When studying non-relativistic charged particles in an electromagnetic field with self-interaction on usually relies on the Schrödinger-Maxwell system
\begin{align}
i\partial_t u = -(\nabla-iA)^2 u + V u\\
-\Delta V -\partial_t \nabla \cdot A = |u|^2 \\
\Box A + \nabla (\partial_t V +\nabla \cdot A) = J(u,A)
\end{align}
where $J(u,A) = 2\Im(\bar{u}(\nabla-iA)u $. Now this system has gauge freedom ($u\mapsto e^{i\lambda} u$, $V \mapsto V-\partial_t \lambda$, $A\mapsto A+\nabla\lambda$) for some function $\lambda\colon \mathbb{R}^{1+3} \rightarrow \mathbb{R}$. One particular gauge fixing is the Coulom gauge $\nabla \cdot A = 0$ which reduces the system above to
\begin{align}
i\partial_t u = -(\nabla-iA)^2 u + V u\\
-\Delta V  = |u|^2 \\
\Box A = \mathbb{P}J(u,A)
\end{align}
where $\mathbb{P}$ is the usual projection on divergence free vector fields (to be precise $\mathbb{P}f = f-\nabla\Delta^{-1} \nabla f$). The gauge $\nabla \cdot A$ is useful in many of the calculations when proving statements about the system, e.g. for reducing $(\nabla-iA)^2u$ to $\Delta u -2iA\cdot \nabla u-|A|^2u$. It is possible to derive a system where the Maxwell equations are approximated in a semi-relativistic fashion by keeping terms only up to order $1/c$ and therefore one would obtain
\begin{align}
i\partial_t u &= -(\nabla-iA)^2 u + V u\\
-\Delta V &= |u|^2 \\
-\Delta A &= J(u,A) \qquad (*)
\end{align}
(cf. [this paper by Masmoudi, Mauser][1]). Using Lorentz gauge the authors set up the wave equation formulation of Maxwell's equations and then do an asymptotic expansion, keeping only terms up to order $\epsilon := 1/c$, here $c$ is the speed of light (I am omitting the fact that the authors actually derive a slightly different system, namely the semi-relativistic approximation of Dirac-Maxwell which gives you an additional spin coupling in the Schrödinger equation but my question focuses on the Poisson equation for $A$).

**My question is:** Does the semi-relativistic system still fix the gauge, namely the Coulomb gauge? That is, can I still impose $\nabla \cdot A = 0$ on $A$, now that it is given by $(\ast)$? Or does $(\ast)$ completely fix the gauge for $A$ and I can no longer expect $\nabla \cdot A= 0$ to hold?

**EDIT** Forgot the minus sign in front of $(\nabla-iA)^2u$.


  [1]: https://www.esi.ac.at/static/esiprpr/esi701.pdf