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Course · Shields 2026c · not a Zenodo paper

The algorithm

The orthodox calculation, one spin, every symbol given a value. A weight of 1/2 is not a click. The update that makes a second measurement agree with the first is not a solution of the Schrödinger equation. The full argument, Einstein’s fork, and the Bell record are in the paper. This page is the sheet.

Read the paper on this page

Assign every symbol

The equation, and only then the example it is evaluated on. This is a spin in a constant field along z. Nothing is left as a letter without a value or a matrix.

i ℏ d|ψ⟩/dt = H |ψ⟩

H = (ℏ ω / 2) σ_z

iImaginary unit. i² = −1. Not a variable.
ℏh / 2π. h = 6.62607015×10⁻³⁴ J·s exactly. ℏ = 1.054571817×10⁻³⁴ J·s.
tTime in seconds. The parameter. Not an operator.
|ψ(t)⟩c↑(t)|↑⟩ + c↓(t)|↓⟩. Two complex numbers, |c↑|² + |c↓|² = 1.
d/dtOrdinary derivative of those two numbers. No spatial derivative in this example.
H(ℏω/2) σ_z = diag(+ℏω/2, −ℏω/2). The two diagonal entries are the two energies.
ωReal angular frequency, s⁻¹, fixed by the magnet. Not an outcome.
σ_zMatrix diag(1, −1). Eigenvalue +1 on |↑⟩ = (1, 0), eigenvalue −1 on |↓⟩ = (0, 1).
|ψ(0)⟩|+x⟩ = (1/√2)(|↑⟩ + |↓⟩). So c↑(0) = c↓(0) = 1/√2.

Componentwise, after cancelling ℏ, which is not zero: i ċ↑ = (ω/2) c↑ and i ċ↓ = −(ω/2) c↓. That is the equation in numbers. No probability has entered.