With all momenta incoming and Fourier convention $e^{-ikx}$, differentiating the scalar contributes $-ik_\mu$. The sole interaction vertex is therefore
$$
\boxed{\bar\psi\psi\phi:\quad -\lambda\gamma^\mu k_\mu=-\lambda\not k,}
$$
where $k$ enters on the scalar line; reversing the convention reverses the irrelevant overall sign. The free internal lines use the <Dirac propagator>
$$
\frac{i(\not p+m)}{p^2-m^2+i\epsilon}
$$
and the scalar <Feynman propagator> $i/(p^2-\mu^2+i\epsilon)$. Momentum is conserved at the vertex.

\Image[/media/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301-derivative-coupling-diagrams.png]
{title=Derivative scalar-current vertex and scalar decay cut diagram}
{description=The scalar momentum enters a derivative vertex on an oriented fermion line. The decay amplitude and its conjugate vanish because the scalar momentum contracts the conserved on-shell Dirac current.}
{height=500}

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301/2/i/solution