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). 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 $(\nabla-iA)^2u$.


You could just try: $A\mapsto A+\nabla \lambda$ leaves the first two equations invariant if I transform $u\mapsto e^{i\lambda}u$ and leave $V$ the same, but then the third equation no longer holds, so the gauge is fixed.

To find which choice of gauge has been made, take the divergence of the third equation $-\Delta A=J$. Within the "semi-relativistic" approximation the divergence of $J$ vanishes, hence $\nabla\cdot A=0$.

  • $\begingroup$ Just to be sure, when taking the divergence you get $\Delta \nabla \cdot A = 0$ because $\nabla \cdot J=0$ but how does that imply $\nabla \cdot A = 0$? $\endgroup$ Feb 4 at 13:08
  • $\begingroup$ let's see what freedom is left: $\nabla\cdot A=0$ satisfies current conservation, if you try $A'=A+\nabla \lambda$ then you need $\Delta\nabla\lambda=0$; that leaves some room, but if you wish $\lambda$ to become constant at infinity only $\lambda\equiv 0$ remains. $\endgroup$ Feb 4 at 16:32
  • $\begingroup$ Sorry I'm still confused, that was not my question: How does taking the divergence of the equation $-\Delta A = J$ imply $\nabla \cdot A =0$? $\endgroup$ Feb 5 at 11:11
  • $\begingroup$ $\nabla\cdot J=0\Rightarrow \nabla\cdot (\Delta A)=0\Rightarrow \Delta(\nabla\cdot A)=0$ and hence either $\nabla\cdot A=0$ or the vector potential is given by $A=A_0+\nabla\lambda$ with $\nabla\cdot A_0=0$ and $\Delta\nabla\lambda=0$; this remaining gauge freedom to choose $\lambda$ will force $\lambda\equiv 0$ if you add the condition that $\lambda\rightarrow 0$ for $r\rightarrow\infty$. $\endgroup$ Feb 5 at 11:53
  • $\begingroup$ Got it, thanks for the clarification! $\endgroup$ Feb 5 at 11:55

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