Must the mean field, in the context of the background field method, satisfy the classical equations of motion?

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?

• Hi @Mr. K. Where in Peskin & Schroeder? Which pages/eqs. in books/links? – Qmechanic Jun 6 at 15:46
• "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. – AccidentalFourierTransform Jun 6 at 15:54
• The answer to the question in the title is — up to possible corrections of order $\mathcal{O}(\hbar)$. – Prof. Legolasov Jun 7 at 9:29