Consider the scalar field transition amplitude $$\tag{1} \mathcal{A} = \int_{\phi_i}^{\phi_f} D\phi e^{iS[\phi]/\hbar}. $$

Let $\phi_{cl}$ solve the classical equation $\frac{\delta S}{\delta\phi}=0$. Denote the stationary phase approximation to (1) by $$ \tag{2} \mathcal{A}^{SP} = e^{iS[\phi_{cl}]/\hbar}\left({\det \frac{S''(\phi_{cl})}{2\pi i \hbar}}\right)^{-1/2}.$$

Denote the tree-level contribution (i.e. sum of all Feynman diagrams with no loops) by $A^{tree}$.

Does $$\mathcal{A}^{SP}=\mathcal{A}^{tree}~?\tag{3}$$

Note: I'm fairly sure the equality holds in the case of a free theory, since then the stationary phase approximation is exact, and there are no loop diagrams. I'm interested in whether it holds in general.

  • $\begingroup$ What prompted the question? Are you following a reference? $\endgroup$
    – Qmechanic
    Commented Jan 29, 2022 at 19:21
  • $\begingroup$ During a calculation I used the stationary phase approx, and wanted to be precise about the size of terms I was neglecting. For a minute I got confused and thought I was precisely throwing away loop terms. But having read your answer, I see that actually it's the interaction terms that I'm throwing away, not the loop ones. In other words: stationary phase approximation is the same as setting all interactions to zero. $\endgroup$ Commented Jan 31, 2022 at 13:25
  • 1
    $\begingroup$ SPA is not the same as the free model. Please reconsider your last comment. $\endgroup$
    – Qmechanic
    Commented Jan 31, 2022 at 13:40
  • $\begingroup$ Ok true. I guess I want to say something like "SPA is the free approx to the theory you get by perturbing around the exact classical solution". I don't know if these words are helpful. In any case, the maths is clear enough. $\endgroup$ Commented Jan 31, 2022 at 16:50
  • $\begingroup$ $\uparrow$ Right. $\endgroup$
    – Qmechanic
    Commented Jan 31, 2022 at 17:14

1 Answer 1

  1. We cannot resist the temptation to include a bulk source $J$. The transition amplitude/overlap is $$\begin{align} \exp&\left\{\frac{i}{\hbar}W^c_{fi}[J]\right\}\cr ~=~& Z_{fi}[J]~=~\langle \phi_f,t_f|\phi_i,t_i\rangle_J\cr ~=~&\int_{\phi(t_i)=\phi_i}^{\phi(t_f)=\phi_f} \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \underbrace{\left(S[\phi]+J_k \phi^k\right)}_{=:~S_J[\phi]}\right\}\cr ~\stackrel{\begin{array}{c}\text{WKB}\cr\text{approx.}\end{array}}{\sim}& {\rm Det}\left(\frac{1}{i}\frac{\delta^2 S[\phi_{fi}[J]]}{\delta \phi^m \delta \phi^n}\right)^{-1/2}\cr &\exp\left\{ \frac{i}{\hbar}\underbrace{\left(S[\phi_{fi}[J]]+J_k \phi_{fi}^k[J]\right)}_{\text{on-shell action}}\right\}\cr &\left(1+ {\cal O}(\hbar)\right) \end{align}\tag{A}$$ in the stationary phase/WKB approximation $\hbar\to 0$. Here $\phi_{fi}^k[J]$ denotes the solution to the Dirichlet boundary value problem $$ \left\{ \begin{array}{rcl} \frac{\delta S[\phi]}{\delta \phi^k}&\approx&-J_k, \cr \phi(t_i)&=&\phi_i,\cr \phi(t_f)&=&\phi_f,\end{array}\right.\tag{B} $$ which we will assume exists and is unique$^1$. It follows from the $\hbar$/loop-expansion that the generator of connected tree diagrams $$\frac{\hbar}{i}\ln Z^{\rm tree}_{fi}[J]~=~W^{c,\rm tree}_{fi}[J]~=~S[\phi]+J_k\phi^k\tag{C}$$ is the Legendre transform of the action $S[\phi]$ between bulk sources $J_k$ and field configurations $\phi^k$ that satisfy the Dirichlet boundary conditions. In particular, $W^{c,\rm tree}_{fi}[J]$ is the on-shell action.

    For more details, see eq. (A8) in my Phys.SE answer here.

    Finally let us return to OPs question. The WKB formula (A) is tree-diagrams $Z^{\rm tree}_{fi}[J]$ [as OP suggests in eq. (3), given by the on-shell action] times a 1-loop functional determinant. So because of the presence of the 1-loop functional determinant, it is strictly speaking not just tree diagrams $Z^{\rm tree}_{fi}[J]$ .

  2. The above is closely related to (but should not be confused with) the 1PI effective/proper action $$\begin{align} \exp&\left\{\frac{i}{\hbar}\Gamma_{fi}[\phi_{\rm cl}]\right\}\cr ~=~&\exp\left\{\frac{i}{\hbar}\left(W^c_{fi}[J[\phi_{\rm cl}]]-J_k[\phi_{\rm cl}]\phi^k_{\rm cl}\right)\right\} \cr ~=~&\int_{\phi(t_i)=\phi_i}^{\phi(t_f)=\phi_f} \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_k[\phi_{\rm cl}] (\phi^k-\phi^k_{\rm cl})\right)\right\}\cr ~\stackrel{\begin{array}{c}\text{WKB}\cr\text{approx.}\end{array}}{\sim}& {\rm Det}\left(\frac{1}{i}\frac{\delta^2 S[\phi_{\rm cl}[J]]}{\delta \phi^m_{\rm cl} \delta \phi^n_{\rm cl}}\right)^{-1/2}\exp\left\{ \frac{i}{\hbar}S[\phi_{\rm cl}]\right\}\left(1+ {\cal O}(\hbar)\right) \end{align}\tag{D}$$ in the stationary phase/WKB approximation $\hbar\to 0$, cf. e.g. this Phys.SE post.


$^1$ If the solution is not unique, we would have to include a sum over different solutions (=instantons).


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