In Matthew D. Schwartz's QFT book, Chapter 28, the author claims when $\hbar \rightarrow 0$, the following equality (eq 28.4) holds:
So how can I see the second "$=$" holds? It seems the method of stationary phase is inapplicable?
UPDATE: Below are my calculations: By definition, \begin{equation} \langle \Omega|\phi(x)|\Omega \rangle=\frac{\int \mathcal{D}\phi \exp\{\frac{i}{\hbar}S[\phi]\}\phi(x)}{\int \mathcal{D}\phi \exp\{\frac{i}{\hbar}S[\phi]\}}. \end{equation} Suppose the solution of equation of motion $\delta S=0$ is given by $\phi=v=$ constant. We write $\phi=\eta+v$ and the expectation value is now \begin{equation} \langle \Omega|\phi(x)|\Omega \rangle=v+\frac{\int \mathcal{D}\eta \exp\{\frac{i}{\hbar}S[\eta+v]\}\eta(x)}{\int \mathcal{D}\eta \exp\{\frac{i}{\hbar}S[\eta+v]\}}. \end{equation} We continue to deal with $S$ up to 2nd order: \begin{equation} S[\eta+v]=S[v]+\mbox{vanishing linear term}+\frac{1}{2}S''[v]\eta^2. \end{equation} After some functional algebra, we get something like \begin{equation} \frac{\int \mathcal{D}\eta \exp\{\frac{i}{\hbar}S[\eta+v]\}\eta(x)}{\int \mathcal{D}\eta \exp\{\frac{i}{\hbar}S[\eta+v]\}}=(-i\hbar)\frac{\partial}{\partial J(x)}\exp\{\int dx' dy\frac{i}{\hbar}J(x')[-2S''(v)]^{-1}J(y)\}|_{J=0}. \end{equation} So the righthand side is vanishing under limit $\hbar \rightarrow 0$? I am not sure whether or not my calculation is correct.
