# Photon propagator in path integral

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 $$(1)=(3).$$

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.