Polarization: ellipse, Jones & Stokes

Keep the chapter-02 convention exp(+jωt−jkz) and fix one observation point z. The physical field components are E_x(t)=A_x cos(ωt+φ_x) and E_y(t)=A_y cos(ωt+φ_y), with non-negative real amplitudes A_x,A_y. Only the relative phase δ=φ_y−φ_x affects the shape traced by the field tip; multiplying both complex amplitudes by the same phase factor shifts the time origin and leaves the polarization state unchanged.

This is why polarization is a projective property of the complex two-component amplitude rather than of an overall complex phase. Absolute field scale controls amplitude, while the ratio of component magnitudes and their relative phase control the shape.

Assume first A_xA_y≠0 and sinδ≠0, and choose the common phase so that φ_x=0; this does not change polarization. Then E_x/A_x=cosψ and E_y/A_y=cos(ψ+δ), where ψ=ωt−kz at the chosen point. Expand the second cosine, isolate sinψ, square once, and use sin²ψ+cos²ψ=1. The zero-amplitude and sinδ=0 cases are handled separately as linear limits, so no division by zero is hidden in the derivation.

The resulting quadratic curve is the polarization ellipse. It degenerates to a straight line when sinδ=0. If both amplitudes are nonzero and sinδ≠0, the enclosed curve has nonzero area. The equation itself contains no arbitrary picture convention: every term comes directly from the two real harmonic components.

Linear polarization occurs when the orthogonal components are in phase or antiphase: δ=0 or π. The field vector then moves back and forth on one fixed line. If one component amplitude is zero, the same conclusion holds regardless of the unused phase.

Circular polarization requires equal nonzero amplitudes and quadrature phase: A_x=A_y and |δ|=π/2 modulo 2π. The field magnitude is constant while its direction rotates. Every other nondegenerate combination produces elliptical polarization.

These are not three unrelated wave types; they are regions and boundaries of the same two-amplitude parameter space.

For +z propagation with AoN's exp(+jωt) convention, let A_x=A_y=A and set φ_x=0. If δ=+π/2, then E_x=A cosωt and E_y=−A sinωt. At t=0 the vector points +x; for a small positive time it moves toward −y. If δ=−π/2, it moves from +x toward +y.

This coordinate statement is unambiguous. The labels right-hand circular polarization and left-hand circular polarization are not safe unless the reference specifies the observer's viewing direction and convention; antenna and optics literature have historically used different descriptions. AoN therefore always gives propagation direction, time convention, and coordinate rotation before attaching any handedness label.

Reversing propagation while keeping the same complex transverse components also changes how a remote observer perceives rotation. Never infer handedness from a static ellipse alone.

For a deterministic monochromatic field, collect the two complex amplitudes into a Jones vector. A nonzero common complex scalar changes overall amplitude and/or common phase but not the normalized polarization state, so two Jones vectors proportional by a nonzero complex number represent the same polarization when intensity scale is ignored.

A convenient normalized representative satisfies |e_x|²+|e_y|²=1. The dagger symbol † denotes conjugate transpose. Examples are x-linear [1,0]^T, y-linear [0,1]^T, +45° linear [1,1]^T/√2, and the two quadrature states [1,±j]^T/√2. Under AoN exp(+jωt), the +j state moves +x→−y at fixed z and the −j state moves +x→+y.

Jones calculus describes fully coherent polarization. A statistical mixture or partially polarized field requires a coherency/Stokes or density-matrix-like description instead of one deterministic Jones vector.

The same transverse field can be expanded in the linear basis {x̂,ŷ} or in the two quadrature basis vectors e_±=(x̂±jŷ)/√2. A basis change is a unitary coordinate transformation, not a physical change of polarization.

Solving E=E_+e_++E_-e_- gives E_+=(E_x−jE_y)/√2 and E_-=(E_x+jE_y)/√2. A pure e_+ state has E_-=0, while a pure e_- state has E_+=0. A linear state generally contains equal-magnitude contributions from the two quadrature basis states.

This basis viewpoint becomes useful at interfaces and in anisotropic systems because a device may couple more naturally to one basis than another.

For the fully polarized deterministic field, AoN defines Stokes quadratic parameters from the complex amplitudes. Here * denotes complex conjugation, so E_x*E_y means the conjugate of E_x multiplied by E_y. We choose S₃=2 Im(E_x*E_y), so under exp(+jωt), positive S₃ corresponds to positive relative phase δ and therefore to the coordinate motion +x→−y in the equal-amplitude circular limit. Some references adopt the opposite sign for S₃; the definition must travel with the number.

S₀ measures total squared field amplitude, S₁ the x/y imbalance, S₂ the ±45° correlation, and S₃ the quadrature correlation under the stated sign convention. For a fully polarized monochromatic state, S₀²=S₁²+S₂²+S₃².

The major-axis azimuth α and ellipticity angle χ can then be obtained with quadrant-aware functions: 2α=atan2(S₂,S₁), while sin(2χ)=S₃/S₀ in this AoN convention. |χ|=45° is circular and χ=0 is linear.

Take E_x=3 V/m and E_y=4 exp(j60°) V/m for a +z monochromatic wave. The relative phase is δ=60°. Because amplitudes are unequal and sinδ≠0, the state is elliptical rather than linear or circular.

The AoN Stokes values are S₀=25, S₁=−7, S₂=12 and S₃=12√3≈20.785 in squared-field units. The identity gives S₁²+S₂²+S₃²=625=S₀². The ellipse azimuth follows 2α=atan2(12,−7), giving α≈60.1° modulo 90°, and sin2χ≈0.8314 gives χ≈28.1°.

At a fixed point with φ_x=0, positive δ means the y component leads in complex phase but, under exp(+jωt), the real vector initially moves toward negative y when starting from a positive-x maximum only in the equal-amplitude quadrature case. Do not generalize circular-motion shorthand to an arbitrary ellipse without evaluating the actual components.

1. Show explicitly that multiplying both Eₓ and Eᵧ by exp(jφ₀) leaves the polarization ellipse unchanged.
2. Derive X²+Y²−2XYcosδ=sin²δ from X=cosψ and Y=cos(ψ+δ), showing the algebraic elimination step.
3. Classify E₀=5x̂+5ŷ V/m and give the line direction.
4. Classify E₀=2x̂+2jŷ V/m. Under AoN exp(+jωt) and +z propagation, state the coordinate motion from t=0 rather than using RH/LH shorthand.
5. Classify E₀=3x̂−4ŷ V/m and determine the linear-polarization angle from +x.
6. For J=[1, e^{jπ/3}]^T, calculate normalized S₀,S₁,S₂,S₃ using AoN's S₃ definition.
7. Transform J=[1,0]^T into the e_± basis and show that the two quadrature-basis amplitudes have equal magnitude.
8. Show that |Eₓ|²+|Eᵧ|²=|E₊|²+|E₋|² using the basis-change equations.
9. For Eₓ=3 and Eᵧ=4e^{jπ/3}, verify the fully polarized Stokes identity and calculate α and χ.
10. A paper says only 'right circular polarization' without declaring propagation direction, viewing direction or harmonic convention. What minimum information should you recover before comparing its complex Jones vector with AoN?

Chapter 04 will apply these plane-wave and polarization tools at an interface: derive tangential boundary conditions, Fresnel reflection/transmission amplitudes, phase changes, power coefficients and the distinction between TE/s and TM/p polarization.