I want to understand a trick in the derivation of the Schwinger-Dyson equations

Question

In the book of Ashok Das, Field theory-path integral approach, he begin the demonstration of the Schwinger-Dyson equation using the fact that the $\delta Z[J]=0$, so \begin{equation} \delta Z[J]=\int \mathcal{D} \phi \frac{\delta S[\phi,J]}{\delta \phi(x)} e^{iS[\phi,J]}=0, \end{equation} but we already know that \begin{equation} \frac{\delta S[\phi, J]}{\delta \phi(x)}=F(\phi(x))-J(x), \end{equation} where $F(\phi(x))$ is the equation of motion.

So if we go back to the first equation and use the identification \begin{equation} \phi(x)\rightarrow -i\frac{\delta}{\delta J(x)}, \end{equation} we conclude that $$ \int \mathcal{D}\phi\left(F(\phi(x))-J(x)\right)e^{iS[\phi,J]}=\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)\int \mathcal{D}\phi e^{iS[\phi,J]} $$ $$ \left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)Z[J]=\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)e^{iW[J]}=0. $$

But is here where I get lost, how did he pass from the above equation for $$ e^{-iW[J]}\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)e^{iW[J]}=F\left(\frac{\delta W[J]}{\delta J(x)}-i\frac{\delta}{\delta J(x)}\right)-J(x)=0. $$

A more important question is: what does this last equation mean at all, mathematically speaking, because the functional derivative now is acting on nothing.

Answer 1

The steps that you already understand showed that $$ \left(F\left(-i\frac{\delta}{\delta J(x)}\right) -J(x)\right)e^{i W[J]}=0. \tag{1} $$ This clearly implies $$ \left(F\left(-i\frac{\delta}{\delta J(x)}\right) -J(x)\right)e^{i W[J]}c=0, \tag{2} $$ where $c$ is any constant. Now use the identity $$ -i\frac{\delta}{\delta J(x)}e^{iW[J]}h[J] = e^{iW[J]}\left(\frac{\delta W[J]}{\delta J(x)} -i\frac{\delta}{\delta J(x)}\right)h[J] \tag{3} $$ to move the factor of $e^{iW[J]}$ from the right-hand side of equation (2) to the left-hand side, where $h[J]$ is an arbitrary functional. The result is the last equation shown in the question, except that here I've written it with an arbitrary constant $c$ on the right-hand side, so that the variational derivatives always have something to act on, even if it's something trivial. The book apparently just didn't bother writing this arbitrary constant.

The remaining functional derivatives $\delta/\delta J$ are still important, because they're inside the argument of $F(\cdots)$, so they still act on the $J$-dependent factors that are also inside the argument of $F(\cdots)$. (For an example, suppose $F[X]=X^2$.) This detail is exactly what makes the Schwinger-Dyson equations different than the classical equation of motion for $\phi(x) := \delta W[J]/\delta J(x)$.

Answer 2

OP's last identity is a functional version of the following identity $$e^{-g(x)} f(\partial_x)e^{g(x)}~=~f\left(e^{-g(x)} \partial_xe^{g(x)}\right)1~=~f\left(e^{-[g(x),\cdot]} \partial_x\right)1~=~f\left( \partial_x-[g(x),\partial_x]\right)1~=~f\left(\partial_x+g^{\prime}(x)\right)1,$$ where $f,g$ are two sufficiently nice functions. The derivative $\partial_x$ acts to the right until the constant function $1$. The constant function $1$ is implicitly implied in Ref. 1.

References:

1. A. Das, Lectures on QFT, 2008; section 16.3 p. 690-691 eq. (16.50).