Break Whittaker's impossibility proof

1 · Whittaker's bit

On your bench, each photon carries Whittaker's hidden bit, fixed as it leaves the source for a polarizer at 30°: 0 at the top, it will pass; 1 at the bottom, it will be blocked. For now, three photons in four get through, as Malus' law requires.

Beginner

According to Whittaker, what dooms this hidden bit?

A photon cannot carry any hidden information You can turn the polarizer while the photons are in flight: the bits already fixed do not change, yet Malus' law demands a different share Malus' law does not hold for photons taken one at a time

2 · Schrödinger's number

Each photon now carries a real number λ between 0 and 1, its height in the beam. It passes if λ ≥ sin²φ.

Intermediate

The polarizer is at 60°. Which photons pass, and what share of the beam?

Those with λ ≥ 0.75: one in four Those with λ ≥ 0.25: three in four Those with λ ≥ 0.67, that is 60/90: one in three

3 · A second polarizer

The first polarizer stays at 30°. Look at the beam just after it: the survivors, which all had λ ≥ 0.25, spread out again over the full height.

Intermediate

A photon with λ = 0.5 crosses the first polarizer, at 30°. What is its hidden number on the way out?

Still 0.5: a photon's number never changes 1/3, that is (0.5 − 0.25)/0.75 A new number drawn at random between 0 and 1

4 · Slip a polarizer between two crossed ones

First polarizer at 0°, last one at 90°: crossed, they let nothing through.

Expert

You slip a third polarizer between them, at 45°. With Schrödinger's rule at every crossing, what share of the photons leaving the first one comes out of the last?

None: the first and the last are still crossed A quarter: half pass at 45°, then half of those at the last one Half: only the middle polarizer stops any

Impossibility proof broken

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