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The Green's function of the Klein-Gordon equation: $$\phi_s(x_\mu-y_\mu) = \int \frac{d^4k}{(2\pi)^4} \; \frac{e^{-i k^\mu (x_\mu -y_\mu)}}{-k_\mu k^\mu + m^2}$$ is the solution to the equation $$ \left(\partial_\mu \partial ^{\mu} + m^2\right) \phi_s = \delta(x_\mu-y_\mu) \, .$$ What's the corresponding equation and solution for a massless gauge field $A_\mu$? The free equation of motion reads $$\partial_\sigma ( \partial^\sigma A^\rho - \partial^\rho A^\sigma) =0 \, $$ but I'm unsure what we write here on the right-hand side to define the Green's function since there is a free index $\rho$ in the equation of motion. Moreover, although I checked several textbooks, I wasn't able to find the expression for the Green's function for photons.

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In momentum space the equation of motion for photon is: $$(-p^2g_{\mu\nu}+p_\mu p_\nu)A_\mu= J_\nu$$ where $J_\nu$ is the source term. Now the term in the parentheses is not invertible since the determinant is zero (no mass term). This is also a signature of Gauge freedom of the photon fields. To remedy this, a gauge term is added to the original Lagrangian. $$\mathcal{L}= -\frac{1}{4} F_{\mu\nu}^2-\frac{1}{2\xi}(\partial_\mu A_\mu)^2 - J_\mu A_\mu.$$ Here $\xi$ is the gauge field which satisfies the equation of motions $\partial_\mu A_\mu=0$ (Lorenz gauge). With this Lagrangian, the equation of motion becomes $$\left[-p^2g_{\mu\nu}+\left(1-\frac{1}{\xi}\right)p_\mu p_\nu\right]A_\mu= J_\nu.$$ Now, one can check that the propagator is given by the following expression $$\Pi_{\mu\nu}=-\frac{g_{\mu\nu}-(1-\xi)p_\mu p_\nu/p^2}{p^2}.$$ To check that this is indeed the propagator, it should satisfy the equation $\left[-p^2g_{\mu\nu}+\left(1-\frac{1}{\xi}\right)p_\mu p_\nu\right]\Pi_{\nu\beta} = g_{\mu\beta}$.

So, the expression for propagator or Green's function is dependent on the gauge choice as it should be but all the physical observables should be independent of the gauge field $\xi$. Some most used choices of gauge fields in Quantum Electrodynamics are:

  1. Feynman gauge: $\xi=1\implies \Pi_{\mu\nu}=-\frac{g_{\mu\nu}}{p^2+i\epsilon}$.

  2. Lorenz gauge: $\xi=0 \implies \Pi_{\mu\nu}=-\frac{g_{\mu\nu}-p_\mu p_\nu/p^2}{p^2+i\epsilon}$

The $i\epsilon$ part is added to choose the causal Green's function and in the final expression of an observable $\epsilon\rightarrow 0$ limit is taken.

To read more about photon propagators you can have a look at any QED book. The discussion of the propagator should be independent of whether this is a classical or quantum field theory.

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