Optics 073 · Polarization, anisotropy, and modulation

Polarization State Explorer

A real rotating polarizer bench: Malus’s law, the third-polarizer effect and count-derived Stokes measurements with handedness ambiguity.

Expected detector / source power

I_D/I_0

$0.50$

Incident polarization / absorbed power

p,I_{\rm abs}

$50\%$

Measurement / projection check

\mathcal V

$0.00\pi$

Model regime $\text{valid model regime}$

Validated scientific apparatus

Polarization State Explorer

Apparatus (named buttons below)

m

0

Azimuth

\psi

0^{\circ}

Ellipticity angle (not axis ratio)

\chi

0^{\circ}

Incident polarized-power fraction

p

1

Final analyzer transmission axis

\alpha

35^{\circ}

Middle transmission axis

\theta_2

45^{\circ}

Middle: zero removed, one inserted

j_2

1

Circular analysis: zero removed, one included

j_C

1

Incident expected electrons per equal exposure

N_{\rm in}

10000

Zero: expectations; one: Poisson counts

n

0

Inspected analysis-record index

j

0

Two crossed polarizers go dark. Insert a diagonal third plate—and light returns. Rotate the actual transmission axes, compare fixed-scale field ellipses, then measure Stokes from separate intensity exposures. Remove circular analysis to discover what linear measurements cannot know.

Ideal absorptive polarizers, monochromatic coherency model and fixed calibrated circular analyzer. Equal exposures; optional Poisson. No extinction leakage, alignment errors or self-calibration.

Physics tutorial

Audit the physics of Polarization State Explorer

BackgroundA linear polarizer transmits a field projection, not a direction-shaped hole. For a general polarization mixture its power is $I_\alpha=\tfrac12(S_0+S_1\cos2\alpha+S_2\sin2\alpha)$ . For purely linear input this becomes Malus’s law. For unpolarized or circular input every ideal linear analyzer transmits one half.

Why it mattersHow do intensity, ellipse shape, Jones vectors, and Stokes measurements encode the same polarization state?

Start with the essentials

Focus question: How do intensity, ellipse shape, Jones vectors, and Stokes measurements encode the same polarization state?
One-sentence intuition: Linear projections cannot see handedness: $I_\alpha(S_3)=I_\alpha(-S_3)$ . Add the two calibrated circular-analysis exposures to determine the missing component. Never infer it from a flat analyzer curve.

Core mathematical model

Jones field and the partial mixture

\begin{aligned}\mathbf j&=\cos\chi\,\mathbf a_\psi+i\sin\chi\,\mathbf b_\psi\\\mathbf a_\psi&=(\cos\psi,\sin\psi)^{\mathsf T}\\\mathbf b_\psi&=(-\sin\psi,\cos\psi)^{\mathsf T}\\\mathbf C&=p\mathbf j\mathbf j^\dagger+\tfrac{1-p}{2}\mathbf1\\\mathbf E(t)&=\Re\{\mathbf j e^{-i\omega t}\}\end{aligned}

Unit source power. Horizontal basis points opposite scene Z, vertical along scene Y; their cross product follows the beam along scene X. Angles are counterclockwise looking back toward the source. Positive ellipticity is left-handed in the stated NIST convention.

Ideal projection and Malus’s limit

\begin{aligned}\mathbf C'&=\mathbf P_\alpha\mathbf C\mathbf P_\alpha^\dagger\\\mathbf P_\alpha&=\mathbf a_\alpha\mathbf a_\alpha^{\mathsf T}\\\mathbf a_\alpha&=(\cos\alpha,\sin\alpha)^{\mathsf T}\\I_{\rm linear}/I_0&=\cos^2(\alpha-\psi)\end{aligned}

The rejected orthogonal component is absorbed. Pure linear output follows the transmission axis; the mount itself does not become transparent.

Three projections, not added energy

\begin{aligned}I_D/I_0&=\tfrac12\cos^2\theta_2\cos^2(\alpha-\theta_2)\\\alpha&=\tfrac\pi2,\quad\theta_2=\tfrac\pi4\\I_D/I_0&=\tfrac18\end{aligned}

This expression assumes unpolarized input and a fixed horizontal first plate. Remove the middle plate and crossed end plates transmit zero.

Counts, not a copied Stokes answer

\begin{aligned}C_b&\sim{\rm Poisson}(N_{\rm in}I_b)\\\widehat S_0&=\tfrac{\sum_b C_b}{3N_{\rm in}}\\\widehat S_1&=\tfrac{C_H-C_V}{N_{\rm in}}\\\widehat S_2&=\tfrac{C_D-C_A}{N_{\rm in}}\\\widehat S_3&=\tfrac{C_L-C_R}{N_{\rm in}}\end{aligned}

Six independent equal exposures, calibrated incident expectation and unit efficiency. With four linear records, total intensity uses two pair sums and circular Stokes stays unknown. Turning noise off records fractional expectations.

Physical states and noisy estimates

p=\frac{\sqrt{S_1^2+S_2^2+S_3^2}}{S_0}\le1

This inequality holds for true coherency matrices. An unconstrained finite-count estimate can violate it. A flat linear-analyzer scan can be circular, unpolarized or a mixture: it does not identify one state.

Common difficulties

More polarizers must mean less light?

Typical misconceptionInserting a middle plate into crossed end plates cannot brighten the detector.

Better mental modelThe middle projection changes the state reaching the final plate. A diagonal middle plate gives one eighth of an unpolarized source, versus zero without it. Each individual plate still absorbs power; no energy is created.

Run the experiment

01
Scene 1: Malus-law analyzer
Start with a single analyzer. Sweep its axis through a half-turn, then compare linear, circular and unpolarized input.
What to observe: A Jones vector describes only the polarized component. The unpolarized remainder is propagated with a coherency matrix, not a fictitious deterministic ellipse.
02
Scene 2: Three-polarizer paradox
Choose three polarizers. Keep the final plate crossed, remove and insert the diagonal middle plate. Watch the physical mount move out of the beam.
What to observe: All ellipse panels share a fixed amplitude scale. Local phase is freely referenced; slow field animation is not photon travel or measured optical phase.
03
Scene 3: Stokes polarimetry
Measure Stokes. Select each exposure to inspect its real analyzer setting. Remove circular records, reverse ellipticity, then add them back. Try low Poisson counts.
What to observe: The six records are separate equal exposures of the unchanged input, not six output branches. The count estimator never receives the reference state.