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?


2 Answers 2


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.


$$\mid 1 \rangle \,\rightarrow\,\, \mid 0 \rangle - i \mid 1 \rangle$$

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
  • $\begingroup$ Thank you! About the phase shift oft the state: The thing is, not the global phase changes, but also the relative ones. Secondly, which matrix do I have to take for the beam splitter? Can I take either of both? $\endgroup$ Commented Jul 9, 2015 at 12:28
  • $\begingroup$ Yes, you're right. The relative phase shift matters. I realized that bullet point was confusing. I just meant that you can always multiply a ket by a global phase factor without changing the state, which I stated because I used that in showing that $\mid 1 \rangle \,\rightarrow\,\, i \mid 0 \rangle + \mid 1 \rangle$ is the same as $\mid 1 \rangle \,\rightarrow\,\, \mid 0 \rangle - i \mid 1 \rangle$. That wasn't clear at all, sorry. I'm removing that point from the summary. I hope the rest makes sense. $\endgroup$ Commented Jul 9, 2015 at 12:33
  • $\begingroup$ The rest makes perfectly sense, but still it´s not clear why the relative phase is shifted when using the Jones matrix and the phase is left unchanged when using the Hadamard gate. $\endgroup$ Commented Jul 9, 2015 at 12:40
  • $\begingroup$ That's because they're two different rotations - different axes. Remember, the spin has complex coefficients. That's why the Bloch sphere is a sphere. By convention, the $xz$-plane is taken to be real, so any spin in that plane has real coefficients. However, any spin with a $y$ component has complex coefficients. So a rotation of a spin in the $xz$-plane about $y$ doesn't change the phase, while a rotation about $x$ or $z$, in general, would. The Hadamard gate is performing the rotation about the $y$-axis, so no phase shift. The other rotates about the $x$-axis, so there is a phase shift. $\endgroup$ Commented Jul 9, 2015 at 13:05
  • $\begingroup$ So both the Jones and the Hadamard Transform would be equally suitable for describing the bs? $\endgroup$ Commented Jul 10, 2015 at 15:09

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.