The interaction is linear in the Gaussian field $\chi$, so completing the functional square makes its order-$\lambda^2$ contribution exact:
$$
W[\phi]=S_1[\phi]-\frac{\lambda^2}{2}
\int d^dx\,d^dy\,\phi^2(x)\Delta_M(x-y)\phi^2(y).
$$
In momentum space, $\Delta_M(p)=(p^2+M^2)^{-1}$. If every external momentum satisfies $p^2\ll M^2$, the <derivative expansion>
$$
\Delta_M(p)=\frac1{M^2}-\frac{p^2}{M^4}+O(M^{-6}p^4)
$$
starts with the local effective interaction
$$
-\frac{\lambda^2}{2M^2}\int d^dx\,\phi^4(x),
$$
which is the field-theory version of the zero-dimensional result.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 3 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-304/3/b/ii/solution