Spin states and eigenspinors along any axis

Take n̂ = (sinθ cosφ, sinθ sinφ, cosθ) and assemble n̂⋅σ = nₓ σₓ + n_y σ_y + n_z σ_z in the basis where S_z is diagonal. The off-diagonal entries collect into sinθ e^(∓iφ) and the diagonal is ±cosθ, so n̂⋅σ = [[cosθ, sinθ e^(−iφ)], [sinθ e^(iφ), −cosθ]]. Three facts are visible without a calculation. It is Hermitian, so its eigenvalues are real and its eigenvectors orthogonal. Its trace is zero, so the eigenvalues sum to zero. And the Pauli identity gives (n̂⋅σ)² = nᵢ nⱼ σᵢ σⱼ = (n̂⋅n̂) I = I, so every eigenvalue squares to 1. Together: the spectrum is exactly (+1, −1), one of each, and S⋅n̂ = (ħ/2) n̂⋅σ returns ±ħ/2 along any axis whatever. For θ = 45°, φ = 90° the matrix is [[0.707, −0.707i], [0.707i, −0.707]]; check its trace and its square before you trust an eigenvector.

Put (n̂⋅σ − I)χ = 0 with χ = (a, b). The first row reads (cosθ − 1) a + sinθ e^(−iφ) b = 0, so b/a = e^(iφ)(1 − cosθ)/sinθ. The half-angle identities 1 − cosθ = 2 sin²(θ/2) and sinθ = 2 sin(θ/2) cos(θ/2) reduce that ratio to e^(iφ) tan(θ/2). The second row gives the same ratio — it must, since the determinant vanishes — so one equation fixes the direction of χ in C² and normalisation fixes its length: a = cos(θ/2), b = e^(iφ) sin(θ/2), with the free global phase spent on making a real and non-negative. That is |n,+⟩. For |n,−⟩ either repeat with +I, or take the vector orthogonal to |n,+⟩: (sin(θ/2), −e^(iφ) cos(θ/2)). At θ = 45° the ratio is tan 22.5° = 0.414 and the spinor is (0.924, 0.383 e^(iφ)). The half-angle has entered through a trigonometric identity, not through a postulate.

A spinor prepared along m̂ and measured along n̂ has probabilities |⟨n,±|m,+⟩|², and the whole art is expanding in the analyser's basis first. Simplest case, m̂ = ẑ: ⟨n,+|↑⟩ = cos(θ/2) and ⟨n,−|↑⟩ = sin(θ/2), so P₊ = cos²(θ/2) and P₋ = sin²(θ/2), summing to one. For two general directions the same algebra collapses to a single angle: |⟨m,+|n,+⟩|² = (1 + n̂⋅m̂)/2 = cos²(γ/2), with γ the real-space angle between the two axes. Try m̂ = x̂ (θ = 90°, φ = 0) measured along n̂ at θ = 45°, φ = 0. Component by component, ⟨n,+|x,+⟩ = 0.924 × 0.707 + 0.383 × 0.707 = 0.924, squared 0.854; and n̂⋅x̂ = sin 45° = 0.707 gives (1 + 0.707)/2 = 0.854 by the shortcut. Squaring the z-basis components of the state to get 0.5 and 0.5 answers a different question — what a z analyser would do — and is the standard way to get this wrong.

With α = cos(θ/2) and β = e^(iφ) sin(θ/2), the three expectation values follow from the Pauli matrices directly: ⟨σ_z⟩ = |α|² − |β|² = cos²(θ/2) − sin²(θ/2) = cosθ; ⟨σₓ⟩ = 2 Re(α*β) = 2 cos(θ/2) sin(θ/2) cosφ = sinθ cosφ; ⟨σ_y⟩ = 2 Im(α*β) = sinθ sinφ. So ⟨n,+|σ|n,+⟩ = n̂ and ⟨S⟩ = (ħ/2) n̂, a vector of full length ħ/2 pointing along the axis the state was built on. Notice what appears where: the amplitudes carry θ/2, but the expectation values carry the full θ, because they are quadratic in the amplitudes. The state (|↑⟩ + |↓⟩)/√2, so often called 'half up and half down', has θ = 90°, φ = 0 and hence ⟨S⟩ = (ħ/2) x̂: it is |x,+⟩, transmitted with certainty by an x analyser. A genuinely unpolarised beam has ⟨S⟩ = 0, and no ket can produce that; it needs the mixture ρ = I/2 that arrives two lessons from now.

Every real-space angle is halved on the way into C². Rotate |↑⟩ about ŷ through θ and the components become (cos(θ/2), sin(θ/2)): at θ = 180° that is (0, 1) = |↓⟩, so a half-turn in the room is a quarter-turn in Hilbert space, and antipodal directions are orthogonal states — |n,−⟩ is |−n̂,+⟩ up to phase. At θ = 360° the components are (−1, 0) = −|↑⟩: the direction is back where it started and the spinor is not. That sign is the double cover, SU(2) → SO(3) two-to-one, unobservable on a state alone (a global phase) yet observable as a relative phase when only one arm of an interferometer is rotated. The census closes the argument. Normalisation and the global phase remove two of the four real numbers in (α, β), leaving θ = 2 arccos|α| and φ = arg β − arg α, the coordinates of a point on a sphere. Every pure spinor is therefore some |n,+⟩. The Bloch sphere is not a picture of the pure-state space; it is the pure-state space.

In NumPy, build sig = [σₓ, σ_y, σ_z], form M = sum(n[i]*sig[i]), and call w, v = np.linalg.eigh(M). Two things need care. eigh sorts eigenvalues ascending, so w = [−1, +1] and the +1 eigenvector is the second column, v[:, 1]. And an eigenvector is defined only up to a phase: the routine may return it with a negative or complex first component, which is still correct and still disagrees with (cos(θ/2), e^(iφ) sin(θ/2)) entry by entry. Impose the convention yourself, v *= exp(−i arg v[0]), and only then compare. At θ = π the first component is zero and the convention has nothing to act on: that is the coordinate singularity at the pole, not a bug. Then run the checks that are basis-free: v†Mv = +1, and |⟨m,+|n,+⟩|² against (1 + n̂⋅m̂)/2 for a few random pairs of axes. Agreement to 10⁻¹⁵ is what a correct spinor looks like; agreement of the components alone is what a phase convention looks like.