I am trying to draw the Feynman diagram for the following scattering amplitude (f a fermion)
$$
i\mathcal{M}(f\overline{f}\phi\phi\phi)
$$
Given the following interaction term in the Lagrangian:
$$
\mathcal{L}_I = -g\phi\psi\overline{\psi}
$$
Now, I am trying to construct a feynman diagram from the rules of the theory, but I am having trouble recognising which diagrams are not allowed.
Which symmetries should I be respecting at each vertex?
Should I think of this as fermion/anti-fermion annihilation which produces two scalars, one of which then decays, producing another scalar?
Best Answer
Assuming that $g\phi\bar{\psi}\psi$ is the only interaction term you will need at least 3 vertices to produce 3 external $\phi$-legs. Therefore, the amplitude at leading order in perturbation theory is going to be of the form $$ \mathcal{M}\propto \bar{v}(p_1)\Delta_F(p_1-p_3)\Delta_F(p_1-p_3-p_4)u(p_2)+\ldots $$ where $\Delta_F$ is the fermion propagator, and the $\ldots$ refers to the permutations of the three bosons momenta $p_3$, $p_4$ and $p_5$. It comes from the Feynmann diagrams where 3 boson legs attach to the internal fermion line in t-channel (corresponding i.e. to 2 fermionic propagators).