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I'm learning the Lagrangian for Yukawa theory, where $L_{int} = \phi\bar{\psi}\psi$. For the fermion-fermion scattering, we can draw the Feynman diagrams as

enter image description here

My question is why we can't have $s$-channel here? If it exists, it still seems like we can have a diagram with the appropriate vertices. Also, does the direction of the arrow matter for that internal propagator?

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2 Answers 2

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Because the interaction has to have fermion going in and fermion coming out, or anti-fermion in/anti-fermion out. The arrow shows the direction of fermion propagation. There is no interaction like

interaction

so there is no s-channel.

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  • $\begingroup$ Thanks for the answer! Why it's not possible for the scalar to decay into two fermions at another vertex? $\endgroup$
    – IGY
    Dec 4, 2022 at 19:44
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    $\begingroup$ @IGY Because $\psi \sim a + b^{\dagger}$ while $\psi^{\dagger} \sim a^{\dagger} + b$, this means that $\psi$ can either destroy a fermion or create an anti-fermion, while $\psi^{\dagger}$ can either destroy an anti-fermion or create an fermion. If your vertex is $\phi \psi^{\dagger} \psi$ you can't create (or destroy) two fermions at once, because you will need $\psi \psi$ (or $\psi^{\dagger} \psi^{\dagger}$) in the vertex. $\endgroup$ Dec 4, 2022 at 19:53
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For $s$-channel, the fermions need to annihilate, so:

$$ e^+ e^- \rightarrow e^+ + e^- $$

is $s$-channel. There is still a $t$-channel, of course, but no $u$-channel as the electron and positron are not identical particles.

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  • $\begingroup$ Thanks for the answer, is $e^-e^+$ fermion-antifermion scattering? $\endgroup$
    – IGY
    Dec 4, 2022 at 19:42
  • $\begingroup$ yes, but so is $\mu^+ e^-$, which lacks an $s$-channel. I think $e-+\bar\nu_e \rightarrow W^- \rightarrow X$ works to. $\endgroup$
    – JEB
    Dec 5, 2022 at 1:25

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