# Follow Up Questions To Presentation on Codirectional Coupling

Revisting the initial value problem of $\begin{bmatrix} \tilde{A}(z)\\ \tilde{B}(z)\\ \end{bmatrix} = F(z;z_0) \begin{bmatrix} \tilde{A}(z_0)\\ \tilde{B}(z_0) \\ \end{bmatrix}$ The question is to solve this subject to the initial condition of when power is launched only into the mode a at $$z_0=0$$. This means $$\tilde{B}(0)=0$$ and $$\tilde{A}(0)\neq 0$$. $F(z;z_0)= \begin{bmatrix} \frac{\beta_c cos \beta_c(z-z_0)-i\delta sin \beta_c(z-z_0)}{\beta_c}e^{i \delta (z+z_0)} & \frac{i \kappa_{ab}}{\beta_c}sin\beta_c(z-z_0)e^{i \delta (z+z_0)}\\ \frac{i \kappa_{ba}}{\beta_c}sin\beta_c(z-z_0)e^{-i \delta (z+z_0)} &\frac{\beta_c cos \beta_c(z-z_0)+i\delta sin \beta_c(z-z_0)}{\beta_c}e^{-i \delta (z+z_0)} \\ \end{bmatrix}$ Subject to our initial conditions then $\begin{bmatrix} \tilde{A}(z)\\ \tilde{B}(z)\\ \end{bmatrix} = F(z;0) \begin{bmatrix} \tilde{A}(0)\\ 0 \\ \end{bmatrix}$ Carrying the matrix multiplication step by step $\begin{bmatrix} \tilde{A}(z)\\ \tilde{B}(z)\\ \end{bmatrix}=\begin{bmatrix} \frac{\beta_c cos \beta_c(z)-i\delta sin \beta_c(z)}{\beta_c}e^{i \delta (z)} & \frac{i \kappa_{ab}}{\beta_c}sin\beta_c(z)e^{i \delta (z)}\\ \frac{i \kappa_{ba}}{\beta_c}sin\beta_c(z)e^{-i \delta (z)} &\frac{\beta_c cos \beta_c(z)+i\delta sin \beta_c(z)}{\beta_c}e^{-i \delta (z)} \\ \end{bmatrix} \begin{bmatrix} \tilde{A}(0)\\ 0 \\ \end{bmatrix}$ $\begin{bmatrix} \tilde{A}(z)\\ \tilde{B}(z)\\ \end{bmatrix}=\begin{bmatrix} \frac{\tilde{A}(0) \beta_c cos \beta_c(z)-i\delta sin \beta_c(z)}{\beta_c}e^{i \delta (z)} & 0\\ \tilde{A}(0) \frac{i \kappa_{ba}}{\beta_c}sin\beta_c(z)e^{-i \delta (z)} & 0\\ \end{bmatrix}$ This leaves us then with $\tilde{A}(z)=\tilde{A}(0)\left( cos \beta_c z -\frac{i \delta}{\beta_c} sin \beta_c z \right) e^{i \delta z}$ $\tilde{B}(z)=\tilde{A}(0)\left(\frac{i \kappa_{ba}}{\beta_c} sin \beta_c z \right) e^{-i \delta z}$ Looking back at my slides I wrote that $$\tilde{B}(z)=\tilde{B}(0)\left(\frac{i \kappa_{ba}}{\beta_c} sin \beta_c z \right) e^{-i \delta z}$$, which given my initial conditions would be wrong since $$\tilde{B}(0)=0$$. Thankfully, this error doesn’t propagate through the rest of the presentation as that error only shows up on slide 8 and the rest of the figures I made are still accurate. I want to point out another error I made in the paper in regards to figure 1 here. The error is that in the figure in the paper, I wrote the same error for the power, corresponding to the blue line in figure one. I wrote $$\frac{P_a(z)}{P_b(o)}$$, when in fact it should be $$\frac{P_a(z)}{P_a(o)}$$, It was corrected for the presentation and here.