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?

  • $\begingroup$ Could you perhaps try to indicate more clearly what the initial and final particles are? Do you mean $i\mathcal M (f\bar f \to \phi\phi\phi)$? $\endgroup$
    – Danu
    Oct 6, 2015 at 16:44
  • $\begingroup$ I guess so, but actually I wouldn't mind knowing if any other initial/final states are even allowed by the theory. For example, I assume that $f\phi\longrightarrow\overline{f}\phi\phi$ isn't allowed, but not exactly certain on the reasons why. $\endgroup$
    – user94752
    Oct 6, 2015 at 17:30
  • 2
    $\begingroup$ when in doubt go back to wick contractions and commutation relations and derive the Feynman rules by your self! $\endgroup$
    – Ali Moh
    Oct 6, 2015 at 18:32

1 Answer 1


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).


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