University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.2

Rotating a spin-half, and the sign change under 2π

Here the rotation operator stops being a formal exponential and becomes a 2 × 2 matrix you can write down. Use it to turn any spinor, to see Larmor precession as a rotation in time, and to meet the one thing no classical arrow does: after a full turn a spin-½ comes back as minus itself, and a neutron interferometer can tell.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 5.1 wrote a rotation of any state as exp(−i n̂⋅J φ/ħ). For spin ½, J = (ħ/2)σ, and one fact from Quantum Mechanics I makes the exponential summable: (σ⋅n̂)² = I. Even powers collapse to I, odd powers to σ⋅n̂, and the series becomes cos(φ/2) I − i sin(φ/2) σ⋅n̂, a unitary matrix of determinant one built from half angles.

The half angles are not a flaw in the description, because nothing measurable sees D alone. Probabilities and expectation values contain D and D† together, the half angles pair into whole ones, and ⟨S⟩ turns through exactly φ, as a classical vector would; the same matrix turns |+z⟩ into the eigenspinor along any axis, and Larmor precession turns out to be this rotation through the angle ω_L t. The half angles show in one place only: at φ = 2π the operator is −I, so every rotation is represented by two matrices, ±D, whose sign cancels from any measurement on one beam.

It survives as a relative phase. Split a neutron into two paths, precess the spin in one, recombine, and the fringes repeat only every 4π of precession, which the 1975 interferometer experiments saw and this lesson quotes. The model stops at its own assumptions: the sign belongs to half-integer j alone, it means something only against a reference that did not rotate, and because a field rotates each neutron by an angle set by its speed, a spread of speeds washes the fringes out at large φ.

Simple definition: The spin-½ rotation operator D(n̂, φ) = cos(φ/2) I − i sin(φ/2) σ⋅n̂ is the 2 × 2 unitary matrix that turns a spinor through the angle φ about the unit axis n̂; it rotates ⟨S⟩ through φ, yet equals −I, not I, at φ = 2π.
Example: A 90° turn about x is D = (I − iσₓ)/√2, which takes |+z⟩ to (1, −i)/√2 = |−y⟩, so ⟨S⟩ swings from +z to −y as a right-handed quarter turn about x requires; four such turns give D⁴ = −I, and only eight give I.

Closed form of a spin-½ rotationD(n̂, φ) = exp(−iσ⋅n̂ φ/2) = cos(φ/2) I − i sin(φ/2) σ⋅n̂

The series collapses because (σ⋅n̂)² = I: even powers make the cosine, odd powers the sine, and any spin-½ rotation becomes two numbers and one Pauli matrix.

n̂ a unit vector, φ the rotation angle in rad; D is a dimensionless 2 × 2 unitary matrix with det D = 1

Conjugation turns S as an ordinary vectorD†S D = S cos φ + (n̂ × S) sin φ + n̂(n̂⋅S)(1 − cos φ)

So ⟨S⟩ after the rotation is the old ⟨S⟩ turned through the whole angle φ. D enters twice, which doubles its half angles and cancels its sign.

S = (ħ/2)σ in J s; the right side is S rotated through φ about n̂ in the right-hand sense

Every eigenspinor from |+z⟩D(ẑ, φ) D(ŷ, θ)|+z⟩ = e^(−iφ/2)[cos(θ/2)|+z⟩ + e^(iφ) sin(θ/2)|−z⟩]

Two rotations replace an eigenvalue problem. The leftover e^(−iφ/2) is a global phase, and it is −1 after one full circuit in φ.

θ and φ the polar and azimuthal angles of n̂ in rad; the bracket is the +ħ/2 eigenspinor of n̂⋅S from Quantum Mechanics I

Larmor evolution is a rotation in timeU(t) = exp(iγt B⋅S/ħ) = D(b̂, ω_L t), with ω_L = −γB

H = −γB⋅S generates a turn through ω_L t about the field: ⟨S⟩ repeats every 2π/|ω_L|, the spinor only every 4π/|ω_L|.

γ in rad s⁻¹ T⁻¹, B in T, b̂ = B/B, ω_L in rad s⁻¹; γₙ = −1.832 × 10⁸ rad s⁻¹ T⁻¹ for the neutron

A full turn on any multipletexp(−2πi n̂⋅J/ħ) = (−1)^(2j) I

Each |j, m⟩ gains e^(−2πim), and m − j is an integer. Orbital states, with integer l, never change sign; every half-integer j does.

j the angular-momentum quantum number of the multiplet: j = ½ gives −I, integer j gives +I

Fringe from a rotation in one armI/I₀ = [1 + cos(φ/2)]/2

Re⟨ψ|D|ψ⟩ = cos(φ/2) for every spin state, so even an unpolarised beam shows fringes, with period 4π: a 2π precession swaps the bright and dark ports.

equal arms, full contrast, no other phase difference; φ the precession angle in one arm in rad, I₀ the intensity at φ = 0

Summing the exponential: (σ⋅n̂)² = I does the work

Topic 5.1 gave D(n̂, φ) = exp(−i n̂⋅J φ/ħ). For spin ½, J = S = (ħ/2)σ, so the exponent is −i(σ⋅n̂)φ/2, dimensionless because S⋅n̂ φ/ħ is J s × rad ÷ J s. The Pauli product rule σᵢσⱼ = δᵢⱼI + iεᵢⱼₖσₖ, summed against the symmetric nᵢnⱼ, loses its ε term and leaves (σ⋅n̂)² = (n̂⋅n̂)I = I. Every even power of σ⋅n̂ is therefore I and every odd power is σ⋅n̂, so the series splits: the even terms 1 − (φ/2)²/2! + … sum to cos(φ/2), the odd terms, each carrying σ⋅n̂, to −i[(φ/2) − (φ/2)³/3! + …]σ⋅n̂ = −i sin(φ/2) σ⋅n̂. Hence D = cos(φ/2) I − i sin(φ/2) σ⋅n̂, derived, with D†D = I and det D = cos²(φ/2) + sin²(φ/2) = 1. Two checks. For small φ it reduces to I − i(S⋅n̂/ħ)φ, the generator form of 5.1. And a half turn about y is −iσ_y, with rows (0, −1) and (1, 0): it takes |+z⟩ to |−z⟩, as flipping the z axis should, and applying it twice gives (−iσ_y)² = −I, the first sight of the sign.

Half angles in D, whole angles in ⟨S⟩

Nothing measurable sees D alone. After the rotation ⟨S⟩ = ⟨ψ|D†SD|ψ⟩, so the object to compute is D†SD. Take n̂ = ẑ. Because σₓ anticommutes with σ_z, carrying σₓ through D flips the sign of the exponent, σₓe^{(−iσ_zφ/2)} = e^(iσ_zφ/2)σₓ, and the two half angles meet: D†σₓD = e^(iσ_zφ)σₓ = σₓ cos φ + iσ_zσₓ sin φ = σₓ cos φ − σ_y sin φ, using σ_zσₓ = iσ_y. The same steps give D†σ_yD = σ_y cos φ + σₓ sin φ, and σ_z, which commutes with D, is untouched. That is the matrix of an ordinary right-handed rotation through φ about z; for a general axis the same algebra gives the Rodrigues form on the formula card. So ⟨S⟩ turns through the whole angle φ, exactly as a classical angular momentum would, and keeps its length ħ/2. Because D appears twice, this step erases both its half angles and its overall sign. Check with |+x⟩ turned 90° about z: D|+x⟩ = (e^(−iπ/4), e^(iπ/4))/√2 = e^(−iπ/4)|+y⟩, and ⟨S⟩ has moved from x to y.

Building every eigenspinor by rotating |+z⟩

Quantum Mechanics I found the +ħ/2 eigenspinor of n̂⋅S by solving a 2 × 2 eigenvalue problem. Rotation builds it instead: tip ẑ to polar angle θ with a turn about y, then swing it to azimuth φ with a turn about z. About y the operator is real, D(ŷ, θ) = cos(θ/2) I − i sin(θ/2)σ_y, with rows (cos(θ/2), −sin(θ/2)) and (sin(θ/2), cos(θ/2)), so |+z⟩ becomes (cos(θ/2), sin(θ/2)). About z it is diagonal, with entries e^(−iφ/2) and e^(iφ/2), and the result is e^(−iφ/2)[cos(θ/2)|+z⟩ + e^(iφ) sin(θ/2)|−z⟩]: the old eigenspinor times a phase. Since D†SD rotates S, this state has ⟨S⟩ = (ħ/2)n̂ with no further work. The phase is unavoidable, because many rotations carry ẑ to n̂: add a turn χ about z first and |+z⟩ gains only e^(−iχ/2). For θ = 60° and φ = 90° the state is (0.866 e^(−iπ/4), 0.500 e^(iπ/4)). Now let φ run once round to 2π at fixed θ: the Bloch point comes home, and the spinor comes back multiplied by −1.

Larmor precession is the same rotation, played in time

Unit 3 put a moment μ = γS in a uniform field and found H = −γB⋅S. Writing B⋅S = B(b̂⋅S), the evolution operator exp(−iHt/ħ) = exp(−i(b̂⋅S)(−γBt)/ħ) is D(b̂, ω_L t) with ω_L = −γB, derived: evolution in a uniform field is a rotation about the field through ω_L t, and by the previous section ⟨S⟩ precesses at ω_L. For the neutron, γₙ = −1.832 × 10⁸ rad s⁻¹ T⁻¹ (quoted from the tables), so in 10.0 mT, ω_L = 1.83 × 10⁶ rad s⁻¹: ⟨S⟩ repeats after 3.43 μs, the spinor only after 6.86 μs. The rate checks against the energies: the levels ±ħω_L/2 are split by ħω_L = 1.21 neV, which is 2|μₙ|B with |μₙ| = 6.03 × 10⁻⁸ eV T⁻¹. One caution. Adding a constant E₀ to H multiplies U by e^{(−iE₀t/ħ)}, which could hide or fake a sign, so the −1 at t = 2π/ω_L is meaningful only because the field adds the traceless −γB⋅S alone; in an interferometer the other arm, with no field, fixes the zero of phase.

Minus one at 2π, and two matrices for every rotation

Put φ = 2π in the closed form: D = cos π I − i sin π σ⋅n̂ = −I for every axis, and only at 4π is D back to I. The rotation of ⟨S⟩ at 2π is the identity, so D and −D perform the same rotation; conjugation, D†SD, cannot see the sign. The 2 × 2 unitary matrices of unit determinant form the group SU(2), and they cover the rotations twice: φ and φ + 2π are one rotation but two operators. On any multiplet, exp(−2πiJ_z/ħ) gives |j, m⟩ the factor e^(−2πim) = (−1)^(2j), derived, so integer spins and every orbital state, whose Yₗₘ carry e^(imφ) and are single-valued, return unchanged, while every half-integer j flips. No experiment on one beam can see it: a probability |⟨χ|Dψ⟩|² and a mean ⟨ψ|D†AD|ψ⟩ each contain D beside its conjugate, and (−1)(−1)* = 1. Nor does the sign carry any power of the fine-structure constant: it follows from the algebra of rotations alone, whereas the rate at which a field produces the rotation depends on the particle's g, which for the electron QED shifts from Dirac's 2 by α/π at leading order, a result quoted here.

Seeing the sign: split, precess, recombine

A perfect-crystal neutron interferometer splits each neutron into two paths centimetres apart and recombines them. Put a field region of length L in one path. A neutron of speed v spends L/v there and is rotated through φ = |γₙ|BL/v about the field. In an ideal balanced interferometer the forward amplitude is (ψ + Dψ)/2, so I/I₀ = [1 + Re⟨ψ|D|ψ⟩]/2, and ⟨ψ|D|ψ⟩ = cos(φ/2) − i sin(φ/2)⟨σ⋅n̂⟩ has real part cos(φ/2) for every ψ, polarised or not. Hence I/I₀ = [1 + cos(φ/2)]/2, derived: dark at 2π, bright again only at 4π. With v = 2.20 km s⁻¹ and L = 2.0 cm, one full fringe needs B = 7.5 mT. The 1975 experiments of Rauch's group and of Werner, Colella, Overhauser and Eagen, quoted here, used unpolarised beams; Werner's group reported 704° ± 38° of precession per fringe, and a 1978 repeat by Rauch's group 716.8° ± 3.8°, against 720° for a spinor and 360° for a vector. In practice the model fails before it does in principle: a speed spread Δv/v smears φ by φΔv/v and washes the fringes out at large φ.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotation angle φ360 °

Spin tilt from axis θ60 °

Set θ to 0: the state only gains a phase, so the dashed curve never leaves 1, yet the arrow swings to −1 at 360° and the fringe goes dark. Then change θ: the dashed curve and the ellipse move, the solid fringe does not, which is why unpolarised neutrons sufficed in 1975.

ONE BEAM |A|²1.00

Re A = cos(φ/2)-1.00

FRINGE I/I₀0.00

⟨S⟩ TURNED, MOD 360°0 °

Live interpretationONE BEAM |A|²: 1.00. Re A = cos(φ/2): −1.00. FRINGE I/I₀: 0.00. ⟨S⟩ TURNED, MOD 360°: 0 °

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading of the half angles is that a spin-½ must be turned through 720° before it points the same way again, so that after 360° some measurement on the particle would come out differently. None does. After the rotation ⟨S⟩ is built from D†SD, in which D appears twice: its half angles add to the full φ and its sign cancels, so the spin vector, every analyser probability and every expectation value repeat exactly after 360°, as for a classical arrow. What does not repeat is the state vector, D(2π)ψ = −ψ, a global phase that one beam cannot detect because every observable carries ψ and ψ* together. The −1 becomes physical only as a relative phase between a rotated and an unrotated part of one wavefunction, which is why it took an interferometer. The mirror-image slip is to call a sign no single beam can see a mere convention; fringes dark at 2π and bright at 4π rule that out.

A neutron interferometer has no phase difference between its two paths until a field region is added to one of them, precessing the spin there through an angle φ about the field. With the field off, the forward port carries intensity I₀. What does that port carry at φ = 360°, and at φ = 180°?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA rotation of 120° about the body diagonal n̂ = (1, 1, 1)/√3 carries the x axis to y, y to z and z to x. Write its spin-½ matrix D, apply it to |+z⟩, check that ⟨S⟩ lands where the rotation sends the z axis, and find D³.

Closed form with φ/2 = 60°: cos 60° = 0.500 and sin 60°/√3 = 0.500, so D = ½[I − i(σₓ + σ_y + σ_z)]. The axis is a unit vector, n̂⋅n̂ = 3 × ⅓ = 1, so (σ⋅n̂)² = I holds and the closed form applies.
σₓ + σ_y + σ_z has rows (1, 1 − i) and (1 + i, −1). Multiplying by −i and adding I gives rows (1 − i, −1 − i) and (1 − i, 1 + i), and D is half of that. Its determinant is ¼[(1 − i)(1 + i) + (1 + i)(1 − i)] = ¼(2 + 2) = 1, as SU(2) requires.
D|+z⟩ is the first column of D: ½(1 − i, 1 − i) = [(1 − i)/√2](1, 1)/√2 = e^(−iπ/4)|+x⟩. So ⟨S⟩ = (ħ/2)x̂ = 5.27 × 10⁻³⁵ J s along x, exactly where the rotation sends the z axis; the factor e^(−iπ/4) is a global phase.
Three turns of 120° make one of 360° about the same axis, so D³ = cos 180° I − i sin 180° σ⋅n̂ = −I.
Cross-check by stepping: D takes |+z⟩ to e^(−iπ/4)|+x⟩, then to e^(−3iπ/4)|+y⟩, then to e^(−iπ)|+z⟩ = −|+z⟩. Every axis is back where it started; the spinor is not.

AnswerD = ½[I − i(σₓ + σ_y + σ_z)]; D|+z⟩ = e^(−iπ/4)|+x⟩, so ⟨S⟩ = (ħ/2)x̂ = 5.27 × 10⁻³⁵ J s along x; D³ = −I.

MediumAn electron starts in |+x⟩ in a uniform field of 25.0 mT along +z. Taking γₑ = −1.7609 × 10¹¹ rad s⁻¹ T⁻¹, find the Larmor angular frequency, the first time at which ⟨S⟩ points along +y and the spinor then, and the state after one Larmor period, with what an x analyser reads there.

ω_L = −γₑB = 1.7609 × 10¹¹ × 0.0250 = 4.40 × 10⁹ rad s⁻¹. Units: (rad s⁻¹ T⁻¹)(T) = rad s⁻¹. ω_L is positive, so U(t) = D(ẑ, ω_L t) is a right-handed turn about +z, with diagonal entries e^{(−iω_L t/2)} and e^{(iω_L t/2)}.
ψ(t) = U(t)|+x⟩ = e^{(−iω_L t/2)}(1, e^{(iω_L t)})/√2, the equatorial spinor at azimuth ω_L t, so ⟨S⟩ = (ħ/2)(cos ω_L t, sin ω_L t, 0).
⟨S⟩ first points along +y when ω_L t = π/2: t = π/(2 × 4.40 × 10⁹ rad s⁻¹) = 3.57 × 10⁻¹⁰ s = 0.357 ns, where ψ = e^(−iπ/4)|+y⟩.
One period is T = 2π/ω_L = 1.43 ns, and there ψ = D(ẑ, 2π)|+x⟩ = −|+x⟩. An x analyser passes every electron, |⟨+x|ψ⟩|² = 1, exactly as at t = 0; the spinor itself repeats only at 2T = 2.85 ns.
Checks: ħω_L = 4.64 × 10⁻²⁵ J = 2.90 μeV equals the Zeeman splitting gμ_B B = 2.0023 × 57.88 μeV T⁻¹ × 0.0250 T = 2.90 μeV. With Dirac's g = 2 the period would be 1.429 ns; the QED anomaly (g − 2)/2, α/2π at leading order, shortens it by 0.116%, 1.66 ps, and changes neither the 2π return of ⟨S⟩ nor the 4π return of ψ.

Answerω_L = 4.40 × 10⁹ rad s⁻¹; ⟨S⟩ reaches +y at 0.357 ns with ψ = e^(−iπ/4)|+y⟩; after T = 1.43 ns, ψ = −|+x⟩ and an x analyser passes every electron, as at t = 0; ψ itself repeats only at 2.85 ns.

HardNeutrons of wavelength 0.235 nm cross an interferometer whose one path holds a field region of effective length 1.50 cm, the field perpendicular to the beam. Find the precession per millitesla, and the forward-port fringe period in B predicted with and without the spinor sign. Scanning B, adjacent minima sit at 2.35 mT and 9.98 mT, each located to ±0.08 mT. What precession does one fringe correspond to, and which prediction survives? Take λ and L as exact.

Speed and transit time: v = h/(mₙλ) = 6.626 × 10⁻³⁴ J s ÷ (1.675 × 10⁻²⁷ kg × 2.35 × 10⁻¹⁰ m) = 1683.3 m s⁻¹, and τ = L/v = 0.0150 m ÷ 1683.3 m s⁻¹ = 8.911 × 10⁻⁶ s, one guard digit carried in each.
Precession per unit field: φ/B = |γₙ|τ = 1.832 × 10⁸ rad s⁻¹ T⁻¹ × 8.911 × 10⁻⁶ s = 1632.5 rad T⁻¹, which is 1.6325 rad or 93.5° per millitesla.
Predicted period: the spinor fringe [1 + cos(φ/2)]/2 repeats every 4π, so ΔB = 4π ÷ 1.6325 rad mT⁻¹ = 7.70 mT. A spin that returned to itself after 2π would repeat every 2π ÷ 1.6325 rad mT⁻¹ = 3.85 mT.
Measured period: ΔB = 9.98 − 2.35 = 7.63 mT, with u = √2 × 0.08 = 0.113 mT, quoted as 0.11 mT, from the two independent positions. The minima need not sit at φ = 2π and 6π, because the arms also differ by an unrelated phase; only their spacing is tested.
Angle per fringe: 1.6325 rad mT⁻¹ × 7.63 mT = 12.456 rad = 713.7°, or 714°, with u = 714° × 0.113/7.63 = 10.6°, quoted as 11°.
Compare: 714° ± 11° lies (720 − 714)/10.6 = 0.6 standard uncertainties from 720° and (714 − 360)/10.6 = 33 from 360°. The sign is confirmed, in line with the quoted 704° ± 38° of 1975.

Answer93.5° (1.6325 rad) of precession per mT; predicted period 7.70 mT with the sign, 3.85 mT without; measured 7.63 ± 0.11 mT, i.e. 714° ± 11° per fringe, consistent with 720° and excluding 360°.