A question on Schwartz's derivation of the Euler-Heisenberg Lagrangian

Question

In Subsection 33.2.2. of Schwartz's Quantum Field Theory and the Standard Model, he starts to derive the Euler-Heisenberg effective Lagrangian by "replacing" the field which is being integrated out with a corresponding classical expectation value.

Specifically, suppose that $|A\rangle$ is a state corresponding to some fixed background configuration of the potential $A^\mu$. Schwartz starts with the standard QED Lagrangian $$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}^2 + \overline{\psi}(i\gamma^\mu\partial_\mu - m)\psi - eA_\mu\overline{\psi}\gamma^\mu\psi,\tag{33.31}$$ and then simply declares that the result of integrating out the fields $(\psi,\overline{\psi})$ is given by $$\mathcal{L}_{\mathrm{eff}} = -\frac{1}{4}F_{\mu\nu}^2 - eA_\mu J_A^\mu,\tag{33.32}$$ where $$J_A^\mu = \langle A|\overline{\psi}\gamma^\mu\psi|A\rangle\tag{33.33}.$$ What is the justification for this step? What happened to quadratic term $\overline{\psi}(i\gamma^\mu\partial_\mu - m)\psi$?

Note that Schwartz has, throughout his book, a rather idiosyncratic way of integrating out fields by solving their equations of motion (see for example the discussion here). This does not seem to be what's happening here however.

Also, just to be clear, I am not asking about the usual process of integrating out fields by performing the marginal path integrals. Nor am I really interested in the derivation of the Euler-Heisenberg Lagrangian itself. I am primarily just curious about whether the "shortcut" performed by Schwartz here to integrate out the $\psi$ field has any real justification. Something akin to Adam's answer (point 1) in the linked question would be a perfect example of the sort of thing I'm looking for.

Answer 1

OP has a point. The approach in Subsection 33.2.2 seems ad hoc at best.

• Eq. (33.32) is certainly only valid for infinitesimal $A_{\mu}$. If we write the effective action as $$S_{\rm eff}[A]~\equiv~\Gamma[A]~=~S_{\rm Maxwell}[A] + \Delta\Gamma[A],$$ then eq. (33.32) should be replaced by $$ \frac{\delta}{\delta A_{\mu}} \Delta\Gamma[A]~=~eJ^{\mu}_A. \tag{33.32'}$$

• $J^{\mu}_A$ is more clearly defined via eq. (33.35) [which in turn relies on eq. (33.27)] than eq. (33.33), whose notation seems not properly explained.

• Eqs. (33.35) and the modified eq. (33.32') are consistent with the later functional determinant approach in Subsection 33.3.1, which integrates out the fermions: $$ i\Delta\Gamma[A]~=~{\rm Tr}\ln(i\not{\!\!D}-m) + {\rm const}. \tag{33.54}$$

References:

1. M.D. Schwartz, QFT & the standard model, 2014; Subsection 33.2.2.

Answer 2

There is a formal process to integrate out fields, based on the path integral formulation of the theory. (Not sure whether it is discussed in Schwartz's book. I don't have that book.) In case the field that is to be integrated out has interaction terms with another field, one would add source terms and pull the interaction term out of the integral by replacing the fields with functional derivatives. The integral can then be evaluated, producing a propagator from the kinetic term dressed by the sources. When the interaction term is allowed to operate on this dressed propagator it produces a term that can be interpreted as a current with the other field with which it interacted. I suspect that this is what Schwartz is doing.