2
$\begingroup$

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?)

$\endgroup$
2
  • $\begingroup$ How did the $J\phi$ term in eq. (1) become $J\phi_{\rm cl}$ in eq. (2)? $\endgroup$ – Qmechanic Mar 29 at 12:20
  • $\begingroup$ @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 $J\phi_\text{cl.}$. $\endgroup$ – Jeanbaptiste Roux Mar 29 at 13:02
2
$\begingroup$
  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.

$\endgroup$
0
$\begingroup$

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)

$\endgroup$
5
  • $\begingroup$ 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 $\frac{\delta S}{\delta \phi}=0$? $\endgroup$ – Jeanbaptiste Roux Mar 29 at 13:26
  • $\begingroup$ 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. $\Delta \phi$. $\endgroup$ – ohneVal Mar 29 at 13:52
  • $\begingroup$ Why will $Z[J]$ still depend on $J$ if $e^{i\int J\phi_\text{cl.}}$ disappear, as you said? $\endgroup$ – Jeanbaptiste Roux Mar 29 at 14:52
  • $\begingroup$ 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 $0$. $\endgroup$ – Jeanbaptiste Roux Mar 29 at 15:24
  • $\begingroup$ 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 $\endgroup$ – ohneVal Mar 29 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.