Interaction of charges in gauge

Question

Let's consider Maxwell theory:

$$ \mathcal{L} = -F_{\mu\nu}F^{\mu\nu} = 2 A_\mu (\Box \eta^{\mu\nu} - \partial^\mu \partial^\nu) A_\nu $$

Is it possible to fix gauge $A_0 = 0 $ and concider Lagrangian:

$$ \mathcal{L} = 2 A_i (- \Box \delta^{ij} - \partial^i \partial^j) A_j $$

And do quntization of such theory in Lorentz non-invariant manner? Or there are some other barriers?

More concreatly, is it possible to calculate integration between 2 external charges due to gauge field?

I see obstruction in interaction therm for statical charges $J^\mu = (q_1\delta(\vec r - \vec r_1)+ q_2\delta(\vec r - \vec r_2), 0 , 0, 0)$ interaction therm $A_\mu J^\mu = 0$. Is possible to do such calculation in this gauge?

Answer 1

Even in classical field theory calculation, one cannot just substitute $A_0=0$ in the Lagrangian. Before gauge fixing, $A_0$, being non-dynamical Lagrange-multiplier-like variable, imposes with its EOM a (gauge-invariant) constraint like $div\, \vec{E} = J^0$ (which is the Gauss law). One should remember about this constraint after fixing the gauge to get sensible answers.

I believe that same applies to the quantum case: one should impose Gauss law constraint by hand as a requirement on physical states of the system. As another example of the same problem, recall how for Polyakov action in string theory after fixing the gauge for the $g_{\mu\nu}$ field we get just a system of $d$ massless free bosonic fields. Yet there, this is not the full story, as one should remember about EOMs for the metric, which is $T=0$; only after imposing this constraint we get correct answers.