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When deriving the effective action $\Gamma$ in the background field method, one splits the field $\phi = \phi_b + \phi_f$ into a background (or mean field) $\phi_b$ and fluctuations $\phi_f$, then proceed to integrate out $\phi_f$:

$$ e^{-\Gamma[\phi_b]} = \int\mathcal{D}\phi_f e^{-S[\phi_b+\phi_f]-\frac{\delta \Gamma}{\delta \phi_b} \phi_f}. $$

The background field is a functional of the source $J$:

$$ \phi_b = \frac{\delta W}{\delta J} $$

and, in general, it doesn't coincide with the classical field $\phi_{cl}$ (which satisfies the classical EOM). Nonetheless, it is sometimes assumed (see eg Peskin, Schroeder, eqs 11.55 and 11.58) that $\phi_b$ satisfies the classical EOM so that one can get rid of the linear term in $\phi_f$ in the Taylor expansion of $S[\phi_b+\phi_f]$:

$$ S[\phi_b+\phi_f] = S[\phi_b] + \int\frac{\delta S[\phi_b]}{\delta{\phi_f}}\phi_f + \cdots $$

Other references (see eg https://inspirehep.net/record/345318?ln=en, eq. 2.91) seem not to assume that $\phi_b$ satisfies the classical EOM.

So here are my questions (for what it's worth, I'm particularly interested in the case of gravity):

  1. When do I have to choose $\phi_b=\phi_{cl}$, i.e. when do I have to assume that $\phi_b$ satisfies the classical EOM?

  2. The importance of the effective action $\Gamma$ is two-fold. For one, it is the generator of 1PI correlation functions. But it can also be used to study the dynamical evolution of $\phi_b$ under the influence of quantum fields via the quantum EOM $\frac{\delta \Gamma}{\delta \phi_b}=0$. If $\phi_b$ also satisfies the classical EOM, then $\phi_b$ is overdetermined and, in particular, there is no new solution to the quantum EOM. What is the meaning of the quantum EOM then?

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  • $\begingroup$ Hi @Mr. K. Where in Peskin & Schroeder? Which pages/eqs. in books/links? $\endgroup$ – Qmechanic Jun 6 at 15:46
  • $\begingroup$ "When do I have to..." we never have to do anything. We do whatever we want. We do what is most useful in every situation. $\endgroup$ – AccidentalFourierTransform Jun 6 at 15:54
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    $\begingroup$ The answer to the question in the title is — up to possible corrections of order $\mathcal{O}(\hbar)$. $\endgroup$ – Prof. Legolasov Jun 7 at 9:29

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