# Calculating generating functional with stationary phase approximation

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?)

• How did the $J\phi$ term in eq. (1) become $J\phi_{\rm cl}$ in eq. (2)? – Qmechanic Mar 29 at 12:20
• @Qmechanic Since $J\phi=J\phi_\text{cl.}+J\Delta\phi$, $J\Delta\phi$ is absorbed by the classical equation of motion $\frac{\delta S}{\delta \phi}\Delta\phi+J\Delta\phi$ when expanding the action around the classical solution. Then the only term surviving this is $J\phi_\text{cl.}$. – Jeanbaptiste Roux Mar 29 at 13:02

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.

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)

• So, in order to have $Z[J]$ that still depends on $J$, I have to take a $\phi_\text{cl.}$ that satisfies the free equation of motion $\frac{\delta S}{\delta \phi}=0$? – Jeanbaptiste Roux Mar 29 at 13:26
• Depends on what you are looking into... I guess. If you want only to approximate $Z[J]$, the saddle point corresponds to the EOM above. And you still have $Z[J]$ where $J$ appears in higher orders together with the fluctuations. $\Delta \phi$. – ohneVal Mar 29 at 13:52
• Why will $Z[J]$ still depend on $J$ if $e^{i\int J\phi_\text{cl.}}$ disappear, as you said? – Jeanbaptiste Roux Mar 29 at 14:52
• But with those terms, in the expansion of the action I have terms like $\left(\left.\frac{\delta S}{\delta\phi}\right|_{\phi_\text{cl.}}+J\right)\Delta\phi$, and this gives $0$. – Jeanbaptiste Roux Mar 29 at 15:24
• True, the one that stays is indeed $J\phi_{cl}$, so that is already a $J$ dependence, but you will still have higher order terms in $F$. I will expand the answer – ohneVal Mar 29 at 17:20