Gauge invariant Green's function for electrodynamics Varying the electromagnetic action
\begin{equation}
S=-m c \int d s\left(\dot{z}^{2}\right)^{\frac{1}{2}}-\frac{e}{c} \int d s A_{\mu} \dot{z}^{\mu}-\frac{1}{16 \pi c} \int d^{4} x F_{\mu \nu} F^{\mu \nu}
\end{equation}
we get the two equations of motion
\begin{equation}
m \ddot{z}^{\mu}=\frac{e}{c^{2}} F^{\mu \nu} \dot{z}_{\nu}
\end{equation}
\begin{equation}
\square A^{\mu}-\partial^{\mu}\left(\partial_{\nu} A^{\nu}\right)=\frac{4 \pi}{c} j^{\mu}
\end{equation}
If we use the Lorenz gauge, the solutions to the second equation are the  Liénard–Wiechert potentials.
Is there a general solution to the second equation without fixing a gauge? By solution I mean a closed expression for $A^{\mu}$ in terms of $j^{\mu}$. Or is it possible to proof that such a solution does not exist.
 A: As my2cts said the equation is not invertible. Here's a quick proof. First we go to momentum space by Fourier transforming both sides of the equation
$$ (-k^2 \eta^{\mu\nu}+ k^\mu k^\nu )\tilde A_\nu(k) = \frac{4\pi}{c} \tilde j(k)^\mu$$
Now the solution for $\tilde A(k)$ would be simply given by solving the equation algebrically and then doing an inverse Fourier transform. Thing is, that the tensor operator on the LHS is not invertible.
You can see this easily by noticing that that operator is zero on the vector space spanned by $k^\mu$, since
$$ (-k^2 \eta^{\mu\nu}+ k^\mu k^\nu ) k_\nu = -k^2 k^\mu + k^\mu k^\nu k_\nu = 0$$
The reason why you have to fix the gauge before solving these equations is to get rid of this "null direction". For example imposing $\partial^\mu A_\mu = 0$ is equivalent, in momentum space to $k^\mu A_\mu = 0$. This condition prevents explicitly any configuration that would make the above operator zero and makes it invertible.
A: If your question is, as the title suggests, whether the second equation can be inverted, the answer is no. Inversion requires 'gauge fixing'. An operator can be inverted only if it establishes a one to one relation between solution and source. The operator in your second equation assigns many potentials to a single source and hence cannot be inverted.
