Modified photon propagator with a fluctuating current

Question

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!

Answer 1

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).