Polarization: The Photon's Hidden Compass

Polarization: the photon's hidden compass

Here's a question that should feel a bit unsettling. We said \(\mathbf{E}\) must lie in the plane perpendicular to \(\hat{k}\). But that plane is two-dimensional, and it contains infinitely many possible directions. The wave equations don't pick one — they're symmetric with respect to all of them. So which direction does \(\mathbf{E}\) actually point?

The answer is polarization, an intrinsic quantum property of the photon itself. It's extra information that the photon carries, encoding the orientation of its electric field (and thus magnetic field) in the transverse plane. Mathematically, it's described by a Jones vector:^[6] a pair of complex numbers,

\[ \boldsymbol{\varepsilon} = \begin{pmatrix} \varepsilon_x \\ \varepsilon_y \end{pmatrix} \in \mathbb{C}^2, \qquad |\varepsilon_x|^2 + |\varepsilon_y|^2 = 1, \]

Let \(\hat{x}\) and \(\hat{y}\) be any two orthogonal axes in the transverse plane. Given \(\boldsymbol{\varepsilon}\) and the propagation direction \(\hat{k}\), the electric field at a point along the propagation axis is:

\[ \mathbf{E}(t) = E_0\, \operatorname{Re}\!\left[(\varepsilon_x\,\hat{x} + \varepsilon_y\,\hat{y})\, e^{-i\omega t}\right]. \]

The real and imaginary parts of \(\varepsilon_x\) and \(\varepsilon_y\) control what the field arrow does over time. When both components are real — say \(\boldsymbol{\varepsilon} = (1,\,0)^T\) — the field oscillates back and forth along \(\hat{x}\): that's linear polarization. When the magnitudes are equal and there's a 90° phase shift between them — \(\boldsymbol{\varepsilon} = (1,\, i)^T/\!\sqrt{2}\) — the field components are \(E_x = (E_0/\!\sqrt{2})\cos\omega t\) and \(E_y = (E_0/\!\sqrt{2})\sin\omega t\), so the tip of \(\mathbf{E}\) traces a circle over time: that's circular polarization. In general, you get an ellipse.

Angular momentum from the polarization state

The polarization vector \(\boldsymbol{\varepsilon} = (\varepsilon_x, \varepsilon_y)^T \in \mathbb{C}^2\) we introduced above — the pair of complex numbers encoding which direction in the transverse plane \(\mathbf{E}\) oscillates, encodes a physically measurable quantity: the angular momentum of the photon. For a photon travelling along \(\hat{k}\), the component of spin angular momentum in that direction is:^[7]

\[ J_k = 2\hbar\, \operatorname{Im}(\varepsilon_x^*\, \varepsilon_y). \]

Let's compute this quantity on our two circular cases. For \(\boldsymbol{\varepsilon} = (1,\,i)^T/\!\sqrt{2}\) — the case shown in the animation, where \(\mathbf{E}\) traces a counterclockwise circle when viewed looking in the direction of propagation:

\[ J_k = 2\hbar\, \operatorname{Im}\!\left(\frac{1}{\sqrt{2}} \cdot \frac{i}{\sqrt{2}}\right) = 2\hbar\cdot\operatorname{Im}\!\left(\frac{i}{2}\right) = +\hbar. \]

For \(\boldsymbol{\varepsilon} = (1,\,-i)^T/\!\sqrt{2}\) (clockwise when viewed looking in the direction of propagation), \(J_k = -\hbar\). The labelling adopted here — LCP for \(+\hbar\), RCP for \(-\hbar\) — follows the helicity convention standard in atomic and quantum optics, where the sign of the angular momentum determines the name.^[7] Note that classical optics texts (IEEE, SPIE) sometimes use the opposite labelling, where handedness is defined by the rotation direction seen by the receiver rather than the angular momentum sign; always verify the convention when reading a source.

For linear polarization, \(\varepsilon_y\) is real so \(\operatorname{Im}(\varepsilon_x^*\varepsilon_y) = 0\) and \(J_k = 0\). A linearly polarised photon carries no net spin — it's a superposition of one LCP (\(+\hbar\)) and one RCP (\(-\hbar\)) photon with equal weight.

As a vector in \(\mathbb{R}^3\):

\[ \mathbf{J} = J_k\,\hat{k} = 2\hbar\, \operatorname{Im}(\varepsilon_x^*\, \varepsilon_y)\,\hat{k}. \]

It always points along or against the propagation direction. The polarization state and the angular momentum are in exact one-to-one correspondence: \(\boldsymbol{\varepsilon}\) determines \(\mathbf{J}\) via the formula above, and knowing \(J_k\) plus the beam's ellipticity lets you reconstruct \(\boldsymbol{\varepsilon}\) up to an overall phase. They are two descriptions of the same physical reality.

Why call it "angular momentum" if a photon doesn't spin?

In classical mechanics, the angular momentum of a point particle is \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\) — position cross momentum. Its SI unit is metres times kg·m/s:

\[ [\mathbf{L}] = [\mathbf{r}][\mathbf{p}] = \mathrm{m} \cdot \frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{s}} = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}}. \]

Now consider the scalar of the photon's angular momentum: \(\hbar = h/(2\pi)\), where \(h\) is the Planck constant introduced by Planck to explain blackbody radiation via the relation \(E = hf\) (energy equals frequency times \(h\)). So its units are energy divided by frequency:

\[ [h] = \frac{[E]}{[f]} = \frac{\mathrm{J}}{\mathrm{Hz}} = \mathrm{J}\cdot\mathrm{s} = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}\cdot\mathrm{s} = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}}. \]

That's identical to the unit of classical angular momentum, even though a photon has no volume and doesn't spin like a sphere. The units work out, but does the concept?

The reason we call \(J_k\) angular momentum — rather than just saying "it happens to have the right units" — is that it satisfies the defining property of angular momentum: it is the conserved quantity associated with rotational symmetry. This is Noether's theorem: whenever a physical system is symmetric under rotations about some axis, there is a corresponding conserved quantity, and that quantity is what we call angular momentum about that axis. A photon propagating through empty space — which is rotationally symmetric — cannot spontaneously change its \(J_k\). When it is absorbed by an atom, the atom's total angular momentum changes by exactly \(\pm\hbar\), because angular momentum is conserved in the interaction. This is directly measurable (it is the origin of the selection rules we meet in the applications section below).

Helicity: why the photon's spin is permanently locked to its direction of travel

The fact that the angular momentum \(\mathbf{J}\) of a photon always points either in the same or in the opposite direction of the photon's direction of propagation has a name — helicity — and its Lorentz invariance is a consequence of the photon being massless. Here is the argument.

Along the way we will also see why the photon, despite being a spin-1 particle, carries only helicity \(\pm 1\) and never \(0\). Massive spin-1 particles (like the W boson) have three helicity states \(-1, 0, +1\); the zero mode is absent for the photon, which is why light has exactly two independent polarisation states rather than three.^[9]

What "Lorentz invariant" means

Special relativity tells us that the laws of physics are the same in every inertial frame — any frame in which force-free particles move at constant velocity in straight lines. Two such frames that are in constant relative-velocity motion are related by a Lorentz boost. Most physical quantities — energy, momentum, velocity — take different numerical values in different frames. A quantity is called Lorentz invariant if every observer in every inertial frame measures the same value for it.

The archetypal Lorentz invariant is the rest mass \(m\). No matter how fast you move relative to a particle, the combination

\[ m^2c^2 = \left(\frac{E}{c}\right)^{\!2} - |\mathbf{p}|^2 \]

always gives the same number. Energy \(E\) and momentum \(\mathbf{p}\) each depend on the frame; their combination in this formula does not.

The Lorentz boost matrix

For an observer B moving at speed \(v = \beta c\) relative to A along the \(z\)-axis, the energy and momentum of any particle transform as:

\[ \begin{pmatrix} E'/c \\ p_x' \\ p_y' \\ p_z' \end{pmatrix} = \underbrace{\begin{pmatrix} \phantom{-}\gamma & 0 & 0 & -\gamma\beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma\beta & 0 & 0 & \phantom{-}\gamma \end{pmatrix}}_{\Lambda(\beta)} \begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix}, \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}. \]

The transverse components \(p_x, p_y\) are untouched — only energy and the parallel momentum mix. Every quantity that transforms via this matrix is called a Lorentz four-vector; the invariant mass formula above is just the statement that the "length" of the energy-momentum four-vector is preserved.

For massive particles, helicity can flip

Consider a spinning baseball moving in \(+z\) with its spin aligned with its motion: helicity \(h = +\hbar\). Now suppose observer B runs faster than the ball in the same direction. In B's frame the ball moves backward — \(p_z\) has reversed sign — while the ball's own spin axis is unchanged. The helicity, \(h = J^{12} \cdot \operatorname{sign}(p_z)\), has flipped to \(-\hbar\).

For any massive particle this is always possible: it moves at some \(v < c\), so a valid observer can always travel faster, overtake it, and see its momentum reversed. Helicity is therefore not Lorentz invariant for massive particles.

For photons, no observer can overtake light

A photon satisfies \(E = |\mathbf{p}|\,c\). For a photon moving in \(+z\), \(E = p_z c\). Substituting into the boost:

\[ p_z' = \gamma\!\left(p_z - \beta\frac{E}{c}\right) = \gamma\!\left(p_z - \beta\, p_z\right) = \gamma\, p_z(1 - \beta). \]

Since no valid observer has \(\beta \ge 1\), the factor \((1-\beta)\) is strictly positive: the sign of \(p_z\) is preserved in every inertial frame. The spin component \(J^{12}\) is equally preserved — the \(x\) and \(y\) rows of the boost matrix are always the identity, so they cannot alter the spin about the \(z\)-axis. Both factors in \(h = J^{12} \cdot \operatorname{sign}(p_z)\) are invariant, and so is the helicity.

The generalisation to boosts in any direction — not just along \(\hat{z}\) — follows from the Wigner little group: the subgroup of Lorentz transformations that leaves a given momentum vector fixed. For massless particles this group is ISO(2), whose only finite unitary representations are singlet helicity states \(|p, \lambda\rangle\). No continuous Lorentz transformation can mix states of different helicity, which is also why the \(\lambda = 0\) mode is absent for the photon altogether.

What photon angular momentum does in the world

If \(\mathbf{J}\) just points along the axis the photon is already travelling, you might wonder what it actually accomplishes. The answer: quite a lot.

Selection rules: angular momentum conservation forces atomic transitions

When an atom absorbs a photon, every conserved quantity — energy, momentum, angular momentum — must pass from the photon to the atom. Atomic electrons occupy states labelled by a magnetic quantum number \(m_J\) describing their angular momentum along a chosen axis. Absorbing a \(+\hbar\) photon (LCP) forces \(\Delta m_J = +1\); absorbing a \(-\hbar\) photon (RCP) forces \(\Delta m_J = -1\). A linearly polarised photon, being a superposition of \(\pm\hbar\), drives both transitions equally. These are the electric dipole selection rules, and they are purely a statement of angular momentum conservation.

Their consequences are enormous: selection rules are the engine behind optical pumping (preparing atomic spin states in atomic clocks and quantum computers), laser cooling (reducing atomic motion near absolute zero by repeated kicks from circularly polarised photons), and the read-out of spin qubits. Without photon angular momentum, the SI second — defined by hyperfine transitions in caesium atoms — could not be measured.

Circular dichroism: probing chirality with opposite spins

Many biological molecules — all naturally occurring amino acids (except glycine), DNA's right-handed double helix, most proteins — are chiral: they come in two mirror-image forms that cannot be superimposed, like a left and a right hand. LCP and RCP carry opposite angular momenta, and a chiral molecule interacts with them differently, absorbing each at a slightly different rate. The difference — called circular dichroism — is a sensitive fingerprint of molecular structure. It reveals whether a protein is folded correctly (alpha-helices and beta-sheets produce characteristic spectra), verifies the enantiomeric purity of pharmaceutical compounds, and maps how small drug molecules bind to DNA.

Optical tweezers: angular momentum as a mechanical torque

A tightly focused laser beam creates a spatial intensity gradient that pulls a microscopic object — a bead, a bacterium, even a single DNA strand — toward its focus. This is an optical tweezer (Arthur Ashkin, Nobel Prize in Physics 2018). When the trapped object is birefringent (its refractive index differs for different polarisation directions, as it does for many crystals and biological fibres), photons passing through it shift their polarisation state. By conservation of angular momentum, \(\pm\hbar\) per photon is transferred as a mechanical torque, spinning the particle about its own axis — an optical spanner.

This torque is real and measurable: it has been used to rotate DNA strands to study how they uncoil under torsional stress, spin organelles inside living cells without puncturing the membrane, and drive microscopic fluid pumps by rotating particles in microfluidic channels.

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References

1. Wikipedia contributors. Jones calculus. Wikipedia, The Free Encyclopedia. en.wikipedia.org/wiki/Jones_calculus
2. Wikipedia contributors. Spin angular momentum of light. Wikipedia, The Free Encyclopedia. en.wikipedia.org/wiki/Spin_angular_momentum_of_light
3. Wikipedia contributors. Helicity (particle physics). Wikipedia, The Free Encyclopedia. en.wikipedia.org/wiki/Helicity_(particle_physics)