University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.7

Berry curvature and the monopole at a degeneracy

Topic 14.6 found the spin's phase to be minus half the solid angle, which looks like a lucky property of one eigenspinor. It is not: the phase is the flux of a field in parameter space, computable from matrix elements alone, and all of that flux has its source at the points where two levels touch. Find those points and you know the phase of every loop.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Berry's phase for a closed loop, γₙ = ∮ Aₙ ⋅ dR with Aₙ = i⟨n|∇_R n⟩, has the form of the circulation of a vector potential, and Stokes' theorem takes the analogy literally: γₙ is the flux of the curvature Fₙ = ∇_R × Aₙ through any surface the loop bounds. The connection depends on how the eigenstates were phased; the curvature does not, because re-phasing adds a gradient and a gradient has no curl. Differentiating H|n⟩ = Eₙ|n⟩ writes Fₙ as a sum over the other states with (Eₙ − Eₘ)² underneath, so curvature piles up wherever a level nearly meets a neighbour and diverges where they meet.

For a spin in a field the sum has one term and gives F = −B̂/(2B²): the field of a point source of strength −½ at B = 0, whose flux through a loop's cap is minus half the solid angle and whose total, −2π, is one whole unit of 2π, as a phase defined modulo 2π requires. Two levels of a complex Hermitian matrix cross only where three real conditions hold, so in three parameters crossings are isolated points, and they are the only sources: a loop far from any of them still collects their flux. The model is only as good as the adiabatic theorem that defines γₙ.

Close to the monopole the gap shrinks, the curvature grows as 1/B², the loop must slow down without limit, and at the point itself no phase is defined.

Simple definition: The Berry curvature of a level is the curl, in parameter space, of its Berry connection i⟨n|∇_R n⟩; its flux through a surface bounded by a closed loop is the geometric phase the level picks up when the Hamiltonian is carried slowly around that loop.
Example: On the sphere |B| = 2.0 mT the aligned spin's curvature has magnitude 1/(2 × (2.0 mT)²) = 0.125 mT⁻²; a loop enclosing 1.0 sr of directions bounds 1.0 × (2.0 mT)² = 4.0 mT² of that sphere, so γ = −0.125 × 4.0 = −0.50 rad.

Berry connectionAₙ(R) = i⟨n(R)|∇_R n(R)⟩

Its loop integral is γₙ (topic 14.5). Re-phasing |n⟩ by e^(iχ(R)) shifts it by −∇_R χ, so Aₙ on its own is a convention, like a vector potential.

R the varied parameters (for a spin, B in T); Aₙ in 1/[R], so T⁻¹; real, since ⟨n|∇_R n⟩ is imaginary

Phase as flux, by Stokes' theoremγₙ = ∮_C Aₙ ⋅ dR = ∬_S Fₙ ⋅ dS, with Fₙ = ∇_R × Aₙ

Derived. The curl of −∇_R χ is zero, so Fₙ is gauge-invariant: the phase is set by what the loop encloses, not by the phases chosen along it.

S any surface bounded by C; for a field, Fₙ is in T⁻² and dS in T², so γₙ is in rad

Curvature as a sum over the other statesFₙ = i Σ_(m≠n) ⟨n|∇_R H|m⟩ × ⟨m|∇_R H|n⟩ / (Eₙ − Eₘ)²

Derived from ⟨m|∇_R n⟩ = ⟨m|∇_R H|n⟩/(Eₙ − Eₘ). No state is differentiated and no phase chosen; the squared gap puts the curvature where levels nearly meet.

∇_R H in J T⁻¹ and energies in J, so Fₙ in T⁻²; m runs over every other eigenstate

The spin's monopoleF_± = ∓B̂/(2B²)

Derived from the sum with the Pauli matrices; the moment and ħ cancel. Through a cone's cap the flux is ∓Ω/2, Ω the solid angle, which is 14.6's phase; through a sphere, ∓2π.

± for spin along or against B̂; B̂ the unit vector along B, B its magnitude in T; F in T⁻²

Flux through a closed surface∯ Fₙ ⋅ dS = 2π Cₙ, with Cₙ an integer

Derived here for the spin's sphere by joining two phase conventions; quoted in general as Chern's theorem. Cₙ can change only when a degeneracy crosses the surface.

Cₙ, the Chern number, is dimensionless; the surface must avoid every degeneracy; C₊ = −1 for the aligned spin

When two levels can crossE = ε₀ ± |h|, so the two levels meet only where h = 0

Derived for the 2 × 2 block: three real conditions, so crossings are isolated points in three parameters and generically missed in fewer. A real symmetric H has Im V = 0 and needs only two.

H = ε₀ + h⋅σ near the crossing; h = (Re V, −Im V, (E₁ − E₂)/2), in J, has three real components

A connection whose curl does not depend on the phases

Topic 14.5 wrote the geometric phase as γₙ = ∮ Aₙ ⋅ dR with the Berry connection Aₙ = i⟨n|∇_R n⟩, real because normalisation makes ⟨n|∇_R n⟩ imaginary. Re-phase the eigenstates, |n⟩ → e^(iχ(R))|n⟩, and Aₙ → Aₙ − ∇_R χ: exactly the freedom of a magnetic vector potential. A curl annihilates a gradient, so the Berry curvature Fₙ = ∇_R × Aₙ is unchanged, and Stokes' theorem makes it the physical object: γₙ = ∬ Fₙ ⋅ dS over any surface the loop bounds. The units keep the bookkeeping honest. With the field B in tesla as the parameter, Aₙ is in T⁻¹, Fₙ in T⁻² and dS in T², so γₙ comes out in radians, as a phase must. For a loop small enough that Fₙ is uniform across it, γₙ ≈ (the component of Fₙ normal to the loop) × (the area it encloses): the curvature is geometric phase per unit area of parameter space. With more than three parameters the curl becomes the antisymmetric array ∂ᵢAⱼ − ∂ⱼAᵢ, but three are enough for everything in this unit.

Curvature without differentiating a state

Written out, ∇_R × Aₙ = i⟨∇n| × |∇n⟩, and inserting Σₘ |m⟩⟨m| = 1 turns it into i Σₘ ⟨∇n|m⟩ × ⟨m|∇n⟩. The m = n term vanishes: ⟨n|∇n⟩ is imaginary, so ⟨∇n|n⟩ is its negative and a vector crossed with a multiple of itself is zero. For m ≠ n, take the gradient of H|n⟩ = Eₙ|n⟩ and project on ⟨m|; the ∇Eₙ term dies by orthogonality, leaving ⟨m|∇n⟩ = ⟨m|∇_R H|n⟩/(Eₙ − Eₘ). That gives Fₙ = i Σ_(m≠n) ⟨n|∇_R H|m⟩ × ⟨m|∇_R H|n⟩/(Eₙ − Eₘ)², derived here. Every term is gauge-invariant as it stands, since |m⟩ appears only in |m⟩⟨m| and |n⟩ only in ⟨n| … |n⟩. Swapping n and m reverses the cross product and leaves the denominator alone, so Σₙ Fₙ = 0, and for two levels F₊ = −F₋. The denominator is the square of the same gap that appears in topic 14.2's adiabatic condition: curvature concentrates exactly where following an eigenstate is hardest.

For a spin the curvature is a monopole at B = 0

Take H = −μσ⋅B, so E_± = ∓μB and ∇_B H = −μσ. With B along z the aligned state is |↑⟩, the only other state is |↓⟩, and the Pauli matrices give ⟨↑|σ|↓⟩ = (1, −i, 0) and ⟨↓|σ|↑⟩ = (1, i, 0), whose cross product is (0, 0, 2i). With (E₊ − E₋)² = 4μ²B², the aligned state has F₊ = i μ²(2i)ẑ/(4μ²B²) = −ẑ/(2B²), and because rotating the field rotates the states, F₊ = −B̂/(2B²) at every B. That is the field of a point source of strength −½ at B = 0. The moment μ has cancelled, and so has ħ: the curvature knows neither the size nor the sign of the coupling, and no power of the fine-structure constant enters. Checks: the units are T⁻². The flux through the cap of the sphere of radius B cut off by a cone of solid angle Ω is −(1/(2B²))(B²Ω) = −Ω/2, which is 14.6's −π(1 − cos θ) with no phase convention chosen along the way; the anti-aligned state gets +Ω/2. The 1987 polarised-neutron measurement quoted in topic 14.6 therefore tests this curvature too.

Why a closed surface carries whole units of 2π

Topic 14.6's eigenspinor (cos(θ/2), e^(iφ) sin(θ/2)) is smooth everywhere except the south pole, where it tends to (0, e^(iφ)) and its phase depends on the direction of approach. Multiplying it by e^(−iφ) gives a second choice, smooth except at the north pole. Use the first on the northern hemisphere and the second on the southern. By Stokes' theorem each hemisphere's flux is a line integral round the equator, taken in opposite senses, so the total is ∮(A_N − A_S) ⋅ dR = ∮ ∇χ ⋅ dR, where χ = −φ is the phase linking the two choices. A single-valued e^(iχ) can wind round the equator only a whole number of times, so the total flux is 2π times an integer: here the winding is −1 and the flux −2π, matching −½ × 4π from the monopole. The integer is the level's Chern number. The same fact reads physically: a tiny loop near the south pole encloses almost no flux through its own small cap, and −2π through the rest of the sphere, and both must give the same e^(iγ). That the flux is quantised for every closed surface and every family of Hamiltonians is Chern's theorem, quoted.

Degeneracies are points, and they are the only sources

Near a crossing, two levels obey a 2 × 2 Hamiltonian ε₀ + h⋅σ with h = (Re V, −Im V, (E₁ − E₂)/2), whose eigenvalues ε₀ ± |h| meet only when all three real components vanish. One varied parameter, as in Quantum Mechanics I's avoided crossing, generically misses: the gap bottoms out at 2|V|. Two parameters still generically miss. Three parameters can satisfy three conditions at isolated points, which is why the spin's only crossing is the single point B = 0 in field space. Away from such points Fₙ is the curl of a smooth local Aₙ, so ∇ ⋅ Fₙ = 0: flux cannot begin or end in empty parameter space, and a closed surface carries non-zero flux only if it encloses a crossing. If H is real and symmetric, Im V is zero for every R, one condition disappears, crossings become points in two parameters, and the curvature collapses onto them, leaving topic 14.5's phases of 0 or π. That a generic crossing, where h is linear in R − R₀, carries a monopole of strength ±½ is quoted; the spin is the example worked here.

Loops far from the source, and the price of coming close

Put the sphere of fields off-centre, B = B₁n̂ + dẑ with B₁ = 2.0 mT. A loop at fixed latitude of n̂ still has constant |B| and traces a cone about ẑ, now of half-angle α as seen from B = 0, so γ = −π(1 − cos α): only the solid angle at the degeneracy counts, not the loop's size or distance. Sweep the latitude from pole to pole and the cap's flux runs from 0 to −2π when |d| is less than B₁, but out and back to 0 when |d| exceeds B₁. The integer can change only as the sphere passes through B = 0, and there the adiabatic theorem fails: for an electron the gap is gμ_B|B|, which is h × 28.0 MHz per millitesla, the curvature grows as 1/|B|², and the condition ω sin α ≪ 2ω_L, with ω_L the gap over ħ, forces the loop to slow in proportion to |B|. Superconducting-qubit experiments published in 2014 by Schroer and colleagues and by Roushan and colleagues measured the curvature from the small lag of a spin swept at finite speed, integrated it over exactly this sphere, and found a Chern number close to 1 in magnitude with B = 0 inside and close to 0 with it outside: the jump predicted here.

Change one variable at a time

Make the relationship visible.

Interactive model

Sphere radius B₁1.8 mT

Offset field d along z0.5 mT

Loop latitude θ60 °

Keep d = 0.5 mT and drag θ to 180°: γ falls all the way to −2π. Then push d beyond B₁: the curve turns back and ends at 0, because the sphere no longer holds B = 0. The count can change only as the sphere sweeps through the degeneracy, where the smallest-gap readout drops toward zero.

CONE HALF-ANGLE α FROM B = 047.8 °

BERRY PHASE γ (CAP FLUX)-1.03 rad

WHOLE-SPHERE FLUX ÷ 2π-1

SMALLEST GAP ÷ h, ELECTRON35.0 MHz

Live interpretationCONE HALF-ANGLE α FROM B = 0: 47.8 °. BERRY PHASE γ (CAP FLUX): −1.03 rad. WHOLE-SPHERE FLUX ÷ 2π: −1. SMALLEST GAP ÷ h, ELECTRON: 35.0 MHz

Catch the common trap

Explain before calculating.

Misconception check

It is tempting to read the curvature as a local cause: |F| is small where |B| is large, so a loop far from B = 0 should collect little phase and a loop hugging the degeneracy a lot. The phase is not a property of the points on the path; it is a flux through a surface the path bounds. For the spin, |F| = 1/(2B²) falls as B⁻² while the cap the loop bounds on the sphere of radius B grows as B², so the two cancel and γ = −½ × (the solid angle seen from B = 0) at any distance: a loop round the equator of directions gives −π at 40 mT and at 4.0 mT alike. What distance does change is the adiabatic condition, because the gap 2μB shrinks with B: the near loop must be run ten times more slowly to stay adiabatic. The local value of F fixes γ on its own only for a loop so small that F is uniform across it.

A spin-½ stays in its state aligned with the field while the field's direction is carried slowly once around a cone subtending 0.80 sr at a field strength of 4.0 mT, and γ = −0.40 rad is found. The run is repeated around the same cone of directions at 1.0 mT. What changes?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA proton spin stays in its lower state, along B̂, while the field's magnitude is held at 50 mT and its direction is carried slowly once around a cone of half-angle 60°, azimuth increasing. Using F = −B̂/(2B²), find the curvature's magnitude on the loop, the area of the cap the loop bounds on the sphere |B| = 50 mT, and the Berry phase. Repeat at 5.0 mT.

On the sphere |B| = 0.050 T, |F| = 1/(2B²) = 1/(2 × 2.5 × 10⁻³ T²) = 200 T⁻². It points along −B̂, so it is normal to the sphere at every point of the cap.
The cap above a cone of half-angle θ on a sphere of radius B has area 2πB²(1 − cos θ) = 2π × 2.5 × 10⁻³ T² × 0.500 = 7.85 × 10⁻³ T².
Increasing azimuth gives, by the right-hand rule, the outward normal on this cap, so the flux is −200 T⁻² × 7.85 × 10⁻³ T² = −1.57 rad: dimensionless, as a phase must be, and equal to −½ × 2π(1 − cos 60°) = −π/2, topic 14.6's result.
At 5.0 mT, |F| = 1/(2 × 2.5 × 10⁻⁵ T²) = 2.0 × 10⁴ T⁻², a hundred times larger, while the cap's area, 7.85 × 10⁻⁵ T², is a hundred times smaller. The flux is again −1.57 rad.
Limits: θ → 0 leaves no area and no phase; θ → 180° takes in the whole sphere, −2π, which is no phase at all modulo 2π, as a loop shrunk onto the south pole must give.

Answer|F| = 200 T⁻² and the cap is 7.85 × 10⁻³ T² at 50 mT (2.0 × 10⁴ T⁻² and 7.85 × 10⁻⁵ T² at 5.0 mT); γ = −1.57 rad, that is −π/2, at both field strengths.

MediumA spin-½ is held in the state along B̂ while B = B₁n̂ + dẑ, with B₁ = 2.0 mT and n̂ = (sin θ cos φ, sin θ sin φ, cos θ). The loop is θ = 120° with φ run once from 0 to 2π. Find its Berry phase for d = 1.0 mT and for d = 3.0 mT, give the flux through the whole sphere of n̂ in each case, and for an electron at d = 1.0 mT find how long the loop must take to be adiabatic.

On the loop B = (ρ cos φ, ρ sin φ, z), with ρ = B₁ sin 120° = 1.732 mT and z = d + B₁ cos 120° = d − 1.0 mT. |B| is constant, so B̂ runs round a cone about ẑ of half-angle α, cos α = z/√(z² + ρ²), and the flux of F through its cap is γ = −π(1 − cos α).
d = 1.0 mT: z = 0, so α = 90° and γ = −π = −3.14 rad. Since |d| is less than B₁, B = 0 lies inside the sphere, and the flux through the whole sphere is −2π.
d = 3.0 mT: z = 2.0 mT, |B| = √(4.00 + 3.00) mT = 2.646 mT, cos α = 0.756 (α = 40.9°), and γ = −π × 0.244 = −0.767 rad. Now |d| exceeds B₁, the sphere does not enclose B = 0, and its total flux is 0: the curvature's flux in through the upper part of the sphere equals its flux out through the lower part.
Adiabatic check at d = 1.0 mT: |B| = 1.732 mT on the loop, so ω_L = gμ_B|B|/ħ = (1.761 × 10¹¹ s⁻¹ T⁻¹)(1.732 × 10⁻³ T) = 3.05 × 10⁸ s⁻¹. With |⟨−|dH/dt|+⟩| = μω|B| sin α and a gap of 2μ|B|, the condition ħ|⟨−|dH/dt|+⟩| ≪ (gap)² reads ω sin α ≪ 2ω_L = 6.1 × 10⁸ s⁻¹.
Here sin α = 1, so the loop period T = 2π/ω must be much longer than 2π/(6.1 × 10⁸ s⁻¹) = 1.0 × 10⁻⁸ s. A loop taking 1 μs is a hundred times slower than that limit.

Answerd = 1.0 mT: γ = −3.14 rad, whole-sphere flux −2π. d = 3.0 mT: γ = −0.767 rad, whole-sphere flux 0. The loop at d = 1.0 mT needs a period much longer than 10 ns.

HardA spin-s particle has H = −κB⋅S, with κ its gyromagnetic ratio. Use the sum-over-states formula and the ladder-operator matrix elements to find the Berry curvature of the state |m⟩ quantised along B̂. Then take a deuteron (s = 1) whose field direction is carried slowly once around a cone of half-angle 45°: find the phase of each of m = +1, 0 and −1, and the relative phase between m = +1 and m = −1.

∇_B H = −κS and Eₘ = −κħmB, so each term of Fₘ carries κ²⟨m|S|m′⟩ × ⟨m′|S|m⟩ over κ²ħ²B²(m − m′)², and κ cancels. Put B along z: Sₓ and S_y change m by one, so only m′ = m ± 1 contribute, each with (m − m′)² = 1.
From S_± = Sₓ ± iS_y and the ladder elements of Quantum Mechanics I, ⟨m ± 1|Sₓ|m⟩ = c_±/2 and ⟨m ± 1|S_y|m⟩ = ∓ic_±/2, with c_± = ħ√(s(s + 1) − m(m ± 1)). Writing a = ⟨m′|S|m⟩ = (c_±/2)(1, ∓i, 0), each term needs a* × a = ∓(ic_±²/2)ẑ.
So Fₘ = i Σ_± (∓ic_±²/2)ẑ/(ħ²B²) = (c₊² − c₋²)ẑ/(2ħ²B²). The ladder factors differ by c₊² − c₋² = −2mħ², so Fₘ = −mẑ/B², and by rotational symmetry Fₘ = −mB̂/B² everywhere.
Checks: m = ½ gives −B̂/(2B²), the spin-½ result; the units are T⁻²; the curvatures sum to zero because m and −m pair off; and the flux through a whole sphere is −4πm = 2π × (−2m), a whole multiple of 2π because 2m is an integer.
The cone of half-angle 45° subtends Ω = 2π(1 − cos 45°) = 2π × 0.2929 = 1.840 sr, and the phase is the flux through its cap, γₘ = −mΩ: γ₊₁ = −1.840 rad, γ₀ = 0 and γ₋₁ = +1.840 rad.
An interferometer compares two states, so it sees their relative phase. The geometric part is γ₊₁ − γ₋₁ = −2Ω = −3.68 rad, independent of κ, of |B| and of the traversal time T provided the loop is slow compared with the precession. On top of it sits the dynamic part, (E₋₁ − E₊₁)T/ħ = 2κ|B|T, which grows with T. A second run round the loop in the opposite sense reverses the geometric part and leaves the dynamic part alone, so the difference between the two runs is 2(γ₊₁ − γ₋₁) = −7.36 rad with no dynamic phase in it.

AnswerFₘ = −mB̂/B². For the deuteron, γ₊₁ = −1.84 rad, γ₀ = 0 and γ₋₁ = +1.84 rad, and the relative geometric phase γ₊₁ − γ₋₁ is −3.68 rad, read once the dynamic difference 2κ|B|T has been cancelled.