I’m calculating some Feynmann Amplitudes, in particular the Positron-Photon (Compton) Scattering.

In general the fermion propagator is:

$$iS_F(q) = \frac{i(\gamma^{\mu}q_{\mu}+m)}{q^2-m^2}$$

The Feynmann diagram of type S of the mentioned process is the following:


Since for internal fermion Lines we have that the four-momentum labels on Feynman diagram always represent energy-momentum flow in the SAME direction as the arrows, in our case then the correct expression for the propagator should be:

$$iS_F(q) = \frac{i(\gamma^{\mu}(-p-k)_{\mu}+m)}{(p+k)^2-m^2}$$

Is this correct?


1 Answer 1


If time is left-to-right in your diagram, then momentum conservation implies that the internal fermion has momentum $p+k=p'+k'$. The convention of reversing arrows for antiparticles is just to make it easy to check that the discrete quantum numbers that change sign for antiparticles -- like charge -- are conserved at each node. But conservation of energy and momentum apply the same to positrons as electrons.

  • $\begingroup$ Yes, indeed if q enters the vertice it has to be q=-k-p. However my question was about another thing, it was if we have to use q=-k-p in the propagator when calculating the amplitude. OR if we had to use k+p like in the elettron case. $\endgroup$ Commented Sep 16, 2021 at 10:51

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