# Chris Spencer Boston University EC 770: Guided-Wave Optoelectronics

Coupled-mode theory is concerned with coupling spatial modes of differing polarizations, distributions, or both. To understand codirectional coupling it is useful to have an understand of background material that builds to codirectional coupling. First, consider coupling normal modes in a single waveguide that is affected by a perturbation. Such case is single-waveguide mode coupling. The perturbation in question is spatially dependent and is represented as $$\Delta P(r)$$, a perturbing polarization. Consider the following Maxwell’s equations $\nabla \times E=i \omega \mu_0 H$ $\nabla \times H=-i \omega \epsilon E-i \omega \Delta P$ Consider two sets of fields $$(E_1,H_1)$$ and $$(E_2,H_2)$$, they satisfy the Lorentz reciprocity theorem give by $$\nabla \cdot \left( E_1 \times H_2^{*} +E_2^{*} \times H_1\right)=- i \omega \left( E_1 \cdot \Delta P_2^{*}-E_2^{*} \cdot \Delta P_1 \right)$$ For $$\Delta P_1=\Delta P$$ and $$\Delta P_2=0$$ and integrating over the result for the cross section of the waveguide in question, we get $\sum_{\nu} \frac{ \partial }{\partial z} A_{\nu}(z)e^{i\left(\beta_{\nu}-\beta_{\mu}\right)} z=i \omega e^{-i \beta_{\nu} z} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} E_{\mu}^{*} \cdot \Delta P dxdy$ Evoking orthonormality, we can get the coupled-mode equation $\pm \frac{ \partial A_{\nu} }{\partial z} =i \omega e^{-i \beta_{\mu} z} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} E_{\nu}^{*} \cdot \Delta P dxdy$ The plus sign indicates forward propagating modes when $$B_{\nu} > 0$$ and the minus sign indicates a backward propagating mode with $$B_{\nu} < 0$$
Many applications are concerned with the coupling between two modes. This coupling between two modes can be within the same waveguide or can be coupled between two parallel waveguides. For a system where we are interested in coupling two modes for either the parallel waveguides case or within the same waveguide, the two modes are described by two amplitudes A and B. The coupled equations are given by $\pm \frac{ \partial A}{\partial z}=i \kappa_{aa} A+ i\kappa_{ab} B e^{i(\beta_b-\beta_a)z}$ $\pm \frac{ \partial B}{\partial z}=i \kappa_{bb} B + i\kappa_{ba} A e^{i(\beta_a-\beta_b)z}$ for the two modes we want to solve. The coupling coefficients are given as part of a matrix $$C=[C_{\nu \mu}]$$ and $$\widetilde{\kappa}=[\widetilde{\kappa}_{\nu \mu}]$$ They are given as $C_{\nu \mu}=\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \left( E_{\nu}^{*} \times H_{\mu} + E_{\mu} \times H_\nu \right) \cdot \hat{z} dxdy=c_{\mu \nu}^{*}$ $\widetilde{\kappa}_{\nu \mu}=\omega \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} E_{\nu}^{*} \cdot \Delta \epsilon_{\mu} \cdot E_{\mu} dxdy$ where $$\Delta \epsilon$$ is the perturbation applied to the system. We then simplify the math further by removing the self coupling terms in equations (5) and (6) by expressing our normal mode coefficients by $A(z)=\widetilde{A}(z)e^{\pm i \int_{0}^{z} \kappa_{aa}(z)dz}$ $B(z)=\widetilde{B}(z)e^{\pm i \int_{0}^{z} \kappa_{bb}(z)dz}$ For cases of interest, perturbation will either be independent of z or be a periodic funciton of z. This further reduces our coupling coefficients to $\pm \frac{\partial \widetilde{A}}{\partial z}=i\kappa_{ab} \widetilde{B}e^{i 2 \delta z}$ $\pm \frac{\partial \widetilde{B}}{\partial z}=i\kappa_{ba} \widetilde{A}e^{-i 2 \delta z}$ The parameter of $$2\delta$$ is the phase mismatch between the two coupled modes.

# Codirectional Coupling

Codirectional coupling is when the coupling of two propagating modes are in the same direction, over some length $$l$$, where $$\beta_a$$ and $$\beta_b$$ are both greater than zero. Coupling equations to be used are $\frac{\partial \widetilde{A}}{\partial z}=i\kappa_{ab} \widetilde{B}e^{i 2 \delta z}$ $\frac{\partial \widetilde{B}}{\partial z}=i\kappa_{ba} \widetilde{A}e^{-i 2 \delta z}$ The general solution of this system is solved as an initial value problem in matrix form is given as $\begin{bmatrix} \widetilde{A}(z)\\ \widetilde{B}(z)\\ \end{bmatrix} = F(z;z_0) \begin{bmatrix} \widetilde{A}(z_0)\\ \widetilde{B}(z_0)\\ \end{bmatrix}$ Where $$F(z;z_0)$$ is the forward coupling matrix and it relates field amplitudes at an intial value of $$z_0$$ to those at z. For the most simple case when power is launched into only mode a at $$z=0$$ , giving $$\widetilde{B}(0)=0$$. With $$z=0$$ $\widetilde{A}(z)=\widetilde{A}(0)\left( cos \beta_c z -\frac{i \delta}{\beta_c} sin \beta_c z \right) e^{i \delta z}$ $\widetilde{B}(z)=\widetilde{B}(0)\left(\frac{i \kappa_{ba}}{\beta_c} sin \beta_c z \right) e^{-i \delta z}$ where $$\beta_c=\left( \kappa_{ab}\kappa_{ba}+\delta^2\right)^\frac{1}{2}$$ Then looking at the power of the two modes when they are completely phase matched, that is when $$\delta=0$$ $\frac{P_a(z)}{P_a(0)}=|\frac{\widetilde{A}(z)}{\widetilde{A}(0)}|^2=cos^2 \beta_c z$ $\frac{P_b(z)}{P_a(0)}=|\frac{\widetilde{B}(z)}{\widetilde{A}(0)}|^2=sin^2 \beta_c z$ where $$\kappa_{ab}=\kappa_{ba}^{*}$$ Define the coupling efficiency for a length l as $$\eta=\frac{|\kappa_{ba}|^2}{\beta_c ^2} sin^2 \beta_c z$$ as well as the coupling length $$l_c=\frac{\pi}{2 \beta_c}$$. I have plotted the phase matching condition of $$\delta=0$$ in figure 1. It is seen that we can only have complete power transfer when we are phase matched between the two coupled modes.

# Reference

[1]- Liu, Jia-Ming. “Ch. 4, Coupling of Waves and Modes.” Photonic Devices. Cambridge: Cambridge, 2005. N. pag. Print.