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In Matthew D. Schwartz's QFT book, Chapter 28, the author claims when $\hbar \rightarrow 0$, the following equality (eq 28.4) holds:

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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.

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    $\begingroup$ In what sense do you think the method of stationary phase is inapplicable? $\endgroup$ – J. Murray Jun 19 at 2:38
  • $\begingroup$ @J.Murray I think beacuse a $\phi(x)$ is inserted into the path integral? $\endgroup$ – Sven2009 Jun 19 at 2:52
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    $\begingroup$ If you can articulate why you think that is problematic, you may have better luck finding an answer. $\endgroup$ – J. Murray Jun 19 at 2:57
  • $\begingroup$ @J.Murray I have updated some calculations. Could you have a look? $\endgroup$ – Sven2009 Jun 19 at 4:06
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Hint: Scale the quantum fluctuations $\eta$ with a factor $\sqrt{\hbar}$, i.e. put $$\phi~=~v+\sqrt{\hbar}\eta.$$ This makes it easier to see that the extra terms vanish as $\hbar\to 0$. See also e.g. this related Phys.SE post.

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