Calculating generating functional with stationary phase approximation

Question

Let's say that I have a generating functional $Z[J]$ defined as: \begin{equation*} Z[J]=\int \mathcal{D}\phi\,e^{iS[\phi]+i\int d^4x\,J\phi}.\tag{1} \end{equation*} I want to use the stationary phase approximation, but it gives this (using $\frac{\delta S}{\delta \phi}+J=0$): \begin{equation*} Z[J]=e^{iS_\text{cl.}[\phi]}e^{iF\left[ \frac{-i\delta}{\delta J} \right]}\int\mathcal{D}\Delta\phi\,e^{\frac{i}{2}\int d^4x \int d^4y \left. \frac{\delta^2\mathcal{L}}{\delta \phi(x)\delta\phi(y)}\right|_{\phi_\text{cl.}}\Delta\phi(x)\Delta\phi(y)}e^{i\int d^4x\,J\phi_\text{cl.}},\tag{2} \end{equation*} where $F$ contains all the terms of order $\geq 3$ in $\Delta\phi$. But this expression induces that: \begin{equation*} Z[J]=e^{iS_\text{cl.}[\phi]}e^{iF\left[ \phi_\text{cl.}\right]}\int\mathcal{D}\Delta\phi\,e^{\frac{i}{2}\int d^4x \int d^4y \left. \frac{\delta^2\mathcal{L}}{\delta \phi(x)\delta\phi(y)}\right|_{\phi_\text{cl.}}\Delta\phi(x)\Delta\phi(y)}e^{i\int d^4x\,J\phi_\text{cl.}}.\tag{3} \end{equation*} So an expectation value like $\langle \phi \rangle$ may be written as: $$\begin{align*} \langle \phi \rangle&=\left. \frac{-i\delta}{\delta J} \frac{Z[J]}{Z[0]} \right|_{J=0} \\ &=\phi_\text{cl.} \end{align*}\tag{4}$$ This seems OK but for me, it is problematic because this result does not depend on the approximation we choose for $F$ (the order in $\Delta\phi$ at which $F$ ends). Is this normal or my calculations are wrong or there's a way to make this result dependant on the approximation? (should I really expand the action around the solution with a source?)

Answer 1

1. The first equality in OP's eq. (4) is a general result in Fourier theory that doesn't depend on the stationary phase/WKB approximation.

2. The second equality in OP's eq. (4) is proven in my Phys.SE answer here. Be aware that $\phi_{\rm cl}$ often denotes the Legendre-transformed variable in the effective action $\Gamma[\phi_{\rm cl}]$, as opposed to a classical solution of the Euler-Lagrange (EL) equations. (OP is talking about the latter). The 2 notions agree to ${\cal O}(\hbar)$, i.e. not necessarily at quantum-level.

Answer 2

If we take as you say that $\phi_{cl}$ is a saddle-point and therefore satisfies $$\frac{\delta S}{\delta \phi}\bigg|_{\phi_{cl}} + J = 0$$ for a given $J$, then the first order term is canceled completely on OP's equation (3), that is, no term $e^{i\int J\phi_{cl}}$ should be present.

While on the contrary all the terms in $F$ still depend on the fluctuations $\Delta \phi$ so they cannot be taken out of the path integral. That is where you lose higher order terms. These are generally vanishing in one-loop approximations, but you can see for example if the interaction where $\phi^3$ or anything of higher order that these terms will generate terms which are being neglected, let us make $\phi\rightarrow \phi_{cl} + \Delta\phi$ $$\phi^3 \rightarrow \phi_{cl}^3 + 3\phi_{cl}\Delta\phi + 3\Delta\phi^2\phi_{cl} + \Delta\phi^3$$ So this sort of terms might appear in the Lagrangian and you are taking them out in your Eq.(3). There is no freedom on $F$ as you seem to believe, there is just a truncation up to a certain order and that completely specifies what $F$ is and what one is neglecting (if you want to keep the path integral Gaussian)