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Considering the Yukawa interaction $-g\phi\bar{\psi}\psi$ and $e^{-}e^{-} \rightarrow e^{-}e^{-}$ scattering, the 2nd order term in the numerator of the gell man low formula is

$\frac{(-ig)^2}{2!}$ <0|$T\psi\psi\bar{\psi}\bar{\psi}\psi\bar{\psi}\phi\psi\bar{\psi}\phi$|0>.

In order to get the $t$-channel diagram for the process, we’d have to contract a $\psi$ from an incoming particle to one of the $\bar{\psi}$ from the first vertex. However we’d need to contract the other $\psi$ from the incoming particle to the first vertex (otherwise we’d get the u channel diagram). But the only other fields to contract with are $\phi$ and $\psi$, which give 0 if I’m correct. So how does the t-channel for electron-electron scattering in this theory exist?

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  • $\begingroup$ It's the Yukawa analogue of Møller scattering in QED. You get the same tree diagrams, namely a $t,u$ channel. $\endgroup$
    – LPZ
    Commented Oct 20, 2023 at 11:52
  • $\begingroup$ @knzhou still, how can you contract it to get the t-channel? Which contractions correspond to that? Each part of a Feynman diagram corresponds to a contraction, so you’d still have 2 spinor fields contracted to two other spinor fields with one being the same spinor field. $\endgroup$
    – user310742
    Commented Oct 20, 2023 at 14:22
  • $\begingroup$ @LPZ I know, what I am asking is how to derive them from contractions. $\endgroup$
    – user310742
    Commented Oct 20, 2023 at 14:24
  • $\begingroup$ Do you know how to derive the Feynman rules from contractions? $\endgroup$
    – LPZ
    Commented Oct 20, 2023 at 14:48
  • $\begingroup$ @LPZ yes, however I’m asking which contractions give the t channel $\endgroup$
    – user310742
    Commented Oct 20, 2023 at 20:29

1 Answer 1

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The three channels are

Instead of looking for the t-channel you consider the s-channel. Which is indeed absent in this model.

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