Stokes Parameters and the Poincaré Sphere

Real light rarely obliges the neat two-component amplitude that Jones calculus wants. Sunlight, an LED, a star, the glow off a wet road — their polarization state reshuffles every few femtoseconds, faster than any detector can follow, so no fixed pair of complex amplitudes describes them on any timescale you can measure. What is stable are averaged intensities: how much light a polarizer passes at $0^\circ$ versus $90^\circ$ , at $45^\circ$ versus $135^\circ$ , right-circular versus left. The electromagnetic origin of these quantities — that they are the measurable quadratic averages of the transverse field, that a fully polarized beam saturates $S_1^2+S_2^2+S_3^2=S_0^2$ , and that the whole structure is the classical shadow of the qubit Bloch sphere — is developed in the companion article aq-em-jones-vectors-stokes-parameters and recalled here by name. This article owns the optical working geometry: the six-intensity polarimetry recipe, the map from a polarization ellipse to a point on the Poincaré sphere, and the reading of every wave plate as a rigid rotation of that sphere. The payoff is a single four-number fingerprint $(S_0,S_1,S_2,S_3)$ that describes any beam — pure, partial, or fully scrambled — using nothing but a photodiode and a few analyzers. Normalize by intensity and the last three numbers become a point in a ball: fully polarized states on the surface (the Poincaré sphere), unpolarized light dead centre, everything real in between, at a radius equal to the degree of polarization. Stokes built this in 1852, decades before Maxwell showed light was electromagnetic, from pure operational cunning; Poincaré wrapped it onto a sphere in 1892; and a century later quantum information reused the entire apparatus, renamed, for the qubit.

Key takeaways

The Stokes parameters $(S_0,S_1,S_2,S_3)$ are four real, directly measurable intensities that completely describe any beam — fully polarized, partially polarized, or unpolarized — where a Jones vector cannot.
Each is a difference of intensity readings: $S_0=I_{0}+I_{90}$ , $S_1=I_{0}-I_{90}$ , $S_2=I_{45}-I_{135}$ , $S_3=I_R-I_L$ — six measurements, three differences plus the total.
Fully polarized light saturates $S_1^2+S_2^2+S_3^2=S_0^2$ ; in general $S_1^2+S_2^2+S_3^2\le S_0^2$ , and the degree of polarization is $\text{DoP}=\sqrt{S_1^2+S_2^2+S_3^2}/S_0$ .
The normalized vector $(S_1,S_2,S_3)/S_0$ is a point on (pure) or inside (partial) the Poincaré sphere: equator = linear, poles = circular, latitude $=2\chi$ , longitude $=2\psi$ .
Every lossless wave plate is a rigid rotation of the sphere through its retardance — polarization engineering becomes navigation on a globe.
Key idea: any beam is a real four-vector of measurable intensities; its normalized spatial part is a point in a ball whose surface is the Poincaré sphere and whose radius is the degree of polarization.

The core insight

Any state of light — pure, partial, or unpolarized — is captured by four real, directly measurable intensities $(S_0,S_1,S_2,S_3)$ . Normalized, $(S_1,S_2,S_3)/S_0$ is a point in the unit ball whose surface is the Poincaré sphere: latitude fixes ellipticity, longitude fixes orientation, the radius is the degree of polarization, and every wave plate acts as a rotation.

Where this lives today

Stokes polarimetry is the workhorse wherever polarization carries information and the light is not a clean laser. A rotating quarter-wave plate in front of a fixed polarizer, its transmitted signal Fourier-analyzed over one turn, returns all four parameters — and hence orientation, ellipticity, and degree of polarization — from a single scan. Solar physicists map the Sun's magnetic field this way, reading the Zeeman-induced polarization of spectral lines into vector magnetograms; NASA and ESA fly imaging Stokes polarimeters to retrieve aerosol size and cloud microphysics from sunlight scattered by planetary atmospheres, precisely because a difference of intensities is robust against absolute calibration drift. The same four-vector is now standard in quantum optics, where a photon's polarization is a qubit: pure states on the Poincaré/Bloch surface, mixed states inside, and decoherence measured as a shrinking radius. Stokes-operator tomography calibrates polarization qubits in quantum key distribution and characterizes entangled-photon sources, so the 1852 recipe — measure six intensities, take three differences — is today read out on single photons.

The physical picture

Polarization has exactly two continuous degrees of freedom plus a brightness, and the Stokes parameters lay them out as three "imbalances" over a total. Ask three yes/no questions of the beam — is it more horizontal or vertical, more $+45^\circ$ or $-45^\circ$ , more right- or left-circular — and measure the bias of each. Those biases are $S_1, S_2, S_3$ , each running from $-S_0$ to $+S_0$ ; $S_0$ is the total power, the size of the pot. A perfectly predictable beam maxes out its total bias and lands on a sphere of radius $S_0$ ; a perfectly scrambled beam has no bias in any basis and sits at the centre; real light lies in between.

Figure

The Poincaré sphere: linear states on the equator (H, +45°, V, −45°), the two circular states at the poles (R, L), a general elliptical state at latitude 2χ and longitude 2ψ

The angles carry the ellipse directly. The longitude of a point is twice the orientation $\psi$ of the polarization ellipse's major axis; the latitude is twice the ellipticity angle $\chi$ (where $\tan\chi=\pm b/a$ ). The doubling is not a nuisance — it is the reason orthogonal polarizations sit at antipodes: rotating a physical ellipse by $180^\circ$ returns the identical state, so a full trip around the equator corresponds to only half a turn of real hardware, and the H/V, $\pm45^\circ$ , R/L orthogonal pairs land at opposite poles of the three coordinate axes. A birefringent retarder touches neither $S_0$ nor the radius; it simply spins the point about an axis fixed by its own fast axis, through an angle equal to its retardance. A quarter-wave plate is a $90^\circ$ rotation, a half-wave plate a $180^\circ$ rotation, and all of compensator design, depolarization budgeting, and fibre polarization control becomes rigid-body navigation on this globe.

Figure

Punchline: a state at ψ = 30°, χ = 15° lands at S = (1, 0.43, 0.75, 0.50) on the surface (DoP 1); a partially polarized beam of the same orientation sits INSIDE at radius DoP = 0.70

The second figure is the whole formalism in one glance. Put a beam's normalized Stokes vector as an arrow from the centre: its length is the degree of polarization and its direction is which polarization the ordered part prefers. The pure elliptical state at $\psi=30^\circ,\chi=15^\circ$ carries $S=(1,0.43,0.75,0.50)$ , and $\sqrt{0.43^2+0.75^2+0.50^2}=1$ puts it exactly on the surface. A partially polarized beam of the same ellipse but only $70\%$ ordered has the identical direction and a radius of $0.70$ — the arrow shrinks straight in toward the centre without turning. That single geometric fact, radius equals degree of polarization, is why the sphere organizes everything from a wave plate to a depolarizing fibre.

The core equations

Everything starts operationally, the way Stokes intended: point a polarizer and a circular analyzer at the beam and read a power meter. The four parameters are built so each is literally a subtraction of two such readings, which is what gives them their experimental immediacy — every one is something you measure, never something you infer from an unobservable absolute phase.

S_0 = I_{0}+I_{90}, \quad S_1 = I_{0}-I_{90}, \quad S_2 = I_{45}-I_{135}, \quad S_3 = I_{R}-I_{L} \qquad (1)

where: $I_\theta$ — intensity through a linear polarizer at angle $\theta$ ; $I_R,I_L$ — intensities of the right- and left-circular components (a quarter-wave plate followed by a polarizer). $S_0$ is the total power; $S_1$ is the horizontal-minus-vertical excess, $S_2$ the diagonal excess, $S_3$ the circular excess. A fully scrambled beam shows no excess anywhere ( $S_1=S_2=S_3=0$ ); a pure state maxes out its total excess. Note $S_2$ is not redundant with $S_1$ : light perfectly balanced in H/V can still be fully polarized at $45^\circ$ . For a monochromatic (pure) beam the same four numbers are quadratic in the field amplitudes, and the polarization ellipse's own angles parametrize them on the sphere of radius $S_0$ . This is the bridge from a measured ellipse to a point.

S_1 = S_0\cos 2\chi\cos 2\psi,\quad S_2 = S_0\cos 2\chi\sin 2\psi,\quad S_3 = S_0\sin 2\chi \qquad (2)

where: $\psi$ — orientation (azimuth) of the ellipse's major axis, $0\le\psi<\pi$ ; $\chi$ — ellipticity angle, $\tan\chi=\pm b/a$ , $-\tfrac\pi4\le\chi\le\tfrac\pi4$ . Read it geometrically: $2\psi$ is longitude, $2\chi$ latitude. Setting $\chi=0$ gives $S_3=0$ — the equator, all linear states; $\chi=\pm\tfrac\pi4$ gives $S_1=S_2=0,\ S_3=\pm S_0$ — the two poles, the circular states. The doubled angles are what make orthogonal ellipses antipodal. Because every real beam is a statistical mixture, the polarized fraction is a single number read straight off the vector length, and the interior of the sphere is exactly the partially polarized light.

\text{DoP} = \frac{\sqrt{S_1^2+S_2^2+S_3^2}}{S_0}, \qquad 0\le\text{DoP}\le 1,\qquad S_1^2+S_2^2+S_3^2 \le S_0^2 \qquad (3)

where: $\text{DoP}=1$ is fully polarized (surface), $\text{DoP}=0$ unpolarized (centre). The inequality — recalled from the coherency-matrix positivity in aq-em-jones-vectors-stokes-parameters — says no physical beam can lie outside the sphere. Inverting the ellipse from data: $\tan 2\psi = S_2/S_1$ and $\sin 2\chi = S_3/\sqrt{S_1^2+S_2^2+S_3^2}$ , valid once you have confirmed $\text{DoP}\approx1$ for the part you call polarized.

Recalled from electromagnetism. The field-theoretic content — that these four intensities are the measurable quadratic averages $\langle E_iE_j^*\rangle$ of the transverse field, that a pure state necessarily saturates $S_1^2+S_2^2+S_3^2=S_0^2$ , and that the sphere is verbatim the qubit Bloch sphere with $(S_1,S_2,S_3)/S_0$ the Bloch vector — belongs to aq-em-jones-vectors-stokes-parameters and is recalled here by name. This article takes the definitions as given vocabulary and owns what optics does with them: the polarimetry recipe, the ellipse-to-sphere map, and the retarder-as-rotation calculus below.

Derivation AThe measurement recipe and why a pure state lands on the sphere

The aim is to show that the six laboratory intensities of Equation (1) really do return the four Stokes parameters, and that a fully polarized beam, parametrized by its own ellipse angles, satisfies $S_1^2+S_2^2+S_3^2=S_0^2$ — so pure states fill the surface and nothing more. The route is direct: project the field through each analyzer to confirm the differences, then substitute the ellipse form of Equation (2) and collapse the trigonometry. Watch the identity fall out as a plain Pythagorean sum once the double angles are in place.

Step 1. A polarizer at $0^\circ$ transmits $I_0=\langle E_x^2\rangle$ and at $90^\circ$ transmits $I_{90}=\langle E_y^2\rangle$ , so $S_0=I_0+I_{90}=\langle E_x^2\rangle+\langle E_y^2\rangle$ is the total intensity and $S_1=I_0-I_{90}$ the horizontal excess. Why: the two orthogonal linear readings partition all the power, so their sum is total intensity and their difference is the first imbalance.

Step 2. Rotating the analyzer to $\pm45^\circ$ passes $E_{\pm45}=\tfrac1{\sqrt2}(E_x\pm E_y)$ , whose intensity difference is $I_{45}-I_{135}=\langle 2E_xE_y\cos\delta\rangle=S_2$ ; inserting a quarter-wave plate before the polarizer converts the circular basis to linear and yields $I_R-I_L=\langle 2E_xE_y\sin\delta\rangle=S_3$ . Why: the diagonal analyzer is sensitive to the real part of the cross-correlation and the circular analyzer to its imaginary part — together they capture the full relative phase $\delta$ .

Step 3. For a pure state write $a=|E_x|,\ b=|E_y|$ with relative phase $\delta$ , so $S_1=a^2-b^2$ , $S_2=2ab\cos\delta$ , $S_3=2ab\sin\delta$ . Then $S_1^2+S_2^2+S_3^2=(a^2-b^2)^2+4a^2b^2=(a^2+b^2)^2=S_0^2$ . Why: the cross terms conspire into a perfect square — this is the algebraic reason every monochromatic beam sits exactly on the sphere of radius $S_0$ , never inside it.

Step 4. Recast the pure state in its principal (ellipse) axes and the amplitudes become $\cos\chi,\sin\chi$ rotated by $\psi$ , giving Equation (2): $S_1=S_0\cos2\chi\cos2\psi$ , $S_2=S_0\cos2\chi\sin2\psi$ , $S_3=S_0\sin2\chi$ , which again squares to $S_0^2$ by $\cos^2+\sin^2=1$ applied twice. Why: the same identity, now read as latitude-and-longitude on a sphere, is what turns the abstract state space into the Poincaré globe. *Verified symbolically ( $S_1^2+S_2^2+S_3^2=S_0^2$ from the ellipse parametrization) and numerically (the state $\psi=30^\circ,\chi=15^\circ\Rightarrow S=(1,0.433,0.75,0.500)$ , DoP $=1$ , and the six-intensity recipe reproducing $S$ ) in verify_stokes-parameters-poincare-sphere.py.*

Go deeper · Derivation B · 2 minPartial polarization, the interior, and the retarder as a sphere rotation

Derivation BPartial polarization, the interior, and the retarder as a sphere rotation

Here the goal is twofold: show that any beam inside the sphere splits uniquely into an unpolarized part plus a fully polarized part, so the radius really is the ordered fraction; and show that a lossless retarder moves a point by pure rotation, leaving the radius — hence the degree of polarization — untouched. The strategy uses the one rule that makes Stokes parameters special: for incoherent beams, intensities add, so Stokes vectors add component by component.

Step 1. Two mutually incoherent beams superpose in intensity, so their Stokes vectors add: $\mathbf S=\mathbf S^{(1)}+\mathbf S^{(2)}$ . Adding equal-power H and V gives $S_0$ doubled while $S_1,S_2,S_3$ cancel — the centre, unpolarized. Why: incoherent light has no fixed relative phase to interfere, so only the measurable powers combine, and equal orthogonal mixtures erase every imbalance.

Step 2. Any allowed vector splits as $\mathbf S=(\text{DoP}\,S_0,\,S_1,S_2,S_3)+((1-\text{DoP})S_0,\,0,0,0)$ : the first piece has spatial length equal to its own intensity ( $\text{DoP}=1$ , on the surface), the second is pure unpolarized. Both intensities are non-negative because $0\le\text{DoP}\le1$ , and the direction of $(S_1,S_2,S_3)$ fixes the polarized ellipse uniquely. Why: this is what "70% polarized" physically means — a 0.70-power fully polarized beam riding on a 0.30-power scrambled background, not a single strange intermediate wave.

Step 3. A retarder with fast axis along $x$ and retardance $\Gamma$ sends $E_x\to E_x,\ E_y\to E_ye^{i\Gamma}$ , so $\delta\to\delta+\Gamma$ ; since $S_2\propto\cos\delta$ and $S_3\propto\sin\delta$ , the pair $(S_2,S_3)$ rotates by $\Gamma$ while $S_0,S_1$ are untouched. Why: delaying one component only advances the relative phase, which is a rotation in the $S_2$ – $S_3$ plane — a rigid turn of the sphere about the $S_1$ axis.

Step 4. That rotation preserves $S_1^2+S_2^2+S_3^2$ , so both $S_0$ and the radius are invariant and $\text{DoP}$ cannot change — a lossless element never depolarizes. A quarter-wave plate ( $\Gamma=\tfrac\pi2$ ) turns a $+45^\circ$ linear state ( $S=(1,0,1,0)$ ) into circular ( $S=(1,0,0,\pm1)$ ); a half-wave plate ( $\Gamma=\pi$ ) reverses handedness. Why: this is the geometric heart of the Poincaré picture — the entire wave-plate toolbox is rotations, and no rotation moves a point off its sphere. *Verified: the retarder rotation preserving $|\vec S|$ , the QWP mapping $+45^\circ$ linear to circular, and the partial-state decomposition (DoP $=0.70$ : polarized power $0.70$ , unpolarized $0.30$ ) in verify_stokes-parameters-poincare-sphere.py.*

Limiting cases

Case	Result	Meaning
Fully polarized	$S_1^2+S_2^2+S_3^2=S_0^2$ , DoP $=1$	on the sphere's surface; a pure Jones vector exists
Unpolarized	$(S_0,0,0,0)$ , DoP $=0$	centre of the sphere; no analyzer orientation is preferred, no wave plate changes any reading
H / V linear	$(1,\pm1,0,0)$	equator, $\pm S_1$ axis; $\psi=0^\circ/90^\circ$ , $\chi=0$
$\pm45^\circ$ linear	$(1,0,\pm1,0)$	equator, $\pm S_2$ axis; the diagonal states
R / L circular	$(1,0,0,\pm1)$	north/south poles; $\chi=\pm45^\circ$ , orientation undefined (a pole has no longitude)
Lossless retarder	rotation of $(S_1,S_2,S_3)$ ; $S_0$ , DoP fixed	wave plates never depolarize — they only spin the point

Common trap“You can just read off the Jones vector from the Stokes parameters”

Only if the light is fully polarized. The Stokes parameters are defined through time averages; drop the averaging and you silently collapse every beam onto the sphere's surface, discarding all partially polarized and unpolarized light. Recovering a Jones vector from measured $S_i$ is legitimate only after you have checked $\text{DoP}\approx1$ — fit a pure state to $\text{DoP}=0.85$ data and you misassign $15\%$ of the power to a phase that genuinely is not there. A second landmine lives inside the convention itself: which diagonal is "positive $S_2$ " and which circular sense is " $R$ " vary between textbooks and instruments, so a Poincaré sphere is only unambiguous once its axis labels — and the handedness of $S_3$ — are pinned down. Always record the convention with the data.

Worked examples

1 · From a polarization ellipse to a point on the sphere (confidence builder)

A fully polarized beam has its ellipse oriented at $\psi=30^\circ$ with ellipticity angle $\chi=15^\circ$ . Find its normalized Stokes vector, its degree of polarization, and where it sits on the Poincaré sphere.

Step 1. Use Equation (2) with $S_0=1$ : $S_1=\cos(2\cdot15^\circ)\cos(2\cdot30^\circ)=\cos30^\circ\cos60^\circ=0.8660\times0.5=\mathbf{0.433}$ . The major axis at $30^\circ$ and a modest ellipticity leave a horizontal excess just under a half.

Step 2. $S_2=\cos30^\circ\sin60^\circ=0.8660\times0.8660=\mathbf{0.750}$ and $S_3=\sin(2\cdot15^\circ)=\sin30^\circ=\mathbf{0.500}$ . The diagonal excess dominates because $2\psi=60^\circ$ is near the $+45^\circ$ axis, and the circular content is set by the ellipticity.

Step 3. Degree of polarization: $\text{DoP}=\sqrt{0.433^2+0.750^2+0.500^2}=\sqrt{0.1875+0.5625+0.2500}=\sqrt{1.000}=\mathbf{1.00}$ — the state is on the surface, as any pure state must be.

Step 4. Location: longitude $2\psi=60^\circ$ , latitude $2\chi=30^\circ$ north — an elliptical state in the front-upper octant, tilting toward $+45^\circ$ and toward the R pole. Inverting confirms the round trip: $\tan2\psi=S_2/S_1=0.750/0.433=1.732\Rightarrow\psi=30^\circ$ ; $\sin2\chi=S_3/1=0.500\Rightarrow\chi=15^\circ$ . S = (1, 0.433, 0.750, 0.500), DoP = 1.

2 · Degree of polarization from six intensity readings (exam level)

A polarimeter reports (in arbitrary units) $I_0=0.50,\ I_{90}=0.50,\ I_{45}=0.85,\ I_{135}=0.15,\ I_R=0.50,\ I_L=0.50$ . Find the Stokes vector, the degree of polarization, and the polarized/unpolarized power split.

Step 1. Total and imbalances from Equation (1): $S_0=I_0+I_{90}=1.00$ ; $S_1=I_0-I_{90}=\mathbf{0}$ ; $S_2=I_{45}-I_{135}=0.85-0.15=\mathbf{0.70}$ ; $S_3=I_R-I_L=\mathbf{0}$ . So $S=(1,0,0.70,0)$ : no H/V or circular bias, a pure diagonal excess.

Step 2. Degree of polarization: $\text{DoP}=\sqrt{0^2+0.70^2+0^2}/1.00=\mathbf{0.70}$ — the beam is $70\%$ polarized and sits inside the sphere at radius $0.70$ , on the $+S_2$ axis.

Step 3. Decompose: the polarized part carries power $\text{DoP}\cdot S_0=\mathbf{0.70}$ and is pure $+45^\circ$ linear (its unit vector is $+S_2$ ); the remainder is unpolarized with power $(1-\text{DoP})S_0=\mathbf{0.30}$ . The measured $45^\circ/135^\circ$ contrast is exactly the fingerprint: $I_{45}/I_{135}=0.85/0.15\approx5.7$ , and $(I_{45}-I_{135})/(I_{45}+I_{135})=0.70$ recovers the DoP directly as a fringe visibility.

Step 4. Consistency: had the beam been fully polarized $+45^\circ$ , we would have measured $I_{45}=1,\ I_{135}=0$ ; the finite $I_{135}=0.15$ is the unpolarized floor leaking through every analyzer at half its power. S = (1, 0, 0.70, 0), DoP = 0.70, split 0.70 polarized + 0.30 unpolarized.

Physics of the portrait

The four Stokes parameters are one of physics' great acts of operational nerve: a complete description of a wave built entirely from things you can put a meter on, written before anyone knew what the wave was.

Origins · 1 minstokes parameters and how this was discovered

stokes parameters

operational optics, 1852 → 1892 · Stokes at Cambridge, Poincaré at the Sorbonne

The parameters carry George Gabriel Stokes's name from his 1852 memoir On the composition and resolution of streams of polarized light from different sources, read to the Cambridge Philosophical Society. Stokes — Lucasian Professor at Cambridge, later Master of Pembroke and President of the Royal Society, and the same Stokes of viscous drag and fluorescence — wanted to describe partially polarized light, the everyday kind that neither the wave theory's amplitudes nor any single ellipse could capture. His stroke of genius was to define the state by four numbers that are each a difference of intensity measurements, decades before Maxwell (1865) identified light as an electromagnetic wave: the description needs no model of what is waving, only a polarizer and a detector, so it survived every later revolution in what light "is". Forty years on, Henri Poincaré — in his 1892 lectures Théorie mathématique de la lumière — gave the normalized parameters their geometry, mapping every polarization state to a point on a sphere so that wave plates became rotations and orthogonal states antipodes. The construction was so natural that quantum mechanics reused it wholesale: the Poincaré sphere is the Bloch sphere, the coherency matrix is the density matrix, and a photon's polarization is the textbook qubit. Polarization optics is simply where that machinery was first built, a century early, out of calcite and lenses.

History

1852

— Stokes defines the four parameters and the description of partially polarized light (Cambridge Phil. Soc.)

1865

— Maxwell identifies light as an electromagnetic wave, later grounding the parameters in field averages

1892

— Poincaré introduces the sphere in his lectures on the mathematical theory of light

1943

— Mueller extends the scheme to 4×4 intensity matrices, the Stokes-vector calculus for depolarizing elements

1946

— Chandrasekhar reintroduces the Stokes parameters for radiative transfer in stellar and planetary atmospheres

Going further

Texts. Hecht, Optics, 5th ed., §8.13 (Stokes parameters, Poincaré sphere) · Born & Wolf, Principles of Optics, 7th ed., §1.4 and §10.9 (partial polarization, coherency matrix) · Goldstein, Polarized Light, 3rd ed., ch. 4–6 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §6.1 · Loudon, The Quantum Theory of Light, 3rd ed. (Stokes operators and the photon Bloch sphere).

Provenance. Definitions and sign conventions follow Hecht §8.13 and Born & Wolf §1.4/§10.9; the pure-state identity $S_1^2+S_2^2+S_3^2=S_0^2$ , the ellipse-to-sphere relations $(2\psi,2\chi)$ , and the polarized/unpolarized decomposition are standard (Born & Wolf, Goldstein). The field-theoretic origin of the parameters, the coherency-matrix positivity behind the fundamental inequality, and the Bloch-sphere isomorphism are recalled by name from the companion electromagnetism article aq-em-jones-vectors-stokes-parameters rather than re-derived. A verifier recomputes: the symbolic identity from the ellipse parametrization; the worked state $\psi=30^\circ,\chi=15^\circ\Rightarrow(1,0.433,0.75,0.5)$ ; and the six-intensity partial-state recipe giving DoP $=0.70$ — all in verify_stokes-parameters-poincare-sphere.py.

Prerequisites. Polarization of Light · Elliptical Polarization · Jones Calculus. Leads to. Mueller Calculus · Degree of Polarization and Partial Polarization · Wave Plates and Retarders.

Why four parameters — isn't polarization just two degrees of freedom?

A fully polarized beam does have only two continuous degrees of freedom (orientation $\psi$ and ellipticity $\chi$ ), plus brightness — three numbers. The fourth appears because real light can be partially polarized: you also need to say what fraction is ordered. So $S_0$ is intensity, the direction of $(S_1,S_2,S_3)$ gives the two ellipse angles, and its length gives the degree of polarization. For a pure state the four collapse back to three via $S_1^2+S_2^2+S_3^2=S_0^2$ , which is exactly the statement that a pure state sits on the sphere and has no independent "how polarized" freedom left.

What is the difference between the Poincaré sphere and the Bloch sphere?

Geometrically, nothing — they are the same unit sphere. The Poincaré sphere is the optician's name: points are polarization states, the equator is linear light, the poles are circular, and wave plates are rotations. The Bloch sphere is the quantum-information name for a qubit: poles are the two basis states, the surface is pure states, the interior is mixed states. The dictionary is exact: $(S_1,S_2,S_3)/S_0$ is the Bloch vector, the coherency matrix is the density matrix, and the degree of polarization is the purity. Polarization optics simply realized the geometry a century before qubits, which is why a single photon's polarization is the cheapest qubit to build and measure.

How is unpolarized light "at the centre" if it still has intensity?

The centre refers only to the normalized spatial vector $(S_1,S_2,S_3)/S_0$ , which is zero for unpolarized light because every imbalance averages away. The intensity $S_0$ is separate and can be anything — a blazing unpolarized beam and a dim one both sit at the centre of their (differently scaled) spheres. It helps to picture the four-vector as (total power) plus (a direction-and-length inside a ball): unpolarized light has full power but zero length, so it is bright yet dead-centre.

If I put unpolarized light through a wave plate, why does nothing change?

Because a lossless retarder is a pure rotation of the Poincaré sphere, and rotating the centre point leaves it at the centre. Physically, unpolarized light has no fixed relative phase between its components for the plate to advance — delaying "one component" is meaningless when the two are uncorrelated — so every measured intensity is unchanged and the light stays unpolarized. Wave plates only reshape light that already has a definite polarization to work on; they can never create order from a scrambled beam. To polarize unpolarized light you need a polarizer (a projection toward the equator), which throws away power, not a retarder.

Can two different beams have the same Stokes vector?

Yes — and that is the point. The Stokes vector is a complete description of everything a polarization measurement can distinguish, but two beams that differ only in an unobservable (absolute phase, or the detailed statistics of how the polarization wanders, as long as the averages match) share the same four numbers. This is exactly the classical analogue of two quantum ensembles with the same density matrix being physically indistinguishable. If any polarization experiment could tell two beams apart, their Stokes vectors would differ; if the vectors are equal, no analyzer, wave plate, or power meter can separate them.

Why does a partially polarized beam decompose uniquely into polarized plus unpolarized?

Because the split is forced by the geometry. The unpolarized part must have zero spatial vector, so it can only be $((1-\text{DoP})S_0,0,0,0)$ ; whatever is left, $(\text{DoP}\,S_0,S_1,S_2,S_3)$ , has spatial length equal to its own intensity and is therefore fully polarized. The direction of $(S_1,S_2,S_3)$ fixes which ellipse the polarized part traces, and only the intensity split remains — and that is pinned by the requirement that both pieces carry non-negative power. There is exactly one way to do it, which is why " $70\%$ polarized, oriented at $+45^\circ$ " is an unambiguous physical statement.