Modified photon propagator with a fluctuating current I am studying a paper on the effects in bound-state QED coming from nuclear dynamics, and I am struggling to understand one basic derivation.
The authors introduce a modified photon propagator
$$
\tag{1}
i\mathcal{D}_{\mu\nu}(x,x') = \langle 0|T[\hat{A}^{\mathrm{rad}}_{\mu}(x) \hat{A}^{\mathrm{rad}}_{\nu}(x')]|0 \rangle,
$$
where the total radiation field $\hat{A}^{\mathrm{rad}}_{\mu}(x)$ is expressed as the sum $$
\hat{A}^{\mathrm{rad}}_{\mu}(x) = \hat{A}^{\mathrm{free}}_{\mu}(x) + \hat{A}^{\mathrm{fluc}}_{\mu}(x).$$
Here, $\hat{A}^{\mathrm{free}}_{\mu}(x)$ is the usual free photon field, and the fluctuating part $\hat{A}^{\mathrm{fluc}}_{\mu}(x)$ is generated by the nuclear transition current $\hat{j}^{\mathrm{fluc}}_{\mu}(x)$.
The authors say that "one easily verifies" that Eq. (1) can be rewritten as
$$
\tag{2}
\mathcal{D}_{\mu\nu}(x,x') = D_{\mu\nu}(x-x') + \int d^4 x_1 \, d^4 x_2 \, D_{\mu \alpha}(x-x_1) \Pi^{\alpha \beta}(x_1,x_2) D_{\beta \nu}(x_2-x'), 
$$
with the free photon propagator $$iD_{\mu\nu}(x-x') = \langle 0|T[\hat{A}^{\mathrm{free}}_{\mu}(x) \hat{A}^{\mathrm{free}}_{\nu}(x')]|0 \rangle$$ and the so-called nuclear-polarization tensor $$i\Pi^{\alpha \beta}(x,x')=\langle 0|T[\hat{j}^{\alpha}_{\mathrm{fluc}}(x) \hat{j}^{\beta}_{\mathrm{fluc}}(x')]|0 \rangle.$$
This is probably something very simple but I cannot seem to understand how one would go from $\hat{A}^{\mathrm{fluc}}_{\mu}(x)$ in Eq. (1) to $\hat{j}^{\mathrm{fluc}}_{\mu}(x)$ in Eq. (2). Any hints/ideas would be greatly appreciated!
 A: Inspired by the answers to this question (thanks to @Nihar Karve for pointing it out to me!), I think I have figured it out.
Using the equation of motion for the photon field:
$$
\partial^2 A^{\mu}_{\mathrm{fluc}}(x) = j^{\mu}_{\mathrm{fluc}}(x),
$$
and the fact that the free photon propagator is the Green's function of this equation:
$$
\partial^2 D^{\mu\nu}(x) = \eta^{\mu\nu} \delta^{(4)}(x),
$$
one can recover Eq. (1) from Eq. (2) in the following way:
$$
\begin{align}
& \int d^4 x_1 \, d^4 x_2 \, D_{\mu \alpha}(x-x_1) \Pi^{\alpha \beta}(x_1,x_2) D_{\beta \nu}(x_2-x') \\
& = (-i)\int d^4 x_1 \, d^4 x_2 \, D_{\mu \alpha}(x-x_1) \langle 0|T[\hat{j}^{\alpha}_{\mathrm{fluc}}(x_1) \hat{j}^{\beta}_{\mathrm{fluc}}(x_2)]|0 \rangle D_{\beta \nu}(x_2-x') \\
& = (-i)\int d^4 x_1 \, d^4 x_2 \, D_{\mu \alpha}(x-x_1) \langle 0|T[\{\partial^2_{x_1} \hat{A}^{\alpha}_{\mathrm{fluc}}(x_1)\} \{\partial^2_{x_2} \hat{A}^{\beta}_{\mathrm{fluc}}(x_2)\}]|0 \rangle D_{\beta \nu}(x_2-x') \\
& = (-i)\int d^4 x_1 \, d^4 x_2 \, \{\partial^2_{x_1} D_{\mu \alpha}(x-x_1)\} \langle 0|T[\hat{A}^{\alpha}_{\mathrm{fluc}}(x_1) \hat{A}^{\beta}_{\mathrm{fluc}}(x_2)]|0 \rangle \{\partial^2_{x_2} D_{\beta \nu}(x_2-x')\} \\
& = (-i)\int d^4 x_1 \, d^4 x_2 \, \eta_{\mu\alpha} \delta^{(4)}(x-x_1) \langle 0|T[\hat{A}^{\alpha}_{\mathrm{fluc}}(x_1) \hat{A}^{\beta}_{\mathrm{fluc}}(x_2)]|0 \rangle \eta_{\beta\nu} \delta^{(4)}(x_2-x') \\
& = (-i) \langle 0|T[\hat{A}^{\mathrm{fluc}}_{\mu}(x) \hat{A}^{\mathrm{fluc}}_{\nu}(x')]|0 \rangle,
\end{align}
$$
where integrations by parts with zero boundary terms are used in between lines 3 and 4.
By running this kind of argument backwards, it can also be seen that the mixed terms
$$\langle 0|T[\hat{A}^{\mathrm{fluc}}_{\mu}(x) \hat{A}^{\mathrm{free}}_{\nu}(x')]|0 \rangle = \langle 0|T[\hat{A}^{\mathrm{free}}_{\mu}(x) \hat{A}^{\mathrm{fluc}}_{\nu}(x')]|0 \rangle = 0,
$$
which completes the proof of Eq. (2).
