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Consider a massless Dirac or Majorana fermion $\psi$ and a massless scalar $\phi$. They interact through a Lagrangian $\mathcal{L}_I(\phi,\psi)$. I would like to understand what are the cubic interactions that contribute to the $2\to2$ scattering of fermions. The general term takes the form $$ \mathcal{L}_I \supset (\partial^m \bar\psi \, M \, \partial^n \psi) \, \partial^p \phi, $$ where in this schematic notation $m,n,p$ are spacetime indices to be contracted in such a way that the final result is a Lorentz scalar and $M$ is a matrix given by (products of) the identity $4\times4$ matrix, the Dirac matrices $\gamma^\mu$ and $\gamma^5$. I'd like to order the Lagrangian terms according to their mass dimension, i.e. with the number of derivatives. Importantly, I don't care about renormalizability and so I can have an arbitrary number of derivatives.

Of course there is the case with $m=n=p=0$, $M=1_{4\times4}$, which is just the ordinary Yukawa term $\bar\psi\psi \phi$ for a scalar (let's not consider for simplicity the case in which the scalar is a pseudo-scalar).

As far as I understand, a cubic vertex produces a four-point amplitude only when the it's non-zero on-shell. For a theory of scalars only, one can argue (see section 3.1.1 of https://arxiv.org/abs/1708.03872) that only $\phi^3$ survives, because all other terms $(\partial^m \phi \partial^n \phi \partial^p \phi)$ vanish on-shell. Is a similar analysis as trivial for a system of fermions and scalars?

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  • $\begingroup$ $m,n,p$ are space time indices, but then you set them to zero to find the Yukawa coupling? $\endgroup$ Commented Apr 24 at 11:49
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    $\begingroup$ It's not that you set them to zero. It was just a schematic notation. $m$ can be a set of spacetime indices. $m=0$ means no derivative at all. $\endgroup$
    – Rubilax96
    Commented Apr 24 at 13:50

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