# Quantum Eigenstates of Curved and Varying Cross-Sectional Waveguides
## 2 The General Equations
### Constant Orientation
We furthermore have that  
\[\mathcal{R}^{2} =v^{2}\frac{\partial^{2}}{\partial v^{2}}+2 vw\frac{\partial^{2}} {\partial v\partial w}+w^{2}\frac{\partial^{2}}{\partial w^{2}}+v\frac{ \partial}{\partial v}+w\frac{\partial}{\partial w}\,,\] \[\left(\frac{\partial}{\partial s}-\frac{h^{\prime}}{h}\mathcal{ R}\right)^{2} =\frac{\partial^{2}}{\partial s^{2}}-2\frac{h^{\prime}}{h}\mathcal{ R}\frac{\partial}{\partial s}+\left(\frac{h^{\prime}}{h}\right)^{2}(\mathcal{R}^{2}+ \mathcal{R})-\frac{h^{\prime\prime}}{h}\mathcal{R}\,.\]  
Denoting the coordinates \((s,v,w)\) by \((u^{1},u^{2},u^{3})\) we can now write the Laplace operator as  
\[\triangle =G^{k\ell}\frac{\partial}{\partial u^{k}}\frac{\partial}{ \partial u^{\ell}}+\left(\frac{G^{k\ell}}{2G}\frac{\partial G}{\partial u^{k }}+\frac{\partial G^{k\ell}}{\partial u^{k}}\right)\frac{\partial}{\partial u^ {\ell}}\] \[=\frac{1}{e^{2}h^{2}}\left(\frac{\partial^{2}}{\partial v^{2}}+ \frac{\partial^{2}}{\partial w^{2}}\right)+\frac{1}{F^{2}}\left(\frac{ \partial}{\partial s}-\frac{h^{\prime}}{h}\mathcal{R}\right)^{2}\] \[\quad-\frac{1}{ehr}\left(\kappa_{1}\frac{\partial}{\partial v}+ \kappa_{2}\frac{\partial}{\partial w}\right)+\frac{\epsilon h(\kappa_{1}^{ \prime}v+\kappa_{2}^{\prime}w)}{F^{3}}\left(\frac{\partial}{\partial s}-\frac {h^{\prime}}{h}\mathcal{R}\right)\,. \tag{11}\]  
As usual [15], we let \(\chi=\sqrt{F}\psi\) and have that  
\[\int_{\Omega}\psi_{1}\psi_{2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\int_{ \Omega_{0}}\int_{0}^{L}\psi_{1}\psi_{2}F\,\mathrm{d}s\,\mathrm{d}v\,\mathrm{d }w=\int_{\Omega_{0}}^{1}\int_{0}^{L}\chi_{1}\chi_{2}\,\mathrm{d}s\,\mathrm{d}v \,\mathrm{d}w\,, \tag{12}\]  
such that the Helmholtz equation reads  
\[-\triangle\psi=\lambda\psi\iff-\triangle\frac{\chi}{\sqrt{F}}=\lambda\frac{ \chi}{\sqrt{F}}\iff-\sqrt{F}\triangle\frac{\chi}{\sqrt{F}}=\lambda\chi. \tag{13}\]  
Later, we shall exemplify the present analysis to the Schrodinger equation for a quantum-mechanical particle. For now, we focus on the Helmholtz equation. To zeroth order in \(\epsilon\), we have  
\[\sqrt{F}\triangle\frac{\chi}{\sqrt{F}}\approx\frac{1}{\epsilon^{2}h^{2}}\left( \frac{\partial^{2}\chi}{\partial v^{2}}+\frac{\partial^{2}\chi}{\partial w^{2} }\right)+\left(\frac{\partial}{\partial s}-\frac{h^{\prime}}{h}\mathcal{R} \right)^{2}\chi+\frac{\kappa^{2}}{4}\chi\,. \tag{14}\]  
Unless \(h^{\prime}=0\), i.e., the thickness is constant, we cannot separate the variables. Instead we will use perturbation theory. We have  
\[\left(\frac{\partial}{\partial s}-\frac{h^{\prime}}{h}\mathcal{R}\right)^{2}= \frac{\partial^{2}}{\partial s^{2}}\underbrace{-2\frac{h^{\prime}}{h} \mathcal{R}\frac{\partial}{\partial s}+\left(\frac{h^{\prime}}{h}\right)^{2} \mathcal{R}^{2}-\left(\frac{h^{\prime\prime}}{h}-\left(\frac{h^{\prime}}{h} \right)^{2}\right)\mathcal{R}}_{\text{treated as a perturbation}}\,.\]The last three terms are considered as perturbations and first order perturbation theory yields  
\[\mathcal{R}^{2} =v^{2}\frac{\partial^{2}}{\partial v^{2}}+2 vw\frac{\partial^{2}} {\partial v\partial w}+w^{2}\frac{\partial^{2}}{\partial w^{2}}+v\frac{ \partial}{\partial v}+w\frac{\partial}{\partial w}\,,\]0 \[\mathcal{R}^{2} =v^{2}\frac{\partial^{2}}{\partial v^{2}}+2 vw\frac{\partial^{2}} {\partial v\partial w}+w^{2}\frac{\partial^{2}}{\partial w^{2}}+v\frac{ \partial}{\partial v}+w\frac{\partial}{\partial w}\,,\]1  
where \(\langle\mathcal{R}\rangle\) and \(\langle\mathcal{R}^{2}\rangle\) are the expectation values of \(\mathcal{R}\) and \(\mathcal{R}^{2}\), respectively. \[\mathcal{R}^{2} =v^{2}\frac{\partial^{2}}{\partial v^{2}}+2 vw\frac{\partial^{2}} {\partial v\partial w}+w^{2}\frac{\partial^{2}}{\partial w^{2}}+v\frac{ \partial}{\partial v}+w\frac{\partial}{\partial w}\,,\]2  
where \(\psi_{0}\) is a transversal eigenstate. This procedure is easy in the case of non-degenerate transverse eigenstates such as for the transverse groundstate. However, in case of degenerate transverse eigenstates, standard degenerate perturbation theory must be used.