University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.2

The adiabatic theorem and its validity condition

Unit 12 let the Hamiltonian change a little at any speed. Here it changes a lot, slowly, and one ratio decides whether a system rides along in its eigenstate: the rate of change of H set against the square of the gap. Learn to compute that ratio, find where along a change it is worst, and name what the theorem leaves behind: a phase.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The model is a Hamiltonian that changes by a large amount but slowly, and the tool is its own eigenbasis, recomputed at every instant; the whole content of the theorem is a precise meaning for “slowly”. Expand the state on the eigenstates of H(t) at each instant, each carrying the dynamic phase −(1/ħ)∫Eₙ dt, and the Schrödinger equation becomes exact coefficient equations whose only couplings are ⟨m|ṅ⟩ = ⟨m|Ḣ|n⟩/(Eₙ − Eₘ), the rate at which the basis itself turns. Each coupling arrives multiplied by a phase that spins at the Bohr frequency (Eₙ − Eₘ)/ħ.

When the coupling hardly changes during one Bohr period, successive half-cycles cancel, and the amplitude leaked into m stays of order ħ|⟨m|Ḣ|n⟩|/(Eₙ − Eₘ)² instead of growing. Make that small for every m and a system started in |n⟩ rides along in |n(t)⟩: a well widened slowly from w to 2w leaves the particle in its nth state at a quarter of its energy. That argument is an outline, derived here; the theorem proper is quoted, and it rests on one hypothesis, a gap that never closes.

At a level crossing or the edge of a continuum no speed is slow enough, and a change completed in a time short compared with ħ/ΔE is the opposite, sudden limit. Even where the theorem holds it fixes the state only up to a phase: the diagonal term ⟨n|ṅ⟩ leaves a second phase it does not determine.

Simple definition: The adiabatic theorem says that a system starting in a non-degenerate eigenstate of a slowly changing Hamiltonian ends in the matching eigenstate |n(t)⟩ of the Hamiltonian at that moment, changed only by a phase, where slowly means ħ|⟨m|Ḣ|n⟩| ≪ (Eₙ − Eₘ)² for every other state m.
Example: An electron in the ground state of a 1.0 nm well whose wall recedes at a steady speed to 2.0 nm over 10 ps ends in the ground state of the wider well, its energy down from 0.376 eV to 0.094 eV, because the adiabatic ratio never exceeds 1.6 × 10⁻⁴.

Expansion on instantaneous eigenstates|Ψ(t)⟩ = Σₙ cₙ(t) e^(iθₙ(t)) |n(t)⟩, with θₙ(t) = −(1/ħ) ∫₀ᵗ Eₙ(t′) dt′

Removing the phase a stationary state would run up leaves coefficients that change only because the basis turns. A definition, so exact.

H(t)|n(t)⟩ = Eₙ(t)|n(t)⟩ at each instant; Eₙ in J, ħ in J s, θₙ in rad, cₙ dimensionless

The coefficient equationsdcₘ/dt = −Σₙ cₙ ⟨m|ṅ⟩ e^{(i(θₙ − θₘ))}

Unit 12's form, but the coupling is the rate at which the basis rotates, not a small H′/ħ, so H itself may change by any amount.

|ṅ⟩ = d|n(t)⟩/dt; ⟨m|ṅ⟩ and dcₘ/dt are in s⁻¹. Derived here with no approximation

The coupling in terms of dH/dt⟨m|ṅ⟩ = ⟨m|Ḣ|n⟩ / (Eₙ − Eₘ), for m ≠ n

The numerator carries the selection rules; the denominator makes the same Ḣ turn the basis faster and faster as two levels approach.

Ḣ = dH/dt in J s⁻¹ over an energy in J gives s⁻¹. Derived by differentiating H|n⟩ = Eₙ|n⟩

The adiabatic conditionħ |⟨m|Ḣ|n⟩| ≪ (Eₙ − Eₘ)², for every m ≠ n

Derived in outline; the rigorous theorem, quoted, needs a gap that never closes. No power of α enters: the only scales are the rate and the gap.

Both sides in J². Their ratio sets the size of the leaked amplitude, and its square the population lost

Coupling in a well with one wall moving⟨m|ṅ⟩ = (−1)^(m+n) [2mn / (m² − n²)] ẇ/w

With the gap (m² − n²)π²ħ²/(2Mw²) it gives the ratio 8Mwẇ/(9π²ħ) from n = 1 to m = 2, which grows as the well widens.

w is the width in m and ẇ = dw/dt in m s⁻¹; m ≠ n. Derived from ψₙ = √(2/w) sin(nπx/w)

The phase the theorem leaves openγₙ(t) = i ∫₀ᵗ ⟨n|ṅ⟩ dt′

It depends on the path through parameter space, not on how long the change takes, unlike θₙ. On an open path a re-phasing of |n(t)⟩ changes it; only round a closed loop is it fixed, and topic 14.05 derives it there as Berry's phase.

Real, in rad, because ⟨n|ṅ⟩ is purely imaginary; zero when the |n(t)⟩ are chosen real

Expand on the eigenstates of the moment

Unit 12 kept a fixed basis and let a small H′ move population. Here H(t) may change by a factor of four, so use the basis that moves with it: at each instant solve H(t)|n(t)⟩ = Eₙ(t)|n(t)⟩ and write |Ψ⟩ = Σₙ cₙ e^(iθₙ)|n(t)⟩, with θₙ = −(1/ħ)∫₀ᵗ Eₙ dt′, the phase a stationary state would have run up. Put this into iħ ∂|Ψ⟩/∂t = H|Ψ⟩. The H terms cancel the derivatives of θₙ exactly, and what survives is Σₙ (dcₙ/dt |n⟩ + cₙ|ṅ⟩) e^(iθₙ) = 0. Project onto ⟨m| and dcₘ/dt = −Σₙ cₙ⟨m|ṅ⟩ e^{(i(θₙ − θₘ))}. Nothing has been approximated. The only thing that can move population is |ṅ⟩, the rate at which the basis turns, so if H stood still every cₙ would stay constant, as Quantum Mechanics I requires of stationary states.

Trade the turning basis for the rate of change of H

|ṅ⟩ is awkward to compute, so differentiate the eigenvalue equation instead: Ḣ|n⟩ + H|ṅ⟩ = (dEₙ/dt)|n⟩ + Eₙ|ṅ⟩. Project onto ⟨m| with m ≠ n: H acts to the left to give Eₘ, ⟨m|n⟩ = 0 removes dEₙ/dt, and ⟨m|ṅ⟩ = ⟨m|Ḣ|n⟩/(Eₙ − Eₘ). The units check: J s⁻¹ over J is s⁻¹, a rate, as a term in dcₘ/dt must be. Two lessons sit in this line. The numerator is how fast H changes in the direction that joins n to m, so symmetry can switch a coupling off: for an oscillator whose frequency changes, Ḣ = Mω(dω/dt)x², and x² joins n only to n ± 2, never to n ± 1. The denominator is the gap, so the same Ḣ turns the basis faster as levels approach, and without limit at a degeneracy.

Why a fast phase starves the other states

Start in n, so cₙ ≈ 1 and every other cₘ is small. To first order, cₘ(t) = −∫₀ᵗ κ e^{(−iωₙₘt′)} dt′, where κ = ⟨m|ṅ⟩ and ωₙₘ = (Eₙ − Eₘ)/ħ, held constant for the outline. If κ changes little during one Bohr period 2π/|ωₙₘ|, the integrand is a slow amplitude times a fast rotation, and successive half-cycles cancel. Integrating by parts leaves the boundary term κ e^{(−iωₙₘt′)}/(iωₙₘ), evaluated between 0 and t, plus a remainder smaller by a further factor of |dκ/dt|/|κωₙₘ|. Its size is |κ/ωₙₘ| = ħ|⟨m|Ḣ|n⟩|/(Eₙ − Eₘ)², the adiabatic ratio, and keeping it small for every m is the validity condition. Dimensions: J s × J s⁻¹ = J², matching the squared gap. In words, H must change by much less than ΔE during a time ħ/ΔE. The leaked population is of order the ratio squared: a ratio of 0.01 costs about 10⁻⁴.

What the rigorous theorem adds, and what it needs

The outline assumed a constant gap and stopped at first order. The theorem, quoted and not derived here, does neither: Born and Fock proved it in 1928 for a discrete spectrum, and Kato in 1950 in the form now used. Let H(t/T) run along a fixed smooth path of Hamiltonians. If Eₙ stays separated from the rest of the spectrum by a gap that never closes, a state starting in |n(0)⟩ ends within O(1/T) of a phase times |n(T)⟩ as T → ∞, and faster than any power of 1/T if the change also starts and stops smoothly. Every hypothesis matters. Close the gap, as at a true crossing, and no finite T suffices; a narrow avoided crossing is the Landau-Zener topic. A continuum touching Eₙ has no gap at all. A measured case: the electrons of H₂ follow the nuclei because the vibrational quantum, 0.516 eV from the measured 4161 cm⁻¹ band, is at most about 5% of the gap of 10 eV or more to the nearest excited electronic state, so the nuclear motion is slow on the electrons' clock.

A well widened slowly keeps its quantum number

Quote Quantum Mechanics I: in a well of width w, Eₙ = n²π²ħ²/(2Mw²) and ψₙ = √(2/w) sin(nπx/w). Move one wall out slowly from w to 2w. The levels never cross, so the theorem applies and the particle ends in the nth state of the wider well at a quarter of the energy. An electron in the ground state of a 1.0 nm well goes from 0.376 eV to 0.094 eV, the missing 0.282 eV being work done on the wall. Now derive the coupling. Here |ṅ⟩ = ẇ ∂ψₙ/∂w, and ∂ψₙ/∂w = −ψₙ/(2w) − √(2/w)(nπx/w²)cos(nπx/w). The first term is orthogonal to ψₘ; with u = πx/w the second gives ⟨m|∂n/∂w⟩ = −(2n/(πw))∫₀^π u sin(μ) cos(ν) du = −(2n/(πw)) × (−1)^(m+n+1) πm/(m² − n²), so ⟨m|ṅ⟩ = (−1)^(m+n)[2mn/(m² − n²)] ẇ/w. Moving one wall breaks the mirror symmetry, so every m couples, but m = 2 dominates, and the ratio for it, 8Mwẇ/(9π²ħ), grows as the well widens because the gap falls as 1/w² faster than the coupling falls as 1/w. Limiting case: a classical ball bouncing off a wall receding at speed u loses 2u per bounce, which keeps vw fixed and so E ∝ 1/w², the same quarter.

The phase the theorem leaves open

Keep the m = n term of the coefficient equations: dcₙ/dt = −cₙ⟨n|ṅ⟩ plus couplings the theorem makes negligible. Differentiating ⟨n|n⟩ = 1 gives ⟨ṅ|n⟩ + ⟨n|ṅ⟩ = 0, so ⟨n|ṅ⟩ is purely imaginary, and cₙ(t) = cₙ(0) e^(iγₙ) with γₙ = i∫₀ᵗ⟨n|ṅ⟩dt′ real. The state therefore carries two phases. The dynamic phase θₙ depends on how long the change takes: for a linear widening from w₀ to 2w₀ in a time T it is −Eₙ(w₀)T/(2ħ), about −2.9 × 10³ rad for the 1.0 nm ground state with T = 10 ps, and it doubles if T doubles. The second does not depend on T at all, because ⟨n|ṅ⟩dt = ⟨n|∂n/∂R⟩dR involves only the path through the parameters R. On an open path it can be changed by re-phasing the |n(t)⟩, and it vanishes when the eigenstates are chosen real along the path, as for the well and the oscillator. When no real, single-valued choice exists, as for a spin whose field sweeps a cone, it survives a closed loop, and topic 14.05 derives it as Berry's phase.

Change one variable at a time

Make the relationship visible.

Interactive model

Gap frequency ω₀/2π1.00 MHz

Turning rate Ω/2π0.50 MHz

Halve the turning rate from 0.50 to 0.25 MHz: the exact peaks fall from 0.20 to 0.059 and the dashed estimate closes onto them. Then raise the gap to 4 MHz and the leakage all but vanishes, though the eigenbasis turns just as fast.

ADIABATIC RATIO Ω/(2ω₀)0.250

PEAK LEAKAGE, EXACT0.2000

PEAK, FIRST ORDER (Ω/ω₀)²0.2500

LEAKAGE PERIOD0.894 μs

Live interpretationADIABATIC RATIO Ω/(2ω₀): 0.250. PEAK LEAKAGE, EXACT: 0.2000. PEAK, FIRST ORDER (Ω/ω₀)²: 0.2500. LEAKAGE PERIOD: 0.894 μs

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that “slowly” means a small Ḣ on a human clock, so anything taking seconds is adiabatic and anything taking femtoseconds is not. The condition compares Ḣ with the square of the gap, so the same rate is gentle for one system and violent for another. For a change in H comparable to the gap itself it reduces to T ≫ ħ/ΔE: 0.66 fs for a 1 eV gap, but 0.66 ns for a 1 μeV gap. The wall of an electron's nanometre well can recede at 6 km s⁻¹ and the electron still follows, while two levels a microelectronvolt apart are knocked out of step by a change lasting a nanosecond. The other half of the error is about what is followed: the system stays in |n(t)⟩, the nth eigenstate of the present H, not in the wavefunction it started with. Keeping the old wavefunction is the sudden limit, the opposite answer.

An electron starts in the ground state of an infinite well, and one wall moves out at constant speed until the width has doubled from w to 2w. How does the adiabatic ratio ħ|⟨2|Ḣ|1⟩|/(E₂ − E₁)² at the end compare with its value at the start?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is in the n = 2 state of an infinite well 0.50 nm wide. One wall is moved out slowly until the well is 1.50 nm wide. Which state does the electron end in, what are its energies before and after, and how much work does it do on the moving wall?

Eₙ ∝ n²/w² keeps the levels in the same order at every width, so the gaps to n = 1 and n = 3 never close and the adiabatic theorem, quoted, applies: the electron ends in n = 2 of the wide well, its one interior node moving from 0.25 nm to 0.75 nm.
Quote E₁ = π²ħ²/(2mₑw²) from Quantum Mechanics I: at w = 0.50 nm, E₁ = 1.504 eV, so E₂ = 4E₁ = 6.02 eV.
Eₙ ∝ 1/w² and the width triples, so the energy falls by a factor of 9: E₂ = 6.02 eV ÷ 9 = 0.668 eV.
Only the wall exchanges energy with the electron, so the 6.02 − 0.668 = 5.35 eV it loses is work done on the receding wall.
Limiting-case check: Ew² = 6.02 eV × (0.50 nm)² = 1.50 eV nm² before and 0.668 eV × (1.50 nm)² = 1.50 eV nm² after, the same invariant a classical ball bouncing in a slowly widening box keeps, since vw is constant there.

AnswerIt stays in n = 2; its energy falls from 6.02 eV to 0.668 eV, and it does 5.35 eV of work on the wall.

MediumAn electron sits in the ground state of an infinite well 1.0 nm wide, and one wall is moved out at constant speed until the width is 2.0 nm. How short can the move be if the adiabatic ratio for every other state must stay at or below 0.010 throughout? Compare that time with the Bohr period of the final well.

For each m, the ratio is rₘ = ħ|⟨m|Ḣ|1⟩|/(Eₘ − E₁)² = ħ|⟨m|ṅ⟩|/(Eₘ − E₁), with |⟨m|ṅ⟩| = [2m/(m² − 1)] ẇ/w for n = 1 and Eₘ − E₁ = (m² − 1)E₁.
So rₘ ∝ 2m/(m² − 1)²: 0.444, 0.094 and 0.036 for m = 2, 3 and 4. The m = 2 term governs, and with E₁ = π²ħ²/(2mₑw²) it becomes r₂ = 8mₑwẇ/(9π²ħ).
At constant ẇ, r₂ ∝ w, so the worst moment is the end, w = 2.0 nm: r₂ = 8 × 9.109 × 10⁻³¹ kg × 2.0 × 10⁻⁹ m × ẇ/(9π² × 1.055 × 10⁻³⁴ J s) = 1.56 × 10⁻⁶ × ẇ/(m s⁻¹).
r₂ ≤ 0.010 needs ẇ ≤ 6.4 × 10³ m s⁻¹, so the 1.0 nm move takes T ≥ 1.0 × 10⁻⁹ m ÷ 6.4 × 10³ m s⁻¹ = 1.6 × 10⁻¹³ s.
Final well: E₂ − E₁ = 3 × 0.0940 eV = 0.282 eV, a Bohr period h/ΔE = 4.136 × 10⁻¹⁵ eV s ÷ 0.282 eV = 14.7 fs. The shortest move is about 11 Bohr periods, and it leaves roughly r² ≈ 10⁻⁴ of the population outside the ground state.

AnswerT ≥ 0.16 ps, a wall speed of at most 6.4 × 10³ m s⁻¹, about 11 Bohr periods of the final well; the end of the move, not the start, sets the limit.

HardA trapped ion starts in the motional ground state of a harmonic trap whose frequency is lowered linearly from ω/2π = 2.00 MHz to 0.500 MHz in a time T. Find the shortest T that keeps the adiabatic ratio at or below 0.010 throughout, estimate the population that ends in an excited state, and give the final energy and the dynamic and geometric phases.

Ḣ = Mω(dω/dt)x², and Quantum Mechanics I gives ⟨n + 2|x²|n⟩ = (ħ/(2Mω))√((n + 1)(n + 2)) with ⟨n + 1|x²|n⟩ = 0. From |0⟩ only |2⟩ couples: ⟨2|Ḣ|0⟩ = √2 (ħ/2)(dω/dt), across a gap 2ħω.
Adiabatic ratio: r = ħ|⟨2|Ḣ|0⟩|/(2ħω)² = √2 |dω/dt|/(8ω²). M and ħ cancel, so the ion's mass never enters: r is the fractional change of ω per radian of oscillation, times √2/8.
|dω/dt| = 2π × 1.50 MHz ÷ T = 9.42 × 10⁶ s⁻¹ ÷ T is constant, so r ∝ 1/ω² is 16 times larger at the end than at the start. Setting r = 0.010 at ω = 3.14 × 10⁶ s⁻¹ gives T = √2 × 9.42 × 10⁶ s⁻¹ ÷ (8 × 0.010 × (3.14 × 10⁶ s⁻¹)²) = 16.9 μs.
The leaked amplitude is set by r at the end, where the ramp stops abruptly, so P₂ ≈ r² = 1.0 × 10⁻⁴. Integrating the coefficient equations numerically at T = 16.9 μs gives P₂ = 9.6 × 10⁻⁵, and the exact equations for ⟨x²⟩ and ⟨p²⟩ give a mean phonon number 1.93 × 10⁻⁴ = 2P₂, in agreement.
Final energy: ½ħω = ½ × 1.055 × 10⁻³⁴ J s × 3.14 × 10⁶ s⁻¹ = 1.66 × 10⁻²⁸ J = 1.03 neV, a quarter of the initial 4.14 neV. E/ω = ħ/2 is unchanged, the quantum form of the pendulum's invariant in topic 14.01.
Phases: θ₀ = −½∫ω dt = −½ × 2π × 1.25 MHz × 16.9 μs = −66 rad, which doubles if T doubles. The oscillator eigenfunctions are real, so ⟨n|ṅ⟩ = 0 for n = 0 and the geometric phase is zero.

AnswerT ≥ 16.9 μs; about 1 × 10⁻⁴ of the population ends in |2⟩ (9.6 × 10⁻⁵ numerically); the final energy is 1.03 neV, with a dynamic phase of −66 rad and no geometric phase.