Optics 076 · Polarization, anisotropy, and modulation

Waveplate & Poincaré Sphere Lab

How does a waveplate move a polarization state on the Poincaré sphere? An ideal waveplate turns the polarization point rigidly around an axis on the sphere's equator. The axis sits at twice the fast-axis angle, and the turn equals the plate's retardance. A quarter-wave plate takes light polarized at 45 degrees to its axis into circular; a half-wave plate mirrors the polarization about its fast axis. Order matters, because turns about different axes do not commute.

Stack three ideal waveplates, applied in the correct Jones-matrix order, and set each fast axis. Watch the state move on a rotatable Poincaré sphere, compare field ellipses drawn at equal scale and read the analyzer power.

Output normalized Stokes

\mathbf s

$0.50$

Analyzer / incident power

I_D/I_0

$50\%$

Reversed-order angular separation

\gamma_{\rm reverse}

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

Waveplate & Poincaré Sphere Lab

Plate A retardance

\delta_A

0.50\pi\,\mathrm{rad}

Plate A fast axis from u toward v

\alpha_A

0^{\circ}

Plate B retardance

\delta_B

1\pi\,\mathrm{rad}

Plate B fast axis

\alpha_B

22.50^{\circ}

Plate C retardance

\delta_C

0.50\pi\,\mathrm{rad}

Plate C fast axis

\alpha_C

0^{\circ}

Number of plates: A, then B, then C

N

1

Order: zero forward, one reversed

j_O

0

Input ellipse azimuth

\psi_0

45^{\circ}

Input signed ellipticity angle

\chi_0

0^{\circ}

Analyzer axis from u toward v

\beta

0^{\circ}

A quarter-wave plate can turn linear polarization circular, and a half-wave plate rotates a linear state. Combine three plates, then reverse their order: the same settings need not give the same final state. Follow each transformation on the sphere and check it with the analyzer below.

The light is fully polarized and monochromatic, the plates are lossless effective retarders, and the analyzer is ideal. Plate thickness is fixed and schematic. Material dispersion, walk-off and reconstructing Stokes parameters from measured data are not included.

Physics tutorial

Audit the physics of Waveplate & Poincaré Sphere Lab

BackgroundA waveplate does not remove a polarization component. It changes the phase between two orthogonal components: $\mathbf J=\mathbf f\mathbf f^{\mathsf T}+e^{i\delta}\mathbf g\mathbf g^{\mathsf T}$ . The fast axis is the one with the smaller phase index. With the time dependence of the phasor written as a negative exponent, the slow component gains a positive phase relative to the fast one. So a quarter-wave plate with its fast axis along the first transverse axis turns diagonal linear light into circular light.

Why it mattersHow does a waveplate move a polarization state on the Poincaré sphere?

Start with the essentials

Focus question: How does a waveplate move a polarization state on the Poincaré sphere?
One-sentence intuition: A single linear retarder rigidly rotates the normalized Stokes vector: $\mathbf s_{\rm out}=\mathcal R_{\mathbf n}(\delta)\mathbf s_{\rm in}$ . The rotation axis lies on the equator, at twice the angle of the physical fast axis. The colored path traces the retardance as it builds up inside the plate. Successive rotations generally do not commute, so reversing the order of the plates can change the final state even though no plate setting changes.

Core mathematical model

Define the axes before the matrix

\begin{aligned}\mathbf f&=(\cos\alpha,\sin\alpha)^{\mathsf T}\\\mathbf g&=(-\sin\alpha,\cos\alpha)^{\mathsf T}\\\delta&=\frac{2\pi(n_s-n_f)d}{\lambda_0}\\\mathbf E(x,t)&=\Re\{\mathbf J_E e^{i(kx-\omega t)}\}\end{aligned}

The first vector is the fast axis and the second the slow axis. The overall phase along the fast axis is dropped, because this Lab looks at polarization and intensity, not at interference between separate arms. You set the effective retardance directly; material dispersion and the real optical thickness are not inferred from the drawing.

Jones spinor to the sphere

\begin{aligned}s_1&=|u|^2-|v|^2\\s_2&=2\Re(u^*v)\\s_3&=2\Im(u^*v)\\\mathbf n&=(\cos2\alpha,\sin2\alpha,0)\\|u|^2+|v|^2&=1\end{aligned}

With this sign convention, a positive retardance rotates the Stokes vector by a positive right-handed angle about the doubled fast-axis vector. The paths need not be great circles. Every ideal retarder keeps the power unchanged and keeps the state on the unit sphere.

A detector tests projection, not phase

\begin{aligned}\mathbf J_{\rm total}&=\mathbf J_C\mathbf J_B\mathbf J_A\\\frac{I_D(\beta)}{I_0}&=\frac{1+s_1\cos2\beta+s_2\sin2\beta}{2}\\\gamma_{\rm reverse}&=\arccos(\mathbf s\cdot\mathbf s_{\rm reverse})\end{aligned}

The rightmost plate in the product acts first. The analyzer formula is checked independently against a time integral of the projected electric field. It has no circular Stokes term, so opposite circular states give identical scans with a linear analyzer. The reversal angle compares two predictions; it is not a recorded measurement.

Common difficulties

Every quarter-wave plate makes circular light?

Typical misconceptionQuarter-wave retardance is sufficient regardless of the incoming polarization and fast-axis angle.

Better mental modelCircular output needs equal field amplitudes along the two eigenaxes and a quarter-cycle phase difference between them. Linear light aligned with either axis stays linear. Circular input comes out linear. A scan with a linear analyzer alone cannot tell the two signs of circular polarization apart.

Run the experiment

01
Scene 1: Quarter-wave conversion
Begin with quarter-wave conversion. The end point reaches a pole, the field ellipse becomes a circle, and the analyzer passes half the power at every angle.
What to observe: The sphere is a map of polarization states, not a lens or detector in the beam. Its poles stand for circular polarization of each sign. Opposite points on the sphere are orthogonal polarization states, not light travelling in opposite directions.
02
Scene 2: Half-wave rotation
Turn the fast axis of plate A to the input azimuth. The circle collapses to a line, because the light no longer has a component along the slow axis.
What to observe: The axes form a right-handed frame: light travels along positive X, the first field axis points along positive Y and the second along positive Z. Angles are measured from the first field axis toward the second. The ellipse plot draws these two axes to the right and upward.
03
Scene 3: Arbitrary retarder sequence
Choose the half-wave preset. Rotate the fast axis and watch the linear output turn through twice the angle, while the power before the analyzer stays the same.
What to observe: A positive circular Stokes value means that, at a fixed point, the electric field turns from the first axis toward the second as time goes on. This operational sign is used instead of right or left labels, which are ambiguous. At a circular pole the ellipse azimuth is undefined.