# How do I know if the propagator in my Feynman diagram is a photon or a fermion?

When going over QED Feynman diagrams I came across the interaction:

$$e^+ (p_1) +e^- (p_2) \rightarrow \gamma (p_3) + \gamma (p_4)$$

I am trying to draw all the tree-level diagrams contributing to this interaction.

I am aware of the s-channel and t-channel forms for the diagram. (Using the s-channel diagram as an example) I have two incoming fermions connecting\annihilating at a vertex (electron and positron) and a propagator from this initial vertex to a final one where two photons are generated.

How do I know if the propagator between the two vertices is a photon propagator (curved line) or a fermion/antifermion propagator (straight line)?

• There isn't an s-channel diagram for this process, there is however a u-channel and a t-channel – Triatticus Dec 27 '19 at 19:36

There is no trivial way to start with a list of initial-state and final-state particles and determine which channels are involved. You just have to start enumerating diagrams, and working out if they are allowed or not.

To do that you use a small set of tools

• Each fundamental interaction has an allowed set of diagrams. You have to know or look these up.
• Each vertex must conserve four-momentum.
• Each exernal line has to be on-shell (have the correct mass) but internal lines can be off-shell.

Then you start drawing and discard any diagram that fails to meet one of those rquirements.

In general it is difficult to enumerate all possible diagrams, but you have a two-in-two-out reaction and have specified "tree level", so there are only three topologies and only one internal line (where you get a choice of identity) in each of them. You've also specified "QED" so I'm going to limit myself to electromagnetic and weak interactions (strict QED doesn't inlude the weak interactions, but I did weak physics for most of my career, so I include it by default).

You just go to town.

Take the s-channel mentioned in the comments for an example. On the incident side the only diagrams with two charged lepton legs have either a photon or a $$Z^0$$ on the third leg, and that must be the particle that participates in the vertex with the two out-going photons. Can either of those bosons do that?

Doing the t- and u-channels is left as an exercise, but I'll give you a hint: the internal line will be off-shell or you won't be able to conserve four-momentum at the vertices.