Bloch state + Photonic crystals master equation

[carlosbgois]

Hello all!

I've been trying to go from the second to the third equation shown in the image.
Here, $\vec{H}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}\vec{u}_\vec{k}(\vec{r})$ is the Bloch state for some periodic dielectric arrangement.

I have tried using the identity that $\vec{\nabla}\times\phi\vec{F}=\nabla\phi\times\vec{F}+\phi\vec{\nabla} \times \vec{F}$, and got to the result $\vec{\nabla}\left(\frac{1}{\epsilon}\left[i\vec{k}\times\vec{u}+\vec{\nabla}\times\vec{u}\right]\right)=\frac{\omega^2}{c^2}\vec{u}$, but can't see where to go from here.

Any help is appreciated.
Thanks in advance.

 

[eljose79]

Thank you for your question. It seems like you are trying to derive the third equation from the second equation in the image provided. The second equation represents the Bloch state for a periodic dielectric arrangement, where \vec{H}(\vec{r}) is the magnetic field and e^{i\vec{k}\cdot\vec{r}}\vec{u}_\vec{k}(\vec{r}) is the periodic part of the Bloch state. In order to derive the third equation, you can use the Maxwell's equations in the form of \vec{\nabla}\times\vec{E}=-\frac{1}{c}\frac{\partial\vec{B}}{\partial t} and \vec{\nabla}\times\vec{H}=\frac{1}{c}\frac{\partial\vec{D}}{\partial t}, where \vec{E} is the electric field and \vec{D} is the electric displacement field. From these equations, you can derive the relation \vec{\nabla}\times\vec{H}=\frac{\omega}{c}\vec{D}, where \omega is the frequency of the electromagnetic wave. Since \vec{D}=\epsilon\vec{E}, where \epsilon is the dielectric constant, you can substitute this into the previous equation to get \vec{\nabla}\times\vec{H}=\frac{\omega}{c}\epsilon\vec{E}. Using the identity you mentioned, \vec{\nabla}\times\phi\vec{F}=\nabla\phi\times\vec{F}+\phi\vec{\nabla} \times \vec{F}, you can rewrite this equation as \vec{\nabla}\times\vec{H}=\frac{\omega}{c}\epsilon\vec{E}=\frac{\omega}{c}\left(\frac{1}{\epsilon}\epsilon\right)\vec{E}=\frac{\omega}{c}\left(\frac{1}{\epsilon}\right)\left[\vec{\nabla}\times\left(i\vec{k}\times\vec{u}+\vec{\nabla}\times\vec{u}\right)\right]. From here, you can equate the coefficients of \vec{E} and \vec{H} to get the desired result \vec{\