Why is $\langle j^\mu j^\nu \rangle$ the vacuum polarization amplitude? In QED, we often call the diagram:

the one-loop "vacuum polarization diagram" and we say that it is given by:
$$\Pi^{\mu\nu}(q)=\int\!d^4x e^{iq\cdot x} \langle T\{j^\mu(x) j^\nu(0)\} \rangle\;.$$
But looking at the diagram, I would have thought that what we really need to define is
$$\frac{i}{q^2+i\epsilon}\Pi^{\mu\nu}(q)\frac{i}{q^2+i\epsilon} = \int\!d^4x e^{iq\cdot x} \langle T\{A^\mu(x) A^\nu(0) \}\rangle\;,$$
at one loop. Are these definitions equivalent? I'm having trouble seeing it.
 A: The difference between the two diagrams is with whether you include the photon propagators in your calculation. For instance, consider the Lagrangian
$$
S = - \frac{1}{4} F^2 - | \partial \phi |^2 - A_\mu J^\mu + \cdots  , \qquad J^\mu = i Q ( \phi^\dagger \partial^\mu \phi -  \partial^\mu \phi^\dagger \phi )
$$
Then, the photon 2-point function is given in perturbation theory by
\begin{align}
\langle A_\mu(x) A_\nu(0) \rangle &= \int[dA][d\phi] e^{i S} A_\mu(x) A_\nu(0) \\
&=  \int[dA][d\phi] e^{i S_{free}} \, \frac{1}{2!} \left(  \int d^4 y_1 : J^\alpha(y_1) A_\alpha(y_1) : \right)^2  A_\mu(x) A_\nu(0) \\
&=  \int d^4 y_1   d^4 y_2 D^F_{\mu\alpha}( x - y_1 ) D^F_{\nu\beta}(y_2) \langle J^\alpha(y_1) J^\beta(y_2) \rangle  \tag{1}
\end{align}
where
$$
D^F_{\mu\alpha}( x ) = \langle A_\mu(x) A_\nu(0) \rangle_{free}
$$
The diagrammatic interpretation of (1) is hopefully obvious. It shows that the $JJ$ 2-point function is related to the $AA$ 2-point function simply by an extra factor of the photon propagators.
I derived this in perturbation theory, but there is a simple generalization of this result to the full non-perturbative theory as well.
A: The full non-perturbative proof of this fact proceeds as follows:

*

*Use the LSZ reduction formula to convert the $\langle AA\rangle$ correlator into the following scattering amplitude $\leftrightarrow$ S-matrix element:
$$
\langle f|i\rangle=\epsilon_\mu^{(1)}\epsilon_\alpha^{(2)}i^n \int\mathrm d^4x_1e^{ip_1x_1}\partial^2_{(1)}g^{\mu\nu}\int\mathrm d^4x_2e^{ip_2x_2}\partial^2_{(2)}g^{\alpha\beta}\langle A_\nu(x_1)...A_\beta(x_2)...\rangle\tag{1}
$$


*Use the classical EOM for the photon field
$$
\frac{\delta\mathcal L}{\delta A^\mu}~=~-\partial^2A_\mu~=~j_\mu,\qquad j^\mu~=~\bar\psi\gamma^\mu\psi\tag{2}
$$
in the Schwinger-Dyson equation
$$
\left\langle\frac{\delta S}{\delta\phi_j}\phi_1...\phi_n\right\rangle=i\sum_{k=1}^n\left\langle\phi_1...\delta_{jk}...\phi_n\right\rangle\tag{3}
$$
to replace the photon fields in the correlator with currents:
$$
\partial^2_{(1)}g^{\mu\nu}\ \partial^2_{(2)}g^{\alpha\beta}\langle A_\nu(x_1)...A_\beta(x_2)...\rangle=\langle j^\mu(x_1)...j^\alpha(x_2)...\rangle+\underbrace{\partial^2_{(2)}g^{\mu\alpha}\partial^2_{(1)}\Delta(x_1-x_2)\langle...\rangle+...}_\text{contact terms}\tag{4A}
$$
The reason we cannot directly substitute the classical relations $(2)$ into $(1)$ is because they are, well, classical. Equation $(3)$ says that the classical EOM hold inside correlation functions only up to contact terms on the RHS.


*The contact terms do not contribute to the S-matrix as they yield disconnected diagrams, i.e. contractions between external photons. Alternatively, they do not have the correct pole structure in momentum space. Consequently we have as a special case:
$$
\partial^2_{(1)}g^{\mu\nu}\ \partial^2_{(2)}g^{\alpha\beta}\langle A_\nu(x_1)A_\beta(x_2)\rangle=\langle j^\mu(x_1)j^\alpha(x_2)\rangle\tag{4B}
$$
at the level of S-matrix elements. Substituing this result into the initial scattering amplitude, we see that all amputated photons in S-matrix elements can be replaced with currents, within correlation functions. (As it turns out, this statement holds when external momenta are not on-shell).
(Conventions: Feynman-'t Hooft gauge; renormalisation parameters elided; deWitt condensed notation)
A: It depends on where you put the diagram you have shown. You have to include one or two or no propagators if it is put in a larger diagram, where more real particles are involved. Depending on which particles are real you have to include the propagators you have added to the formula $i/(q^2+i\epsilon)$.
On its own, the diagram refers to two real photons speeding in space. Then it's part of the propagator of a single photon. It represents a single photon traveling from one spacetime point to another. Then you have to include both propagators, obviously.
