quantum mechanics - 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?

Answer

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 $-i$.

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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