# Confusions on expectation value for $\hbar$ going to zero

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.

• In what sense do you think the method of stationary phase is inapplicable? – J. Murray Jun 19 at 2:38
• @J.Murray I think beacuse a $\phi(x)$ is inserted into the path integral? – Sven2009 Jun 19 at 2:52
• If you can articulate why you think that is problematic, you may have better luck finding an answer. – J. Murray Jun 19 at 2:57
• @J.Murray I have updated some calculations. Could you have a look? – Sven2009 Jun 19 at 4:06

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.