So, I study photonics and there is a theme "Modal decomposition for 1D Phc". It has the following part:
Maxwell equation for both polarizations read:
$$\eta(z)\frac{d}{dz}\left(\frac{1}{\eta(z)}\frac{d\psi(z)}{dz}\right) + \omega^2\epsilon(z)\mu(z)\psi(z) = \beta^2\psi(z)$$
We denote a second order operator:
$$A(z) = \eta(z)\frac{d}{dz}\left(\frac{1}{\eta(z)}\frac{d}{dz}\right) + \omega^2\epsilon(z)\mu(z)$$
$$A(z)\psi(z)=\beta^2\psi(z)$$
In case of a pure dielectric material this operator is self-conjugate, having real eigenvalues and eigenfunctions which form a complete orthonormal set:
$$A\psi_m(z)=\beta_m^2\psi_m(z)$$
$$\frac{1}{\Lambda}\int\limits_0^\Lambda \frac{\psi_m(z)\psi_n^*(z)}{\omega \eta(z)}dz = \delta_{mn}$$
Here, we have a 1D Phc consisting of 2 layers, and $\eta(z)$ is either $\epsilon$ or $\mu$, depending on polarization. First layer has $\epsilon_1,\mu_1$ and $\eta_1$ for $z=[0,d_1]$, and second layer has $\epsilon_2,\mu_2$ and $\eta_2$ for $z=[d_1,d_2]$, $\Lambda = d_1+d_2$ is the period. And I wonder how do I get this exact orthogonality condition?
I tried to use property of this operator $$\int \psi_m^*A\psi_n dV = \int (A^*\psi_m^*)\psi_ndV$$ But it just leads to $\int \psi_m^* \psi_n dz = \delta_{mn}$, which looks like is not the case here.