Photon polarisation

Photon polarisation is the quantum mechanical treatment of light polarisation. Photoelectric effect experiments have shown that the direction of emission of electrons is dependent on the polarisation of the incident light. In quantum mechanics, we refer to classical polarisation directions as quantum states, with the states $\vert z\rangle$ and $\vert x\rangle$ representing linear polarisation parallel to the $z$ -axis and the $x$ -axis respectively (see diagram below).

If we consider a unit circle, the state of a photon $\vert\psi\rangle$ polarised at $\theta$ can therefore be written as a linear combination of $\vert z\rangle$ and $\vert x\rangle$ :

$\vert\psi\rangle=cos\theta\vert z\rangle+sin\theta\vert x\rangle\; \; \; \; \; \; \; \; 237$

where $\vert z\rangle$ and $\vert x\rangle$ become basis state vectors, which we can define them as unit vectors $\vert z\rangle=\begin{pmatrix} 1\\0 \end{pmatrix}$ and $\vert x\rangle=\begin{pmatrix} 0\\1 \end{pmatrix}$ , with $\vert \psi\rangle=\begin{pmatrix} cos\theta\\sin\theta \end{pmatrix}$ .

Question

Can a single photon be unpolarised?

Answer

Each photon in a string of photons (e.g. from the sun) has a definite polarisation, even though one photon’s oscillation direction may be randomly oriented relatively to another photon’s oscillation direction.

As per any normalised quantum state of $\vert\psi\rangle=\sum_{i=1}^{2}c_i\vert\phi_i\rangle$ , the square of each coefficient $\vert c_i\vert^{2}$ is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the corresponding eigenstate $\phi_i$ , i.e. $cos^{2}\theta +sin^{2}\theta=1$ . If we measure the polarisation of a string of photons, whose generic states are given by eq237, through a calcite crystal (see diagram above; the optic axis is in the plane of the viewer’s screen), photons with oscillations that are perpendicular to the plane of the viewer’s screen emerge as part of the o-ray with the state $\vert\psi\rangle=\vert x\rangle$ , while photons with oscillations that are parallel to the plane of the viewer’s screen emerge as part of the e-ray with the state $\vert\psi\rangle=\vert z\rangle$ . A photon oscillating at $\theta=\pm45^{\circ}$ ( $\vert\psi\rangle=\frac{1}{\sqrt{2}}[\vert z\rangle\pm\vert x\rangle]$ ) has 50% probability of emerging as part of the o-ray and 50% probability emerging as part of the e-ray.