Split the field into disjoint ranges of <Fourier modes>,
$$
\phi_{\rm tot}=\phi+\phi^+,
\qquad
\widetilde\phi(p)=0\ (p^2>\Lambda^2),
\qquad
\widetilde{\phi^+}(p)=0\ \text{unless }\Lambda^2<p^2\leq\Lambda_0^2.
$$
The <Wilsonian effective action> is defined by
$$
e^{-S_\Lambda^{\rm eff}[\phi]}
=\int_\Lambda^{\Lambda_0}\mathcal D\phi^+\,
e^{-S_{\Lambda_0}[\phi+\phi^+]}.
$$
Writing $\Delta S[\phi,\phi^+]=S_{\Lambda_0}[\phi+\phi^+]-S_{\Lambda_0}[\phi]$ immediately gives
$$
\boxed{S_\Lambda^{\rm eff}[\phi]=S_{\Lambda_0}[\phi]
-\log\int_\Lambda^{\Lambda_0}\mathcal D\phi^+e^{-\Delta S[\phi,\phi^+]}.}
$$
Quadratic cross terms vanish because the momentum supports do not overlap. For $h_0\ne0$,
$$
\begin{aligned}
\Delta S={}&S_{0,>}[\phi^+]
+\int d^4x\left\{
\frac{h_0}{3!}\left[3\phi^2\phi^++3\phi(\phi^+)^2+(\phi^+)^3\right]\right.\\
&\left.\hspace{31mm}
+\frac{g_0}{4!}\left[4\phi^3\phi^++6\phi^2(\phi^+)^2
+4\phi(\phi^+)^3+(\phi^+)^4\right]\right\},
\end{aligned}
$$
where
$$
S_{0,>}[\phi^+]=\frac12\int d^4x\,
\left[(\partial\phi^+)^2+m_0^2(\phi^+)^2\right].
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 2 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-304/2/a/i/solution