Riemann–Silberstein vector

Remarks

The pair $(E, B)$ solves the source-free Maxwell's equations exactly when $$\nabla \cdot F = 0, \qquad i \, \partial_t F = c \, \nabla \times F.$$ Indeed, the real and imaginary parts of the second equation are $\partial_t B = -\nabla \times E$ and $\partial_t E = c^2 \nabla \times B$, and the first equation contains both Gauss laws. A duality rotation by the angle $\theta$ maps $F$ to $e^{-i\theta} F$, and $\frac{\varepsilon_0}{2} \abs{F}^2 = \frac{\varepsilon_0}{2} (\abs{E}^2 + c^2 \abs{B}^2)$ is the energy density. Some authors use $E - icB$ or a normalizing factor such as $\sqrt{\varepsilon_0 / 2}$.

Example: for the plane wave $E = (0, E_0 \cos(kx - \omega t), 0)$, $B = (0, 0, \frac{E_0}{c} \cos(kx - \omega t))$ with $\omega = ck$, one gets $F = E_0 \cos(kx - \omega t) \, (0, 1, i)$. With $s = \sin(kx - \omega t)$, both $i \partial_t F$ and $c \nabla \times F$ equal $(0, i \omega E_0 s, -\omega E_0 s)$, and $\nabla \cdot F = 0$.

Non-example: $F(x, y, z, t) = (-y, x, 0)$ is a Riemann–Silberstein vector, with $E = (-y, x, 0)$ and $B = 0$, but it does not come from a solution of the source-free equations, since $i \partial_t F = 0$ while $c \nabla \times F = (0, 0, 2c)$.