University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.1

Adiabatic invariants: a pendulum of slowly changing length

This is the classical rehearsal for the whole unit, on an apparatus you can test with a stopwatch. Draw a pendulum's string in slowly and its energy changes by an amount fixed in advance, whatever the schedule; jerk it and the result depends on the instant you chose. What slow change protects turns out to be the quantum number.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Hang a bob on a string that runs up through a small hole in a fixed plate and draw the string in, so the length L(t) shrinks. Energy is not conserved, because the hand does work, and angular momentum about the hole is not conserved, because gravity exerts a torque. One quantity survives all the same, provided L changes by only a small fraction during each swing. Averaged over a swing the tension exceeds mg by E/(2L), where E is the energy of swinging, so drawing the string in by a length |dL| costs E|dL|/(2L) beyond lifting the bob, and dE/E = −dL/(2L).

E therefore grows as L⁻¹⁄², exactly as ω = √(g/L) does, and E/ω stays fixed. That ratio is the area of the orbit in the plane of θ against p_θ = mL²θ̇, divided by 2π: slow change reshapes the orbit without changing its area. It was Einstein's answer when Lorentz asked about this pendulum at the 1911 Solvay conference. A sudden jerk keeps the bob's angle and angular momentum instead, and E/ω after it depends on where in the swing the jerk fell.

The argument fails when L changes appreciably within one swing, at amplitudes large enough for ω to depend on θ₀, and when the pulls are timed to the swing, which is how a swing is pumped. For a quantum oscillator E/ω = (n + ½)ħ, so the invariant is the quantum number, and the question left for the next lesson is how slow a change must be to keep a state in its level.

Simple definition: An adiabatic invariant is a quantity that stays fixed when a parameter of a periodic system changes slowly compared with one period, even though the energy changes; for an oscillator it is E/ω, the area of its phase-space orbit divided by 2π.
Example: Draw a 1.00 m pendulum slowly up to 0.25 m: ω rises from 3.13 to 6.26 rad s⁻¹, E doubles, E/ω is unchanged, and a 2.0° swing widens to 4³⁄⁴ × 2.0° = 5.7°.

Tension averaged over one swing⟨T⟩ = mg + E/(2L)

Derived here from T = mg cos θ + mLθ̇² with ⟨θ²⟩ = θ₀²/2 and ⟨θ̇²⟩ = ω²θ₀²/2. The excess E/(2L) is the only force beyond the weight that the hand works against, so it alone changes the swing energy.

T tension in N, m bob mass in kg, g = 9.81 m s⁻², L length in m, E = ½mgLθ₀² the swing energy in J

Swing energy under a slow pulldE/E = −dL/(2L), so E ∝ L⁻¹⁄²

Derived here from dE − mg dL = −⟨T⟩ dL. Quartering the length doubles E, and the rate never enters, provided it is slow.

E in J, L in m; valid while L changes by a small fraction of itself per radian of swing

The adiabatic invariant of the swingI = E/ω = (1/2π)∮ p_θ dθ = constant

Derived here: E and ω both go as L⁻¹⁄². I is the area of the phase-space ellipse over 2π, the quantity Einstein told Lorentz slow change keeps (1911, quoted).

ω = √(g/L) in rad s⁻¹ and p_θ = mL²θ̇ in J s, so I is an action, in J s

Angular amplitude under a slow pullθ₀ ∝ L⁻³⁄⁴

Derived here from E = ½mgLθ₀² ∝ L⁻¹⁄². A string drawn in to a quarter of its length swings 2.83 times as wide in angle, over 0.71 times the sideways distance.

θ₀ in rad, small swings; the bob's sideways reach Lθ₀ goes as L¹⁄⁴, so it shrinks

A sudden pull at phase φ of the swingI₂/I₁ = r³⁄² cos²φ + r⁻³⁄² sin²φ, with r = L₂/L₁

Derived here from θ and p_θ = mL²θ̇ both surviving a jerk along the string. At r = 0.25 it runs from 0.125 to 8 and averages 4.06 over φ; at r = 1 it is 1.

θ = θ₀ cos φ at the pull, so φ = 0 at a turning point and 90° at the bottom; r and I₂/I₁ are pure numbers

Quantum reading of the invariantE/ω = (n + ½)ħ, with n = 0, 1, 2, …

Quoted from the Quantum Mechanics I oscillator spectrum. If slow change keeps I it keeps n; whether a quantum state really stays in its level is the next lesson's question.

ħ = 1.055 × 10⁻³⁴ J s; a 1.00 kg bob on 1.00 m swinging 0.10 rad has n ≈ 1.5 × 10³²

Lorentz's pendulum, where no conservation law helps

The bob, mass m, hangs on a string that passes up through a small hole in a fixed plate, and a hand draws the string in, so the length L(t) shrinks. The tension acts along the string, through the hole, so it has no torque about the hole, and gravity's torque gives the exact equation of motion d/dt(mL²θ̇) = −mgL sin θ. Neither familiar conservation law survives. Energy is not conserved, because the hand does work on the bob, and p_θ = mL²θ̇ is not conserved, because gravity has a torque. Call E = ½mL²θ̇² + mgL(1 − cos θ) the energy of swinging at the current length. Lorentz asked in 1911 whether a slowly shortened pendulum keeps its energy a whole multiple of hν (quoted history). First-order perturbation theory cannot answer, because the change in L is not small; only its rate is. Measure the rate by ε = |dL/dt|/(Lω), the fraction by which L changes per radian of swing. At ε ≪ 1 each swing is a harmonic oscillation at ω = √(g/L), and the question becomes the accumulated effect of many nearly identical swings.

Average the tension over a swing

The tension supports the weight component mg cos θ and supplies the centripetal force mLθ̇², so T = mg cos θ + mLθ̇², with d²L/dt² negligible. Over one swing θ = θ₀ cos ωt gives ⟨θ²⟩ = θ₀²/2 and ⟨θ̇²⟩ = ω²θ₀²/2, so ⟨T⟩ = mg(1 − θ₀²/4) + mgθ₀²/2 = mg + E/(2L), with E = ½mgLθ₀². Changing the length by dL, the hand does work −⟨T⟩ dL, and the bob's total mechanical energy measured from the plate is E − mgL, dropping ½m(dL/dt)², which is second order in the pull speed. So dE − mg dL = −⟨T⟩ dL, the mg terms cancel, and dE/E = −dL/(2L), which integrates to E ∝ L⁻¹⁄². Numbers: a 1.00 kg bob on 1.00 m swinging at 0.10 rad has E = 0.0491 J and an average tension 0.0245 N above its weight, so drawing in 1.0 cm costs 2.45 × 10⁻⁴ J beyond the lift, a 0.50% rise in E. Checks: E/(2L) is J m⁻¹ = N; as θ₀ → 0 the excess vanishes and a bob at rest stays at rest; and no power of α appears, because this is classical mechanics.

What stays fixed is an area

Since E and ω = √(g/L) both scale as L⁻¹⁄², E/ω is constant. The ratio has a geometric meaning. In the plane of θ against p_θ = mL²θ̇ a swing traces an ellipse with semi-axes θ₀ and mL²ωθ₀, whose area is πmL²ωθ₀² = 2πE/ω. So I = E/ω = (1/2π)∮ p_θ dθ, an action in J s. Slow change reshapes the ellipse and keeps its area: θ₀ grows as L⁻³⁄⁴ while the peak p_θ shrinks as L³⁄⁴. For the 1.00 kg bob, I = 0.0491 J ÷ 3.132 rad s⁻¹ = 0.0157 J s, and the ellipse encloses 0.0984 J s at whatever length the string is slowly drawn to. Liouville's theorem, quoted from classical mechanics, explains why area is what survives: any Hamiltonian flow, time-dependent or not, carries a closed ring of phase points into a ring enclosing the same area. Put the ring on the old orbit, one point for each phase. A slow change keeps all its points on one orbit of the current Hamiltonian, to the accuracy of the averaging, so that orbit has the old area. A fast change carries the ring intact but strands its points on different orbits.

A sudden jerk keeps θ and p_θ, so timing decides

Now draw the string from L₁ to L₂ = rL₁ in a time much shorter than a swing. The pull is along the string, so p_θ = mL²θ̇ is kept, gravity having no time to act, and θ has no time to change. The phase-space point stays put, but the new orbit through it has a different shape, because the ratio of peak p_θ to θ₀ is mL²ω, which scales as L³⁄². With θ = θ₀ cos φ at the jerk, placing the unchanged point on the new ellipse gives I₂/I₁ = r³⁄² cos²φ + r⁻³⁄² sin²φ, which is 1 for every φ at r = 1. At r = 0.25 a jerk at a turning point, where the bob is at rest, gives 0.125; one at the bottom, where p_θ peaks, gives 8. It equals 1 only where |tan φ| = r³⁄⁴, at φ = 19.5° and the mirror phases 160.5°, 199.5° and 340.5°, by coincidence. Averaged over a random phase it is (r³⁄² + r⁻³⁄²)/2 = 4.06, never below 1: sudden changes at random moments heat an oscillator. Timed ones pump it. Shorten at each bottom and lengthen at each turning point, as a child on a swing does by standing and squatting, and every jerk multiplies I by r⁻³⁄².

How slow is slow, checked by integration

The averaging needed ε = |dL/dt|/(Lω) ≪ 1, equivalently |dω/dt| ≪ ω², since dω/dt = −ω(dL/dt)/(2L). Integrating d/dt(mL²θ̇) = −mgL sin θ numerically, for a 1.00 m string drawn in to 0.25 m from 24 starting phases, measures how well I survives. A pull lasting 0.001 of a swing reproduces the jerk formula, I₂/I₁ from 0.125 to 8.00. A constant-speed pull over one swing leaves I between 0.79 and 1.27 of its start value; over 10 swings, within 2.4%; over 100, within 0.23%. The same change with a smooth start and stop holds I to 2 parts in 10⁵ over 10 swings. The constant-speed residue comes from the abrupt start and finish, where dL/dt jumps, and is of order ε there; for a change smooth to every order it falls faster than any power of ε (quoted). The other limits are physical. At large amplitude ω depends on θ₀, and the kept quantity is the true area ∮ p_θ dθ/2π, which exceeds E/√(g/L), the small-swing E/ω, by a fraction θ₀²/32 and falls short of E divided by the true, amplitude-dependent frequency by about the same fraction (checked numerically): 0.03% at 0.10 rad, 3% at 1 rad. And pulls repeated twice per swing are not slow at all.

E/ω read as a quantum number

For a quantum oscillator E = (n + ½)ħω (quoted from Quantum Mechanics I), so E/ω = (n + ½)ħ, and the classical invariant is the quantum number in disguise: the 1.00 kg pendulum has n ≈ 0.0157 J s ÷ 1.055 × 10⁻³⁴ J s = 1.5 × 10³². That is why Ehrenfest made adiabatic invariants the quantities the old quantum theory quantised (quoted history): a quantum number that slow change could move continuously would mean nothing. The reading passes a second test. In the infinite well Eₙ = n²π²ħ²/(2mL²); a classical particle bouncing between the walls has action 2pL, which slow change keeps, so E ∝ L⁻², exactly what fixed n gives. Squeeze a cube of monatomic gas evenly and slowly and every atom keeps its action, so E ∝ V⁻²⁄³, and PV = 2E/3 makes PV⁵⁄³ constant: γ = 5/3 = 1.67, the measured heat-capacity ratio of helium and argon (tables). The quantum question can now be posed. Multiplied by ħ², the classical condition |dω/dt| ≪ ω² reads ħ d(ħω)/dt ≪ (ħω)², the rate of change of a level spacing against its square. Whether that keeps a state in level n is the next lesson's theorem.

Change one variable at a time

Make the relationship visible.

Interactive model

Length ratio L₂/L₁0.50

Phase φ of the sudden pull60 °

Hold L₂/L₁ at 0.50 and walk φ round the swing: at the turning points, 0° and 180°, the dashed orbit's area falls to 0.35 of the circle's; at the bottom, 90° and 270°, it rises to 2.83. At 30° it almost matches the solid orbit, by luck of timing rather than slowness.

SLOW PULL E₂/E₁1.41

SUDDEN PULL E₂/E₁3.12

SUDDEN PULL I₂/I₁2.21

MEAN SUDDEN I₂/I₁ OVER φ1.59

Live interpretationSLOW PULL E₂/E₁: 1.41. SUDDEN PULL E₂/E₁: 3.12. SUDDEN PULL I₂/I₁: 2.21. MEAN SUDDEN I₂/I₁ OVER φ: 1.59

Catch the common trap

Explain before calculating.

Misconception check

Slow is not the same as small. The tempting argument is that a gentle pull barely disturbs the pendulum, so E, or at least θ₀, should come out nearly unchanged. That is first-order perturbation thinking, and it fails because the change in L is large even though its rate is small. The work done per swing is tiny, but the number of swings is huge, and their product is finite and definite: drawing the string from 1.00 m to 0.25 m doubles E and nearly triples θ₀, however slowly it is done. What slowness buys is not a small effect but a predictable one, independent of the schedule and of the phase of the swing, which is exactly what a sudden jerk lacks. The thermodynamic word misleads the same way: an adiabatic compression exchanges no heat, yet the work done on the gas raises its energy.

A pendulum on a 1.00 m string swings with angular amplitude 2.0°. Over a hundred swings the string is drawn up through a hole in the ceiling until it is 0.36 m long. What is the new angular amplitude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 0.40 kg bob hangs on a string 1.60 m long that runs up through a hole in a fixed plate, and swings with angular amplitude 3.0°. The string is drawn up steadily over one minute until the length is 0.90 m. Find ω, E and E/ω before and after, the new angular amplitude, and the work the hand does on the swing beyond lifting the bob.

Check the pull is slow. Drawing in 0.70 m in 60 s is 0.0117 m s⁻¹, so ε = |dL/dt|/(Lω) is 0.0029 at the start and 0.0039 at the end, far below 1; the bob makes some 24 to 32 swings during the pull.
Before: ω₁ = √(9.81/1.60) = 2.476 rad s⁻¹ and θ₀ = 3.0° = 0.05236 rad, so E₁ = ½mgL₁θ₀² = ½ × 0.40 × 9.81 × 1.60 × 0.05236² = 8.61 × 10⁻³ J and E₁/ω₁ = 3.48 × 10⁻³ J s, in J ÷ s⁻¹ = J s, the units of an action.
The length ratio is r = 0.90/1.60 = 9/16, so r⁻¹⁄² = 4/3. Then ω₂ = 3.302 rad s⁻¹ and E₂ = (4/3)E₁ = 1.148 × 10⁻² J, and E₂/ω₂ = 3.48 × 10⁻³ J s, unchanged, as it must be.
Amplitude: θ₀ ∝ L⁻³⁄⁴, so θ₀₂ = 3.0° × (16/9)³⁄⁴ = 3.0° × 1.540 = 4.62° = 0.08061 rad. Check: ½mgL₂θ₀₂² = ½ × 0.40 × 9.81 × 0.90 × 0.08061² = 1.148 × 10⁻² J, matching E₂. The sideways reach Lθ₀ falls from 83.8 mm to 72.6 mm, a factor r¹⁄⁴ = 0.866.
Work: lifting the bob 0.70 m costs mgΔL = 2.75 J, and the swing takes E₂ − E₁ = E₁/3 = 2.87 × 10⁻³ J more, about 0.1% of the total, paid against the excess tension E/(2L). At 4.6° the small-angle error in sin θ is 0.11%, so the model holds.

Answerω rises from 2.48 to 3.30 rad s⁻¹ and E from 8.61 to 11.5 mJ, with E/ω = 3.48 × 10⁻³ J s throughout; θ₀ grows from 3.0° to 4.6°, and the hand does 2.87 mJ of work on the swing beyond the 2.75 J lift.

MediumA 0.20 kg bob on a 1.00 m string through a hole swings with amplitude 5.0°. When it is 3.0° from the vertical and moving towards it, the string is jerked up to 0.64 m in a few milliseconds. Find E/ω before and after and the new amplitude, and compare with a slow pull to the same length. At what angle from the vertical would the same jerk have left E/ω unchanged?

Before: ω₁ = √(9.81/1.00) = 3.132 rad s⁻¹ and E₁ = ½ × 0.20 × 9.81 × 1.00 × 0.08727² = 7.47 × 10⁻³ J, so I₁ = E₁/ω₁ = 2.385 × 10⁻³ J s. At the jerk θ = θ₀ cos φ = 3.0°, so cos φ = 0.60, sin φ = 0.80, and θ̇ = −ω₁θ₀ sin φ = −0.2187 rad s⁻¹.
A jerk along the string keeps θ and p_θ = mL²θ̇. With r = 0.64, θ̇ becomes −0.2187/0.64² = −0.5338 rad s⁻¹ while θ stays 0.05236 rad. The new swing energy is ½mL₂²θ̇² + ½mgL₂θ² = 1.167 × 10⁻² J + 1.72 × 10⁻³ J = 1.339 × 10⁻² J.
ω₂ = √(9.81/0.64) = 3.915 rad s⁻¹, so I₂ = 3.421 × 10⁻³ J s and I₂/I₁ = 1.434. The jerk formula agrees: 0.64³⁄² × 0.36 + 0.64⁻³⁄² × 0.64 = 0.184 + 1.250 = 1.434.
New amplitude: θ₀₂² = θ² + (θ̇/ω₂)² = 0.05236² + (0.5338/3.915)² = 0.02133 rad², so θ₀₂ = 0.1461 rad = 8.37°. A slow pull would give 5.0° × 0.64⁻³⁄⁴ = 6.99° and E = 1.25E₁ = 9.34 × 10⁻³ J, with I₂/I₁ = 1.
A jerk keeps I only where r³⁄² cos²φ + r⁻³⁄² sin²φ = 1, which rearranges to tan²φ = r³⁄², so |tan φ| = r³⁄⁴ = 0.7155, φ = 35.6° and θ = 5.0° × cos 35.6° = 4.07°, on either side and moving either way. That is a coincidence of timing, not adiabaticity: jerk a moment earlier or later and I changes.

AnswerI rises from 2.39 × 10⁻³ to 3.42 × 10⁻³ J s, a factor 1.43, and the amplitude from 5.0° to 8.4°, against 7.0° with I unchanged for a slow pull. A jerk at 4.07° from the vertical would have left I unchanged.

HardA motor moves the string of a 0.50 kg pendulum between 1.00 m and 0.90 m, starting from a 2.0° swing at 1.00 m. In run (a) it cycles the length slowly five times, over many swings per cycle; in run (b) it jerks the string in to 0.90 m whenever the bob passes the bottom and lets it out to 1.00 m at each turning point. Find the amplitude and swing energy at 1.00 m after five full periods of run (b), compare run (a), and say where the energy comes from.

Run (a): slow change keeps I = E/ω at every stage, and at a given length E/ω fixes E and θ₀. So each time L returns to 1.00 m, E and θ₀ return too: after five slow cycles the swing is 2.0° again, and the net work done on the swing is zero.
Run (b), at a bottom pass: θ = 0, so a jerk to r = 0.90 keeps θ = 0 and multiplies θ̇ by r⁻² = 1.235. The new amplitude is θ̇/ω₂ with ω₂ = ω₁r⁻¹⁄², so θ₀ grows by r⁻² × r¹⁄² = r⁻³⁄² = 1.1712, and E/ω by the same factor.
At a turning point the bob is at rest, so letting out to 1.00 m keeps θ = θ₀ and θ̇ = 0: the amplitude is unchanged, E grows by 1/r = 1.111 and ω falls by r¹⁄², so I again grows by r⁻³⁄² = 1.1712. Four jerks per period multiply I by 1.1712⁴ = 1.882.
After five periods, ten bottom passes: θ₀ = 2.0° × 1.1712¹⁰ = 2.0° × 4.857 = 9.71°. Integrating the exact equation d/dt(mL²θ̇) = −mgL sin θ, with each jerk lasting 0.2% of a period, gives 9.72°. At 1.00 m, E ∝ θ₀², so E rises by 4.857² = 23.59, from ½ × 0.50 × 9.81 × 1.00 × 0.03491² = 2.988 mJ to 70.5 mJ.
The motor supplies the 67.5 mJ: it pulls in against the largest tension, mg + mLθ̇², at the bottom, and lets out under the smallest, mg cos θ₀, at the turning points, so it puts in more work than it gets back. Swap the timing, shortening at turning points and lengthening at the bottom, and each jerk multiplies I by r³⁄²: the swing dies to 2.0° × 0.9¹⁵ = 0.41° in five periods. Sudden change makes timing everything; slow change makes it nothing.

AnswerRun (a) returns to 2.0° with no net work on the swing. Run (b) reaches 9.71° after five periods, E at 1.00 m rising from 2.99 mJ to 70.5 mJ, the 67.5 mJ supplied by the motor; reversing its timing would damp the swing to 0.41°.