# Three level Feynman diagramas lagrangian density

1. Nov 17, 2014

### pstq

Hi,
I am trying to figure out how to draw all the three level Feynmann diagrams corresponding to this lagrangian density $$L = \frac{1}{2} \partial _{\mu} \phi \partial^{\mu} \phi - \frac{\mu^2}{2}\phi^2- \frac{\eta}{3!}\phi^3-\frac{\lambda}{4!} \phi^4+i \bar{\psi} \gamma _{\mu} \partial^{\mu} \psi \phi -m \bar{\psi} \psi+ig \bar{\psi} \gamma^{5} \psi \phi$$
for this process $$F+ \bar{F} → F+ \bar{F}$$
and φ is the field associated to this particle F.

So i was thinking on drawing the 3 Feynman diagram (i.e. u, s,t channels ) for every interaction term . I mean

for the interaction $$\phi^3$$ three Feynman diagrams, whose vertex are proportional to
$$\eta^2$$
for $$\phi^4$$ another three , is that right ?

the problem is that I think that we don't have u channel in the $$\phi^3$$ case, but I am not sure why . So if someone could enlighten me about this as well, you will make another fellow human interested in particle physics very happy today.

and another question, $$\psi$$ is the dirac spinor for another particle X which is not F, would i need to take into account the last term of the above Lagrangian which is interaction term between the particles F and the others , if I am considering only the above process $$F+ \bar{F} → F+ \bar{F}$$ or not?

Any help with any question/ or any remark would be highly appreciated

thanks !

2. Nov 17, 2014

### pstq

this is not coursework questions nor homework per se, so I hope this is the right place to post this. In case Im wrong , my sincerest apologies

3. Nov 17, 2014

### The_Duck

I guess you mean *tree* level?

From your Lagrangian, it looks like $\phi$ is a real scalar field. So the F particle is its own antiparticle. So I guess you are considering $F + F \to F + F$?

The $\phi^3$ interaction produces three tree-level diagrams, corresponding to the u, s, and t channels. However the $\phi^4$ interaction produces only one tree-level diagram for the process you are interested in.

There is definitely a u channel diagram.

The X particle will only matter if you consider loop diagrams. Since there are no X's in the initial and final states, they can only appear in loops.

4. Nov 17, 2014

### ChrisVer

It's better to ask yourself why don't you see a u-channel for $\phi$? (however you saw a t-channel)

For $\phi^4$ you will only get the diagram that looks like this: X
for 4 external legs...

For the last term, #2 post is totally right...

5. Nov 22, 2014

### pstq

thanks a lot :):):)