If we start in the gauge

\begin{align*} \textbf{E}=-\nabla\phi-\frac{\partial\textbf{A}}{\partial t}, \end{align*}

\begin{align*} \textbf{B}=\nabla\times\textbf{A} \end{align*}

We can express everything in terms of the vector potential by performing the gauge transformation

\begin{align*} \textbf{A}\rightarrow\textbf{A}'=\textbf{A}+\nabla\chi \end{align*} \begin{align*} \phi\rightarrow\phi'=\phi-\frac{\partial\chi}{\partial t} \end{align*}

Where $\nabla\phi=\nabla\partial_{t}\chi$, then we have a gauge where both the electric and magnetic fields are expressed solely in terms of vector potential:

\begin{align*} \textbf{E}=-\frac{\partial\textbf{A}'}{\partial t}, \end{align*}

\begin{align*} \textbf{B}=\nabla\times\textbf{A}' \end{align*}

Often, in textbooks on quantum optics, I see them start with this definition for both fields before choosing the Coulomb gauge. To transform to the Coulomb gauge we require $\nabla\cdot\textbf{A}=0$. Thus repeating the gauge transformation to enforce this

\begin{align*} \textbf{A}'\rightarrow\textbf{A}''&=\textbf{A}'+\nabla\chi'\\ &=\textbf{A}+\nabla\chi+\nabla\chi' \end{align*}

\begin{align*} \phi'\rightarrow\phi''&=\phi'-\frac{\partial\chi'}{\partial t}\\ &=\phi-\frac{\partial\chi}{\partial t}-\frac{\partial\chi'}{\partial t} \end{align*}

As long as

\begin{align*} \nabla^{2}\chi'=-\nabla\cdot\textbf{A}' \end{align*}

we're in the Coulomb gauge. However is it possible to do this without changing the value of $\nabla\phi$? Intuitively, if

\begin{align*} \nabla\frac{\partial\chi'}{\partial t}=0 \end{align*}

then it's fine, but that implies that $\nabla^{2}\chi'$ has no time dependence, and therefore that $\nabla\cdot\textbf{A}'$ has no time dependence. Feels like theres a loss of generality in doing this, for instance if the fields are functions of time. So my question is, is it okay to take both these gauges in conjunction for a completely general pair of electric and magnetic fields? And if not, what conditions is this okay in?

  • $\begingroup$ A few points for clarification: "If we start in the gauge" -- the first two equations are not a gauge, they are the general expression for fields in terms of potentials. "are expressed solely in terms of vector potential" -- this is an indirect way of saying that $\phi = 0$ (Weyl gauge or temporal gauge). "To transform to the Coulomb gauge we require $\boldsymbol{\nabla}\phi = \mathbf{0}$" -- did you mean $\boldsymbol{\nabla} \cdot \mathbf{A} = 0$? $\endgroup$
    – nanoman
    Commented Sep 27, 2022 at 8:26
  • $\begingroup$ Yeh this was a typo on my behalf, apologies $\endgroup$ Commented Sep 27, 2022 at 10:11

1 Answer 1


You're asking whether we can impose both $\phi = 0$ and $\nabla \cdot \mathbf{A} = 0$ simultaneously. This will not be possible in any situation where $\rho \neq 0$, since if both conditions on the potentials hold we necessarily have $$ \frac{\rho}{\epsilon_0} = \nabla \cdot \mathbf{E} = - \nabla^2 \phi - \frac{\partial (\nabla \cdot \mathbf{A})}{\partial t} = 0. $$

However, in the absence of any charge (which is usually the case in optics), it is possible to impose both of these conditions simultaneously. Specifically, we want to construct a $\chi(\mathbf{r},t)$ so that (given our original $\phi$ and $\mathbf{A}$) we simultaneously have $$ \nabla^2 \chi = - \nabla \cdot \mathbf{A}, \qquad \frac{\partial \chi}{\partial t} = \phi. \tag{1} $$ The second condition implies that we can write $$ \chi(\mathbf{r},t) = \int^t_{t_0} \phi(\mathbf{r},t')\,dt' + f(\mathbf{r}) $$ where $t_0$ is an arbitrary initial time and $f(\mathbf{r})$ is an as-yet-undetermined function of $\mathbf{r}$ only. Assuming that there is no charge present, we then have \begin{align*} \nabla^2 \chi(\mathbf{r},t) &= \int^t_{t_0} \nabla^2 \phi(\mathbf{r},t')\,dt' + \nabla^2 f(\mathbf{r})\\ &= - \int^t_{t_0} \left[ \underbrace{\nabla \cdot\mathbf{E}}_{{}=\,0} + \nabla \cdot \frac{\partial \mathbf{A}}{\partial t'} \right]_{\mathbf{r},t'}\,dt' + \nabla^2 f(\mathbf{r})\\ &= - \nabla\cdot \mathbf{A}(\mathbf{r},t) + \nabla\cdot \mathbf{A}(\mathbf{r},t_0) + \nabla^2 f(\mathbf{r}) \end{align*} Choosing $f$ to satisfy $$ \nabla^2 f(\mathbf{r}) = - \nabla \cdot \mathbf{A} (\mathbf{r},t_0) $$ completes the proof: the $\chi(\mathbf{r},t)$ that we have constructed satisfies both of the conditions in (1) simultaneously.

  • $\begingroup$ thanks, did think it was a bit suspicious $\endgroup$ Commented Sep 27, 2022 at 10:10

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