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

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.

• Which paper? Which page? Commented Feb 16, 2021 at 22:02
• Apologies I mis-spoke it is a PhD thesis called "Quantum and Classical Optics of Dispersive and Absorptive Structured Media" by Navin Bhat that I have been told to work through. It is on pages 21 to 22. As I said I don't think it appropriate to promulgate the paper as it was sent to me by my supervisor and the question has been answered anyway both below and subsequently by my supervisor. Commented Feb 17, 2021 at 14:21

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....
• I am surprised that the system can be shown to be Hermitean because it has been known since 1960's that a lossless (yes, $\epsilon, \mu$ are real) but inhomogeneously filled waveguide can have complex eigenvalues and with that propagating modes $e^{\mathfrak{j}(\omega t - \beta z)}$ such that the wavenumber $\beta$ is complex number. See, e.g., Clarricoats "COMPLEX MODES OF PROPAGATION IN DIELECTRIC LOADED CIRCULAR WAVEGUIDE", ELECTRONICS LETTERS July 1965 Vol. 1 No. 5, and Rozzi : "General Constraints on the Propagation of Complex Waves..." IEEE MTT VOL. 46, NO. 5, MAY 1998 Commented Feb 16, 2021 at 21:17
• As I said to hyportnex, I did not consider the BC's required to make the boundary terms vanish. For a perfect conductor and uniform $\epsilon$, they are the usual BC's of Maxwell fields at a metallic interface, and would imagine that they still work for non uniform $\epsilon$, but I do not know. For the uniform case you can look at page 252 in our book goldbart.gatech.edu/PostScript/MS_PG_book/bookmaster.pdf Commented Feb 16, 2021 at 22:39