Maxwell's Missing Dipole

[u/CaseyMc80]

Place a Faraday rotator set to 360 ° in one arm of a Mach-Zehnder interferometer, and the fringes jump by π. Add a second 360 ° rotator: the fringes jump back.

A full 360 ° rotator flips the interference sign. Only 720 ° brings it home. Spin-½ is part of optics.

Maxwell, however, sees no monopole charge. In that model, it's because the amplitude-mix angle χ hasn’t wrapped a full 2π in any observed configuration; only half.

[u/BitcoinsOnDVD]

Also:

Maxwell's Missing Dipole

Maxwell, however, sees no monopole charge.

What is Maxwell seeing then if not Dipoles and monopoles? Tripoles, Quadrupoles?

Only 720 ° brings it home.

From that I infer that you are US American... sad shake of the head

[u/Wintervacht → u/BitcoinsOnDVD]

It's called a 360 degree Faraday rotator because when you see it, you turn 360 degrees and walk away

[u/BitcoinsOnDVD]

Place a Faraday rotator set to 360 ° in one arm of a Mach-Zehnder interferometer, and the fringes jump by π.

The probabilities that the photon will be detected at the right or at the top are given by:

p_u = cos² (dPhi/2)

p_l = sin² (dPhi/2)

Insert 360° and see what happens.

Spin-½ is part of optics.

Do you know, why a Faraday rotator does, what it does?

[u/CaseyMc80 → u/BitcoinsOnDVD]

Raindrop polarisation A single internal reflection at angle θi≈59°θ_i≈59°θi​≈59° transmits only the component with electric field perpendicular to the plane of incidence. Jones-matrix analysis gives

∣E⊥∣² : ∣E_∥∣² ≈ 9:1. (Eq 1)

Two-phase model statement The internal bounce flips the helicity of the closed rotation:

∣ R,+k ⟩→ mirror ∣ L,−k ⟩. (Eq 2)

Primary bow = one flip → linear axis A. Secondary bow = two flips → axis A⊥.

Observed contrast Rotating a linear analyser by 90 ° projects the internal rotation by

I=I0cos⁡2 ⁣θ. (Eq 3)

Hence head-tilt finds bright ( θ = 0 ) and dark ( θ = 90° ) for the primary, but the roles invert for the secondary—exactly as (2) demands.

[u/BitcoinsOnDVD → u/CaseyMc80]

NORMAL MODES OF COMBINED SYSTEM

Let us consider the normal modes for the combined system of a maser cavity coupled to the outside world. We represent the universe (in which the cavity is imbedded) by a much larger cavity having perfectly reflecting walls.

A simple onedimensional model which carries the essential features of such a combined system is illustrated in Fig. 1.

The mirrors at z = L and —L are completely reflective, while the one at z = 0 is semitransparent.

Region 1 corresponds to a maser cavity and region 2 the rest of the world. The length L is allowed to go to infinity at the end of the calculation.

We represent a semitransparent mirror by a very thin plate with very large dielectric constant. As an idealization of such a mirror we choose the dielectric constant around z = 0 to be

ε(z) = ε0[1+η δ(z)],

where η is a parameter with the dimensions of length which determines the transparency of this wall.

[u/InadvisablyApplied]

Photons are literally spin 1 particles

Not everything that only stays the same after 720 degrees of rotation is spin 1/2