Spin measurement: projectors, Born rule and cascades

Start from the eigenvalue problem solved in UPV-10.04: n̂⋅σ has eigenvalues ±1 with eigenspinors |n,±⟩. A function of a Hermitian operator is defined on its eigenvalues, and the function returning 1 on +1 and 0 on −1 is (1 + λ)/2, so P₊ = (I + n̂⋅σ)/2 and P₋ = (I − n̂⋅σ)/2. Check the algebra without writing a component. The Pauli identity σᵢσⱼ = δᵢⱼ I + iεᵢⱼₖ σₖ gives (n̂⋅σ)² = nᵢ nⱼ δᵢⱼ I + iεᵢⱼₖ nᵢ nⱼ σₖ = I, because n̂ is a unit vector and the antisymmetric ε kills the symmetric product nᵢ nⱼ. Then P±² = (I ± 2n̂⋅σ + I)/4 = P±, P₊P₋ = (I − I)/4 = 0, and P₊ + P₋ = I. Each has trace 1, so each is rank one: P± = |n,±⟩⟨n,±|. For n̂ = (sin θ, 0, cos θ) the matrix is P₊ = ½[[1 + cos θ, sin θ], [sin θ, 1 − cos θ]] = [[cos²(θ/2), sin(θ/2)cos(θ/2)], [sin(θ/2)cos(θ/2), sin²(θ/2)]], visibly the outer product of (cos(θ/2), sin(θ/2)) with itself.

The Born rule for a two-outcome measurement is p± = ⟨ψ|P±|ψ⟩. Because P± is rank one this equals |⟨n,±|ψ⟩|², the squared overlap with the eigenspinor; because P± is linear in n̂⋅σ it also equals (1 ± ⟨n̂⋅σ⟩)/2 = (1 ± n̂⋅a)/2, with a the Bloch vector. The three forms are one number; use whichever is cheapest. For |ψ⟩ = |z,+⟩ = (1, 0) and n̂ at polar angle θ, the overlap is ⟨n,+|z,+⟩ = cos(θ/2), so p₊ = cos²(θ/2) and p₋ = sin²(θ/2); from the Bloch form, a = ẑ and n̂⋅a = cos θ, so p₊ = (1 + cos θ)/2, the same thing by the half-angle identity. The expectation value of the observable follows for free: ⟨n̂⋅S⟩ = (ℏ/2)(p₊ − p₋) = (ℏ/2) cos θ. What you must never do is square the z-basis components of |ψ⟩ and read them as probabilities for an n̂ measurement: |α|² and |β|² answer the z question only. Expand in the eigenbasis of the operator you are measuring, then square.

After the + outcome the state is P₊|ψ⟩ divided by its norm, and the norm is not decoration: ‖P₊|ψ⟩‖² = ⟨ψ|P₊²|ψ⟩ = ⟨ψ|P₊|ψ⟩ = p₊, so the unnormalised vector carries the probability in its length. Dividing by √p₊ leaves |n,+⟩ times the phase of ⟨n,+|ψ⟩, which no single-spin experiment can see. That is the whole content of collapse for one spin: the output is the analyser's eigenspinor, and nothing about the input survives except the fraction that got through. Two consequences drop straight out of the algebra. Measuring along the same n̂ again gives + with certainty, because P₊P₊ = P₊; and blocking the − port after a + result blocks nothing, because P₋P₊ = 0. The map |ψ⟩ → P₊|ψ⟩/√p₊ is not unitary and not even linear, since the denominator depends on ψ, so it cannot be the solution of the Schrödinger equation for any Hamiltonian. The general form of the rule, p± = Tr(ρP±) and ρ → P±ρP±/p± for a mixed input, and its extension to POVMs and Kraus operators for measurements that are not projective, belong to UPV-20.03, Measurement: projectors, the Born rule, and POVMs; for one spin in a pure state the projector is all you need.

A cascade is a product of projectors in time order, read right to left. A z-up source, analyser A at angle θ_A keeping its + port, then analyser B at θ_B keeping +: the unnormalised final state is P₊(B)P₊(A)|z,+⟩ and its squared norm is the joint probability. Because P₊(A) collapses to |A,+⟩, the norm factorises: p(A+, B+) = |⟨A,+|z,+⟩|² |⟨B,+|A,+⟩|² = cos²(θ_A/2) cos²((θ_B − θ_A)/2), and the second factor depends only on the angle between the two analysers. Take θ_A = 90° (an x analyser) and θ_B = 180° (a z analyser read at its − port): cos²45° × cos²45° = ½ × ½ = ¼ of the source arrives. Remove A and the fraction is |⟨z,−|z,+⟩|² = 0. Inserting a filter increased transmission, which no filter that merely removes atoms carrying a fixed, pre-existing label can do: the x analyser re-prepares every atom it passes, as a 45° polariser between crossed polarisers does for classical light. In operator language P₋(z)P₊(x)P₊(z) ≠ 0 although P₋(z)P₊(z) = 0, because P₊(x) does not commute with P±(z). Draw every cascade as boxes with a fraction on each beam and multiply along the path you keep. If both ports of an analyser are recombined with no record kept, no projector is applied and the state passes unchanged.

Two boundaries deserve a plain statement. First, collapse is postulated. The Schrödinger equation with H = −γ S⋅B carries |z,+⟩ through an x-oriented Stern-Gerlach magnet into a superposition of two separated branches, and carries both branches through the next magnet too; nothing in unitary evolution selects one. The Born rule and the renormalised update are what you add to turn branch amplitudes into beam counts. Second, one qubit's projective statistics say nothing against determinism. Let the hidden variable be a unit vector m drawn uniformly over the sphere, and declare the result of measuring along n̂ to be the sign of (a + m)⋅n̂. Because m⋅n̂ is uniform on [−1, 1] for every n̂ (Archimedes' hat-box theorem), the probability of + is (1 + n̂⋅a)/2, exactly the Born rule, for every axis and every state; and because one m fixes an answer for every axis at once, the answer along −n̂ always opposite to the answer along n̂, the model is deterministic and noncontextual. Bell made this point in 1966 with a model of his own. Excluding local hidden variables needs two spins and a correlation inequality; excluding noncontextual ones needs a Hilbert space of dimension three or more, where Gleason's theorem and Kochen–Specker bite. Neither excludes every hidden-variable theory: Bohm's, nonlocal and contextual, survives both. Everything in this lesson is compatible with classical dice under the hood; the entanglement of the closing units is not compatible with local ones.

The whole lesson fits in NumPy. Build sx, sy, sz as 2×2 complex arrays, set n = (sin θ, 0, cos θ), and P = 0.5*(np.eye(2) + n[0]*sx + n[1]*sy + n[2]*sz). Then np.allclose(P @ P, P) confirms idempotence to machine precision and np.trace(P) returns 1. For a state psi as a length-2 array, p = (ψ.conj() @ P @ ψ).real is the Born probability and ψₒᵤₜ = P @ ψ / np.√(p) the renormalised state; np.allclose(P @ ψₒᵤₜ, ψₒᵤₜ) checks it is an eigenvector. A cascade is a loop that applies the next projector, records the squared norm, and renormalises; with θ_A = 90° and θ_B = 180° the running product returns 0.25. Sweep θ from 0 to 2π and plot p(θ): the curve is cos²(θ/2), with period 2π in the probability though 4π in the amplitude cos(θ/2), which is the double cover made visible.