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?
Best Answer
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.