How to prove that the normal mode eigenvalue problem constitutes that of a Hermitian operator?

Question

I am physics PhD student working on quantisation of electromagnetic fields in a non-homogeneous media. I am working through a paper at the moment and I am struggling with one of the statements. In the paper, there is a time independent eigenvalue problem obtained from inserting a sum of exponential functions of time into the Maxwell equations.

$$\pmb{\nabla}\times\left(\frac{\pmb{\nabla}\times\pmb{B}_m(\pmb{r})}{n^2(\pmb{r})}\right) = \frac{\omega_m^2}{c^2}\pmb{B}_m(\pmb{r})$$

The text calls this a "second order Hermitian eigenvalue problem". I am unsure how to demonstrate that this problem is Hermitian and would like some help, I will attach the paper below for clarity but only the relevant couple of pages as it was sent to me by my supervisor and I am don't think it is appropriate to distribute the whole thing. I first tried establishing an inner product integrating over a large box of volume $V$ with periodic boundary conditions.

$$\hat{O}(\pmb{Y}) = \pmb{\nabla}\times\left(\frac{\pmb{\nabla}\times\pmb{Y}(\pmb{r})}{n^2(\pmb{r})}\right) \\ (\pmb{X},\hat{O}\pmb{Y}) = \iiint_{V}d^3\pmb{r} \pmb{X}\cdot\hat{O}(\pmb{Y}) = \iiint_{V}d^3\pmb{r} \pmb{X}\cdot\pmb{\nabla}\times\left(\frac{\pmb{\nabla}\times\pmb{Y}(\pmb{r})}{n^2(\pmb{r})}\right)$$

I then tried to find the components of the operator $\hat{O}$ acting on $\pmb{Y}$ to determine the inner product. I have used implicit summation over repeated symbols.

$$\left[\pmb{\nabla}\times\left(\frac{\pmb{\nabla}\times\pmb{Y}(\pmb{r})}{n^2(\pmb{r})}\right)\right]_k = \epsilon_{ijk}\frac{\partial}{\partial x_i}\left(\frac{1}{n^2(\pmb{r})}\left[\pmb{\nabla}\times\pmb{Y}(\pmb{r})\right]_j\right) = \epsilon_{ijk}\frac{\partial}{\partial x_i}\left(\frac{1}{n^2(\pmb{r})}\epsilon_{mnj}\frac{\partial}{\partial x_m} Y_n(\pmb{r})\right) \\ = -\epsilon_{jik}\epsilon_{jmn}\left(-\frac{2}{n^3(\pmb{r})}\frac{\partial n(\pmb{r})}{\partial x_i}\frac{\partial Y_n(\pmb{r})}{\partial x_m} + \frac{1}{n^2(\pmb{r})}\frac{\partial^2 Y_n(\pmb{r})}{\partial x_i\partial x_m} \right) \\ = -\left(\delta_{im}\delta_{kn} - \delta_{in}\delta_{km}\right)\left(-\frac{2}{n^3(\pmb{r})}\frac{\partial n(\pmb{r})}{\partial x_i}\frac{\partial Y_n(\pmb{r})}{\partial x_m} + \frac{1}{n^2(\pmb{r})}\frac{\partial^2 Y_n(\pmb{r})}{\partial x_i\partial x_m} \right) \\ = \frac{2}{n^3(\pmb{r})}\frac{\partial n(\pmb{r})}{\partial x_m}\frac{\partial Y_k(\pmb{r})}{\partial x_m} - \frac{1}{n^2(\pmb{r})}\frac{\partial^2 Y_k(\pmb{r})}{\partial x_m\partial x_m} - \frac{2}{n^3(\pmb{r})}\frac{\partial n(\pmb{r})}{\partial x_n}\frac{\partial Y_n(\pmb{r})}{\partial x_k} + \frac{1}{n^2(\pmb{r})}\frac{\partial}{\partial x_k}\frac{\partial Y_n(\pmb{r})}{\partial x_n}$$

From here I tried to integrate the inner product of another vector function with this vector function.

$$(\pmb{X},\hat{O}\pmb{Y}) = \iiint_{V}d^3\pmb{r} \pmb{X}\cdot\pmb{\nabla}\times\left(\frac{\pmb{\nabla}\times\pmb{Y}(\pmb{r})}{n^2(\pmb{r})}\right) \\ = \iiint_{V}d^3\pmb{r} X_k\left(\frac{2}{n^3(\pmb{r})}\frac{\partial n(\pmb{r})}{\partial x_m}\frac{\partial Y_k(\pmb{r})}{\partial x_m} - \frac{1}{n^2(\pmb{r})}\frac{\partial^2 Y_k(\pmb{r})}{\partial x_m\partial x_m} - \frac{2}{n^3(\pmb{r})}\frac{\partial n(\pmb{r})}{\partial x_n}\frac{\partial Y_n(\pmb{r})}{\partial x_k} + \frac{1}{n^2(\pmb{r})}\frac{\partial}{\partial x_k}\frac{\partial Y_n(\pmb{r})}{\partial x_n}\right)$$

I want to demonstrate:

$$(\pmb{X},\hat{O}\pmb{Y}) = (\pmb{Y},\hat{O}^*\pmb{X})$$ So it tried to do this by parts but it won't work due to the reciprocal of the refractive index, $n$. Are there any other ways I can prove this, any help would be appreciated. Thank you.

Answer 1

My original answer was misleading -- at least for the B equation. Consider the vector identity $$ \nabla \cdot ({\bf a}\times {\bf b})= {\bf b}\cdot (\nabla\times {\bf a})- {\bf a} \cdot (\nabla \times {\bf b}) $$ with ${\bf b}=n^{-2} \nabla \times {\bf v}$: $$ \nabla \cdot \left({\bf u} \times \left(\frac 1 {n^2} \nabla\times {\bf v}\right)\right)= \frac 1 {n^2}(\nabla\times {\bf u})\cdot (\nabla\times {\bf v})- {\bf u}\cdot (\nabla \times \frac 1 {n^2}(\nabla \times {\bf v})) $$ The $u,v$ symmetry shows that, discarding boundary terms $$ \int {\bf u}\cdot \left (\nabla \times \frac 1 {n^2}( \nabla \times {\bf v})\right) d^3x = \int{\bf v}\cdot \left(\nabla \times \frac 1 {n^2}( \nabla \times {\bf u})\right). $$
so $$ O \equiv \nabla\times\left( \frac 1{n^2}\nabla\,\,\times\right. $$ is hermitian with the usual inner product. I supect that $$ \nabla\times \left(\nabla\,\times \frac 1{n^2}\right. $$ is hermitian wrt to the inner product with the $1/n^2$, but it's time for my bath.... OK Hyporntex distracted me from my bath.

We also have $$ \nabla \cdot \left(\frac 1{n^2} {\bf D}_1 \times \left( \nabla\times \frac 1{n^2}{\bf D}_1\right)\right)= \frac 1 {n^2}(\nabla\times \frac 1{n^2} {\bf D}_1)\cdot (\nabla\times \frac 1 {n^2}{\bf D}_2)- \frac 1 {n^2} {\bf D}_1\cdot (\nabla \times (\nabla \times \frac 1{n^2}{\bf D}_2)) $$ so the same algebra shows that $$ \int \frac 1{n^2} {\bf D}_1\cdot \left( \nabla\times \nabla\times \frac 1{n^2}{\bf D}_2\right)d^3x =\int \frac 1{n^2} {\bf D}_2\cdot \left( \nabla\times \nabla\times \frac 1{n^2}{\bf D}_1\right)d^3x $$ so that with $$ \langle {\bf u},{\bf v}\rangle_n = \int \frac 1{n^2} {\bf u}\cdot {\bf v} d^3 x, $$ the equation for the ${\bf D}$ is self adjoint. This is why the orthogonality
$$ \int {\bf D}_i^* \frac 1{n^2} {\bf D}_j d^3x = \delta_{ij} $$ needs the $1/n^2$. Now it's time to cook supper....