For momenta much smaller than $M$,
$$
\frac{i}{p^2-M^2+i\epsilon}=-\frac{i}{M^2}+O(p^2/M^4),
\qquad
D_F(x-y)=-\frac{i}{M^2}\delta^{(4)}(x-y)+O(M^{-4}).
$$
Expanding the source functional to order $g^2$, equivalently integrating out the heavy scalar by its field equation, produces
$$
\mathcal L_{\rm eff}=\bar\psi(i\not\partial-m)\psi
+\frac{g^2}{2M^2}(\bar\psi\psi)^2+O(M^{-4},g^4).
$$
Thus 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 $-2$.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-304/2/d/solution