Intrinsic spin, spinors and the Pauli matrices

Run the ladder argument on any triple obeying [Jₓ, J_y] = iħJ_z. Demanding that the ladder terminate at both ends forces the eigenvalues j(j+1)ħ² and mħ with 2j + 1 rungs, and j comes out integer or half-integer. Orbital angular momentum kept only the integers, and not because e^(imφ) must be single-valued — a half-integer m only flips an overall sign, which is unobservable. It kept them because the l = ½ tower fails: L₋ does not annihilate the supposed bottom rung, and lowering again leaves the normalisable functions on the sphere altogether. That veto is a property of the position representation, not of the algebra. A degree of freedom that has no position representation is free to take j = ½ — and the evidence says one exists: two Stern-Gerlach beams from an l = 0 silver atom, doublet lines in sodium, and Zeeman patterns with the wrong number of components.

With s = ½ the space has 2s + 1 = 2 dimensions, so take |↑⟩ = (1,0)ᵀ and |↓⟩ = (0,1)ᵀ as the S_z eigenvectors with eigenvalues ±ħ/2. That alone fixes S_z = (ħ/2)diag(1, −1). The other two come from the ladder: S₊|s, m⟩ = ħ√(s(s+1) − m(m+1))|s, m+1⟩, and for s = ½, m = −½ the root is √(¾ + ¼) = 1, so S₊|↓⟩ = ħ|↑⟩ while S₊|↑⟩ = 0. In matrix form S₊ = ħ[[0,1],[0,0]] and S₋ = ħ[[0,0],[1,0]]. Then Sₓ = (S₊ + S₋)/2 = (ħ/2)[[0,1],[1,0]] and S_y = (S₊ − S₋)/2i = (ħ/2)[[0,−i],[i,0]]. Nothing was assumed except the commutators and the dimension: the Pauli matrices are the output of the derivation, not an input to it. Writing S = (ħ/2)σ strips off the units and leaves three dimensionless matrices to work with.

Everything about a spin-half follows from σᵢσⱼ = δᵢⱼ1 + iεᵢⱼₖσₖ. Set j = i: σᵢ² = 1, so each Pauli matrix has eigenvalues ±1 and every spin component, along every axis, measures ±ħ/2 — no state and no direction gives anything else. Symmetrise: σᵢσⱼ + σⱼσᵢ = 2δᵢⱼ1, so different components anticommute, the sharpest possible incompatibility. Antisymmetrise: [σᵢ, σⱼ] = 2iεᵢⱼₖσₖ, which is [Sₓ, S_y] = iħS_z restored. Sum the squares: S² = (ħ²/4)(σₓ² + σ_y² + σ_z²) = ¾ħ²1, proportional to the identity, so every spinor is an S² eigenstate and s = ½ is a property of the electron rather than a quantum number of its state. Finally 1, σₓ, σ_y and σ_z span all 2×2 Hermitian matrices, so every observable of a spin-half is a1 + b⋅σ and there is nothing else left to find.

Write n̂ = (sinθ cosφ, sinθ sinφ, cosθ). Then S⋅n̂ = (ħ/2)[[cosθ, sinθ e^(−iφ)], [sinθ e^(iφ), −cosθ]], whose trace is 0 and determinant −ħ²/4 for every θ and φ — so the eigenvalues are ±ħ/2 on every axis and no direction in space is privileged. Solving for the + eigenvector gives χ₊(n̂) = (cos(θ/2), e^(iφ)sin(θ/2))ᵀ. Check it: θ = 90°, φ = 0 returns (1/√2)(1, 1)ᵀ, the +x spinor. Two features matter. First, ⟨S⟩ = χ₊†Sχ₊ = (ħ/2)n̂ exactly, so the mean vector has length ħ/2 and not √3ħ/2; the shortfall is transverse variance, not a shorter arrow. Second, the angle is halved: turn n̂ through θ and the components turn through θ/2, so χ₊ at θ = 2π is −χ₊ at θ = 0. The half-angle is not a convention — it is the whole reason the period is 4π.

Prepare |↑⟩ and analyse along n̂ at angle Θ from z. The amplitude is ⟨χ₊(n̂)|↑⟩ = cos(Θ/2), so P(+) = cos²(Θ/2) = (1 + cosΘ)/2 and P(−) = sin²(Θ/2). In numbers: Θ = 0 gives 1, Θ = 60° gives 0.750, Θ = 90° gives 0.500, Θ = 120° gives 0.250, Θ = 180° gives 0. Now chain three analysers, z then x then z. The x stage splits the beam 50/50, and its +x output — a state that no longer knows anything about z — splits 50/50 again, so a quarter of an initially pure +z beam ends in the −z channel although the x stage removed nothing along z. No assignment of definite values to every axis, fixed in advance and left unchanged by the x stage, reproduces that. Be precise about what it refutes, though: it kills values that pre-exist along all axes and pass through the other measurements undisturbed, but a model in which the x stage resets the hidden values can still match these counts, so this single-particle result is not a Bell test of local hidden variables.

Expand the rotation operator using (n̂⋅σ)² = 1 and it collapses to U(α, n̂) = cos(α/2)1 − i sin(α/2)(n̂⋅σ). At α = 2π that is −1: every spinor comes back reversed, and only α = 4π restores it. A global sign cannot be measured, but a relative one can — split a neutron beam, rotate the spin in one arm only, and the interference intensity is periodic in 4π of rotation, which is what the 1975 interferometry experiments found. No classical picture survives alongside that. Model the electron as a uniform sphere of the classical electron radius 2.82 × 10⁻¹⁵ m carrying L = ħ/2: the equatorial speed is 5ħ/(4mr) = 5.1 × 10¹⁰ m s⁻¹, about 170c, and keeping it below c would demand a radius above 4.8 × 10⁻¹³ m when experiment bounds the electron below 10⁻¹⁸ m. Spin says how a state transforms under rotations, not how fast anything turns.