University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.9

Measurement modelled as entanglement with a pointer

Quantum Mechanics I let a measurement happen by decree. Here the apparatus is a quantum system too, and the projection postulate's statistics fall out of one unitary and one partial trace. Seeing exactly what that calculation delivers, and what it leaves undone, is the sharpest test of everything this unit has built.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I made measurement a postulate: the state jumps to an eigenstate |sₖ⟩ with probability |cₖ|². Von Neumann's model asks what happens if the apparatus obeys the same Schrödinger equation as the system. It needs one design rule, a unitary that leaves each eigenstate alone and moves a pointer from its ready state |R⟩ to a reading state |Pₖ⟩, and linearity does the rest: a superposition Σ cₖ|sₖ⟩ comes out as the entangled state Σ cₖ|sₖ⟩|Pₖ⟩.

Tracing out the pointer multiplies each coherence cⱼcₖ* by the overlap ⟨Pₖ|Pⱼ⟩ and leaves every population |cₖ|² alone. With orthogonal pointer states the system's density operator is exactly the mixture the projection postulate predicts, and no measurement on the system alone can tell the difference; with overlapping ones, as for a Stern–Gerlach kick comparable with the atom's own momentum spread, the measurement is partial and so is the loss of coherence. All of that is derived.

What the model cannot do is pick an outcome. The joint state stays pure, a coupling that is undone restores the superposition, and the Born rule has only been moved from the system to the pointer. That gap is the measurement problem, and this lesson states it without pretending to close it.

Simple definition: In von Neumann's model a measurement of A is a unitary coupling that leaves each eigenstate |sₖ⟩ of A unchanged and moves a pointer from its ready state |R⟩ to a distinct state |Pₖ⟩, so that reading the pointer reads A.
Example: A silver atom in (|↑⟩ + |↓⟩)/√2 leaves a Stern–Gerlach magnet as (|↑⟩|φ₊⟩ + |↓⟩|φ₋⟩)/√2, any Larmor phase dropped. If its two momentum packets overlap by ⟨φ₋|φ₊⟩ = 0.20, the spin's coherence |ρ↑↓| falls from 0.50 to 0.10 while P(↑) stays 0.50.

Von Neumann couplingU(|sₖ⟩ ⊗ |R⟩) = |sₖ⟩ ⊗ |Pₖ⟩

A modelling choice, not a result: it fixes U on eigenstates only, and unitarity permits it for any normalised pointer states, orthogonal or not.

|sₖ⟩ eigenstates of the measured A, |R⟩ the pointer's ready state, |Pₖ⟩ reading k. Discrete kets and U are dimensionless.

What linearity does to a superpositionU(Σₖ cₖ|sₖ⟩ ⊗ |R⟩) = Σₖ cₖ |sₖ⟩ ⊗ |Pₖ⟩

Derived, not assumed. The output is entangled whenever two cₖ are non-zero, and it is still one pure state of system plus pointer.

cₖ = ⟨sₖ|ψ⟩, dimensionless, Σ|cₖ|² = 1. With ⟨Pⱼ|Pₖ⟩ = δⱼₖ it is a Schmidt form with coefficients |cₖ|.

Reduced state after the couplingρ_S = Σⱼₖ cⱼ cₖ* ⟨Pₖ|Pⱼ⟩ |sⱼ⟩⟨sₖ|

Populations (j = k) never change; each coherence is multiplied by an overlap. Orthogonal pointers leave Σ|cₖ|²|sₖ⟩⟨sₖ|, the projection postulate's mixture.

Partial trace over the pointer. ρ_S is dimensionless with trace 1; the overlap ⟨Pₖ|Pⱼ⟩ is a dimensionless complex number.

Stern–Gerlach momentum kickΔp = (gₛ/2) μ_B |∂B_z/∂z| τ

The electron's moment points against its spin, μ_z = ∓(gₛ/2)μ_B for ↑ and ↓, so with the gradient pointing down, B_z falling with z, spin ↑ is kicked to +Δp and spin ↓ to −Δp: the atom's transverse momentum becomes the pointer. The α/2π inside gₛ/2 is quoted from QED and moves Δp by 0.1%.

μ_B = 9.274 × 10⁻²⁴ J T⁻¹, gradient in T m⁻¹, transit time τ = L/v in s, Δp in kg m s⁻¹; gₛ/2 = 1.00116, measured.

Overlap of two Gaussian pointer packets⟨φ₋|φ₊⟩ = exp(−Δp²/(2σₚ²))

Derived by completing the square. It is 0.61 at Δp = σₚ and 0.011 at 3σₚ: coherence dies once the kick beats the packet's own momentum spread.

φ±(p) = φ(p ∓ Δp) with φ a real, minimum-uncertainty Gaussian; σₚ is the standard deviation of |φ|², in kg m s⁻¹ like Δp. The overlap is real and dimensionless.

Bloch vector of the measured spinr_⊥ → D r_⊥ and r_z → r_z, with D = ⟨φ₋|φ₊⟩

For a spin prepared at polar angle θ the vector slides straight toward the z axis and stops at (0, 0, cos θ), the projection postulate's mixture; a spin-only measurement of σₓ sees its mean shrink by D. A uniform field would add Larmor precession about z, which turns r_⊥ without shrinking it; it is left out here.

r = 2⟨S⟩/ħ is dimensionless with |r| ≤ 1; r_⊥ = (rₓ, r_y) holds the coherence, r_z the populations. D is real here, as for the Gaussian packets; a complex D would also turn r_⊥ about z.

Define the measurement by what it does to eigenstates

Von Neumann makes the apparatus quantum too: a pointer with a ready state |R⟩ and a state |Pₖ⟩ for each reading of A = Σ aₖ|sₖ⟩⟨sₖ|. The one design rule concerns eigenstates. A system that already has the value aₖ is left alone and the pointer moves to reading k: U|sₖ⟩|R⟩ = |sₖ⟩|Pₖ⟩. A concrete U is a short pulse of H = g A ⊗ p̂_Q, with p̂_Q the momentum conjugate to a pointer coordinate Q. On |sₖ⟩ it becomes exp(−iτgaₖp̂_Q/ħ), Quantum Mechanics I's translation operator, which shifts the pointer by gτaₖ. Dimensions: [H] = J and [p̂_Q] = kg m s⁻¹, so gA is a speed and gτaₖ a length. Unitarity asks only that inner products survive, ⟨sⱼ|sₖ⟩⟨R|R⟩ = ⟨sⱼ|sₖ⟩⟨Pⱼ|Pₖ⟩, and both sides are δⱼₖ for any normalised pointer states. Orthogonality is a design goal, not a law: a pointer packet of width σ_Q gives cleanly distinguishable readings only once the shift gτ(aⱼ − aₖ) is many σ_Q.

Linearity turns a superposition into entanglement

Feed in |ψ⟩ = Σ cₖ|sₖ⟩. U is linear, so it acts term by term: U(|ψ⟩|R⟩) = Σ cₖ|sₖ⟩|Pₖ⟩. That is derived, with nothing added. With orthonormal pointer states it is already a Schmidt decomposition with coefficients |cₖ|, so system and pointer are entangled whenever two cₖ are non-zero, with entanglement entropy −Σ|cₖ|² log₂|cₖ|². A spin prepared along +x, c↑ = c↓ = 1/√2, ends one full bit entangled with its pointer, as entangled as the singlet. The limiting case is the reassuring one: an eigenstate input, one |cₖ| = 1, comes out as the product |sₖ⟩|Pₖ⟩, undisturbed and unentangled, which is what the postulate says a repeated measurement should find. Because U is unitary the joint state stays pure, Tr ρ_SP² = 1. A correlation has been written, reading k goes with state |sₖ⟩, but no term has been removed: every branch of the superposition is still present.

Trace out the pointer: coherences shrink, populations stay

The joint density operator is Σⱼₖ cⱼcₖ* |sⱼ⟩⟨sₖ| ⊗ |Pⱼ⟩⟨Pₖ|, and the partial trace replaces each |Pⱼ⟩⟨Pₖ| by the number ⟨Pₖ|Pⱼ⟩. So ρ_S = Σⱼₖ cⱼcₖ*⟨Pₖ|Pⱼ⟩|sⱼ⟩⟨sₖ|. On the diagonal the overlap is 1, so the populations |cₖ|² survive any coupling of this form; off the diagonal each coherence is multiplied by an overlap, and the trace stays 1. For |ψ⟩ = √0.7|↑⟩ + √0.3|↓⟩ the coherence √0.21 = 0.458 becomes 0.115 if the pointers overlap by 0.25, and 0 if they are orthogonal, while the populations stay 0.70 and 0.30. In the orthogonal limit ρ_S = Σ|cₖ|²|sₖ⟩⟨sₖ|, and Tr(ρ_S O) agrees with the projection postulate for every system observable O. Applying the Born rule, quoted from Quantum Mechanics I, to the pointer, it reads k with probability ⟨Ψ|(1 ⊗ |Pₖ⟩⟨Pₖ|)|Ψ⟩ = |cₖ|², and a second pointer coupled the same way agrees with the first every time: repeatability, derived.

Stern–Gerlach: the atom's momentum is the pointer

Unit 3's magnet is this coupling. The energy −μ_z B_z gives a force μ_z ∂B_z/∂z, and the electron's moment points against its spin, μ_z = −(gₛ/2)μ_B for ↑. With the gradient pointing down, ∂B_z/∂z = −b with b positive, the transit time τ = L/v gives spin ↑ a kick +Δp and spin ↓ a kick −Δp, with Δp = (gₛ/2)μ_B b τ. The reading states are the incoming packet shifted, φ±(p) = φ(p ∓ Δp). For a minimum-uncertainty Gaussian of momentum spread σₚ, completing the square in ∫φ(p + Δp)φ(p − Δp) dp derives ⟨φ₋|φ₊⟩ = exp(−Δp²/(2σₚ²)): dimensionless, 1 at Δp = 0, even in Δp. Take numbers of the size Stern and Gerlach used: silver at 550 m s⁻¹ through 3.5 cm of a 10³ T m⁻¹ gradient gets Δp = 5.9 × 10⁻²⁵ kg m s⁻¹ and drifts about 0.1 mm sideways inside the magnet, the scale of the 0.2 mm split measured on their 1922 plate. Even a packet only 1 nm wide, σₚ = ħ/(2σ_z) = 5.3 × 10⁻²⁶ kg m s⁻¹, gives Δp/σₚ = 11 and an overlap of e^(−62.8) ≈ 5 × 10⁻²⁸, and the sideways drift only lowers it further. The first-order QED term α/2π = 0.00116 in gₛ/2, quoted, changes none of this.

The overlap that counts is fixed at the magnet

Two facts about ⟨φ₋|φ₊⟩ head off a common misreading. After the magnet both branches drift under the same spin-independent Hamiltonian, so one unitary U_f acts on each and ⟨φ₋|U_f†U_f|φ₊⟩ = ⟨φ₋|φ₊⟩: the spots on the screen move apart, but the coherence neither shrinks further nor recovers in flight. And at the exit of a short magnet the two position packets still lie on top of each other, yet their overlap is already exp(−Δp²/(2σₚ²)), because the pointer states differ in momentum. What enters ρ_S is the Hilbert-space inner product, not whether two spots touch. The same formula prices undoing the measurement. A reversed gradient that cancels each kick to within a residual δ leaves D = exp(−δ²/(2σₚ²)). Counting momentum alone, recovering 90% of the coherence needs δ ≤ 0.459σₚ; for a packet 1 μm wide in the magnet above that is 4 parts in 10⁵ of the kick, and the 0.1 mm drift must be undone too: reversible in principle, hopeless in practice.

What the model does not deliver: one outcome

Every step so far was unitary evolution plus a partial trace, and neither can select a reading. Suppose a coupling left the +x spin in |↑⟩|P↑⟩ on one run and |↓⟩|P↓⟩ on another: linearity has already fixed the output as (|↑⟩|P↑⟩ + |↓⟩|P↓⟩)/√2, the same on every run. Adding a camera or an observer only adds factors, Σ cₖ|sₖ⟩|Pₖ⟩|Cₖ⟩. Nor is ρ_S ignorance of a fixed result, since the joint state is pure and U† restores |+x⟩. The entangled state does not even name the measured observable: with |P±⟩ = (|P↑⟩ ± |P↓⟩)/√2 it equals (|+x⟩|P₊⟩ + |−x⟩|P₋⟩)/√2 just as well. And the probabilities |cₖ|² came from applying the Born rule to the pointer, so the postulate has been moved, not derived. That gap, a unitary theory that writes correlations but never a single result, is the measurement problem, stated here and not solved. Decoherence, the pointer's own entanglement with an environment, is named here only.

Change one variable at a time

Make the relationship visible.

Interactive model

Initial spin angle θ from +z90 °

Kick κ = Δp/σₚ1.2

Keep θ = 90° and drag κ from 0 to 3: the peaks separate, each keeping area 0.50, and the arrow shrinks from the rim to the centre. Then set θ = 30°: the arrow now stops on the z axis at cos θ = 0.87, not at the centre.

P(↑), EVERY KICK0.500

POINTER OVERLAP ⟨φ₋|φ₊⟩0.487

COHERENCE |ρ↑↓|0.243

PURITY Tr ρ_S²0.618

Live interpretationP(↑), EVERY KICK: 0.500. POINTER OVERLAP ⟨φ₋|φ₊⟩: 0.487. COHERENCE |ρ↑↓|: 0.243. PURITY Tr ρ_S²: 0.618

Catch the common trap

Explain before calculating.

Misconception check

The partial trace is not a collapse. After an ideal coupling ρ_S = Σ|cₖ|²|sₖ⟩⟨sₖ| looks exactly like the density operator of systems each secretly in one |sₖ⟩, and it gives identical predictions for every measurement on the system alone, so it is tempting to read it as ignorance of an outcome that has already happened. It is not. The joint state Σ cₖ|sₖ⟩|Pₖ⟩ is still one pure state with no reading singled out. For the +x spin with orthogonal pointer states, the joint observable σₓ ⊗ X_P, with X_P = |P↑⟩⟨P↓| + |P↓⟩⟨P↑|, has mean 1 in the entangled state and 0 in the ignorance mixture, and running U backwards restores |+x⟩, which no ignorance mixture allows. ρ_S is an improper mixture: a statement about what the system alone can show, not about which outcome occurred.

A spin-½ atom in |+x⟩ = (|↑⟩ + |↓⟩)/√2 crosses a weak field gradient (zero field on its path, so no Larmor precession) that leaves its two momentum packets with overlap ⟨φ₋|φ₊⟩ = 0.30. The spin alone is then measured along x by a device that never looks at the momentum. What is P(+x)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-½ is prepared in |ψ⟩ = (√3|↑⟩ + i|↓⟩)/2 and coupled to a pointer by U|↑⟩|R⟩ = |↑⟩|P↑⟩ and U|↓⟩|R⟩ = |↓⟩|P↓⟩, with ⟨P↑|P↓⟩ = 0. Find the spin's density matrix and Bloch vector before and after the coupling, and the entanglement entropy of spin and pointer.

Linearity gives U|ψ⟩|R⟩ = (√3/2)|↑⟩|P↑⟩ + (i/2)|↓⟩|P↓⟩, already in Schmidt form with coefficients √3/2 = 0.866 and 1/2.
Before: ρ = |ψ⟩⟨ψ| has ρ↑↑ = 3/4, ρ↓↓ = 1/4 and ρ↑↓ = c↑c↓* = (√3/2)(−i/2) = −0.433i. From ρ↑↓ = (rₓ − i r_y)/2: rₓ = 0, r_y = 0.866, and r_z = ρ↑↑ − ρ↓↓ = 0.500, so |r| = 1, a pure state.
After: the partial trace multiplies ρ↑↓ by ⟨P↓|P↑⟩ = 0, leaving ρ_S = diag(0.75, 0.25) and r = (0, 0, 0.500). The populations, and so P(↑) = 0.75, have not moved.
Purity ½(1 + |r|²) = ½(1 + 0.25) = 0.625. Entanglement entropy −0.75 log₂ 0.75 − 0.25 log₂ 0.25 = 0.311 + 0.500 = 0.811 bits.
Limit check: had ψ been |↑⟩, the output would be the product |↑⟩|P↑⟩ with entropy 0; an equal superposition would give the maximum, 1 bit.

AnswerBefore: ρ = [[0.75, −0.433i], [0.433i, 0.25]], r = (0, 0.866, 0.500). After: ρ_S = diag(0.75, 0.25), r = (0, 0, 0.500), purity 0.625; entanglement entropy 0.811 bits.

MediumA slow silver atom (m = 1.79 × 10⁻²⁵ kg, magnetic moment of size μ_B) is in |+x⟩ with a minimum-uncertainty transverse packet of width σ_z = 2.0 μm. A field-gradient pulse lasting τ = 1.0 ms, with zero field on the atom's path so that the spin does not precess, kicks the two spin components by ±Δp. What gradient leaves the spin's coherence at half its initial value, and what is P(+x) afterwards? Check that the atom barely moves during the pulse.

Momentum spread of a minimum-uncertainty packet: σₚ = ħ/(2σ_z) = 1.055 × 10⁻³⁴ J s ÷ 4.0 × 10⁻⁶ m = 2.64 × 10⁻²⁹ kg m s⁻¹.
The coherence is ½⟨φ₋|φ₊⟩ = ½ exp(−Δp²/(2σₚ²)). Halving it needs exp(−Δp²/(2σₚ²)) = 0.5, so Δp = σₚ√(2 ln 2) = 1.177σₚ = 3.10 × 10⁻²⁹ kg m s⁻¹.
With Δp = μ_B b τ and b = |∂B_z/∂z|: b = 3.10 × 10⁻²⁹ ÷ (9.274 × 10⁻²⁴ J T⁻¹ × 1.0 × 10⁻³ s) = 3.3 × 10⁻³ T m⁻¹. Units: kg m s⁻¹ ÷ (J T⁻¹ s) = T m⁻¹. Using gₛ/2 = 1.00116 instead of 1 lowers b by 0.12%, an order-α correction quoted from QED and invisible at two figures.
Now rₓ = 0.50, so P(+x) = ½(1 + rₓ) = 0.75, between the 1.00 of no coupling and the 0.50 of a complete measurement.
Impulse check: under a constant force the atom moves ½(Δp/m)τ = ½ × (3.10 × 10⁻²⁹ ÷ 1.79 × 10⁻²⁵) × 1.0 × 10⁻³ = 8.7 × 10⁻⁸ m during the pulse, 4% of σ_z. The extra overlap factor from that shift, exp(−(0.043)²/2) = 0.999, is negligible.

Answerb ≈ 3.3 × 10⁻³ T m⁻¹ (about 3.3 mT m⁻¹), after which P(+x) = 0.75; the atom moves only 0.087 μm during the pulse, so the kick is effectively impulsive.

HardA spin-1 atom is prepared in the Sₓ = +ħ eigenstate, |ψ⟩ = ½|+1⟩ + (1/√2)|0⟩ + ½|−1⟩, and crosses a pure field gradient, with no Larmor precession, that kicks the component m by mΔp, with Δp = 2.0σₚ for a minimum-uncertainty Gaussian packet. Find the spin's reduced density matrix, ⟨Sₓ⟩ and the purity afterwards. Which coherence dies first, and why?

Linearity gives Σ cₘ|m⟩|φₘ⟩ with φₘ(p) = φ(p − mΔp). Two Gaussian packets whose centres are d apart overlap by exp(−d²/(8σₚ²)); the spin-½ result is the case d = 2Δp.
Neighbours (d = Δp = 2σₚ) overlap by e^(−0.5) = 0.607; the outer pair (d = 2Δp = 4σₚ) by e⁽⁻²⁾ = 0.135.
Populations stay ¼, ½, ¼. Coherences: ρ₊₀ = ρ₀₋ = (½)(1/√2)(0.607) = 0.214, and ρ₊₋ = (½)(½)(0.135) = 0.0338, all real.
With Sₓ = (ħ/√2)[[0, 1, 0], [1, 0, 1], [0, 1, 0]] from Quantum Mechanics I, ⟨Sₓ⟩ = √2ħ(ρ₊₀ + ρ₀₋) = √2 × 0.429ħ = 0.607ħ, down from ħ; the outer coherence does not enter.
Purity Σ|ρᵢⱼ|² = (1/16 + 1/4 + 1/16) + 4(0.2144)² + 2(0.0338)² = 0.375 + 0.184 + 0.002 = 0.561. Limits: Δp → 0 gives 1 and ⟨Sₓ⟩ = ħ; Δp → ∞ gives 0.375 and ⟨Sₓ⟩ = 0.
The m = ±1 coherence dies first: those two pointer states are twice as far apart, so its exponent is four times larger. The readings easiest to tell apart lose their coherence soonest.

Answerρ_S = [[0.250, 0.214, 0.034], [0.214, 0.500, 0.214], [0.034, 0.214, 0.250]]; ⟨Sₓ⟩ = 0.607ħ; purity 0.561. The ±1 coherence, at 0.135 of its value against 0.607 for neighbours, goes first.