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There is a difference between the classical field $\phi(x)$ (which appears in the classical action $S[\phi]$) and the quantity $\phi_c$ defined as $$\phi_c(x)\equiv\langle 0|\hat{\phi}(x)|0\rangle_J$$ which appears in the effective action. Even though $\phi_c(x)$ is referred to as the "classical field", I don't see why $\phi(x)$ and $\phi_c$ should be the same.

In what sense, therefore, is the effective action $\Gamma[\phi_c]$ a quantum-corrected classical action $S[\phi]$? How can we compare the functionals of two different objects (namely, $\phi(x)$ and $\phi_c(x)$) and claim that $\Gamma[\phi_c]$ is a correction over $S[\phi]$?

I apologize for any lack of clarity in the question and the confusion I'm hoping to clear up.

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We want to calculate the path integral $$ Z = \int \mathcal{D}{\phi}\, e^{i \hbar^{-1} S[\phi]} $$ which encodes a transition amplitude between initial and final quantum states.

If we had the effective action $\Gamma[\phi]$ at our disposal, we would have calculated the same result by solving for $$ \phi_c(x):\quad \left. \frac{\delta \Gamma}{\delta \phi} \right|_{\phi=\phi_c} = 0 $$ and plugging it back in the effective action: $$ Z = e^{i \hbar^{-1} \Gamma[\phi_c]}. $$

This is the definition of $\Gamma$.

Note that no path integrals are required at this point. Boundary conditions are implicitly present throughout this answer, encoding the exact states between which the quantum transition occurs. Their existence ensures that there is only one solution $\phi_c$.

Now to why $\phi_c$ is called classical: it solves the e.o.m. given by the action $\Gamma$.

Think of $\Gamma$ as of an object in which all the short-scale properties of the integration measure $\mathcal{D}\phi$ (including renormalization-related issues) are already accounted for. You simply solve the e.o.m. and plug the solution in the exponential and you are done: here is your transition amplitude.

That being said, $\Gamma$ is not classical in the sense that it still describes dynamics of a quantum theory. Only in a different fashion. Simple algebraic manipulations instead of path integrals.

Finally, note how if the path integral is Gaussian, $$\Gamma[\phi] = S[\phi] + \text{const},$$ where $\text{const}$ accounts for the path integral normalization constant. There are no quantum corrections.

In classical theory, however, we solve the e.o.m. w.r.t. $\phi = \phi_c$ for $S[\phi]$, not $\Gamma[\phi]$. Plugging it back into $S[\phi_c]$ gives us the Hamilton function. When the path integral is Gaussian, it doesn't matter if we use $S$ or $\Gamma$, and exponentiating the Hamilton function gives you the transition amplitude. However if we are dealing with an interacting theory, the correct way to do this would be to use $\Gamma$ instead of $S$. In this sense, $\Gamma$ is the quantum-corrected version of $S$.

And yes, it is always true (can be shown using the saddle point approximation formula) that $$ \Gamma[\phi] = S[\phi] + \mathcal{O}(\hbar). $$

Why wouldn't we just use $\Gamma[\phi]$ to define the quantum theory and forget about $S[\phi]$ alltogether? Because $\Gamma$ is non-local and contains infinitely many adjustable parameters. These can be determined from the form of $S[\phi]$ by, well, quantization. That's why it is $S[\phi]$ which defines the theory, not $\Gamma$. $\Gamma$ is to be calculated via path integrals.

UPDATE: It is also important to understand that in naive QFT $\Gamma$ contains divergences, while $S$ doesn't. However, the actual situation is opposite. It is $S$ which contains divergences (divergent bare couplings), which cancel out against the divergences coming from the path integral, rendering a finite (i.e. renormalized) $\Gamma$. That $\Gamma$ should be finite is evident from how we use it to calculate physical properties: we only solve the e.o.m. and plug the result back in $\Gamma$.

Actually, the whole point of renormalization is to make $\Gamma$ finite and well-defined while adjusting only a finite number of diverging couplings in the bare action $S$.

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  • $\begingroup$ So, sorry for making a question in an old question, but would appreciate an insight: when you say that appropriate boundary conditions are assumed throughout the answer, do you mean something to make the volume of the phase space resulting from the 'integration of $D\phi$' unity? $\endgroup$ – GaloisFan Apr 2 at 0:31
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    $\begingroup$ @GaloisFan nah, I assume that the integrand field (be it $x(t)$ or $\phi(x)$) has fixed values at the $t = t_I$ and $t = t_F$ which correspond to quantum state. That is standard procedure in path integrals. Sorry about the confusing term "boundary conditions" which could mean many things. $\endgroup$ – Prof. Legolasov Apr 2 at 1:12
  • $\begingroup$ I see! And why is the intuition from defining the effective action as an analogous to the action but without the functional integration? Just an analogous to the classical partition function? I mean, intuitively it is easy to accept 'the path integral represents the quantum effects' and be ok with it but I'd like to have a deeper understanding. $\endgroup$ – GaloisFan Apr 2 at 1:29
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    $\begingroup$ @GaloisFan intuition is given by the stationary phase approximation formula. Path integrals can be approximately evaluated by looking at the extremum of the action — the classical solutions. So we choose to redefine the action to accommodate the difference between the path integral value and the approximation while retaining the stationary phase formula, now exact. $\endgroup$ – Prof. Legolasov Apr 4 at 2:19
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There is already a good answer by Solenodon Paradoxus. Here we provide a formal proof (via the stationary phase/WKB approximation).

  1. To fix notation, we define the effective/proper action $$ \Gamma[\phi_{\rm cl}]~=~W_c[J]-J_k \phi_{\rm cl}^k, \tag{1}$$ as the Legendre transformation of the generating functional $W_c[J]$ for connected diagrams. We assume that the Legendre transformation is regular, i.e. the formula $$ \phi_{\rm cl}^k~=~\frac{\delta W_c[J]}{\delta J_k} \qquad \Leftrightarrow \qquad J_k~=~-\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k} \tag{2}$$ is invertible. Here $J_k$ are the sources and $\phi_{\rm cl}^k$ are the so-called classical fields. (The latter terminology is a bit of a misnormer as $\phi_{\rm cl}^k[J]$ as a function of the sources $J_{\ell}$ could depend explicitly on $\hbar$. See also section 8 below.)

  2. The partition function/path integral is $$ \exp\left\{ \frac{i}{\hbar} W_c[J]\right\}~=~Z[J] ~:=~\int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_k \phi^k\right)\right\} . \tag{3}$$ The first equality in eq. (3) is the linked cluster theorem, cf. e.g. this Phys.SE post.

  3. At this place it is customary to mention some elementary facts. The 1-pt function/quantum averaged field is by definition $$\begin{align} \langle \phi^k \rangle_J &~:=~\frac{1}{Z[J]} \int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\phi^k\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_{\ell} \phi^{\ell}\right)\right\}\cr &~=~\frac{1}{Z[J]} \frac{\hbar}{i} \frac{\delta }{\delta J_k}\int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_{\ell} \phi^{\ell}\right)\right\}\cr &~\stackrel{(3)}{=}~\frac{1}{Z[J]} \frac{\hbar}{i}\frac{\delta Z[J]}{\delta J_k}~\stackrel{(3)}{=}~\frac{\delta W_c[J]}{\delta J_k}~\stackrel{(2)}{=}~\phi_{\rm cl}^k, \end{align} \tag{4}$$

  4. The 2-pt function is by definition $$\begin{align} \langle \phi^k \phi^{\ell}\rangle_J &~:=~\frac{1}{Z[J]} \int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\phi^k\phi^{\ell}\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_m \phi^m\right)\right\}\cr &~\stackrel{(3)}{=}~\frac{1}{Z[J]} \left(\frac{\hbar}{i}\right)^2\frac{\delta^2 Z[J]}{\delta J_k\delta J_{\ell}}~\stackrel{(3)}{=}~\frac{1}{Z[J]} \frac{\hbar}{i} \frac{\delta}{\delta J_k} \left(Z[J]\frac{\delta W_c[J]}{\delta J_{\ell}}\right)\cr &~\stackrel{(4)}{=}~\frac{\hbar}{i} \frac{\delta^2 W_c[J]}{\delta J_k\delta J_{\ell}} + \langle \phi^k \rangle_J \langle \phi^{\ell} \rangle_J,\end{align} \tag{5}$$ i.e. the connected 2-pt function plus a disconnected piece.

  5. Now let us return to OP's question. By formal Fourier transformation of the path integral (3), we get $$ \exp\left\{ \frac{i}{\hbar}S[\phi_{\rm cl}]\right\} ~\stackrel{(3)}{=}~\int \! {\cal D}\frac{J}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \left(W_c[J]-J_k \phi_{\rm cl}^k\right)\right\} $$ $$~\stackrel{(1)}{\sim}~ {\rm Det}\left(\frac{1}{i}\frac{\delta^2 W_c[J[\phi_{\rm cl}]]}{\delta J_k \delta J_{\ell}}\right)^{-1/2} \exp\left\{ \frac{i}{\hbar}\Gamma[\phi_{\rm cl}]\right\}\left(1+ {\cal O}(\hbar)\right) $$ $$~\stackrel{(8)}{=}~ {\rm Det}\left(\frac{1}{i}\frac{\delta^2 \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k \delta \phi_{\rm cl}^{\ell}}\right)^{1/2} \exp\left\{ \frac{i}{\hbar}\Gamma[\phi_{\rm cl}]\right\}\left(1+ {\cal O}(\hbar)\right) \quad\text{for}\quad\hbar~\to~0 \tag{6}$$ in the stationary phase/WKB approximation. In the last equality of eq. (6), we used that $$\begin{align}\delta^k_{\ell} ~=~\frac{\delta \phi_{\rm cl}^k[J[\phi_{\rm cl}]]}{\delta\phi_{\rm cl}^{\ell}} ~=~&\frac{\delta \phi_{\rm cl}^k[J[\phi_{\rm cl}]]}{\delta J_m} \frac{\delta J^m[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\ell}} \cr ~\stackrel{(2)}{=}~& -\frac{\delta^2 W_c[J[\phi_{\rm cl}]]}{\delta J_k\delta J_m} \frac{\delta^2 \Gamma[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^m\delta\phi_{\rm cl}^{\ell}},\end{align} \tag{7}$$ i.e.

    $$\text{The 2-pt functions } \frac{1}{i}\frac{\delta^2 W_c[J]}{\delta J_k\delta J_m} \text{ and } \frac{1}{i}\frac{\delta^2 \Gamma[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^m\delta\phi_{\rm cl}^{\ell}} \text{ are inverses of each other.} \tag{8}$$

  6. Eq. (6) shows that the effective action $$\begin{align} \Gamma[\phi_{\rm cl}] ~\stackrel{(6)}{=}~& S[\phi_{\rm cl}] +\frac{i\hbar}{2}\ln {\rm Det}\left(\frac{1}{i}\frac{\delta^2 \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k \delta \phi_{\rm cl}^{\ell}}\right) +{\cal O}(\hbar^2) \tag{9} \cr ~\stackrel{(9)}{=}~& S[\phi_{\rm cl}] +\frac{i\hbar}{2}\ln {\rm Det}\left(\frac{1}{i} H_{k\ell}[\phi_{\rm cl}]\right) +{\cal O}(\hbar^2), \qquad H_{k\ell}[\phi]~:=~ \frac{\delta^2 S[\phi]}{\delta\phi^k\delta\phi^{\ell}}, \tag{10}\end{align} $$ agrees with the action $S$ up to quantum corrections. (The square root factor in eq. (6) only contributes at one-loop and beyond.)

    In other words, if we assume that the action $S$ has no explicit $\hbar$-dependence, we deduce that to zeroth-order in $\hbar$/tree diagrams in the effective action

    $$ \Gamma_{\text{tree}}[\phi_{\rm cl}] ~\stackrel{(9)}{=}~S[\phi_{\rm cl}] \tag{11}$$

    is equal to the action $S$ itself. Similarly, we deduce that to first-order in $\hbar$/one-loop diagrams in the effective action

    $$ \Gamma_{\text{1-loop}}[\phi_{\rm cl}] ~\stackrel{(10)}{=}~\frac{i\hbar}{2}\ln {\rm Det}\left(\frac{1}{i} H_{k\ell}[\phi_{\rm cl}] \right) \tag{12}$$

    is equal to a functional determinant of the Hessian of the action $S$. Eqs. (10), (11) & (12) answer OP's question. See also this related Phys.SE post.

  7. At this place it is customary to mention some elementary facts. Let there be given fixed sources $J_k$. From$^1$ $$\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k}~\stackrel{(2)}{=}~-J_k~\stackrel{\text{EL eqs.}}{\approx}~\frac{\delta S[\phi_0]}{\delta \phi^k}, \tag{13} $$ we deduce that the so-called classical solution $\phi_{\rm cl}^k$ and the Euler-Lagrange (EL) solution $\phi_0^k$ agree$^1$ $$ \phi_{\rm cl}^k[J]~\stackrel{(9)+(13)}{\approx}~\phi_0^k[J] +{\cal O}(\hbar) \tag{14} $$ up to quantum corrections. Eq. (14) justifies the practice to call $\phi_{\rm cl}^k$ the classical field. (We assume that each solution to eq. (13) is unique, due to pertinent boundary conditions. We have excluded instantons for simplicity.)

  8. Alternatively, from the background field method $$ \underbrace{\phi^k}_{\text{quan. field}} ~=~\overbrace{\underbrace{\phi^k_{\rm cl}}_{\text{clas. field}}}^{\text{backgr. field}}+\underbrace{\eta^k}_{\text{fluctuation}}, \tag{15}$$ the effective action (1) becomes $$\exp\left\{\frac{i}{\hbar}\Gamma[\phi_{\rm cl}]\right\} ~\stackrel{(1)+(3)}{=}~ \int\!{\cal D}\frac{\phi}{\sqrt{\hbar}} ~\exp\left\{\frac{i}{\hbar} \left(S[\phi] +J_k[\phi_{\rm cl}](\phi^k-\phi^k_{\rm cl}) \right) \right\} $$ $$~\stackrel{(15)}{=}~ \int\!{\cal D}\frac{\eta}{\sqrt{\hbar}} ~\exp\left\{\frac{i}{\hbar} \left(S[\phi_{\rm cl}+\eta] +J_k[\phi_{\rm cl}] \eta^k \right)\right\} $$ $$~=~ \int\!{\cal D}\frac{\eta}{\sqrt{\hbar}} ~\exp\left\{\frac{i}{\hbar} \left( S[\phi_{\rm cl}] +\underbrace{\left(\frac{\delta S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k} +J_k[\phi_{\rm cl}]\right)}_{={\cal O}(\hbar)} \eta^k +\frac{1}{2}\eta^k H_{k\ell}[\phi_{\rm cl}] \eta^{\ell} +{\cal O}(\eta^3) \right)\right\} $$ $$~\stackrel{\text{WKB approx.}}{\sim}~ {\rm Det}\left(\frac{1}{i}H_{mn}[\phi_{\rm cl}] \right)^{-1/2}\left(1+ {\cal O}(\hbar)\right) $$ $$ \times \exp\left\{ \frac{i}{\hbar}\left(S[\phi_{\rm cl}] -\frac{1}{2}\underbrace{\left(\frac{\delta S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k} +J_k[\phi_{\rm cl}]\right)}_{={\cal O}(\hbar)} (H^{-1})^{k\ell}[\phi_{\rm cl}] \underbrace{\left(\frac{\delta S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^{\ell}} +J_{\ell}[\phi_{\rm cl}] \right)}_{={\cal O}(\hbar)} \right)\right\} $$ $$~\stackrel{(2)+(14)}{=}~ {\rm Det}\left(\frac{1}{i}H_{mn}[\phi_{\rm cl}]\right)^{-1/2}\exp\left\{ \frac{i}{\hbar}S[\phi_{\rm cl}]\right\}\left(1+ {\cal O}(\hbar)\right) \quad\text{for}\quad\hbar~\to~0 \tag{16}$$ in the stationary phase/WKB approximation. Eq. (16) again leads to the sought-for eq. (10).

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$^1$ The $\approx$ symbol means here equality modulo the Euler-Lagrange (EL) equations.

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