0
$\begingroup$

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:

https://i.sstatic.net/n9Cg4.jpg

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.