Change variables in the defining integral from the fluctuation to the total field, $\varphi=\phi_0+\eta$. Then
$$
\begin{aligned}
e^{-W(J;\phi_0)/\hbar}
&=\int d\varphi\,
e^{-[S(\varphi)+J(\varphi-\phi_0)]/\hbar}\\
&=e^{J\phi_0/\hbar}e^{-W(J;0)/\hbar},
\end{aligned}
$$
and hence
$$
W(J;\phi_0)=W(J;0)-J\phi_0.
$$
It follows nonperturbatively that
$$
\chi=\partial_JW(J;\phi_0)
=\partial_JW(J;0)-\phi_0.
$$
Thus the same source $J_\chi$ corresponds at zero background to the mean field $\Phi=\chi+\phi_0$. Using the source-sign-compatible Legendre transform $\Gamma(\chi;\phi_0)=W(J_\chi;\phi_0)-J_\chi\chi$,
$$
\Gamma(\chi;\phi_0)
=W(J_\chi;0)-J_\chi(\chi+\phi_0)
=\Gamma(\chi+\phi_0;0).
$$
Renaming $\chi$ as $\eta$ gives the requested identity
$$
\boxed{\Gamma(\eta;\phi_0)=\Gamma(\phi_0+\eta;0).}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-304/1/d/solution