University Physics V · Quantum Spin · 10.5

Stern–Gerlach as an Entangling Device

The magnet does not measure anything. It runs a Hamiltonian that ties the spin to where the atom is heading, and the split beam on the screen is one entangled state, not two possibilities. Learn to compute the coherence that survives and you learn when a measurement has actually happened.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put a neutral atom carrying a magnetic moment into a field that varies in space and the coupling term −γB(z)Ŝ_z is an operator on both factors of the Hilbert space at once: it multiplies each S_z eigenvalue by its own function of position. Because [Ŝ_z, H] = 0 in the idealised field the spin never flips; instead each branch acquires its own force γħmₛ ∂B_z/∂z, and an initial product state (α|↑⟩ + β|↓⟩)⊗|χ₀⟩ runs unitarily into α|↑⟩|χ₊⟩ + β|↓⟩|χ₋⟩. That state is entangled the moment the two packets differ, and nothing in it has collapsed: the evolution is reversible and the global state stays pure.

What the experiment calls a measurement is the death of the overlap ⟨χ₋|χ₊⟩, which is precisely the factor multiplying the off-diagonal elements of the reduced spin density matrix. While that overlap is one the spin is still coherently polarised in the x–y plane; when it vanishes the reduced state is diag(|α|², |β|²) and the interference is gone — not destroyed, but exported into correlations with a variable nobody is tracking. Two branches can be orthogonal in momentum long before they are visibly apart in position, so the record is usually written well before the beam looks split.

The model's cost is the field itself: ∇⋅B = 0 forbids a purely axial gradient, so a transverse component always rides along and only fast Larmor precession about a strong holding field averages it away. And the pattern on the screen reports two independent numbers — the deflection fixes |μ_z|, while the number of spots fixes 2s + 1.

Simple definition: A Stern-Gerlach magnet is a unitary device rather than a detector: its inhomogeneous field gives each S_z eigenvalue its own force, so it correlates — entangles — the spin with the atom's transverse motion.
Example: Silver at 500 m s⁻¹ crossing 4.0 cm of ∂B_z/∂z = 800 T m⁻¹: each branch feels μ_B b = 7.4 × 10⁻²¹ N, so the packets leave 0.27 mm apart and land 3.6 mm apart on a screen 25 cm further on. Whether that counts as a measurement is not read off the picture but off |⟨χ₋|χ₊⟩| — and for a σ = 100 nm packet the momentum split alone has already cut that to 1/e after 50 μm of flight, while the centres are still 0.42 nm apart.

The magnet's coupling HamiltonianH = p̂²/2m − γ B(z) Ŝ_z, with [Ŝ_z, H] = 0

One operator acts on both factors, so H is block diagonal in (|↑⟩, |↓⟩): two ordinary one-dimensional problems, one per eigenvalue.

γ in rad s⁻¹ T⁻¹ (γ = −gₛ e/2mₑ < 0 for an electron); B in T; Ŝ_z in J s

Spin-dependent forceU_{mₛ} = −γħ mₛ B(z) → F_z = γħ mₛ ∂B_z/∂z

A uniform field only precesses the spin. The gradient is what converts a spin amplitude into transverse momentum.

mₛ = ±½; |γħ/2| ≈ μ_B for silver, so |F| = μ_B b with b in T m⁻¹ and F in N. γ < 0 sends mₛ = +½ toward weaker field.

The entangled output(α|↑⟩+β|↓⟩)|χ₀⟩ → α|↑⟩|χ₊⟩ + β|↓⟩|χ₋⟩

Entangled whenever αβ ≠ 0 and |⟨χ₋|χ₊⟩| < 1. Schmidt rank two, and the total state is still pure.

|χ_±⟩ are normalised packets displaced by ±½at² and boosted by ±Ft; α, β dimensionless

Reduced spin state and its purityρₛ = [[|α|², αβ*c], [α*βc*, |β|²]], c = ⟨χ₋|χ₊⟩

One number multiplies every off-diagonal element: c is the dial running from pure superposition to plain mixture.

c dimensionless with |c| ≤ 1; Tr ρₛ² = 1 − 2|α|²|β|²(1 − |c|²)

Gaussian overlap of the two branches|c| = exp(−Δz²/8σ² − Δp²σ²/2ħ²)

Two exponents, and the momentum one usually wins first: Δp = √2 ħ/σ alone already costs a factor of e, and Δz = 7.5σ alone drives |c| to 8.8 × 10⁻⁴.

Δz in m and Δp in kg m s⁻¹ are the branch separations; σ is the coherent packet width in m

Maxwell's constraint and adiabatic followingB = (B₀ + bz)ẑ − b x x̂ω_L = |γ|B₀ ≫ v/L

No legal field has a pure z gradient. Only fast precession about the local B averages the transverse piece away, leaving U = 2μ_B mₛ|B| — so mₛ = +½ is the low-field seeker.

∇⋅B = b − b = 0 and ∇×B = 0; ω_L in rad s⁻¹, B₀ the holding field in T, L the pole length in m

Write the Hamiltonian before you say measurement

The atom is neutral, so no Lorentz force acts; the only coupling is between its magnetic moment and the field. The state space is L²(ℝ³) ⊗ ℂ², and H = p̂²/2m ⊗ I − γ B(z) ⊗ Ŝ_z is an operator on both factors at once. Because B points along z in the idealised model, [Ŝ_z, H] = 0: the spin populations are constants of the motion, and H is block diagonal in the basis (|↑⟩, |↓⟩) that diagonalises Ŝ_z. Each block is an ordinary one-dimensional problem, p̂²/2m − γħmₛ(B₀ + bz) — a linear potential, which is free fall with effective acceleration γħmₛ b/m. The boundary condition is only normalisability on the line: no walls, no quantised levels, just a packet accelerating. For silver, with |γħ/2| = μ_B and b = 800 T m⁻¹, that acceleration is 4.14 × 10⁴ m s⁻², about 4200 g, and because γ < 0 it points toward −z for mₛ = +½ and toward +z for mₛ = −½.

Unitary evolution correlates; it does not collapse

Start from a product state, (α|↑⟩ + β|↓⟩) ⊗ |χ₀⟩. Each spin branch is dragged by its own constant force, so after time t the state is α|↑⟩⊗|χ₊(t)⟩ + β|↓⟩⊗|χ₋(t)⟩, where |χ_±⟩ are the same packet displaced by ±½at² and boosted by ±Ft. This is entangled — Schmidt rank two — as soon as αβ ≠ 0 and the two packets differ. Nothing has been measured. The evolution operator is unitary, the global state is still pure, and running the gradient backwards returns the original product state exactly. Structurally the magnet is a controlled-displacement gate: the spin is the control, the centre of mass is the target, and the same circuit drawn with two qubits is how a CNOT copies a bit without anybody reading it. That is the entire content of the beam splitting: a correlation was created, not an outcome.

The overlap is the meter

Nobody watches the atom's centre of mass, so the spin's state is the partial trace: ρₛ carries |α|² and |β|² on the diagonal, unchanged for ever, and αβ*⟨χ₋|χ₊⟩ off it. Every coherence is multiplied by that one number c. The purity Tr ρₛ² = 1 − 2|α|²|β|²(1 − |c|²) runs from 1 when the packets coincide down to ½ for an equal superposition when they are orthogonal. For equal-width Gaussians separated in position only, |c| = exp(−Δz²/8σ²): 0.88 at one width, 0.61 at two, 0.32 at three, and 8.8 × 10⁻⁴ by seven and a half. Halfway, at |c| = 0.5 and still with equal amplitudes, the reduced state has eigenvalues 0.75 and 0.25 and an entanglement entropy of 0.81 bits — a partial record, which is exactly what a partly completed measurement should mean. Note that σ is the coherent packet width, not the collimated beam width: a slit-limited beam is an incoherent ensemble of far narrower packets.

The record is written in momentum first

The two branches differ in mean momentum as well as position, and for a minimum-uncertainty Gaussian both terms count: |c| = exp(−Δz²/8σ² − Δp²σ²/2ħ²). Momentum is the faster channel, and its width σₚ = ħ/2σ is fixed at the source and never changes under free flight or a uniform force. A thermal silver beam has a transverse coherence width near σ = 100 nm, giving σₚ = 5.3 × 10⁻²⁸ kg m s⁻¹, while Δp = 2Ft grows at 1.48 × 10⁻²⁰ kg m s⁻¹ per second of flight. The momentum exponent reaches 1 after t = 0.10 μs — 50 μm into a 40 mm magnet — while the centres have moved apart by only 0.42 nm, four thousandths of a packet width. Coherence is down to 1/e there and, since the exponent grows as t², to 10⁻³ by 132 μm. The which-path information exists long before there is anything to see. That is why the reversible Stern-Gerlach of Wigner, and of Scully, Englert and Schwinger, is so brutally hard: undoing the split means matching a momentum kick to better than ħ/σ.

There is no such thing as a pure z gradient

B = bz ẑ has ∇⋅B = b, so it is not a magnetic field at all. The simplest legal first-order model near the pole gap is B = (B₀ + bz)ẑ − b x x̂, whose divergence is b − b = 0 and whose curl vanishes in the empty gap. The transverse piece tilts the field off the z axis by bx/B₀, and what rescues the simple picture is speed: the spin precesses about the local field at ω_L = |γ|B₀ = 1.8 × 10¹¹ rad s⁻¹ for B₀ = 1.0 T, while the direction seen by the atom turns at only v/L ≈ 1.3 × 10⁴ s⁻¹. With a ratio of 1.4 × 10⁷ the spin follows adiabatically, the energy is U = 2μ_B mₛ|B| ≈ 2μ_B mₛ(B₀ + bz + b²x²/2B₀), and the residual transverse force is bx/B₀ = 8% of the axial one at 0.10 mm off axis — inward for the low-field seeker mₛ = +½, outward for mₛ = −½. Remove the holding field and the argument dies: near a zero of |B| the spin cannot follow, and Majorana flips scramble the branches.

The deflection and the spot count fix different things

Integrating the constant force through the pole gap and the drift gives z = (μ_z b/mv²)(L²/2 + LD) — 1.79 mm for the silver numbers above — so the position of a spot fixes |μ_z| and nothing else. For silver in 1922 that came out at one Bohr magneton. It does not give the spin, because |μ_z| = gₛ μ_B mₛ is one equation in two unknowns. The second equation is the count: the beam divides into 2s + 1 spots, and two spots force s = ½, hence mₛ = ±½ and gₛ = 2. Stern and Gerlach read their two spots as space quantisation of an l = 1 orbital moment, which would in fact have produced three spots including an undeflected one; spin was not proposed until 1925. Notice too that z carries 1/v², so a thermal velocity spread smears each spot along the deflection axis — that width is kinematics, not quantum uncertainty.

Change one variable at a time

Make the relationship visible.

Interactive model

Packet separation d1.5 σ

Input tilt θ from +z90 °

Input phase φ0 °

Hold θ = 90° and push d from 0 to 2σ: the pair still reads as one broad hump, yet the overlap has already fallen to 0.61 and both spin bars with it. Now turn φ — the bars trade length while the density below never moves, which is why the relative phase is invisible in the split pattern.

OVERLAP ⟨χ₋|χ₊⟩0.755

⟨Sx⟩ ÷ (ħ/2)0.755

⟨Sy⟩ ÷ (ħ/2)0.000

PURITY Tr ρ²0.785

Live interpretationOVERLAP ⟨χ₋|χ₊⟩: 0.755. ⟨Sx⟩ ÷ (ħ/2): 0.755. ⟨Sy⟩ ÷ (ħ/2): 0.000. PURITY Tr ρ²: 0.785

Catch the common trap

Explain before calculating.

Misconception check

The standard error is that the magnet measures S_z the instant the field acts. It cannot: [Ŝ_z, H] = 0, so the populations |α|² and |β|² are constants of the motion, and the evolution is unitary and reversible. What the field changes is coherence, and only through the single factor ⟨χ₋|χ₊⟩ multiplying the off-diagonal elements of ρₛ. While the packets overlap in both position and momentum the spin state is still almost pure, and reversing the gradient restores the original superposition. The second half of the error is reading the split as the atom having gone one way. Even at zero overlap the global state is one pure entangled vector; ρₛ is an improper mixture, and branch language becomes safe only once the record has been amplified into something irreversible. Decoherence relocates the phase — it does not delete it.

In the silver beam above (σ = 100 nm, F = 7.42 × 10⁻²¹ N per branch) an atom prepared in |+x⟩ has flown 50 μm into the magnet. The branch centres are 0.42 nm apart — 0.0042σ — and their mean momenta now differ by Δp = 1.49 × 10⁻²⁷ kg m s⁻¹, which is exactly √2 ħ/σ. What is the length of the spin's transverse Bloch vector √(⟨σₓ⟩² + ⟨σ_y⟩²), and hence the largest fraction any transverse analyser can pass?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA silver atom (m = 1.79 × 10⁻²⁵ kg, |μ_z| = μ_B = 9.274 × 10⁻²⁴ J T⁻¹) crosses a 4.0 cm magnet with ∂B_z/∂z = 800 T m⁻¹ at 500 m s⁻¹, then drifts 25 cm to a screen. Find the force on each branch, the separation of the two packets at the magnet exit, and their separation at the screen.

Each S_z eigenvalue gets its own linear potential, U_{mₛ} = −γħmₛ B(z), so F_z = −dU/dz = γħmₛ b. With |γħ/2| = μ_B the two branches feel equal and opposite forces of size F = μ_B b = 9.274 × 10⁻²⁴ × 800 = 7.42 × 10⁻²¹ N, the mₛ = +½ branch pushed toward −z because γ < 0.
Transverse acceleration a = F/m = 7.42 × 10⁻²¹ / 1.79 × 10⁻²⁵ = 4.14 × 10⁴ m s⁻², about 4200 g. The longitudinal motion is untouched, so the time in the field is t₁ = L/v = 0.040/500 = 8.0 × 10⁻⁵ s.
Each centre moves ½at₁² = 0.5 × 4.14 × 10⁴ × (8.0 × 10⁻⁵)² = 1.33 × 10⁻⁴ m, so at the exit the separation is Δz = 2 × 0.133 mm = 0.27 mm.
The exit transverse velocity is at₁ = 3.31 m s⁻¹. Over the drift, t₂ = 0.25/500 = 5.0 × 10⁻⁴ s, each centre adds 3.31 × 5.0 × 10⁻⁴ = 1.66 mm, giving 1.79 mm per branch and Δz = 3.6 mm at the screen.

AnswerF = 7.4 × 10⁻²¹ N on each branch; Δz = 0.27 mm at the magnet exit and 3.6 mm at the screen. Only the gradient does this — a uniform field would give torque and precession, never deflection.

MediumThe same beam now enters in |+x⟩ = (|↑⟩ + |↓⟩)/√2, with a transverse coherent packet of minimum-uncertainty width σ = 100 nm. Write the state, give the reduced spin matrix, and find how far into the magnet an atom must fly for the branch overlap |c| to fall to 1/e, and then to 10⁻³.

The magnet is a controlled displacement: |+x⟩|χ₀⟩ evolves unitarily into |Ψ(t)⟩ = (|↑⟩|χ₊⟩ + |↓⟩|χ₋⟩)/√2, the two branches being the same packet pushed by ±F = ±7.42 × 10⁻²¹ N.
Tracing out the centre of mass gives ρₛ = ½[[1, c], [c*, 1]] with c = ⟨χ₋|χ₊⟩. The populations stay ½ and ½ for ever; only the transverse components ⟨Sₓ⟩ = (ħ/2)Re c and ⟨S_y⟩ = (ħ/2)Im c can move.
The source fixes σₚ = ħ/2σ = 1.055 × 10⁻³⁴/(2 × 10⁻⁷) = 5.3 × 10⁻²⁸ kg m s⁻¹, and neither free flight nor a uniform force changes it. The branches separate in momentum at Δp = 2Ft = 1.48 × 10⁻²⁰ t.
With |c| = exp(−Δz²/8σ² − Δp²σ²/2ħ²), the momentum exponent reaches 1 when Δp = √2 ħ/σ = 1.49 × 10⁻²⁷ kg m s⁻¹, that is at t = 1.49 × 10⁻²⁷ / 1.48 × 10⁻²⁰ = 1.0 × 10⁻⁷ s — just 50 μm of flight at 500 m s⁻¹.
At that instant Δz = at² = 4.14 × 10⁴ × (1.0 × 10⁻⁷)² = 4.2 × 10⁻¹⁰ m = 0.0042σ, so the position exponent is only 2.2 × 10⁻⁶ and |c| = e⁻¹ = 0.368 to five figures. The ballistic spreading term σₚ t/m is 2.96 × 10⁻¹⁰ m, so the width √(σ² + (σₚ t/m)²) is still 100.000 nm: the packet has not spread at all, it has only been kicked.
The exponent grows as t², so |c| = 10⁻³ needs exponent ln 1000 = 6.91, that is t = 1.0 × 10⁻⁷ × √6.91 = 2.6 × 10⁻⁷ s, or 132 μm of flight. By the magnet exit at 40 mm the centres are 0.27 mm apart, the packet has spread to 0.26 μm — about a thousand of those widths — and Δp is 2.3 × 10³ σₚ, an exponent above 10⁶: |c| is zero to any precision worth writing down.

Answer|Ψ⟩ = (|↑⟩|χ₊⟩ + |↓⟩|χ₋⟩)/√2 with ρₛ = ½[[1, c], [c*, 1]]. The overlap falls to 1/e after 50 μm and to 10⁻³ after 132 μm, and momentum does it: at 50 μm the centres are still only 0.42 nm apart, 0.4% of a packet width.

HardModel the pole-gap field as B = (B₀ + bz)ẑ − b x x̂ with B₀ = 1.0 T and b = 800 T m⁻¹, for the same silver beam (v = 500 m s⁻¹, L = 4.0 cm, D = 25 cm). Show the model is forced by Maxwell, find the adiabatic potential and the transverse force 0.10 mm off axis, check that the spin can follow the field, and say what the deflection and the number of spots each fix.

∇⋅B = ∂B_z/∂z + ∂Bₓ/∂x = b + (−b) = 0. The often-quoted B = bz ẑ has ∇⋅B = b ≠ 0 and is not a field at all, so a transverse component is compulsory; the −bx x̂ choice is the one that also keeps ∇×B = 0 in the empty gap.
If the spin tracks the local direction, the energy is U = −γħmₛ|B| = 2μ_B mₛ|B| with |B| = √((B₀+bz)² + b²x²) ≈ B₀ + bz + b²x²/2B₀. Differentiating: F_z = −2μ_B mₛ b, the same size as before, plus a new Fₓ = −2μ_B mₛ b²x/B₀.
At x = 0.10 mm, |Fₓ/F_z| = bx/B₀ = 800 × 1.0 × 10⁻⁴ / 1.0 = 0.080. The low-field seeker mₛ = +½ (U = +μ_B|B|) is pulled back toward the axis and mₛ = −½ is pushed out: one branch focused, the other defocused, at 8% of the splitting force.
Adiabatic following needs the Larmor rate to beat the rate at which B turns in the atom's frame: ω_L = |γ|B₀ = gₛ μ_B B₀/ħ = 2 × 9.274 × 10⁻²⁴ / 1.055 × 10⁻³⁴ = 1.8 × 10¹¹ rad s⁻¹, against v/L = 500/0.040 = 1.3 × 10⁴ s⁻¹ — a ratio of 1.4 × 10⁷.
That margin exists only because B₀ = 1.0 T dwarfs b × (beam radius) = 800 × 10⁻⁴ = 0.08 T. Let B₀ → 0 and |B| has a zero inside the beam, where ω_L → 0 and Majorana flips mix the branches. Every real magnet therefore carries a holding field.
The spot position z = (μ_z b/mv²)(L²/2 + LD) = 0.166 × 0.0108 = 1.79 mm fixes |μ_z| = 1 μ_B and nothing more. Two spots rather than three give 2s + 1 = 2, hence s = ½ and |mₛ| = ½; only then does gₛ = |μ_z|/(μ_B|mₛ|) = 2 follow.

AnswerMaxwell forces the −bx x̂ term; |Fₓ/F_z| = bx/B₀ = 8% at 0.10 mm, focusing the mₛ = +½ low-field seeker; ω_L/(v/L) ≈ 1.4 × 10⁷ so the spin follows adiabatically; and s = ½ comes from the two-spot count, never from the 1.79 mm deflection.