 9, 2021 at 19:27

1 Answer 1


I don't know what you mean with "rigorously" determine the Feynman diagrams that contribute to a process. In general from a theoretical point of view you can have two approaches

1)The hamiltonian approach in interaction picture, thorugh Dyson expansion

2)The path integral formalism. LSZ formula relates the transition amplitude to a green function. The latter can be calculated with the generating functional formalism through a perturbative expansion.

Both these methods, if carried on show, which Feynman diagrams are relevant, i.e. give a non null contribute to the transition amplitude. However that is not how things are done in practice. In practice you use the Feynman rules. To understand which diagram contributes, you just have to keep fixed the external states (ex, $ e^-(p_1)+e^+(p_2) \rightarrow e^-(p_3)+e^+(p_4)$, fixed to the angles of a rectangle) and look for which channels are permitted by the vertices of the theory. The fundamental point are the vertices rules, these shows which coupling are permitted and hence which Feynman diagrams contributes. For example from the covariant derivative term you can get a $e^2\chi^*\chi A_{\mu}A^{\mu}$ term, to which correspond the rule $ie^2g^{\mu\nu}$. So you can see that the seagull channel can contribute to the process with two final photons you wrote, but obviously not to your first process $e^-+e^+ \rightarrow e^-+e^+$. To sum up, check which vertices there are in your interaction lagrangian, fix the external legs of the process you desire, build all the Feynman diagrams with that external legs using the vertices the theory has.

  • $\begingroup$ ok man thanks. So in a process like $e^{+} e^{-} \rightarrow \tilde{e}^{+} \tilde{e}^{-}$ we just have an s and a t diagram?($\tilde{e}^{-}$ is a selectron i.e. spin 0)? $\endgroup$
    – jmstf94
    Commented Oct 13, 2021 at 6:56

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