Optics 073 · Polarization, anisotropy, and modulation

Polarization State Explorer

How can a Jones vector, an ellipse and Stokes measurements all describe the same polarization state? They describe one state in different ways. The Jones vector lists the two field components and their relative phase; the ellipse is the path the field tip traces in one cycle; the Stokes parameters are built from intensities measured through polarizers. Linear polarizers alone cannot show which way the field turns. That handedness needs a circular analysis.

Rotate polarizers on a bench to see Malus’s law and the third-polarizer effect, then measure Stokes parameters from counts and see when the handedness becomes ambiguous.

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 $\text{valid model regime}$

Model checked against analytic results

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

Expected incident electrons per equal exposure

N_{\rm in}

10000

Zero: expectations; one: Poisson counts

n

0

Inspected analysis-record index

j

0

Two crossed polarizers block the light. Insert a third one at a diagonal and light comes through again. Rotate the transmission axes, compare field ellipses drawn at a fixed scale, then measure the Stokes parameters from separate intensity exposures. Remove the circular analyzer to find out what linear measurements alone cannot tell you.

The polarizers are ideal and absorptive, the light is described by a monochromatic coherency matrix, and the circular analyzer is fixed and calibrated. All exposures are equal, and Poisson noise is optional. Leakage through crossed polarizers, alignment errors and self-calibration are not included.

Physics tutorial

Audit the physics of Polarization State Explorer

BackgroundA linear polarizer passes the projection of the field onto its axis; it does not act like a slot shaped for one direction. For a general mixture of polarizations, the transmitted 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 can a Jones vector, an ellipse and Stokes measurements all describe the same polarization state?

Start with the essentials

Focus question: How can a Jones vector, an ellipse and Stokes measurements all describe the same polarization state?
One-sentence intuition: Linear projections cannot see handedness: $I_\alpha(S_3)=I_\alpha(-S_3)$ . Finding the missing component takes the two calibrated exposures through circular analyzers. A flat analyzer curve never tells you what it is.

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}

The source power is one. The horizontal basis vector points along negative Z in the scene and the vertical one along Y, so their cross product points along the beam in the X direction. Angles are counted counterclockwise as you look back toward the source. In the NIST convention used here, positive ellipticity is left-handed.

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 orthogonal component is absorbed. The light that leaves is purely linear along the transmission axis, and the opaque mount around the plate stays opaque.

Three projections add no 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 first plate fixed horizontal. Without the middle plate, the crossed end plates transmit nothing.

Estimating the Stokes parameters from counts

\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}

The estimate uses six independent, equal exposures, a calibrated expected incident count and unit efficiency. With only the four linear records, the total intensity comes from two pair sums and the circular Stokes parameter stays unknown. With noise turned off, the records are expected values and can be fractional.

Physical states and noisy estimates

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

Every true coherency matrix satisfies this inequality, but an unconstrained estimate from finite counts can break it. A flat scan with a linear analyzer could come from circular light, unpolarized light or a mixture, so 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 plate projects the light onto a new direction, which changes the state that reaches the last plate. With an unpolarized source, a diagonal middle plate lets through one eighth of the power; without it, nothing gets through. Every plate still absorbs power, so 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 part. The unpolarized remainder is carried by a coherency matrix, and no definite ellipse is drawn for it.
02
Scene 2: Three-polarizer paradox
Choose three polarizers and keep the final plate crossed. Remove the diagonal middle plate and insert it again, and watch its physical mount move out of the beam.
What to observe: All the ellipse panels share one fixed amplitude scale. The zero of local phase is arbitrary, and the slow field animation shows neither photon travel nor a measured optical phase.
03
Scene 3: Stokes polarimetry
Measure the Stokes parameters. Select each exposure to see the analyzer setting it really used. Remove the circular records, reverse the ellipticity, then add them back. Also try low Poisson counts.
What to observe: The six records are separate, equal exposures of the same input, not six output branches. The count estimator works only from the counts and never sees the reference state.