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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 $\bar{\Psi}\Psi \to \bar{\Psi}\Psi$?

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The interaction is the building block, you need to construct the process by expanding the interaction term into higher terms and write the probability amplitude (See a QFT or particle physics textbook like Griffiths' or Peskin-Schröder's).

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.

So, for the tree level (the most minimal process), one of the incoming fermions should match with one of the outgoing fermions. Since it is from 2 to 2, it matches all. So, the scalar field should be virtual, i.e., one of the scalar field from one incoming-outgoing fermion vertex should also be the scalar for the other pair.

enter image description here Here the vertices represent the interaction term on the question, p's and q are momenta.

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  • $\begingroup$ Aren't you discribing the process $\Psi\Psi\to\bar{\Psi}\bar{\Psi}$ here? I think I must be missing the point... $\endgroup$ – MeMeansMe Jul 15 '18 at 10:27
  • $\begingroup$ @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. $\endgroup$ – Oktay Doğangün Jul 15 '18 at 10:57
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    $\begingroup$ @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 $\Psi \Psi \rightarrow \Psi \Psi$. $\endgroup$ – Prof. Legolasov Jul 15 '18 at 12:40

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