Beam splitter in Q.M. and C.M. - Formalism In Q.M. the beam splitter is represented by the Hadamard transform (at least if the particle is in a state $|\Psi \rangle = \left( \frac{1}{\sqrt2} \right )(|0\rangle + |1\rangle)$ ) 
The Hadamard Matrix is $H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1\end{pmatrix}$
In the classical Jones-Formalism the Matrix for the beam splitter is
$\begin{pmatrix}
E_{3} \\
E_{4}
\end{pmatrix}
= \frac{1}{\sqrt{2}} \begin{pmatrix}
1 & i \\
i & 1 \\
\end{pmatrix}
\begin{pmatrix}
E_{1} \\
E_{2}
\end{pmatrix}
$
where $E_3,E_4$ are the incident beams.
Clearly, in the classical formulism the reflected beam is phase shifted, as opposed to the Q.M. formulism. Why are the matrices different, shouldn`t they be similar?
 A: In QM a phase shift has no effect on the physical state represented by the ket. That is, $\mid \Psi \rangle$ corresponds to the same state as $\mid \Psi \rangle e^{i\phi}$. Also note that any unitary matrix is a valid quantum logic gate (operation). And every operation on a qubit can be represented by a rotation or reflection on the Bloch sphere.
Now keeping all that in mind, let's see what happens when we act on the most general state $\mid \Psi \rangle = \alpha \mid 0 \rangle + \beta \mid 1 \rangle$ (with $\alpha$ and $\beta$ complex) with the operator defined in your classical beam-splitter. 
$$\mid \Psi' \rangle = \frac{1}{\sqrt{2}} 
\begin{pmatrix}
1 & i \\
i & 1 \\
\end{pmatrix}
\begin{pmatrix}
\alpha \\
\beta
\end{pmatrix} =
\frac{ \alpha + i \beta }{\sqrt{2}} \mid 0 \rangle + \frac{i \alpha + \beta}{\sqrt{2}} \mid 1 \rangle 
\\
= \alpha
\frac{ \mid 0 \rangle + i \mid 1 \rangle}{\sqrt{2}}  + \beta \frac{i \mid 0 \rangle + \mid 1 \rangle }{\sqrt{2}}$$
So the operation turns the bit $\mid 0 \rangle$ into $\mid 0 \rangle + i \mid 1 \rangle$ (times a constant). This looks a lot like the operation
$$\mid + \rangle \,\rightarrow\,\, \mid + \rangle + i \mid - \rangle = \,\, \mid S_{y}; + \rangle$$
or the rotation of a $+\mathbf{\hat z}$ spin by $\pi / 2$ about the $x$-axis, yielding a $+\mathbf{\hat y}$ spin.
Indeed, the operation $\mid 0 \rangle \,\rightarrow\,\, \mid 0 \rangle + i \mid 1 \rangle$, defined by the action of
$$\begin{pmatrix}
1 & i \\
i & 1 \\
\end{pmatrix}$$
on the $\mid 0 \rangle$ qubit, yields a $+\mathbf{\hat y}$ vector on the Bloch sphere.
Similarly, 
$$\mid 1 \rangle \,\rightarrow\,\, \mid 0 \rangle - i \mid 1 \rangle$$
corresponds to $-\mathbf{\hat y}$ on the Bloch sphere, where I have multiplied the state by $-i$.
$$\\ 
\\ 
\\$$
Now let's look at what the Hadamard gate does.
$$\mid \Psi' \rangle = \frac{1}{\sqrt{2}} 
\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}
\begin{pmatrix}
\alpha \\
\beta
\end{pmatrix} =
\frac{ \alpha + \beta}{\sqrt{2}} \mid 0 \rangle + \frac{ \alpha - \beta}{\sqrt{2}} \mid 1 \rangle 
\\
= \alpha
\frac{ \mid 0 \rangle + \mid 1 \rangle}{\sqrt{2}}  + \beta \frac{\mid 0 \rangle - \mid 1 \rangle }{\sqrt{2}}$$
Notice the $\mid 0 \rangle \pm \mid 1 \rangle$ states look like the $\mid S_x; \pm \rangle$ states. Again, this corresponds to a rotation on the Bloch sphere. This time we turn a $\pm \mathbf{\hat z}$ state into a $\pm \mathbf{\hat x}$ state.
So to summarize:


*

*Both matrices are valid quantum gates

*They both correspond to rotations of the qubit

A: Both are valid representations for a lossless beam splitter, and it makes no difference which one you use as long as you are consistent and using just one of them.
