# Lossless beam splitter relations

Wikipedia says that for a beam splitter $$\begin{bmatrix}E_{c}\\ E_{d} \end{bmatrix}=\begin{bmatrix}r_{ac} & t_{bc}\\ t_{ad} & r_{bd} \end{bmatrix}\begin{bmatrix}E_{a}\\ E_{b} \end{bmatrix}$$ the demand $$\left|E_{c}\right|^{2}+\left|E_{d}\right|^{2}=\left|E_{a}\right|^{2}+\left|E_{b}\right|^{2}$$ for energy conservation gives $$r_{ac}t_{bc}^{*}+r_{bd}^{*}t_{ad} =0$$ but I fail to see why. If I substitute the relations into the energy conservation demand, I get: $$\left(r_{ac}t_{bc}^{*}+r_{bd}^{*}t_{ad}\right)E_{a}E_{b}^{*}+\left(r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}\right)E_{a}^{*}E_{b}=0$$doesn't it only imply that the real part of the complex number $$r_{ac}t_{bc}^{*}+r_{bd}^{*}t_{ad}$$ is zero? What am I missing?

Write $$\left(\begin{array}{c} E_c\\ E_d\end{array}\right)= U \left(\begin{array}{c} E_a\\ E_b\end{array}\right)\, .$$ If the beamsplitter is lossless, then $$(E_c^*,E_d^*)\left(\begin{array}{c} E_c\\ E_d\end{array}\right) =(E_a^*,E_b^*) \left(\begin{array}{c} E_a\\ E_b\end{array}\right) = (E_a^*,E_b^*)U^\dagger U \left(\begin{array}{c} E_a\\ E_b\end{array}\right)$$ which implies $$U^\dagger U=\mathbb{I}$$, i.e. $$U$$ is unitary. Unitary matrices are row and column orthonormal, v.g the second column is orthogonal to the first column $$(U^*_{12},U_{22}^*) \left(\begin{array}{c} U_{11}\\ U_{21}\end{array}\right)=0 =U_{11}U_{12}^*+U_{21} U_{22}^* = r_{ac}t_{bc}^*+t_{ad} r_{bd}^*\, .$$
$$\left(r_{ac}t_{bc}^{*}+r_{bd}^{*}t_{ad}\right)E_{a}E_{b}^{*}+\left(r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}\right)E_{a}^{*}E_{b}=\Re\left[\left(r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}\right)E_{a}^{*}E_{b}\right]=0$$ Since this should hold for arbitrary complex values of $$E_{a}^{*}E_{b}$$, we have $$r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}=0$$
I use a slightly different notation. This is your beam splitter and this is you system $$\begin{equation*} \begin{cases} \tilde{\boldsymbol{E}}_a = T\boldsymbol{E}_a +R\boldsymbol{E}_b \\ \tilde{\boldsymbol{E}}_b = R\boldsymbol{E}_a +T\boldsymbol{E}_b \\ |\boldsymbol{E}_a|^2 + |\boldsymbol{E}_b|^2 = |\tilde{\boldsymbol{E}}_a|^2 + |\tilde{\boldsymbol{E}}_b|^2 \end{cases} \end{equation*}$$ And solving it you get $$\begin{gather*} |\boldsymbol{E}_a|^2 + |\boldsymbol{E}_b|^2 = |T\boldsymbol{E}_a +R\boldsymbol{E}_b|^2 + |R\boldsymbol{E}_a +T\boldsymbol{E}_b|^2 \\ |\boldsymbol{E}_a|^2 + |\boldsymbol{E}_b|^2 = \left(|R|^2+|T|^2\right) \left(|\boldsymbol{E}_a|^2+|\boldsymbol{E}_b|^2\right) + \left(RT^\ast+R^\ast T\right) \left( \boldsymbol{E}_a\cdot\boldsymbol{E}_b^\ast + \boldsymbol{E}_a^\ast\cdot\boldsymbol{E}_b \right) \\ 1 = |R|^2+|T|^2 + \left(RT^\ast+R^\ast T\right) \frac{ \boldsymbol{E}_a\cdot\boldsymbol{E}_b^\ast + \boldsymbol{E}_a^\ast\cdot\boldsymbol{E}_b }{ |\boldsymbol{E}_a|^2+|\boldsymbol{E}_b|^2 } \end{gather*}$$ The result must be true for every input field, in particular if $$\boldsymbol{E}_a=\pm\boldsymbol{E}_b$$ so that you deduce $$\begin{gather*} \begin{cases} 1 = |R|^2+|T|^2 - RT^\ast - R^\ast T \\ 1 = |R|^2+|T|^2 + RT^\ast+R^\ast T \end{cases} \\ \rightarrow \begin{cases} 1 = |R|^2+|T|^2 \\ 0 = RT^\ast+R^\ast T \end{cases} \end{gather*}$$