University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.6

Berry's phase for a spin in a slowly rotating field

The spin in a turning field is the one Berry phase you can compute in three lines and then check against an exact solution. Work it through and you learn to keep the phase that grows with time apart from the phase that remembers only the path, and to say which of the two an experiment can see.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put a spin-½ in a field of fixed magnitude B₀ and carry the field's direction once round a cone of half-angle θ, slowly compared with the Larmor precession about it. The adiabatic theorem keeps the spin in the instantaneous eigenstate |+n⟩, so when the field returns the state can differ from its start only by a phase. Part of that phase is dynamic, −ω₀T/2, and grows with the duration T.

The rest is Berry's phase, and for the eigenspinor of Quantum Mechanics I it takes three lines: only the azimuthal derivative survives, and γ₊ = −π(1 − cos θ), minus half the solid angle the loop subtends at zero field. Nothing about the particle enters, not B₀, not the size or sign of its moment, not T; only the loop and the spin's projection on it, which is why the anti-aligned state gets +½Ω. Unit 3's exact rotating-frame solution recovers the same number as the T-independent part of its phase, and shows where the picture stops: corrections of order (ω/ω₀) sin²θ grow as the field turns faster, and as B₀ falls toward zero the gap closes and the same loop is no longer adiabatic.

The phase of one state alone is never seen. The relative phase −Ω between the aligned and anti-aligned components turns a transverse polarisation by the solid angle, and that is what polarised neutrons and nuclear magnetic resonance measured in 1987.

Simple definition: Berry's phase for a spin-½ is the phase, beyond the dynamic phase −ET/ħ, gained by the spin state aligned with a field of fixed magnitude when the field's direction is carried slowly once round a closed loop; it equals minus half the solid angle of the loop.
Example: A field turned once round a cone of half-angle 60° encloses Ω = 2π(1 − cos 60°) = π sr, so the aligned spin gains γ₊ = −π/2 = −1.571 rad whether the turn takes 1 ms or 1 s, provided it stays adiabatic.

Berry connection of the aligned spinor⟨+n|∂/∂φ|+n⟩ = i sin²(θ/2), for |+n⟩ = (cos(θ/2), e^(iφ) sin(θ/2))

Derived by differentiating the Quantum Mechanics I spinor. Only the φ-component survives, so on a loop at fixed θ the Berry integral is 2π times this number.

θ and φ the field's polar and azimuthal angles, in rad; the element is dimensionless, per radian of φ; ⟨+n|∂/∂θ|+n⟩ = 0

Berry's phase of the aligned spinγ₊ = −π(1 − cos θ) = −½Ω

Derived from γ₊ = i∮⟨+n|∂/∂φ|+n⟩ dφ. It depends on the loop alone, not on T, B₀, or the size or sign of the moment: −1.571 rad for θ = 60°.

γ₊ in rad, defined mod 2π; θ the cone half-angle; Ω = 2π(1 − cos θ) the solid angle of the cap round +z, in sr, for a loop run with φ increasing (anticlockwise seen from +z)

Any spin projection m along the fieldγₘ = −mΩ

Derived with the rotation operator e^{(−iφS_z/ħ)}. For spin-½ it gives γ₋ = +½Ω, so the two states separate by −Ω: the phase a polarisation experiment reads.

m = +½ for spin along the field and −½ against it; Ω in sr, γₘ in rad; reversing the loop's sense flips the sign

Dynamic phase of the aligned spinΦ_dyn = −E₊T/ħ = −ω₀T/2

Derived as −(1/ħ)∫E₊ dt. It grows with T and usually dwarfs γ₊, which is why a measurement must cancel it against a reference or vary T to separate the two.

ω₀ = 2|μ|B₀/ħ in rad s⁻¹ (tabulated: 1.761 × 10¹¹ rad s⁻¹ T⁻¹ × B₀ for an electron); T the loop time in s

Exact phase after one turn, rotating frameΦ = π − πλ/ω, with λ = √(ω₀² − 2ω₀ω cos θ + ω²)

Derived from Unit 3's static rotating-frame Hamiltonian; the π is the frame's e^(−iπσ_z) = −1. For ω ≪ ω₀ it is −ω₀T/2 + γ₊ − (π/2)(ω/ω₀) sin²θ.

ω₀ the Larmor and ω the field's angular frequency, in rad s⁻¹; Φ in rad, mod 2π, for the state that returns to itself

Adiabaticity: the peak spin-flip probabilityP_flip = (ω sin θ / λ)²

Derived from the tilt of the rotating-frame field. The adiabatic condition reads ω sin θ ≪ 2ω₀; at ω = 0.05ω₀ and θ = 60°, P_flip = 2.0 × 10⁻³.

Exact, for a spin that starts along the field; dimensionless; λ as in the card above, in rad s⁻¹

From the eigenspinor to minus half the solid angle

Write the field as B₀n̂, with n̂ = (sin θ cos φ, sin θ sin φ, cos θ) and H = (ħω₀/2) n̂⋅σ; for an electron or a neutron the spin along n̂ is the upper level, E₊ = +ħω₀/2. Quantum Mechanics I supplies |+n⟩ = (cos(θ/2), e^(iφ) sin(θ/2)), and the previous topic supplies Berry's formula, γ = i∮⟨n|∇n⟩⋅dR for a single-valued eigenstate. Differentiate. ∂/∂θ gives (−½ sin(θ/2), ½ e^(iφ) cos(θ/2)), whose overlap with |+n⟩ is zero. ∂/∂φ gives (0, i e^(iφ) sin(θ/2)), whose overlap is i sin²(θ/2). On a loop at fixed θ only φ changes, from 0 to 2π, so γ₊ = i × i sin²(θ/2) × 2π = −2π sin²(θ/2) = −π(1 − cos θ). The cap above the loop has solid angle Ω, the integral of 2π sin θ′ dθ′ from 0 to θ, which is 2π(1 − cos θ); so γ₊ = −½Ω. Both sides are pure numbers, radians against steradians. For θ = 60°, Ω = π sr and γ₊ = −1.571 rad.

What the answer depends on, and what it does not

The anti-aligned spinor |−n⟩ = (e^(−iφ) sin(θ/2), −cos(θ/2)) gives ⟨−n|∂/∂φ|−n⟩ = −i sin²(θ/2), so γ₋ = +½Ω. The sign follows the spin's projection on the field, not which level is lower: a proton, whose aligned state is its ground state, gets the same γ₊ as an electron. Run the loop the other way and both signs flip. Re-phase the eigenstate, |+n⟩ → e^(−iφ)|+n⟩, and the integrand gains −i, so γ₊ becomes π(1 + cos θ) = −½Ω + 2π: the same phase factor, now read as half the solid angle of the southern cap. The half is what makes that choice harmless, because ½ × 4π = 2π. For any spin, |n, m⟩ = e^(imφ) e^{(−iφS_z/ħ)} e^{(−iθS_y/ħ)}|m⟩ gives γₘ = −mΩ. The moment never enters, so its quantum-electrodynamic anomaly, of order α/2π ≈ 1.16 × 10⁻³ (quoted), shifts ω₀ and the dynamic phase but leaves γ₊ exactly where it was.

The exact solution splits into two phases

Let the field precess at ω, so n̂ = (sin θ cos ωt, sin θ sin ωt, cos θ). Unit 3 solved this exactly, and the result is quoted: in the frame turning with the field the Hamiltonian is static, H′ = (ħ/2)[ω₀ sin θ σₓ + (ω₀ cos θ − ω)σ_z], with eigenvalues ±ħλ/2 and λ = √(ω₀² − 2ω₀ω cos θ + ω²). Its upper eigenstate returns to itself after T = 2π/ω having gained e^(−iλT/2), and undoing the frame's turn multiplies by e^(−iπσ_z) = −1. So Φ = π − πλ/ω, with no Berry formula used. Expand for ω ≪ ω₀: λ ≈ ω₀ − ω cos θ + ω² sin²θ/(2ω₀), and Φ ≈ −ω₀T/2 + π(1 + cos θ) − (π/2)(ω/ω₀) sin²θ. The first term is the dynamic phase and grows with T. The second is γ₊ plus 2π and does not depend on T. The third is the non-adiabatic correction and vanishes as the turning slows: at ω = 0.01ω₀ and θ = 60°, Φ + ω₀T/2 is −1.5826 rad against γ₊ = −1.5708 rad.

How slow is slow enough

This unit's adiabatic condition asks that ħ|⟨−n|∂H/∂t|+n⟩| ≪ (E₊ − E₋)². Here ∂H/∂t = (ħω₀/2)(dn̂/dt)⋅σ, and dn̂/dt, of length ω sin θ, is perpendicular to n̂, so its element between the two eigenstates has magnitude ω sin θ. The condition becomes ħ²ω₀ω sin θ/2 ≪ ħ²ω₀², or ω sin θ ≪ 2ω₀: the field must turn slowly compared with the precession about it, and a field that never moves, θ = 0, causes no transitions at any rate. The exact solution sharpens this. The rotating-frame field is tilted from n̂ by δ, with sin δ = ω sin θ/λ, so a spin that starts along the field nutates about it, and its flip probability peaks at sin²δ. An electron in 10 μT has ω₀ = 1.76 × 10⁶ rad s⁻¹; turning that field round a great circle at 100 Hz gives a peak flip probability of 1.3 × 10⁻⁷. At 10 nT the same loop gives 0.11: as B₀ approaches the degeneracy at zero field the gap closes and the loop stops being adiabatic.

At θ = 90° the spinor sign reappears

Carry the starting state rigidly round the cone with R_z(φ) = e^{(−iφσ_z/2)}. It stays an eigenstate at every φ, but it is not the adiabatically carried state, because the rotation's generator imprints a phase −⟨S_z⟩φ/ħ; apart from the dynamic phase, the adiabatic state is the rotated one with that phase undone. After a full turn R_z(2π) = −1, the spinor sign of Unit 5, so e^(iγ₊) = (−1) × e^{(2πi⟨S_z⟩/ħ)} with ⟨S_z⟩ = (ħ/2) cos θ, giving γ₊ = π + π cos θ, which is −π(1 − cos θ) mod 2π. At θ = 0 the two factors cancel. At θ = 90° the spin lies in the plane of the loop, ⟨S_z⟩ = 0, and the adiabatic state is simply the spinor rotated through 2π: γ₊ = π, a factor of −1. For integer spin the rotation gives +1 and γₘ = −mΩ is a multiple of 2π on the great circle. Both spin-½ states gain π there, so their relative phase, −Ω = −2π, vanishes: the sign shows only against a reference that was never rotated.

What an experiment actually measures

One state's phase is invisible. Prepare the spin across the field instead: it is an equal superposition of |+n⟩ and |−n⟩, and one turn changes their relative phase by −ω₀T − Ω, so the transverse polarisation precesses by ω₀T about the field plus an extra angle Ω. At θ = 45° that extra turn is 1.840 rad, or 105.4°. Bitter and Dubbers (1987, Institut Laue–Langevin) sent polarised neutrons at about 500 m s⁻¹ through a transverse field that turned once about the beam axis, adding a longitudinal field B_z to set tan θ = B_⊥/B_z; the extra rotation followed the solid-angle law as B_z was varied, departing from it only where the passage was least adiabatic (quoted). Suter, Chingas, Harris and Pines saw the same law in nuclear magnetic resonance in the rotating frame, also in 1987 (quoted). These experiments confirm the geometry, but the spin's expectation value obeys the classical Bloch equation, which predicts the same extra turn Ω. The half in −½Ω belongs to the spinor, and only a reference beam that never went round can show it.

Change one variable at a time

Make the relationship visible.

Interactive model

Cone half-angle θ60 °

Turning rate ω/ω₀0.20

Set θ = 90°: the loop is a great circle and γ₊ = −π, the spinor's sign under a full turn. Then drag ω/ω₀ down to 0.02: the dashed spin closes onto the field and the solid curve settles onto −½Ω. Only the correction shrinks; the geometric phase stays.

SOLID ANGLE Ω3.142 sr

BERRY PHASE γ₊ = −½Ω-1.571 rad

EXACT SHIFT VS STATIC FIELD-1.830 rad

PEAK FLIP PROBABILITY0.0357

Live interpretationSOLID ANGLE Ω: 3.142 sr. BERRY PHASE γ₊ = −½Ω: −1.571 rad. EXACT SHIFT VS STATIC FIELD: −1.830 rad. PEAK FLIP PROBABILITY: 0.0357

Catch the common trap

Explain before calculating.

Misconception check

It is tempting to argue that because the spin ends where it began, e^(iγ₊) multiplies the whole state and can never be seen, which would make Berry's phase a bookkeeping device. The same argument would make the dynamic phase unobservable, yet Larmor precession is measured every day. What cannot be seen is a phase common to everything that is later compared; a phase that differs between the parts of a superposition can. The aligned and anti-aligned components gain −½Ω and +½Ω, so a spin prepared across the field comes back turned by an extra Ω, which a polarimeter reads directly. A spin carried round the loop and a reference spin held in a static field of the same magnitude share the dynamic phase and differ by γ₊, which an interferometer reads as a fringe shift. The phase is geometric, not fictitious.

A spin-½ starts along a field of fixed magnitude whose direction is carried once round a cone of half-angle 60° in a time T, slowly enough for the adiabatic theorem. The run is repeated with the same loop traversed in 2T. What is the geometric phase of the aligned state in the second run?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA field of fixed magnitude is carried slowly once round a cone of half-angle 40°. Evaluate the Berry connection of the aligned eigenspinor, then the geometric phases of the aligned and anti-aligned states, and check the result at θ = 0 and θ = 90°.

With |+n⟩ = (cos(θ/2), e^(iφ) sin(θ/2)) only the φ-derivative survives: ⟨+n|∂/∂φ|+n⟩ = i sin²(20°) = i × 0.34202² = 0.11698i per radian.
Integrate once round the loop: γ₊ = i × 0.11698i × 2π = −2π × 0.11698 = −0.7350 rad.
Solid angle: Ω = 2π(1 − cos 40°) = 2π × 0.23396 = 1.4700 sr, so γ₊ = −½Ω as it must be. The anti-aligned state gains γ₋ = +0.7350 rad, and the two separate by −Ω = −1.470 rad, which turns a transverse spin by an extra 84.2°.
Limits: at θ = 0 the field never moves and −π(1 − cos 0) = 0; at θ = 90° the loop is a great circle and γ₊ = −π, the sign a spinor takes under a full turn.

Answerγ₊ = −0.735 rad and γ₋ = +0.735 rad for a loop enclosing Ω = 1.470 sr; a transverse spin would come back turned by an extra 84.2°.

MediumPolarised neutrons at 400 m s⁻¹ cross a 0.60 m region in which a field of fixed magnitude 0.50 mT turns once about the beam axis on a cone of half-angle 50°. The tabulated neutron ratio is ω₀/B₀ = 1.832 × 10⁸ rad s⁻¹ T⁻¹. Is the passage adiabatic? Find the dynamic and geometric phases of the aligned state and the extra turn of a neutron polarised across the field.

Transit time T = 0.60 m ÷ 400 m s⁻¹ = 1.50 ms, so ω = 2π/T = 4.19 × 10³ rad s⁻¹. Larmor: ω₀ = 1.832 × 10⁸ rad s⁻¹ T⁻¹ × 5.0 × 10⁻⁴ T = 9.16 × 10⁴ rad s⁻¹, so ω/ω₀ = 0.0457.
λ/ω₀ = √(1 − 2 × 0.0457 × cos 50° + 0.0457²) = √(1 − 0.0588 + 0.0021) = 0.9712, and the peak flip probability is (0.0457 × sin 50° ÷ 0.9712)² = 0.0361² = 1.30 × 10⁻³. The passage is adiabatic.
Dynamic phase: −ω₀T/2 = −9.16 × 10⁴ rad s⁻¹ × 1.50 × 10⁻³ s ÷ 2 = −68.7 rad. Geometric phase: γ₊ = −π(1 − cos 50°) = −π × 0.3572 = −1.122 rad, about sixty times smaller.
A spin across the field is an equal mix of |+n⟩ and |−n⟩. Its polarisation precesses by ω₀T = 137.4 rad about the field, plus an extra Ω = 2π × 0.3572 = 2.244 rad, or 128.6°, from γ₋ − γ₊.
Doubling the speed halves the dynamic phase to −34.35 rad and leaves γ₊ and Ω unchanged, while the peak flip probability rises to 5.5 × 10⁻³: still adiabatic. Varying the speed is therefore one way to separate the two phases.

AnswerAdiabatic, with peak flip probability 1.3 × 10⁻³; dynamic phase −68.7 rad and geometric phase −1.12 rad; a transverse polarisation turns an extra 2.24 rad (128.6°) on top of 137.4 rad of Larmor precession.

HardA spin-½ with Larmor angular frequency ω₀ sits in a field whose direction turns once round a 60° cone at ω = 0.20ω₀, far too fast for the adiabatic theorem. Using Unit 3's rotating-frame solution, find the exact phase gained in one turn by the state that returns to itself, compare it with dynamic plus Berry, and show that it equals that state's own dynamic phase plus minus half the solid angle of the cone its spin traces.

λ = ω₀√(1 − 2 × 0.20 × 0.500 + 0.040) = ω₀√0.840 = 0.9165ω₀. The state that returns to itself is the upper eigenstate of the rotating-frame Hamiltonian; after T = 2π/ω = 10π/ω₀ it gains Φ = π − πλ/ω = π − 4.5826π = −3.5826π, which is 0.4174π = 1.311 rad mod 2π.
The adiabatic reading gives −ω₀T/2 + γ₊ = −5π − π/2 = −5.5π, which is 0.5π = 1.571 rad mod 2π. It misses by −0.259 rad; the first-order correction −(π/2)(0.20) sin²60° = −0.236 rad accounts for most of the gap.
The returning spin points along the rotating-frame field (ω₀ sin θ, 0, ω₀ cos θ − ω)/λ: cos θ′ = 0.300/0.9165 = 0.3273, so θ′ = 70.9°, lagging 10.9° outside the field's cone. The angle δ between spin and field has cos δ = (ω₀ − ω cos θ)/λ = 0.900/0.9165 = 0.9820.
Its own dynamic phase is −(1/ħ)∫⟨H⟩ dt = −(ω₀T/2) cos δ = −5π × 0.98198 = −15.425 rad. Its cone encloses Ω′ = 2π(1 − 0.3273) = 4.227 sr, worth −½Ω′ = −2.113 rad. The sum, −17.538 rad, is 1.311 rad mod 2π: exactly Φ.
So minus half the solid angle is exact when the solid angle is the one the spin itself traces (the Aharonov–Anandan phase, quoted); the adiabatic theorem's job is to make that cone the field's. Integrating the two amplitude equations numerically over one turn gives the same 1.3114 rad, and a spin started along the field instead nutates, with peak flip probability sin²δ = 0.0357 = 1/28.

AnswerΦ = 1.311 rad (mod 2π), against 1.571 rad from dynamic plus Berry; it equals the returning state's dynamic phase, −15.425 rad, plus −½ × 4.227 sr = −2.113 rad for its own 70.9° cone.