Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 2

With all momenta incoming and Fourier convention $e^{-ikx}$, differentiating the scalar contributes $-i{k}_{\mu }$. The sole interaction vertex is thereforewhere $k$ enters on the scalar line; reversing the convention reverses the irrelevant overall sign. The free internal lines use the Dirac propagatorand the scalar Feynman propagator $i/(p^2-{\mu }^2+iϵ)$. Momentum is conserved at the vertex.

In four spacetime dimensions $[\psi ]=3/2$, $[\varphi ]=1$, and $[{\partial }_{\mu }]=1$. HenceThe coupling is an irrelevant coupling by power counting in quantum field theory, so it is a nonrenormalizable interaction that would ordinarily define only an effective theory with a cutoff. Here it is also a redundant operator: integration by parts gives $\lambda \varphi {\partial }_{\mu }{j}^{\mu }$ up to a boundary term, and the Dirac current $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ is conserved. This derivative coupling to a conserved current is removed exactly by the local phase redefinition $\psi =e^{-i\lambda \varphi }\chi$.

The tree amplitude is, up to an overall convention-dependent sign,The Clifford algebra gives the momentum-space Dirac equation and its adjoint:ThereforeAlthough $\mu >2m$ permits the relativistic two-body decay kinematically, the matrix element vanishes. Thus

Every scalar-fermion vertex contracts the scalar momentum with the Dirac current. Between on-shell external spinors,and the analogous particle-antiparticle identity also vanishes. Equivalently, the field redefinition in the previous part turns the theory into a free theory. Consequently every putative tree channel for $\psi \bar{\psi }\to \psi \bar{\psi }$ has zero amplitude and