Calc Feynman Amp Q: Chi+ Chi- -> Chi+ Chi-

[div curl F= 0]

I'm having a problem calculating the Feynman amplitude for the scalar scattering process \chi^+ \chi^- \to \chi^+ \chi^- for an interaction Lagrangian which is:

\mathcal{L} = - g \chi^\dagger \chi \Phi - \frac{\lambda}{4} (\chi^\dagger \chi)^2

So far I have the 2 Feynman Diagrams for \chi^+ \chi^- \to \Phi \to \chi^+ \chi^- but I can't think/remember how many there should be for the quartic term. I'm thinking there should only be one diagram and hence only one contribution to the Feynman amplitude (which should be -i lambda/4), so the total amplitude becomes:

(-ig)^2 \left(\frac{i}{(p_1 + p_2)^2 - M^2} + \frac{i}{(p_1 - k_1)^2 - M^2} \right) - \frac{i\lambda}{4}

where M is the mass of Phi boson, p_1 and p_2 are the incoming energy-momenta and k_1 and k_2 are the outgoing energy-momenta.

Am I along the right lines?

Thanks

 

[azntoon]

in advance!Yes, you are on the right track. The Feynman amplitude for the scalar scattering process $\chi^+ \chi^- \to \chi^+ \chi^-$ is given by: \begin{align}i\mathcal{M} = (-ig)^2 \left(\frac{i}{(p_1 + p_2)^2 - M^2} + \frac{i}{(p_1 - k_1)^2 - M^2} \right) - \frac{i\lambda}{4}.\end{align}This is due to the fact that there will be two diagrams for the interaction term $g \chi^\dagger \chi \Phi$ and only one diagram for the quartic term $\lambda (\chi^\dagger \chi)^2$.