How do normalised plane wave solutions work? I am probably being really dense in this situation, but I cannot work out why the answer to the question is logical. The question is this:
Employ the definition 
$$\psi=E+icB
$$ to show that $i\frac{d\psi}{dt}=c\nabla \times \psi $. (This bit is fine.) Assuming that $\psi \propto \psi_0e^{ikx-i\omega t}$ find the normalised plane-wave solutions with the magnetic field of $\psi_0$ along the z-axis and calculate their associated currents.
The answer is as follows: 
The plane-wave substitution provides the Eigen-equation 
\begin{equation}
\omega \psi_0 = ick\hat{x} \times \psi_0, 
\end{equation}which is solved by 
\begin{equation}
\psi_0 = \frac{1}{\sqrt{2}}(\hat{y}+i\hat{z}) =\pm \frac{\omega}{c}.
\end{equation}
If anybody could help explain then I would be grateful!
 A: $\hat x \times\vec\psi_0=\frac{\omega}{ick}\vec\psi_0 $ is a matrix eigenvalue equation for $\vec\psi_0$ because the map $\vec a \mapsto \vec b\times\vec a$ is linear in $\vec a$ for a given vector $\vec b$ in the sense $\vec c+\vec d\mapsto\vec b\times\vec c+\vec b\times\vec d$. So the problem is to find an explicit matrix form of the linear operator $\hat x\times(\cdot)$. Write $$\vec\psi_0=(\psi_0^1,\psi_0^2,\psi_0^3)^T.$$So,$$\hat x\times\vec\psi_0=(0,-\psi_0^3,\psi_0^2)^T.$$ One such matrix is, 
$$ \left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0 \end{array} \right).$$
Now standard procedure of calculating eigenvalues and eigenvectors yields a family of "normalized" eigenvectors $\vec\psi_0=e^{i\theta}\frac{1}{\sqrt2}(0,\pm i,1)$ corresponding to the eigenvalues $\frac{\omega}{ick}=\pm i$ respectively. To satisfy the said condition, set the phase $\theta=\pi/2$. Then, for $k=\omega/c$, $\vec\psi_0=\frac{1}{\sqrt2}(0,1,i)$ in which the magnetic field $\mathbf{Im}(\vec\psi_0)=1/\sqrt2\hat z$ points to the $\hat z$ direction. For $k=-\omega/c$, the same happens for $\theta =-\pi/2$. Note that there is another eigenvector $(1,0,0)$ corresponding to $\frac{\omega}{ick}=0$ but you may discard it depending on the actual problem.
