Substitution of $\Phi=\varphi+\lambda\varphi^2$ gives
$$
\mathcal L_{\mathrm{free}}
=\frac12(\partial\varphi)^2-\frac12m^2\varphi^2
$$
and
$$
\mathcal L_{\mathrm{int}}
=2\lambda\varphi(\partial\varphi)^2-m^2\lambda\varphi^3
+2\lambda^2\varphi^2(\partial\varphi)^2
-\frac12m^2\lambda^2\varphi^4.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 301 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-301/4/a/solution