There was a part in the lecture one didn't quite understand. (The charge $e=0$ in this post.) The partition function for massive fields such as scalar fields or the spin-1/2 fields were quite standard. They could both be put into the form like
$$Z[j]=Z_0\exp\left[-\frac{1}{2}\int d^{4}x\int d^{4}y\,j(x) D(x-y) j(y)\right].$$
However, when dealing with the photon fields, i.e. the massless fields, after the gauge fixing,
$$S_\text{gf}=\frac{1}{2} \int d^{4}x\, A_{\mu}(x)\left[\partial^2g^{\mu\nu}-\left(1-\frac{1}{\xi}\right)\partial^\mu\partial^\nu\right]A_\nu(x).$$
"Commonly", one wrote
$$S_\text{gf}=\frac{1}{2} \int \frac{d^{4}k}{(2\pi)^2} A_{\mu}(k)\left[-k^2g^{\mu\nu}-\left(1-\frac{1}{\xi}\right)k^\mu k^\nu\right]A_\nu(-k),$$
where the Fourier form of the photon propagator was written as
$$D_{F}^{\mu\nu}(k)=\frac{-i}{k^2+i\epsilon}\left[g_{\mu\nu}-(1-\xi)\frac{k^\mu k^\nu}{k^2}\right],$$
with
$$Z[J^\mu]=\ldots(\text{gauge stuff}) \int{\cal D}A\,\exp\left[iS_\text{ gf}-i\int d^{4}x J(x)^\mu A_\mu(x)\right].$$
Supposedly $\int {\cal D}A\,\exp[iS_\text{ gf}]$ was the $Z_0$ part of the partition function, and if one wanted to obtain the $\langle A_\mu(x)\rangle $, would the usual approach of $$\frac{\delta }{\delta iJ^\mu(x)} Z[J^\mu(x)]|_{J^\mu=0}$$ apply?
Question 1
However, here's the thing. In $$\frac{\delta }{\delta iJ^\mu(x)} Z[J^\mu(x)]=Z[J^\mu]\cdot (-A_\mu(x))$$ there is no photon propagator showing up.
Question 2
Further, why couldn't the photon propagator be in analogy to that of the free scalar field?
Thus, one attempted to use a hilarious approach by just saying that there's an operator that (notice the minus sign has disappeared)
$$\left[\partial^2 g_{\mu\rho} -\left(1-\frac{1}{2\xi}\right)\partial_\mu\partial_\rho\right] D_F^{\rho\nu}(x-y)=i\delta_\mu^\nu\delta^{(4)} (x-y),$$
and through the same approach in analogy to the free scalar field
$$A^\mu\rightarrow A^\mu+A_0^\mu$$
$$Z[J^\mu]=Z_0\exp\left[-\frac{1}{2} \int d^{4}x\int d^{4}y\, J^\mu(x) D_{F\mu\nu}(x-y) J^\nu(y)\right],$$
and the things seemed to be working out. The "photon propagator" showed up and the math were quite the same as the massive fields.
But then it occurred to ask that whether this was even possible. After all, the massless field was subject to the relativistic constraint. Classical "physically", they didn't just "travel" through the spacetime; they somewhat "were" the spacetime by tracing out the geodesic (though with the interaction it's a different story, but $e=0$ here).
How do we write the spatial photon propagator from the generating function $Z[J]$ in QED?
Best Answer
It's hard to understand where your confusion is. Indeed given a generic quadratic action the generating functional is $$Z[J]=\int\mathfrak{D}\Phi \exp\left[i\int d^4x\ \Phi(x) {\cal O}\Phi(x)+i\int d^4x \ \Phi(x)J(x)\right]\tag{1}$$
where ${\cal O}$ is some differential operator. We can express, up to normalization $$Z[J]\propto \exp\left(-\dfrac{i}{2}\int d^4x d^4y\ J(x)\Delta(x,y)J(y)\right)\tag{2}$$
where $\Delta(x,y)$ is the inverse of ${\cal O}$ in the sense that if we write ${\cal O}$ in terms of a Kernel ${\cal O}(x,y)$ then $$\int d^4y\ \Delta(x,y){\cal O}(y,z)=\delta(x-z)\tag{3}.$$
This applies to scalar, fermionic and gauge-fixed gauge massless vector fields. In the scalar case $\Phi(x)$ is a scalar, in the fermionic case $\Phi(x)$ is a $\mathbb{C}^4$-tuple and ${\cal O}$ is really a matrix of differential operators, and in the vector field case ${\Phi}(x)$ is a $\mathbb{R}^4$-tuple and ${\cal O}$ is also a matrix of differential operators. In any case, from the form (2) one readily obtains the two-point function $$\langle \Phi(x)\Phi(y)\rangle=(-i)^2\dfrac{\delta^2 Z}{\delta J(x)\delta J(y)}\bigg|_{J=0}=\Delta(x,y)\tag{4}.$$