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  • 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.

    Plot made in Mathametica to show the power exchange for two codirectionally coupled modes when \(\delta=0\). Notice that places with maximum power transfer are for integer multiples of \(l_c^{PM}\), where \(l_c=\frac{\pi}{2 \beta_c}\)

    Figure 2 made in Mathametica to show how \(\delta\) changes the maximum efficiency