QFT: Scattering Amplitudes in Yukawa Theory

1. Sep 20, 2013

Sonny Liston

1. The problem statement, all variables and given/known data
I'm working with the Yukawa theory, where the interaction term in the Lagrangian density is $g\varphi\overline{\psi}\psi$. As an exercise for getting used to using the Feynman rules for the theory, I'm asked to show explicitly (i.e. I'm not allowed to invoke charge conservation) that, at tree level, the amplitude for the process $e^-\varphi \rightarrow e^+\varphi$ is zero.

2. Relevant equations

The interaction is $g\varphi\overline{\psi}\psi$. Just set notation, we're using Srednicki's QFT book and the discussion of Feynman rules for the Yukawa theory is chapter 45.

3. The attempt at a solution

This process yields two diagrams at tree level. My first though was that I would write down the diagrams and they would cancel in some nice way, but I'm actually stuck even trying to write down sensible diagrams for this process; writing down diagrams that are in accord with the Feynman rules while also yielding things that look like amplitudes is proving difficult. In particular, there are no barred spinors in the resultant amplitudes; the incoming electron and outgoing positron both contribute unbarred spinors. Is there a way to write down diagrams for this process that make sense? If so, am I on the right track in expecting that they would cancel, or have I totally misunderstood how to go about this?

2. Sep 20, 2013

Chopin

Can you describe the diagrams you've found that describe this process? Are you sure you have all of your arrowheads going in the correct directions?

3. Sep 21, 2013

Sonny Liston

I think I've got my arrows right. I tried to type out the diagrams themselves, but couldn't get it to work so I'll just describe them.

Starting from the left:
Diagram 1: incoming electron, momentum P, with an arrow to the right flowing toward the first vertex. A scalar line with momentum K flowing into the first vertex, and an internal fermion propagator with momentum (P+K) flowing out of the vertex (arrow to the right). The second vertex has a scalar line with momentum K' flowing out of the vertex, and an outgoing positron, with momentum P'. This has an arrow facing to the left, flowing into the second vertex.

Diagram 2: The same as diagram 1, except the order of the scalar lines is swapped: the first vertex now has a scalar line with momentum K' flowing out of the vertex, yielding an internal propagator with momentum (P-K'), while the second vertex has a scalar line with momentum K flowing into the vertex.

Does that sound right, or have I done something stupid with the arrows for the outgoing positron?

4. Sep 21, 2013

Chopin

Well, that's the right direction of arrow for the outgoing positron, but that means you have two arrows flowing into the second vertex, correct? If so, those don't correspond to valid Feynman rules--the rules say that every vertex in Yukawa theory has one arrow flowing in and one flowing out. This is the way that conservation of charge is enforced at the diagram level--it shouldn't be possible for you to make a diagram that violates charge conservation if you follow the rules correctly.

5. Sep 21, 2013

Sonny Liston

Right, that's part of what I was trying to get at (and expressing poorly) when I said I was having trouble writing down diagrams that made sense. Following the rules for electron/positron arrows produces a diagram that violates the rule concerning arrows flowing into and out of vertices.

So I guess that's my question: does "showing explicitly" that the tree level amplitude is zero just mean showing that you can't write down diagrams for the process that are consistent with all the Feynman rules? Or when you say "it shouldn't be possible for you to make a diagram that violates charge conservation if you follow the rules correctly", do you mean I've not been careful enough somewhere in writing down the diagrams?

6. Sep 22, 2013

Chopin

It's an oddly-worded question, so I guess I'm not sure, but if it were me, that's probably what I would go with. I'm not really sure what else you can do with it--charge conservation in QFT holds diagram-by-diagram, so you're not going to find even one single diagram that violates it. That means there aren't any calculations to do, so I'm not sure how you can "show it explicitly" in any way other than to say that the sum of all relevant diagrams is trivially 0.

7. Sep 22, 2013

Sonny Liston

That was my original thought -- you just couldn't write down sensible diagrams for the process. The wording was what then got me all tangled up thinking diagrams should cancel and whatnot.