How to prove that Feynman propagator are equivalent in Coulomb gauge and $R_\zeta$ gauge? (Be more specific, they are same when they contract with external current)
In $R_\zeta$ gauge, the propagator takes the form $$ D_{F\mu\nu}=-\frac{1}{k^2+i\epsilon}\left[g_{\mu\nu}-(1-\zeta)\frac{k_{\mu}k_\nu}{k^2}\right] $$
When $\zeta=1$ it's the Feynman gauge, $D_{F\mu\nu}=-\frac{g_{\mu\nu} }{k^2+i\epsilon} $. It's the usual form we use in Feynman rules. Because the $A_{\mu}$ is always coupled to the conserved current，i.e. $\partial_\mu J^\mu(x)=0$ so $k_\mu J^{\mu}(k)=0$. So it proves that propagator in $R_{\zeta}$ gauge is equivalent to the Feynman gauge.
In Coulomb gauge, the propagator takes the form: $$ \begin{aligned} D^C_{F\mu\nu}&=\frac{1}{k^2+i\epsilon}\left(\sum_{i=1,2}\epsilon_\mu(k,i)\epsilon_\nu(k,i)\right)\\ &= -\frac{g_{\mu\nu} }{k^2+i\epsilon} -\frac{n_{\mu}n_{\nu}}{(k\cdot n)^2-k^2}-\frac{1} {k^2+i\epsilon} \frac{k_\mu k_\nu-(k_\mu n_\nu+k_\nu n_\mu )(k.n)}{(k\cdot n)^2-k^2} \end{aligned} $$
Using the same argument of above, the 3rd term has no contribution. But the second term is an instantaneous Coulomb interaction. If we choose $n^{\mu}=(1,0,0,0)$, the second the term is $$ \frac{\delta_{\mu,0}\delta_{\nu,0}}{|\mathbf{k}|^2} $$
In coordinate space, this term is $$ \delta_{\mu,0}\delta_{\nu,0}\frac{\delta(x_0-y_0)}{4\pi|\mathbf{x}-\mathbf{y}|} $$
I can't see why it's zero when it couples with external current.
