Photon propagator in path integral

Question

In scalar fiel theory the two point function is given by $$\langle{0}| T\phi(x) \phi(y) |0\rangle= \int D[\phi] \phi(x) \phi(y) \exp[i\int d^4x\mathcal{L_{scalar}}]$$ introducing the the generating functional

$$Z=\int D\phi \exp[i\int d^4x(\mathcal{L_{scalar}}+J\phi )]$$ we have

$$\langle{0}| T\phi(x) \phi(y) |0\rangle=\frac{1}{i^2} \frac{\delta Z[J]}{\delta J (x) \delta J (y)}\bigg|_{J=0}$$ Now in QED the interacting photon propagator is given by

$$\langle{0}| TA_\mu(x) A_\nu(x)|0\rangle=\int [DA] A_\mu(x) A_\nu(y) \exp[i\int d^4x \mathcal{L_{qed}}]\tag 1$$

where $$\mathcal{L_{qed}} =\overline{\psi}(-i\gamma^\mu \partial_\mu +m)\psi+\frac{1}{4}F^{\mu \nu}F_{\mu \nu}+J^{\mu}A_\mu$$ with $J^{\mu} = \bar{\psi} \gamma^{\mu} \psi$

Now usually in textbooks they define $$Z=\int DA \exp[i\int d^4x\mathcal{L_{qed}} ] $$

and $$\langle{0}| TA_\mu(x) A_\nu(y)|0\rangle=\frac{1}{i^2} \frac{\delta Z[J]}{\delta J^\mu(x) \delta J^\nu (y)}\bigg|_{J=0} \tag 2$$

But if we put $J=0$ we are turning the interaction off. Shouldn't we have $$\langle{0}| TA_\mu(x) A_\nu(y)|0\rangle=\frac{1}{i^2} \frac{\delta Z[J]}{\delta J^\mu(x) \delta J^\nu (y)}~? \tag 3$$ Then we would have $ (1)=(3).$

Answer 1

In perturbation theory the interaction term is expanded - I.e. The exponential involving J is expanded - and its contribution is determined term by term with respect to what is now the free theory. Finding the contribution requires calculation of correlators of the form you posted (internal lines in Feynman diagrams are precisely photon or electron propagators) and these are fixed by the free theory.

In essence there are two J that are being confused. The physical current, J, that will be expanded term by term and then an artificial $\cal{J} $ that is introduced to allow the correlator to be determined by functional differentiation. This fictitious $\cal{J} $ will be put to 0 at the end of the calculation.