University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.5
The geometric phase for a cyclic change of parameters
Every adiabatic calculation so far kept the dynamic phase and threw the rest away. This lesson keeps the rest: the phase a slowly cycled Hamiltonian leaves on its eigenstate, fixed by the loop rather than the duration, untouched by any legal re-phasing, and, when H is real, a sign that molecules show in their spectra.
Build the model
Connect the measurement to the mechanism.
Steer the parameters R of a Hamiltonian slowly round a closed loop C, keeping a gap to every other level, and the adiabatic theorem promises that a system started in the eigenstate |n(R)⟩ is still in it when R comes home. What the theorem leaves open is the phase. Keep the diagonal term of the coefficient equations and the returning state is e(iθₙ)e(iγₙ)|n⟩: the dynamic phase θₙ = −(1/ħ)∫Eₙ dt, which grows with the duration, and Berry's phase γₙ = i∮⟨n|∇R n⟩⋅dR, in which dt has cancelled against dR, so it depends on the loop and not on the speed.
Born and Fock set this term to zero in 1928 by choosing the phases of the eigenstates, which is legal along an open path. Round a closed loop it is not: a single-valued re-phasing shifts γₙ only by a multiple of 2π, so e(iγₙ) belongs to the loop, and an interferometer with an arm that stayed at home can measure it. When H is real the phase can only be a sign, and it is −1 when the loop encircles an odd number of points where level n touches a neighbouring real level: the sign change Longuet-Higgins found round intersecting molecular surfaces, seen as half-odd-integer pseudorotation in Na₃.
The result stops where its premises do: a loop through or too near a degeneracy breaks the adiabatic condition, an open path has no invariant phase, and a degenerate level needs a matrix of phases rather than one number.
- Simple definition
- Berry's phase is the phase γₙ = i∮⟨n|∇R n⟩⋅dR, over and above the dynamic phase −(1/ħ)∫Eₙ dt, that an eigenstate acquires when the parameters R of its Hamiltonian are carried slowly once round a closed loop; it depends on the loop, not on how long the trip takes.
- Example
- Carry H = Xσz + Yσₓ once round a circle of radius 2.0 meV centred on X = Y = 0, and the real upper eigenvector (cos θ/2, sin θ/2) comes back as its own negative: γ = π, whether the trip takes 1 ns or 1 μs.
Derived: the adiabatic theorem drops the off-diagonal terms, and the diagonal one, ċₙ = −⟨n|∂ₜn⟩cₙ, integrates to e(iγₙ), not to 1.
θₙ = −(1/ħ)∫₀ᵗ Eₙ dt′ and γₙ = i∫₀ᵗ ⟨n|∂ₜ′n⟩ dt′, both in rad; Eₙ in J, ħ in J s
Derived with ∂ₜ = Ṙ ⋅ ∇R, so dt cancels against dR: one loop gives one γₙ whether it takes 1 ns or 1 s, while θₙ grows in proportion to the time.
R in the parameters' own units, ∇R in their inverse, γₙ in rad; real, since ∇⟨n|n⟩ = 0 makes ⟨n|∇R n⟩ imaginary
Derived from Aₙ → Aₙ − ∇R χ, with Aₙ = i⟨n|∇R n⟩. On an open path the shift is χ(start) − χ(end), which is anything: only a closed loop carries a physical phase.
χ real with e(iχ) single-valued on C; k is the integer number of turns χ makes round C
Derived from ⟨n(R)|n(R + ΔR)⟩ ≈ e(−iAₙ⋅ΔR). Each state appears once as a ket and once as a bra, so the arbitrary phases a diagonaliser returns cancel.
|nₖ⟩ the eigenstate at N points taken in order round C, each with any phase; the product is dimensionless
Derived. γₙ is π or 0, set by whether C encircles an odd or even number of conical intersections of level n with a neighbour; it cannot change without crossing one.
real eigenvectors make ⟨n|∇R n⟩ real and imaginary at once, so zero; only an end-to-end sign survives
Derived: θ → θ + 2πw reverses both vectors when w is odd. Both levels flip together, so the relative phase of a superposition of them picks up no geometric part.
X, Y in J or eV; w the winding number of C round X = Y = 0; eigenvectors (cos θ/2, sin θ/2), (−sin θ/2, cos θ/2); θ the polar angle of (X, Y), so X = √(X² + Y²) cos θ and Y = √(X² + Y²) sin θ
The diagonal term the adiabatic theorem leaves open
Expand on the instantaneous eigenstates, |ψ(t)⟩ = Σ cₘ(t) e(iθₘ(t)) |m(R(t))⟩ with θₘ = −(1/ħ)∫Eₘ dt′. The Schrödinger equation gives ċₘ = −cₘ⟨m|∂ₜm⟩ − Σ(n ≠ m) cₙ⟨m|∂ₜn⟩e(i(θₙ − θₘ)). The adiabatic theorem of UQMII-14.02, quoted here, discards the sum when ħ|⟨m|∂ₜH|n⟩| ≪ (Eₙ − Eₘ)², because ⟨m|∂ₜn⟩ = ⟨m|∂ₜH|n⟩/(Eₙ − Eₘ). The diagonal term survives: ċₙ = −⟨n|∂ₜn⟩cₙ, so cₙ(t) = cₙ(0)e(iγₙ(t)) with γₙ(t) = i∫⟨n|∂ₜ′n⟩ dt′. Differentiating ⟨n|n⟩ = 1 gives 2 Re⟨n|∂ₜn⟩ = 0, so ⟨n|∂ₜn⟩ is imaginary, γₙ is real and |cₙ| is constant: the diagonal term turns the state and never drains it. Units: ⟨n|∂ₜn⟩ is in s⁻¹, so γₙ is in radians. Born and Fock removed the term in 1928 by choosing the phase of |n(R(t))⟩ so that ⟨n|∂ₜn⟩ = 0 at every instant. The next sections show when that choice is allowed.
The clock cancels, so the phase belongs to the path
The eigenstate depends on time only through R, so ∂ₜ|n⟩ = ∇R|n⟩ ⋅ Ṙ and γₙ = i∫⟨n|∇R n⟩ ⋅ Ṙ dt = i∫⟨n|∇R n⟩ ⋅ dR. The dt has gone: run the loop in 1 ns or in 1 s and γₙ is the same line integral. Call the integrand Aₙ(R) = i⟨n|∇R n⟩, the Berry connection, a real vector field on parameter space with units of 1/R, and γₙ(C) = ∮Aₙ ⋅ dR. The dynamic phase is the opposite kind of object: a level at 2.0 meV held for 1.0 ns gives θₙ = −EₙT/ħ = −3.04 × 10³ rad, and for 2.0 ns −6.08 × 10³ rad. Two checks keep them apart. Scale H → λH with λ positive and every energy and θₙ grows by λ, while the eigenvectors, and so γₙ, do not change. And θₙ carries 1/ħ, while γₙ carries no ħ, no α and no energy scale: there is no order in α to check, because the phase is fixed by the geometry of the loop, and its test is against a measured phase.
Re-phasing moves the phase but cannot remove it
An eigenstate is defined only up to a phase, so replace |n(R)⟩ by e(iχ(R))|n(R)⟩ with χ real. Then ⟨n|∇R n⟩ gains i∇R χ, Aₙ becomes Aₙ − ∇R χ, and γₙ shifts by −[χ(end) − χ(start)]. Along an open path that difference is whatever you choose, so an open-path γₙ is a convention, and Born and Fock were right to drop it there. On a closed loop the states must be single-valued, so e(iχ) comes back to itself and χ can grow only by 2πk: γₙ moves by a whole multiple of 2π and e(iγₙ) is fixed. To cancel γₙ you would need χ to grow by γₙ round the loop, which makes the eigenstate multi-valued unless γₙ is already a multiple of 2π. The phase can sit in the connection or in a mismatch between the first and last state, but it cannot vanish. The gauge-free way to compute it multiplies overlaps round a chain of points, ⟨n₁|n₂⟩⟨n₂|n₃⟩ … ⟨nN|n₁⟩ → e(−iγₙ): every state enters once as a ket and once as a bra, so the phases a diagonaliser attached cancel.
A real Hamiltonian leaves only a sign
If H(R) is real and level n is non-degenerate, |n⟩* is an eigenvector with the same energy, so it is |n⟩ times a phase, and a compensating phase makes |n⟩ real. For a real normalised vector ⟨n|∇R n⟩ is real, and normalisation makes it imaginary, so it is zero: the connection vanishes, and all that can survive is a sign between the first and last state, e(iγₙ) = ±1. The two-level case shows which sign. For H = Xσz + Yσₓ with X = E cos θ and Y = E sin θ, the levels are ±E = ±√(X² + Y²) and the upper eigenvector is (cos θ/2, sin θ/2). Carry R once round X = Y = 0 and θ gains 2π while θ/2 gains only π: the vector returns as its own negative, γ = π. A loop that misses the origin brings θ back unchanged, γ = 0. A real 2 × 2 H needs two conditions, X = 0 and Y = 0, to be degenerate, so in a plane of two parameters its degeneracies are isolated points where the surfaces ±E meet as a double cone: the conical intersections Longuet-Higgins studied in molecules.
Seeing the phase needs a reference that stayed at home
A phase shared by the whole state is invisible; γₙ is measured against something that did not go round. Split the system, carry one part round C while the other waits at the start for the same time T, and recombine. The fringe phase is Δ = γₙ − (Ē − E₀)T/ħ, with Ē the loop's time-averaged energy and E₀ the energy at the start. Repeat for several T inside the adiabatic range and Δ falls on a straight line: the slope is dynamic, the intercept is γₙ. Or choose a loop of constant energy, such as a circle centred on the degeneracy of H = Xσz + Yσₓ, with the waiting part held at the same energy: the dynamic phases cancel and Δ = π turns a bright output port dark. What fails for that H is superposing its two levels on one path: both flip, γ₊ = γ₋ = π, and the geometric part of their relative phase is zero. A complex Hamiltonian can give γ₊ ≠ γ₋, and the spin in a rotating field of the next lesson is the standard case, quoted here and derived there.
Where the geometric phase stops being well defined
Everything above assumed the adiabatic condition all the way round, and it is hardest to meet where the loop passes nearest a degeneracy. For H = Xσz + Yσₓ at distance r from the origin, ⟨−|∂ₜH|+⟩ = rθ̇ and the gap is 2r, so the condition reads ħθ̇ ≪ 4r: the mixing angle must turn slowly on the scale of the gap. On the figure's default loop, radius 1.75 meV centred at X = 0.50 meV, the closest approach is r = 1.25 meV, where θ̇ = ρΩ/r peaks; a circuit in 1.0 ns gives ħρΩ/(4r²) = 1.2 × 10⁻³ there, and the ratio reaches 1 only near T = 1.2 ps. Let the loop pass through the origin and no T is slow enough: the gap closes, the Landau-Zener jump of UQMII-14.04 takes over, and γ is undefined. That is also the only way a real H can change its answer, because a sign cannot change continuously: γ is π on one side of the crossing and 0 on the other. Degenerate levels are a further limit, where the phase becomes a matrix; this course does not need it.
Change one variable at a time
Make the relationship visible.
With c below ρ, drag φ from 0° to 360°: the dashed line sweeps once round the dot, the arrow only half way, and it ends reversed at overlap −1. Push c above ρ and θ only rocks and returns, so γ = 0. No slider sets the speed, because γ has no clock in it.
MIXING ANGLE θ312 °
OVERLAP WITH START-0.91
γ FOR THE CLOSED LOOP180 °
SMALLEST GAP ON LOOP2.50 meV
Live interpretationMIXING ANGLE θ: 312 °. OVERLAP WITH START: −0.91. γ FOR THE CLOSED LOOP: 180 °. SMALLEST GAP ON LOOP: 2.50 meV
Catch the common trap
Explain before calculating.
The real Hamiltonian H = Xσz + Yσₓ is carried slowly once round a loop that encloses X = Y = 0, in 2.0 ns on one run and in 4.0 ns on a second. Both runs are well inside the adiabatic limit. What is the geometric phase of the upper level on each run?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe real Hamiltonian H = Xσz + Yσₓ, with X and Y in meV, is steered round circles in the (X, Y) plane. Loop 1 has radius 2.0 meV and centre (1.0, 0) meV; loop 2 has radius 2.0 meV and centre (3.0, 0) meV; each circuit takes 1.0 ns. Find the geometric phase of the upper level on each loop, say what changes if loop 1 takes 2.0 ns or is run twice, and check that 1.0 ns is slow enough.
- The degeneracy is at X = Y = 0. The centre of loop 1 is 1.0 meV from it, inside the 2.0 meV radius, so loop 1 winds once round it, w = 1. The centre of loop 2 is 3.0 meV away, outside its radius, so w = 0.
- With γ = πw, loop 1 gives γ = π: the real eigenvector (cos θ/2, sin θ/2) comes home as its own negative. On loop 2, θ only rocks between ±asin(2.0/3.0) = ±41.8° and returns, so γ = 0.
- Duration: γ contains no time, so loop 1 in 2.0 ns still gives π. The dynamic phase does change: averaged numerically round loop 1 the level sits at Ē = 2.13 meV, so −ĒT/ħ goes from −3.23 × 10³ rad to −6.46 × 10³ rad. Running loop 1 twice gives w = 2 and γ = 2π: two turns of the Hamiltonian bring the vector home.
- Adiabaticity: loop 1 passes closest to the degeneracy at r = 2.0 − 1.0 = 1.0 meV, where ħθ̇ = ħρΩ/r is largest, with ρ = 2.0 meV the radius. With Ω = 2π/(1.0 ns), ħΩ = 4.14 × 10⁻³ meV, so ħρΩ/(4r²) = (4.14 × 10⁻³ × 2.0)/(4 × 1.0²) = 2.1 × 10⁻³, far below 1.
Answerγ = π on loop 1, the same in 1.0 ns or 2.0 ns, and 2π if it is run twice; γ = 0 on loop 2. The adiabatic ratio on loop 1 is at most 2.1 × 10⁻³.
MediumA code diagonalises H = Xσz + Yσₓ at (X, Y) = (1, 0), (0, 1), (−1, 0) and (0, −1) meV, in that order round the origin, and returns the upper eigenvectors with whatever signs its routine picks: (1, 0), (0.707, 0.707), (0, 1) and (0.707, −0.707). Extract the Berry phase from these vectors alone, show that the chosen signs do not matter, and compare the square of points (4, 0), (3, 1), (2, 0) and (3, −1) meV, which misses the origin.
- Check the last vector: at (0, −1) meV, H = −σₓ, whose upper eigenvector is (1, −1)/√2 = (0.707, −0.707). Continuing (cos θ/2, sin θ/2) smoothly from (1, 0) would give (−0.707, 0.707) there instead, so the code has flipped a sign, as codes do.
- Form P = ⟨n₁|n₂⟩⟨n₂|n₃⟩⟨n₃|n₄⟩⟨n₄|n₁⟩ = (0.707)(0.707)(−0.707)(0.707) = −0.250. Flipping any one vector changes two factors, once as a bra and once as a ket, so P is unchanged; with the smooth choice the minus sign simply moves into the last factor.
- P tends to e(−iγ) as the points get denser, and P is real and negative, so γ = π. For a real H the discrete product already has the exact phase at N = 4; only its modulus is off.
- The modulus: with N equally spaced points on a circle round the origin, each overlap is cos(π/N), so |P| = cos(π/N)N = 0.250, 0.531 and 0.733 for N = 4, 8 and 16, tending to 1. The sign stays negative for every N.
- The other square has θ = 0, 18.4°, 0 and −18.4°, so its vectors are (1, 0), (0.987, 0.160), (1, 0) and (0.987, −0.160). All four overlaps are 0.987 and P = +0.949, so γ = 0.
AnswerP = −0.250 for the square round the origin, so γ = π whatever signs the code returns; the square that misses the origin gives P = +0.949 and γ = 0.
HardIn a model triangular molecule, two nuclear displacements enter H = Xσz + Yσₓ for two electronic states as X = κQ cos φ and Y = κQ sin φ. The nuclei circulate in a trough of radius Q₀ = 0.25 Å on the lower surface, with effective mass μ = 20.0 u and electronic gap 2κQ₀ = 0.40 eV along it. Treating φ as a slow ring coordinate, find the boundary condition on the nuclear wavefunction χ(φ), the three lowest levels and their signature, and check adiabaticity.
- Here the mixing angle is θ = φ, and the lower electronic state u₋ = (−sin φ/2, cos φ/2) is real and reverses sign when φ gains 2π: the trough encircles the intersection at Q = 0, so γ = π.
- The full wavefunction Ψ = χ(φ)u₋(φ) must be single-valued, so χ must reverse sign as well: χ(φ + 2π) = −χ(φ). Ring states e(ijφ) then need j = ±1/2, ±3/2, ±5/2, … instead of integers.
- Rotor constant B = ħ²/(2μQ₀²). With μQ₀² = 20.0 × 1.6605 × 10⁻²⁷ kg × (0.25 × 10⁻¹⁰ m)² = 2.076 × 10⁻⁴⁷ kg m², B = (1.0546 × 10⁻³⁴ J s)²/(4.151 × 10⁻⁴⁷ kg m²) = 2.68 × 10⁻²² J = 1.67 meV. Dimensions: (J s)²/(kg m²) = J.
- Levels are Bj², plus a shift B/4, from |∂u₋/∂φ|² = 1/4, that every level shares: j = ±1/2, ±3/2, ±5/2 give 0.42, 3.76 and 10.5 meV, each doubly degenerate. Without the sign change, j = 0, ±1, ±2 would give 0, 1.67 and 6.69 meV, with a single ground level.
- Signature: the successive spacings 3.34 and 6.69 meV stand in the ratio 2, against 1.67 and 5.02 meV, ratio 3, for integer j, and the ground level is a degenerate pair. That half-odd-integer pattern is what was resolved in the spectrum of Na₃ in 1986, quoted here as the measured check.
- Adiabaticity: at j = 3/2, ħφ̇ = 2Bj = 5.0 meV, against 4r = 4 × 0.20 eV = 0.80 eV with r the half-gap, a ratio of 6.3 × 10⁻³, so the electrons follow the nuclei.
Answerχ(φ + 2π) = −χ(φ), so j is half-odd: B = 1.67 meV and the lowest levels are 0.42, 3.76 and 10.5 meV above a common shift B/4, a degenerate ground pair with spacings in the ratio 2 rather than 3.