# 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!

## closed as unclear what you're asking by Emilio Pisanty, Kyle Kanos, ZeroTheHero, Yashas, AccidentalFourierTransformApr 11 '17 at 14:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $\vec B\cdot\vec E=0$. – ZeroTheHero Apr 8 '17 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.