Making sense of stationary phase method for the path integral

Question

I am trying to understand this paper/set of notes. I have already seen the following related question: Does the stationary phase approximation equal the tree-level term? but had some trouble following the accepted answer as I am still new to this topic.

They consider a path integral of the form $$\int[d\phi]e^{iS[\phi]}$$ and use the stationary phase method/saddle point method. I am not very familiar with this method, but from what I understand one takes a field $\phi_c$ such that it is an extrema of the classical action: $\frac{\delta S}{\delta \phi}\Big|_{\phi_c} = 0$ if no sources are present or $\frac{\delta S}{\delta \phi}\Big|_{\phi_c} = J$ in the presence of a source $J$. The stationary phase method/saddle point method gives a way to approximate the path integral by using the fact that it is a Gaussian integral. The final approximation involves a functional determinant.

One can rewrite any field by centering it around the $\phi_c$ found above, i.e. $$\phi = \phi_c + \delta \phi. \tag{1}$$

The paper then says that by applying the decomposition (1) to the path integral one gets $$Z = e^{iS[\phi_c]} \int [d\delta \phi]\exp\Big(\frac i2\delta\phi \frac{\delta^2 S}{\delta\phi\delta\phi}\Big|_{\phi_c}\delta\phi\Big). \tag{2}$$

My questions are:

1. Is my understanding of the stationary phase method/saddle point method right or is there more to it?

2. How did they obtain (2)? It seems more was done than just substituting the decomposition (1) into the path integral.

3. How is the integrand in (2) the variation of an amplitude?

4. How does (2) "recover $\hbar$ below the classical action" in any sense if $\hbar$ is not even present?

I apologize for including multiple questions in one post. If the Phys.SE community prefer I can split this into multiple questions.

Answer 1

You want to compute the generating functional $$Z[J]=\int [d\phi] e^{\frac{i}{\hbar} (S[\phi]+\phi \cdot J)}, \quad Z[0]=1, \quad \phi\cdot J:=\int\! dx \, \phi(x) J(x)$$ in the quasiclassical approximation. To this end you perform the transformation of variables $\phi= \phi_c + \sqrt{\hbar} \, \chi$ with the new integration variable $\chi$. The function $\phi_c=\phi_c[J]$ is determined in such a way that the term linear in $\chi$ in the expansion $$\begin{align}S[\phi_c+\sqrt{\hbar} \,\chi]+(\phi_c+\sqrt{\hbar} \, \chi)\cdot J&=S[\phi_c]+ \phi_c \cdot J \\[5pt] &+\sqrt{\hbar}\int \! dx \,\left(\frac{\delta S[\phi]}{\delta \phi(x)} {\huge|}_{\phi=\phi_c}\!\!+J(x)\right) \chi(x) \\[5pt]& {}+ \frac{\hbar}{2} \int \! dx_1 \, dx_2 \, \frac{\delta^2 S[\phi]}{\delta\phi(x_1) \delta \phi(x_2)}{\huge|}_{\phi=\phi_c}\chi(x_1) \chi(x_2) \\[5pt] &+ \mathcal{O}(\hbar^{3/2}), \end{align}$$ vanishes, such that $$\frac{\delta S[\phi]}{\delta \phi(x)} {\huge|}_{\phi=\phi_c}+J(x) =0.$$ Using the translation invariance of the measure in the functional integral, one finds $$Z[J]= e^{\frac{i}{\hbar} (S[\phi_c]+ \phi_c \cdot J)} \int [d\chi] e^{\frac{i}{2} \delta^2 S/\delta \phi^2|_{\phi=\phi_c} \chi^2 } + \ldots$$

Answer 2

Your understanding seems fine (if we can deal with these questions it will be perfect).

To obtain (2) the picture I was taught is to see $\phi$ as a big vector with values in each point of space-time. This means you have to descretize spacetime to understand the path integral. $S[\phi]$ is an operator on this vector space, and $\int \ [d\delta \phi]$, also written as $\int D\phi$, is integrating over all coordinates of the vector from $-\inf$ to $\inf$.

With those definitions of course your expansion expansion (1) up to 2nd order in $\delta \phi$ will lead to (2), and one then goes on to show that it has a limit if the discretization steps of space-time go to zero.

The integrand is the amplitude of the wave functional if we create a Schrodinger-type wave equation for the big vector space, basically it is a Gaussian wave function $\Psi(\phi)$ of very many coordinates. We see that it behaves as a large multi-dimensional harmonic oscillator. In the usual way $\Psi(\phi)$ gives the variance of the value of $\phi$ in all space-time points.

The symbol $\hbar$ and its precise value are only recovered if they are already in the first equation you wrote. But the point is that however you scale this, the result will be this big harmonic oscillator so any non-zero scale factor of the initially defined $S$ will do, and it can later be absorbed in the choice of the scale factor for $\phi$.