# How do normalised plane wave solutions work? [closed]

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!

• I edited your question by inserting a factor of $e$ in front of $i(kx-\omega t)$ otherwise your expression did not make sense. There are some notational issues with your question: are $E$ and $B$ vectors and if so what is their respective direction? This would be helpful since it would pin down $\psi$ and $\psi_0$ as vectors so you could legitimately take the curl and cross-product. Also, it would clarify if your final expression is a vector or a scalar, and if the $\psi_0$ you are looking for is a magnitude or a direction. In particular, in a plane wave $\vec B\cdot\vec E=0$. Apr 8, 2017 at 13:20

$\hat x \times\vec\psi_0=\frac{\omega}{ick}\vec\psi_0$ 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.