How is the polarization of a photon able to hold quantum information?

Question

It is my understanding that light, and its polarization, is used to transfer information in quantum computers, but how can the information encoded in say, an electron also be stored in light? I understand that the spin of an electron or other particle is often used to create the qubit, where its "value" can be any point on the sphere below between zero and one.

However, how is the polarization of a photon able to have that kind of variety of possible values? Is there something I'm missing? Is there a quantized version/explanation of polarizations that I simply haven't seen that allows for the same range of information storage and extraction?

Answer 1

Firstly, that sphere that you've pictured is convenient for thinking about what's going on, but remember that it is not what is actually happening. So the fact that you don't visualise light as having a little arrow pointing somewhere doesn't matter.

The fact of that matter is that for an electron spin, having the two possible states "up" and "down", we associate two distinguishable states, which we label as perhaps $|up\rangle,|down\rangle$ or $|0\rangle,|1\rangle$. Since quantum mechanics is a linear theory, and linear superposition of the two is allowed, $\alpha|0\rangle+\beta|1\rangle$.

Similarly for a photon, there are two distinguishable states of polarisations, horizontal and vertical. You can label these as $|H\rangle,|V\rangle$ or $|0\rangle,|1\rangle$. The labels really are arbitrary. But again, because quantum mechanics is linear, you can have a linear superposition of these, $\alpha|0\rangle+\beta|1\rangle$, and you can capture the same information as you can with the state of an electron.

On the Bloch Sphere, particular points are often highlighted, such as $0\rangle$, $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$ and $(|0\rangle+i|1\rangle)/\sqrt{2}$. If we've decided that $|0\rangle$ is the same as a horizontally polarised photon, then $|+\rangle$ is the same as a photon polarised at $45^{\circ}$, and $(|0\rangle+i|1\rangle)/\sqrt{2}$ is the same as a circularly polarised photon.

Answer 2

In quantum theory, the pure states are associated with the unit vectors of the Hilbert space. A pure state of a quantum bit can be represented as $$| \psi \rangle = \alpha | 0 \rangle + \beta | 1 \rangle$$ where $|\alpha|^2 + |\beta|^2 = 1$. The basis $| 0 \rangle$ and $| 1 \rangle$ can be viewed as two orthogonal polarization directions.

Bloch sphere is a geometric representation of qubits which makes the mathamatics of a qubit more intuitive. To see this, we may rewrite the equation as $$| \psi \rangle = e^{i \gamma} ( \cos \frac \theta 2 | 0 \rangle + e^{i\phi} \sin \frac \theta 2 | 1 \rangle )$$

Since global phase has no observable effects, we can ignore the factor $e^{i\gamma}$. It follows that a pure qubit state $$| \psi \rangle = \cos \frac \theta 2 | 0 \rangle + e^{i\phi } \sin \frac \theta 2 | 1 \rangle$$ 1-to-1 correponds to a point on the Bloch sphere whose sphere coordinate is exactly $(\theta, \phi)$.

More generally, a state can be probability mixture of pure states, which leads to the notion of density operators. A density operator $\rho$ is positive and $\operatorname{Tr}(\rho) = 1$. In a qubit system, a density operator can be decomposed as $$\rho = \frac 1 2 (I+r_1\sigma_1+r_2\sigma_2+r_3\sigma_3) $$ where $r_1,r_2,r_3 \in \mathbb R$, $r_1^2 + r_2^2 + r_3^2 \leq 1$ and $\sigma_i$ are Pauli matrices $$ \sigma_0 = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ $$ \sigma_1 = \sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ $$ \sigma_2 = \sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} $$ $$ \sigma_3 = \sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$

This indicates that the whole Bloch ball can be associated with states. In particular, the states on the sphere is pure and those inside the sphere is mixed.