Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 2

Completing the square in the Gaussian functional integral and choosingnormalizes $Z_0[0]=1$. The result iswhere the Feynman propagator in the convention of the question is

The analogous Grassmann Gaussian integral, normalized by $Z_0[0,0]=1$, giveswith the free Dirac propagatorIndeed $(i/{\partial }_x-m)S_F(x-y)=i{\delta }^{(4)}(x-y)$.

With left functional derivatives for the Grassmann sources, replace fields in the interaction by $\varphi \mapsto i^{-1}\delta /\delta J$, $\psi \mapsto i^{-1}\delta /\delta \bar{\eta }$, and $\bar{\psi }\mapsto -i^{-1}\delta /\delta \eta$. ThusThe ordering displayed fixes the Grassmann signs and makes this an explicit source functional with no dynamical fields.

For momenta much smaller than $M$,Expanding the source functional to order $g^2$, equivalently integrating out the heavy scalar by its field equation, producesThus the four-spinor coefficient is $g^2/(2M^2)$ in this normalization. Since $[\psi ]=3/2$ and $[g]=0$ in four dimensions, $(\bar{\psi }\psi {)}^2$ has dimension six and its coefficient has the required dimension $-2$.