University Physics V · Quantum Spin · 10.7

Spin Measurement: Projectors, Born Rule & Cascades

Stop squaring z-components. A spin measurement is a projector, P± = (I ± n̂⋅σ)/2, and once you can write it down you can price any Stern-Gerlach cascade: which beam survives, in what state, and why inserting an analyser can let atoms through a filter that had blocked them.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A projective measurement of spin along n̂ is a pair of Hermitian operators P± = (I ± n̂⋅σ)/2 = |n,±⟩⟨n,±|. Because (n̂⋅σ)² = I, each squares to itself, the two are orthogonal, and they sum to the identity: the complete set of answers to the question "is the spin up along n̂?" The Born rule then says the probability of + on a state |ψ⟩ is the expectation value of the projector, p₊ = ⟨ψ|P₊|ψ⟩ = |⟨n,+|ψ⟩|², and the projection postulate says the surviving state is P₊|ψ⟩ renormalised, which for a rank-one projector is |n,+⟩ whatever came in. For a z-up spinor and an analyser tilted by θ that gives cos²(θ/2): the half angle of the SU(2) representation, not the cos θ of a classical moment or the cos²θ of Malus.

Cascades follow by multiplication, because the beam leaving each analyser is that analyser's eigenspinor and remembers nothing else: a z-up beam never reaches a z-down port directly, yet an x analyser in between passes ½ × ½ = ¼ of it there. Two things the algebra does not buy. Collapse is not derived from the Schrödinger equation, which entangles spin with position but never selects a branch; and the statistics of one qubit under every projective measurement are reproduced by a deterministic hidden variable, so nothing here rules classical dice out.

That needs two spins, or a larger Hilbert space.

Simple definition: A projective spin measurement along n̂ is the pair of orthogonal projectors P± = (I ± n̂⋅σ)/2; each outcome has probability ⟨ψ|P±|ψ⟩, and the state afterwards is P±|ψ⟩ divided by the square root of that probability.
Example: Send |z,+⟩ into an analyser 60° from +z: p₊ = cos²30° = 0.750, p₋ = sin²30° = 0.250, and every atom in the + port leaves as |n,+⟩ = (cos 30°, sin 30°) = (0.866, 0.500), with no memory that it arrived along z.

Projectors for spin along n̂P± = (I ± n̂⋅σ)/2 = |n,±⟩⟨n,±|

Writing the yes/no question as an operator is what lets you take its expectation value and multiply it through a cascade.

n̂ a unit vector; σ the vector of Pauli matrices; P± are dimensionless 2×2 Hermitian matrices with eigenvalues 0 and 1

Projector algebraP±² = P±, P₊P₋ = 0, P₊ + P₋ = I, tr P± = 1

Idempotence is repeatability: a second measurement along the same axis is certain. Trace one means each projector selects a single spinor.

all from (n̂⋅σ)² = I, which follows from σᵢσⱼ = δᵢⱼ I + iεᵢⱼₖ σₖ with |n̂| = 1

Born rule for one spinp± = ⟨ψ|P±|ψ⟩ = |⟨n,±|ψ⟩|² = (1 ± n̂⋅a)/2

Probability is the expectation value of the projector, so expand in the eigenbasis you are measuring; never square the z-basis components for another axis.

a = ⟨ψ|σ|ψ⟩ is the Bloch vector, dimensionless, |a| = 1 for a pure state; p₊ + p₋ = 1 automatically

Half-angle transmissionp₊ = cos²(θ/2), p₋ = sin²(θ/2) for |ψ⟩ = |z,+⟩

The half angle is SU(2)'s double cover counted in a beam: orthogonal spin states are antiparallel directions, not perpendicular ones.

θ the angle between n̂ and +z; θ = 60° gives 0.750, θ = 90° gives 0.500, θ = 180° gives 0

Post-measurement state (Lüders rule)|ψ'⟩ = P±|ψ⟩ / √p± = |n,±⟩ up to a phase

The update is non-unitary and nonlinear: the beam leaving an analyser remembers that analyser's axis and nothing else about what entered.

the unnormalised P±|ψ⟩ has squared norm p±; the phase is unobservable on a single spin

Two analysers in cascadep(A+ then B+) = cos²(θ_A/2) · cos²((θ_B − θ_A)/2)

Multiply conditional probabilities stage by stage; the second factor depends only on the angle between the analysers, which is collapse doing arithmetic.

θ_A, θ_B the analyser angles from +z in one plane; each factor is an overlap with the previous eigenspinor

Build the projector from the Pauli identity

Start from the eigenvalue problem of the previous topic: 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.

Probability is the expectation value of the projector

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.

Collapse renormalises, and forgets

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 Stern-Gerlach Hamiltonian H = −γ S⋅B with a field gradient is unitary and produces α|↑⟩|χ₊⟩ + β|↓⟩|χ₋⟩, two separated branches; keeping one and renormalising is the projection postulate, an added rule and not a theorem.

Chain the analysers: multiply conditional probabilities

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 classical filter does; 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.

What the algebra does not buy you

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 a hidden variable λ be uniform on [−1, 1] and declare the result of measuring along n̂ to be +1 if λ ≤ n̂⋅a and −1 otherwise. The probability of + is (1 + n̂⋅a)/2, exactly the Born rule, for every axis and every state; Bell made this point in 1966 with a slightly fancier model. Excluding hidden variables needs two spins and a correlation inequality, or a Hilbert space of dimension three or more, where Gleason's theorem and Kochen–Specker bite. Everything in this lesson is compatible with classical dice under the hood; the entanglement of the closing units is not.

Check it numerically

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Analyser A angle θ_A from +z90 deg

Analyser B angle θ_B from +z180 deg

Set A to 90° and B to 180°: with A removed a z-up beam never reaches the z-down port, yet the x analyser lets a quarter through. Then slide A and watch the solid peak follow it — B sees cos²((θ_B − θ_A)/2), so the beam leaving A remembers only θ_A.

A KEEPS cos²(θ_A/2)0.500

B KEEPS cos²((θ_B − θ_A)/2)0.500

REACHES B+ VIA A0.250

REACHES B+ WITHOUT A0.000

Live interpretationA KEEPS cos²(θ_A/2): 0.500. B KEEPS cos²((θ_B − θ_A)/2): 0.500. REACHES B+ VIA A: 0.250. REACHES B+ WITHOUT A: 0.000

Catch the common trap

Explain before calculating.

Misconception check

The trap is to treat the analyser like a polariser and write the transmitted fraction as cos²θ, or like a classical moment and write it as the projection cos θ. Both fail for the same reason. The probability is a squared overlap of eigenspinors, and eigenspinors along n̂ are built from the half angle: ⟨n,+|z,+⟩ = cos(θ/2), so p₊ = cos²(θ/2). At θ = 90° that gives ½ where cos²θ and cos θ both give 0; at θ = 180° it gives 0 where cos²θ gives 1. Perpendicular directions in space are not orthogonal states, antiparallel ones are, and the half angle is exactly the statement that SU(2) turns a spinor through half the angle SO(3) turns a vector. The same error hides the cascade that surprises people: an x analyser between a z-up source and a z-down port transmits ¼, not 0 and not ½.

A |z,+⟩ beam passes an analyser along x whose + port is kept, then a z analyser. What fraction of the source reaches the z-down port, and why?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA |z,+⟩ beam meets an analyser whose axis is tilted 60° from +z in the xz-plane. Write P₊ as a matrix, find the fraction in the + port, and write the normalised state leaving it.

n̂ = (sin 60°, 0, cos 60°) = (0.866, 0, 0.500), so n̂⋅σ = 0.866σₓ + 0.500σ_z = [[0.500, 0.866], [0.866, −0.500]].
P₊ = (I + n̂⋅σ)/2 = [[0.750, 0.433], [0.433, 0.250]]. Check: trace 1, and the diagonal entries are cos²30° = 0.750 and sin²30° = 0.250.
p₊ = ⟨z,+|P₊|z,+⟩ is the top-left entry, 0.750 = cos²(60°/2). Not cos²60° = 0.250.
P₊|z,+⟩ is the first column, (0.750, 0.433); its squared norm is 0.5625 + 0.1875 = 0.750 = p₊, as it must be. Divide by √0.750 = 0.866: (0.866, 0.500) = (cos 30°, sin 30°) = |n,+⟩.

Answerp₊ = cos²30° = 0.750; the + beam leaves as |n,+⟩ = (0.866, 0.500), whatever entered.

MediumA z-up source feeds analyser A at θ_A = 60°, whose + port is kept, then a z analyser B with both ports counted. Find the fraction of the source at each port of B, compare with A removed, and give ⟨S_z⟩ in the beam between A and B.

Stage A: p(A+) = cos²30° = 0.750, and the survivors are |A,+⟩ = (cos 30°, sin 30°) = (0.866, 0.500).
Stage B measures S_z on |A,+⟩: p(z+ | A+) = |⟨z,+|A,+⟩|² = cos²30° = 0.750 and p(z− | A+) = sin²30° = 0.250. Equivalently, B's axis is 60° from A's, and cos²(60°/2) = 0.750.
Joint fractions of the source: z-up port 0.750 × 0.750 = 0.5625; z-down port 0.750 × 0.250 = 0.1875. They sum to 0.750, the A+ fraction, as P₊ + P₋ = I demands.
With A removed the z-up port takes 1 and the z-down port 0. The extra analyser moved 18.75% of the source into a port that was empty.
In the beam between A and B, ⟨S_z⟩ = (ℏ/2)(0.750 − 0.250) = 0.250ℏ, which is (ℏ/2) cos 60° as the Bloch vector along A's axis predicts.

Answerz-up port 0.5625, z-down port 0.1875 of the source (1 and 0 without A); ⟨S_z⟩ = +ℏ/4 between the analysers.

HardTurn a z-up beam into a z-down beam by measurement alone: N analysers in the xz-plane at angles 180°/N, 2⋅180°/N, …, 180°, each keeping its + port, the last one being a z-down port. Find the fraction delivered for N = 1, 2, 3 and 10, the large-N limit, and say why this chain cannot exclude a hidden-variable model.

Stage k receives |n_(k−1),+⟩ and measures along nₖ, which is 180°/N away. By collapse the conditional probability of + is cos²(90°/N), independent of k and of the original source.
Multiply N identical factors: F(N) = [cos²(90°/N)]^N = cos^(2N)(90°/N).
N = 1: cos²90° = 0. N = 2: cos⁴45° = (½)² = 0.250. N = 3: cos⁶30° = 0.750³ = 0.422. N = 10: cos²⁰9° = 0.98769²⁰ = 0.781.
Large N: cos(π/2N) ≈ 1 − π²/(8N²), so F ≈ exp(−π²/(4N)) → 1; N = 100 delivers 0.976. Measurement alone steers the spin from up to down while losing almost nothing, the Zeno flavour of projective collapse.
Every stage is one projective measurement on a known state. A hidden variable λ uniform on [−1, 1], with outcome + when λ ≤ n̂⋅a and λ drawn afresh for each freshly prepared beam, gives cos²(90°/N) at each stage and the same product; no count from this chain separates the two models. That takes two spins and a correlation inequality.

AnswerF(N) = cos^(2N)(90°/N): 0, 0.250, 0.422 and 0.781 for N = 1, 2, 3, 10, tending to 1 as exp(−π²/4N); every number is reproducible by a classical hidden variable.