University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.5

Stern-Gerlach as a dynamical process

Every course shows the two spots; this lesson makes the magnet earn them. Give it a Hamiltonian and watch: a phase written across the packet, an opposite momentum kick for each spin, and a state in which position carries the spin. The payoff: knowing exactly what the Schrödinger equation delivers and where the detector takes over.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Here the Stern-Gerlach magnet stops being a box that sorts spins and becomes a Hamiltonian, so the sorting becomes a calculation. In the gap the field along z is B₀ + αz, so the spin term is H = −γ(B₀ + αz)S_z, and over a pulse of length T the evolution operator multiplies the spin-up part of the packet by e^(iγB₀T/2) e^(iγαTz/2) and the spin-down part by the conjugate phases. The uniform factor is Larmor precession.

The factor linear in z is a translation in momentum, so the two spin components leave with p_z shifted by ±γαħT/2, the impulse the classical force μ_z ∂B_z/∂z delivers in the time T. A spinor a|↑⟩ + b|↓⟩ riding one packet becomes a|↑⟩|φ₊⟩ + b|↓⟩|φ₋⟩, no longer a spin state times a spatial state: position now carries the spin label, and the Born rule for where the atom lands returns |a|² and |b|² with no new postulate. Two simplifications are argued, not assumed.

Dropping p²/2m leaves the kick in a linear field exact, and leaves the coherence result intact while the pulse is too short for the packet to spread; the transverse gradient that ∇⋅B = 0 demands is beaten by a strong B₀, which spins the transverse moment too fast to be pushed. The model stops in three places: the field is classical and the atom non-relativistic, gₑ is quoted from measurement and quantum electrodynamics, and nothing in it says why one detector fires rather than the other. The magnet correlates; it does not decide.

Simple definition: In the dynamical model a Stern-Gerlach magnet is the Hamiltonian −γ(B₀ + αz)S_z acting for a time T; it kicks the spin-up and spin-down parts of one packet by opposite momenta ±γαħT/2, and so ties spin to momentum rather than measuring it.
Example: A silver atom (m = 1.79 × 10⁻²⁵ kg, |μ_z| = 9.28 × 10⁻²⁴ J T⁻¹) that spends 60 μs in a 1000 T m⁻¹ gradient has its two spin parts kicked by ±5.57 × 10⁻²⁵ kg m s⁻¹, so they separate at 2 × 3.11 = 6.22 m s⁻¹.

Spin Hamiltonian inside the magnetH = −γ(B₀ + αz)S_z, for 0 ≤ t ≤ T

The model, postulated: p²/2m dropped for the pulse, and the γαxSₓ term that ∇⋅B = 0 adds argued away, because B₀ turns the transverse spin too fast to be pushed.

γ in s⁻¹ T⁻¹ (−1.761 × 10¹¹ for the electron, quoted), B₀ in T, α = ∂B_z/∂z in T m⁻¹, S_z in J s, H in J.

Phase written across the packetU|↑⟩ψ(r) = e^(iγB₀T/2) e^(iγαTz/2) |↑⟩ψ(r)

Derived from U = exp(−iHT/ħ). The first factor is Larmor precession, uniform over the packet; the second grows with z, and a phase linear in z shifts momentum.

γB₀T and γαTz in rad, so γαT/2 is a wavenumber in m⁻¹. For |↓⟩ both exponents change sign.

Momentum kick on each spin componentp_z → p_z ± Δp, with Δp = γαħT/2

Derived, and exact for a linear field even with p²/2m kept: the Heisenberg equation gives dp_z/dt = γαS_z, a constant. Only the displacement inside the magnet is lost.

Δp in kg m s⁻¹, + for spin up. Since μ_z = ±γħ/2, it equals the classical impulse μ_z(∂B_z/∂z)T.

Spin–momentum correlated state(a|↑⟩ + b|↓⟩)|φ⟩ → a|↑⟩|φ₊⟩ + b|↓⟩|φ₋⟩

Derived by linearity. Not a spin state times a spatial state once ab ≠ 0 and Δp ≠ 0: the first entangled state of the course, taken up again in Unit 4.

|a|² + |b|² = 1, dimensionless; φ±(p) = φ(p ∓ Δp), with the Larmor phases absorbed into a and b.

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

Derived. The transverse spin that survives is sinθ times this factor: at Δp = 3σₚ only 1% is left, and no evolution that treats both spins alike brings it back.

Minimum-uncertainty Gaussian, σₚ = ħ/(2σ_z) in kg m s⁻¹. Equivalently exp(−γ²α²T²σ_z²/2), with σ_z in m. Valid while T ≪ 4mσ_z²/ħ, so the packet does not spread in the magnet.

Splitting at a screenΔz = 2|Δp|D/(mv)

Derived: a constant force across the magnet equals one kick at its midpoint. Two spots appear only if Δz beats the beam's own width there, which the slits set.

Atom mass m in kg, beam speed v in m s⁻¹, D in m measured from the magnet's centre, Δz in m.

The magnet as a Hamiltonian, and what is left out

Start from the field, not from the force μ_z ∂B_z/∂z quoted in Quantum Mechanics I. In the gap take B = (−αx, 0, B₀ + αz): a strong uniform B₀ along z with gradient α = ∂B_z/∂z, plus the transverse gradient ∂Bₓ/∂x = −α that ∇⋅B = 0 makes compulsory, because a B_z that grows along z needs field lines converging from the side. With μ = γS the spin term is −γ(B₀ + αz)S_z + γαxSₓ. The model keeps the first term and drops two things. The kinetic energy p²/2m goes for the pulse of length T, and the next section shows this costs the kick nothing. The Sₓ term goes because B₀ dominates it. At distance x from the axis it tilts the field by about αx/B₀, and a spin that starts along z and precesses about the tilted axis flips with probability at most (αx/B₀)², the zero-frequency case of the Rabi formula. For B₀ = 1.0 T, α = 1000 T m⁻¹ and x = 50 μm that is 2.5 × 10⁻³. The x-force −γαSₓ fails for the same reason: Sₓ precesses at |γ|B₀ = 1.8 × 10¹¹ s⁻¹, so over a 60 μs pulse its push averages to about 10⁻⁷ of the z-kick.

A phase linear in z is a momentum kick

With the model Hamiltonian, U = exp(−iHT/ħ) = exp[iγ(B₀ + αz)TS_z/ħ], and on a spin-up packet S_z is simply ħ/2: U|↑⟩ψ(r) = e^(iγB₀T/2) e^(iγαTz/2)|↑⟩ψ(r). The first factor is the same everywhere; it is the Larmor precession of the last two topics. The second is new. Multiplying a wavefunction by e^(ik₀z) shifts its momentum-space amplitude, ψ̃(p_z) → ψ̃(p_z − ħk₀), so every momentum rises by ħk₀. Spin up therefore leaves with p_z raised by Δp = γαħT/2, and spin down, whose phase has the opposite sign, with p_z lowered by the same amount. Dimensions: s⁻¹ T⁻¹ × T m⁻¹ × J s × s = kg m s⁻¹. Classical limit: μ_z = γħ/2 for spin up, so Δp is the impulse μ_z(∂B_z/∂z)T. And the pulse need not be short. Keep p²/2m and the Heisenberg equation gives dp_z/dt = (i/ħ)[H, p_z] = γαS_z, which commutes with H, so p_z(T) = p_z(0) ± Δp exactly. All that dropping p²/2m loses is the displacement inside the magnet, and one kick at the magnet's centre restores it.

Spin now labels the packet

Linearity does the rest. An atom arriving as (a|↑⟩ + b|↓⟩)|φ⟩ leaves as a|↑⟩|φ₊⟩ + b|↓⟩|φ₋⟩, where φ₊ and φ₋ are the incoming packet displaced in momentum by +Δp and −Δp and the Larmor phases sit in a and b. Try to write this as one spin state times one spatial state, |χ⟩|η⟩. Projecting onto ⟨↑| would give a|φ₊⟩ and onto ⟨↓| would give b|φ₋⟩, and both would have to be multiples of the same |η⟩, which fails once ab ≠ 0 and Δp ≠ 0. The state is entangled, the first of the course, and Unit 4 builds the tensor-product and density-operator tools that say how much. Now ask where the atom lands. Summed over spin, the position density is |a|²|φ₊(z)|² + |b|²|φ₋(z)|² with no cross term, since ⟨↑|↓⟩ = 0: the lobes never show fringes, even where they overlap, because the spin already records which lobe is which. Once the lobes are apart, the Born rule for position gives |a|² for the spin-up lobe. The spin probabilities have come out of dynamics plus the position postulate of Quantum Mechanics I.

Where the transverse spin goes

Take a = cos(θ/2), real, and b = sin(θ/2) times a phase: a spin tilted θ from z, whose transverse Bloch component has length sinθ. After the magnet ⟨σₓ⟩ = 2a Re[b⟨φ₊|φ₋⟩], so that length is multiplied by |⟨φ₊|φ₋⟩|, the size of the packet overlap ⟨φ₊|φ₋⟩ = ∫|φ(z)|² e^(−iγαTz) dz. For a minimum-uncertainty Gaussian of spread σ_z its size is exp(−γ²α²T²σ_z²/2), the same number as exp(−Δp²/(2σₚ²)) with σₚ = ħ/(2σ_z). That needs T ≪ 4mσ_z²/ħ, so that the packet does not spread inside the magnet: one that does also separates there in position, and the overlap falls faster still. Read it two ways. As precession: the spin at height z turns at |γ|(B₀ + αz), so the precession angles across the packet fan out by |γ|ασ_zT, and a fan wider than about a radian averages to nothing. As momentum: the kick has outrun the momentum spread, and any later evolution that treats both spins alike conserves the overlap, so the loss is permanent unless something spin-dependent acts again. The scale is brutal. A silver atom with σ_z = 1 μm, for which 4mσ_z²/ħ = 6.8 ms, in 1000 T m⁻¹ for 60 μs has |γ|ασ_zT ≈ 1.1 × 10⁴, and the factor is exp(−5.6 × 10⁷). The spin has not been randomised; it has been entangled.

From kicks to spots on a screen

After the magnet the atom flies freely and each lobe keeps its momentum. Because a constant force acting across the magnet is equivalent to one kick at its centre, a screen a distance D beyond that centre finds the lobes displaced by ±|Δp|D/(mv), a splitting Δz = 2|Δp|D/(mv). Which lobe is spin up depends on the sign of γ. The electron's moment points against its spin, so γ is negative, and in a silver atom with α positive spin up is pushed towards −z. The spots are resolved only if Δz beats the width of the beam at the screen, and that width is set by the collimating slits, tens of micrometres, not by the quantum packet. For silver, m = 1.79 × 10⁻²⁵ kg at 550 m s⁻¹ through a 3.5 cm magnet with α = 1000 T m⁻¹: T = 64 μs, |Δp| = 5.9 × 10⁻²⁵ kg m s⁻¹, and at the magnet's exit, D = 1.75 cm, the lobes are 0.21 mm apart. Gerlach and Stern's 1922 deposit showed two lobes separated by at most about 0.2 mm, and fixed the silver moment at one Bohr magneton to within about 10 per cent.

Derived, quoted, and not derived

Derived here: the kick, the entangled state, the loss of transverse spin, the splitting and the weights |a|² and |b|². Quoted: the moment. γ = −gₑ e/(2mₑ) takes gₑ = 2.002319304 from measurement; the Dirac equation gives gₑ = 2, and quantum electrodynamics raises it by the fraction α_fs/2π = 1.16 × 10⁻³ at first order in the fine-structure constant α_fs ≈ 1/137, not to be confused with the gradient α. That correction moves the splitting by 0.12 per cent, far below what a beam experiment resolves; Penning-trap measurements pin gₑ to about a part in 10¹³. Not derived at all: which lobe a detector registers. The magnet has only correlated spin with position, unitarily, so a second magnet with the gradient reversed could in principle bring φ₊ and φ₋ back into one packet and restore the transverse spin. Atom-chip experiments have done this, but only by recombining the two paths with great precision, which no laboratory magnet does by accident. Turning the correlation into a record is the detector's job, and the Schrödinger equation alone does not say which lobe it reports.

Change one variable at a time

Make the relationship visible.

Interactive model

Spin tilt θ from +z60 °

Kick Δp in units of σₚ1.5 σₚ

Set θ = 90° and raise the kick from zero: the after bar falls to half near 1.2σₚ while the two lobes still merge into one barely dimpled hump, so most of the spin coherence is gone before the spots are resolved. At θ = 0° there is one lobe and nothing to lose.

SPIN-UP WEIGHT |a|²0.750

SPIN-DOWN WEIGHT |b|²0.250

PACKET OVERLAP0.325

TRANSVERSE SPIN LEFT0.281 ħ/2

Live interpretationSPIN-UP WEIGHT |a|²: 0.750. SPIN-DOWN WEIGHT |b|²: 0.250. PACKET OVERLAP: 0.325. TRANSVERSE SPIN LEFT: 0.281 ħ/2

Catch the common trap

Explain before calculating.

Misconception check

The tempting argument: the force is μ_z ∂B_z/∂z, an atom with its spin along x has ⟨S_z⟩ = 0, so it feels no force and flies straight to the middle of the screen. Ehrenfest's theorem seems to agree, since the packet's centroid does follow ⟨F_z⟩ = γα⟨S_z⟩ = 0. The mean is right; the picture is wrong. The force is the operator γαS_z, with eigenvalues ±γαħ/2, and the dynamics sends each eigencomponent its own way, so an x-polarised atom arrives as two lobes of weight ½ each and essentially nothing lands in the middle. The same error predicts a continuous smear of deflections for an unpolarised beam, which classical moments with random orientations would give and the 1922 deposit did not show. Expectation values locate the centroid of a distribution; they say nothing about its shape.

A silver atom with its spin along +x crosses a Stern-Gerlach magnet whose kick Δp is 30 times the packet's momentum spread σₚ. Before any detector acts, which description of the atom is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium atoms (m = 3.82 × 10⁻²⁶ kg) moving at 800 m s⁻¹ cross a 5.0 cm magnet with gradient 400 T m⁻¹. Treat the atom's moment as the electron's, |μ_z| = 9.28 × 10⁻²⁴ J T⁻¹. Find the time spent in the gradient, the momentum kick on each spin component, and the rate at which the two components separate.

Time in the gradient: T = 0.050 m ÷ 800 m s⁻¹ = 6.25 × 10⁻⁵ s.
The phases e^{(±iγαTz/2)} shift p_z by ±Δp with Δp = |γ|αħT/2, and |γ|ħ/2 = |μ_z|, so Δp = |μ_z|αT = 9.28 × 10⁻²⁴ × 400 × 6.25 × 10⁻⁵ = 2.32 × 10⁻²⁵ kg m s⁻¹.
Units: J T⁻¹ × T m⁻¹ × s = J s m⁻¹ = kg m s⁻¹, a momentum, as it must be.
Each component gains a transverse velocity ±Δp/m = ±2.32 × 10⁻²⁵ ÷ 3.82 × 10⁻²⁶ = ±6.07 m s⁻¹, so they separate at 2Δp/m = 12.1 m s⁻¹, about 1.5% of the forward speed.

AnswerT = 62.5 μs; Δp = ±2.32 × 10⁻²⁵ kg m s⁻¹; the components separate at 12.1 m s⁻¹.

MediumA potassium atom (m = 6.47 × 10⁻²⁶ kg, |μ_z| = 9.28 × 10⁻²⁴ J T⁻¹, moment antiparallel to the electron spin) moves at 600 m s⁻¹ with its spin tilted 60° from +z. It crosses a 4.0 cm magnet with ∂B_z/∂z = +200 T m⁻¹, then flies 30 cm to a screen. Where do the two lobes land relative to the undeflected line, with what probabilities, and where is ⟨z⟩?

T = 0.040 m ÷ 600 m s⁻¹ = 6.667 × 10⁻⁵ s, so Δp = |μ_z|αT = 9.28 × 10⁻²⁴ × 200 × 6.667 × 10⁻⁵ = 1.237 × 10⁻²⁵ kg m s⁻¹, a transverse velocity Δp/m = 1.91 m s⁻¹.
Sign: γ is negative, so spin up has μ_z = −9.28 × 10⁻²⁴ J T⁻¹ and the force μ_z ∂B_z/∂z is negative. Spin up goes towards −z, spin down towards +z.
A constant force across the magnet equals one kick at its centre, so the lever arm is D = 0.020 + 0.30 = 0.32 m, flown in 0.32 ÷ 600 = 5.33 × 10⁻⁴ s. Each lobe moves 1.91 × 5.33 × 10⁻⁴ = 1.02 × 10⁻³ m.
Weights: a = cos 30° and b = sin 30°, so |a|² = 0.75 for the spin-up lobe at −1.02 mm and |b|² = 0.25 for the spin-down lobe at +1.02 mm, 2.04 mm apart.
Ehrenfest: ⟨z⟩ = 0.75 × (−1.02 mm) + 0.25 × (+1.02 mm) = −0.51 mm, halfway between the undeflected line and the spin-up lobe. For a beam narrower than the splitting, no atom lands there.

AnswerSpin-up lobe at −1.02 mm with probability 0.75, spin-down lobe at +1.02 mm with probability 0.25; ⟨z⟩ = −0.51 mm, a point where essentially no atom arrives.

HardA cold silver atom (m = 1.79 × 10⁻²⁵ kg) in a minimum-uncertainty packet with σ_z = 1.0 μm, spin along +x, is exposed to a pulsed gradient α for T = 1.0 μs. Take |γ| = 1.761 × 10¹¹ s⁻¹ T⁻¹ and ħ = 1.055 × 10⁻³⁴ J s. Check that the packet does not spread during the pulse, find the α that leaves half the transverse spin, say whether the two momentum lobes are then resolved, and repeat for a 1000 T m⁻¹ gradient.

Validity: 4mσ_z²/ħ = 4 × 1.79 × 10⁻²⁵ × (1.0 × 10⁻⁶)² ÷ 1.055 × 10⁻³⁴ = 6.8 × 10⁻³ s, nearly 7000 times T, so the packet does not spread during the pulse and dropping p²/2m leaves the overlap formula intact.
With a = b = 1/√2 the transverse spin's length after the magnet is |⟨φ₊|φ₋⟩| = exp(−(|γ|αTσ_z)²/2). It equals ½ when |γ|αTσ_z = √(2 ln 2) = 1.1774.
|γ|Tσ_z = 1.761 × 10¹¹ × 1.0 × 10⁻⁶ × 1.0 × 10⁻⁶ = 0.1761 m T⁻¹, so α = 1.1774 ÷ 0.1761 m T⁻¹ = 6.69 T m⁻¹.
Momentum picture: σₚ = ħ/(2σ_z) = 1.055 × 10⁻³⁴ ÷ 2.0 × 10⁻⁶ = 5.275 × 10⁻²⁹ kg m s⁻¹ and Δp = |γ|αħT/2 = 6.21 × 10⁻²⁹ kg m s⁻¹, so Δp/σₚ = 1.18, the same number, since Δp/σₚ = |γ|αTσ_z identically.
Resolution: with equal weights the momentum density is ½[G(p − Δp) + G(p + Δp)], G a unit-peak Gaussian of width σₚ. At p = 0 it is exp(−1.1774²/2) = 0.50, and its maxima, near ±0.95σₚ, reach only 0.54: a 7% dip, so the lobes are not resolved.
At 1000 T m⁻¹, |γ|αTσ_z = 176.1: lobes centred at ±176σₚ, cleanly apart, and a transverse spin of exp(−176.1²/2) = e^{(−1.55 × 10⁴)} ≈ 10⁻⁶⁷³⁴. Both answers are set by one number, so keeping the coherence and resolving the lobes exclude each other.

AnswerWith 4mσ_z²/ħ = 6.8 ms against a 1.0 μs pulse the packet does not spread; α = 6.69 T m⁻¹ keeps half the transverse spin, but the lobes then show only a 7% dip; at 1000 T m⁻¹ they sit at ±176σₚ and the transverse spin is about 10⁻⁶⁷³⁴.