quantum field theory - The physicality of the photon propagator

The equation for the photon propagator is straightforward

$$D_{ij} = \langle 0 |T \{ A_{i}(x')A_{j}(x) \}|0 \rangle$$ However, $A_{i}(x)$ is gauge-dependent and therefore unphysical (in the arguable sense). Then, since the propagator is dependent on the vector potential, the propagator is unphysical. Sadly, my whole understanding of what amplitudes mean may be skewed, but I would assume the probability amplitude for a photon to propagate between $x$ and $x'$ is something we would want to be gauge-independent.

Edit:

I guess I wasn't clear enough. By computing the probability amplitude for a process, we obtain a complex number that when multiplied by it's complex conjugate we obtain a probability for such a process to occur (when normalized). Here, the physical process is propagation, and the probability is $|\langle 0 |T \{ A_{i}(x')A_{j}(x) \}|0 \rangle|^2$. However, this probability is gauge dependent, and hence, the usual physical interpretation of $|\langle 0 |T \{ A_{i}(x')A_{j}(x) \}|0 \rangle|^2$ is questionable to me. Where has my interpretation gone astray?

Answer

The photon propagator $D_{\mu\nu}(x,y) = \langle 0 | A_\mu(x) A_\nu(y)|0\rangle$ is a building block for amplitudes, but it isn't necessarily an amplitude itself. The source for an electromagnetic field has to be a conserved current, which basically means that you create states from the vacuum using linear combinations of $A_\mu(x)$ operators whose coefficients are conserved currents.

$$|J\rangle = \int J^\mu(x) A_\mu(x) dx |0\rangle$$ where $\partial_\mu J^\mu = 0$.

You can show by direct computation that the amplitude $\langle J_1 | J_2 \rangle = \int\int J_1^\mu(x) D_{\mu\nu}(x,y)J_2^\nu(y)dxdy$ is gauge invariant if the currents $J_1$ and $J_2$ are conserved.