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?
 A: 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
$$
(U^*_{12},U_{22}^*) \left(\begin{array}{c} U_{11}\\ U_{21}\end{array}\right)=0
=U_{11}U_{12}^*+U_{21} U_{22}^*
= r_{ac}t_{bc}^*+t_{ad} r_{bd}^*\, .
$$
A: $$\left(r_{ac}t_{bc}^{*}+r_{bd}^{*}t_{ad}\right)E_{a}E_{b}^{*}+\left(r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}\right)E_{a}^{*}E_{b}=\Re\left[\left(r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}\right)E_{a}^{*}E_{b}\right]=0$$
Since this should hold for arbitrary complex values of $E_{a}^{*}E_{b}$, we have
$$r_{ac}^{*}t_{bc}+r_{bd}t_{ad}^{*}=0$$
A: 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*}
