homework and exercises - Phase of output in beam splitter

In the basis $\begin{pmatrix} H \\ V \end{pmatrix}$, let us take the beam splitter matrix to be

$\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$

This transforms $\begin{pmatrix} 1 \\ 0 \end{pmatrix} \rightarrow \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix} \rightarrow \frac{1}{\sqrt{2}}\begin{pmatrix} i \\ 1 \end{pmatrix}$.

However, real beam splitters e.g. the one shown below (taken from Wikipedia) do not give the same phase shift to the horizontal and vertical inputs.

So is the representation there incorrect (or represents a different physical system) in which case, how do I represent the beam splitter given in the image? I'm guessing we want something that does $\begin{pmatrix} 1 \\ 0 \end{pmatrix} \rightarrow \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix} \rightarrow \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

But candidates like $\begin{pmatrix} 1 & i \\ 1 & 1 \end{pmatrix}$ are not unitary and this seems to be an issue.

Answer

The transformation matrix given by the figure would be $$ U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} . $$

An alternative way to represent a beam splitter that is sometimes also used is to say $$ a \rightarrow \frac{1}{\sqrt{2}} (a+b) $$ $$ b \rightarrow \frac{1}{\sqrt{2}} (a-b) $$

In other words, the transformation is represented by a matrix $$ U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} . $$ It is still unitary in the sense that $U^{\dagger}U=UU^{\dagger}=I$, but det$(U)=-1$.

The difference between these definitions and the one with the $i$'s is simply some global phase changes at the output ports, which does not give an observable effect.