# Fermion-Fermion scattering in Yukawa theory

The interaction term in the Lagrangian for Yukawa theory is given by

$$\mathcal{L}_\text{int} = -g\phi\bar{\Psi}\Psi,$$

where $g$ is the coupling constant, $\phi$ some scalar field and $\Psi$ a fermion field. My question might be a little bit naive but I'm trying to understand how you can see that for a given quantum field theory a particular scattering process is possible.

Consider e.g. fermion-fermion scattering, so $\Psi\Psi\to\Psi\Psi$. How is such a process allowed in Yukawa theory? My point is that there is no term in the interaction Lagrangian which is proportional to something like $\Psi\Psi$. Such a term would probably not be Lorentz invariant, but how do I see that the scattering event I mentioned is allowed nevertheless and that there are not only processes like $\bar{\Psi}\Psi \to \bar{\Psi}\Psi$?

In your case, you actually have incoming fermion $\Psi$, an outgoing fermion $\bar{\Psi}$ and a real scalar field (incoming=outgoing). But you need to see two incoming fermions and two outgoing fermions. Here the vertices represent the interaction term on the question, p's and q are momenta.
• Aren't you discribing the process $\Psi\Psi\to\bar{\Psi}\bar{\Psi}$ here? I think I must be missing the point... – MeMeansMe Jul 15 '18 at 10:27
• @MeMeansMe No. That process can not happen since it violates charge conservation. Note that incoming-to-outgoing fermion, i.e., $\Psi \rightarrow \phi \Psi$ vertex is as same as fermion annihilation, i.e. $\Psi \bar{\Psi} \rightarrow \phi$, the only difference is the direction of time. – Oktay Doğangün Jul 15 '18 at 10:57
• @MeMeansMe if you really want to understand this answer, I suggest you plug the expressions for $\psi$ and $\bar{\psi}$ in terms of creation and annihilation operators and see for yourself that the relevant process is $\Psi \Psi \rightarrow \Psi \Psi$. – Prof. Legolasov Jul 15 '18 at 12:40