# Why there is no $s$-channel for fermion-fermion scattering?

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

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?

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

so there is no s-channel.

• Thanks for the answer! Why it's not possible for the scalar to decay into two fermions at another vertex?
– IGY
Dec 4, 2022 at 19:44
• @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. Dec 4, 2022 at 19:53

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.

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