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\[\widetilde{A}(z)=\widetilde{A}(0)\left( cos \beta_c z -\frac{i \delta}{\beta_c} sin \beta_c z \right) e^{i \delta z} \]  \[\widetilde{B}(z)=\widetilde{B}(0)\left(\frac{i \kappa_{ba}}{\beta_c} sin \beta_c z \right) e^{-i \delta z} \] 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 $\delta=0$   \[\frac{P_a(z)}{P_a(0)}=|\frac{\widetilde{A}(z)}{\widetilde{A}(0)}|^2=cos^2 \beta_c z \] \[\frac{P_b(z)}{P_a(0)}=|\frac{\widetilde{B}(z)}{\widetilde{A}(0)}|^2=sin^2 \beta_c z\]