University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.3

The sudden approximation, the opposite limit

When a Hamiltonian is switched faster than the system can respond, the state has no time to change and every outcome is a squared overlap. This lesson sets the clock that decides “faster”, works the doubled well and the beta decay of tritium, and keeps account of where the energy goes.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The adiabatic theorem says what a slow change of H does; the sudden approximation answers the opposite question, and its answer comes almost free. The Schrödinger equation is first order in time, so a Hamiltonian that stays finite cannot turn the state through a finite angle in a vanishing time: across a switch lasting τ, the amplitude to have left the initial state is about τΔH/ħ, with ΔH the spread of the Hamiltonian in that state. Take τ to zero and the state is simply kept.

What changes is the question you put to it: the new Hamiltonian has new eigenstates |φₙ⟩, and each is found with probability |⟨φₙ|ψ⟩|², summing to one by completeness. Nothing asks the change to be small, which separates this from Unit 12: tritium's nuclear charge doubles under its electron and the rule still holds. Energy is not conserved level by level; only its mean is fixed, shifted by the work ⟨ΔV⟩ done on the frozen state.

The approximation fails when τ grows to ħ over the spacing of the levels that matter, where the Landau-Zener problem of the next lesson takes over; it fails for levels so high that no real switch outruns them; and it needs more physics when the change also kicks the system, as a recoiling nucleus does, or when a slow β electron lingers inside the atom.

Simple definition: The sudden approximation treats a Hamiltonian switched in a time short compared with ħ over the relevant level spacings as switched instantly: the state is unchanged, and each new eigenstate |φₙ⟩ is found with probability |⟨φₙ|ψ⟩|².
Example: An electron in the ground state of a 1.0 nm well, E₁ = 0.376 eV, whose width jumps to 2.0 nm lands in the new ground state, at 0.094 eV, with probability 32/(9π²) = 0.360, and in the n = 2 level, at 0.376 eV, with probability 0.500.

Sudden rule: the state is kept, the basis changesPₙ = |⟨φₙ|ψ(0⁻)⟩|², because |ψ(0⁺)⟩ = |ψ(0⁻)⟩

Derived. Completeness of the new eigenstates makes Σ Pₙ = 1, so the sum checks every overlap; nothing requires the change in H to be small.

|φₙ⟩ are eigenstates of the new H; Pₙ is dimensionless; a 1D wavefunction carries m^(−1/2), a 3D one m^(−3/2)

Leading error of a switch lasting τ1 − |⟨ψ|U(τ)|ψ⟩|² ≈ (τ/ħ)² (⟨H̄²⟩ − ⟨H̄⟩²)

Derived to second order in τ. The mean of H̄ only adds a phase; its spread ΔH̄ moves population, so sudden means τΔH̄ ≪ ħ.

τ in s, ħ = 1.0546 × 10⁻³⁴ J s, H̄ = (1/τ)∫H dt over the switch, in J; the variance is taken in the initial state

Mean energy just after the switch⟨H_new⟩ = Σₙ Pₙ E′ₙ = ⟨ψ|H_old|ψ⟩ + ⟨ψ|ΔV|ψ⟩

Derived. H depended on time, so energy is not conserved level by level; the mean shifts by the work done on the unchanged state, zero where ψ vanishes.

E′ₙ the new eigenvalues in J or eV; ΔV = H_new − H_old, the change of potential, averaged over the frozen state

Infinite well widened suddenly from L to rLPₙ = (4/(π²r)) sin²(nπ/r) / (1 − n²/r²)², and Pₙ = 1/r at n = r

Derived. The level at n = r sits at the old energy E₁ yet takes only 1/r; ⟨H⟩ stays E₁, against E₁/r² for slow widening.

L in m, r = L′/L dimensionless; r = 2 gives P₁ = 32/(9π²) = 0.360, P₂ = 1/2, P₃ = 0.130, other even n zero

Hydrogenic 1s to 1s when Z becomes Z′P₁ₛ = [4ZZ′/(Z + Z′)²]³

Derived, order α⁰: only the charge ratio enters, and Z′ = Z gives 1. For Z to Z + 1 the loss 1 − P₁ₛ approaches 3/(2Z + 1)² at large Z: 0.0178 against 0.0176 at Z = 6, but 0.333 against 0.298 at Z = 1.

Z, Z′ nuclear charges in units of e; 1s radius a/Z with a = 5.29 × 10⁻¹¹ m; Z = 1 to Z′ = 2 gives 512/729 = 0.702

Is β decay sudden? The timing ratioτΔE/ħ ≈ (3/2) α/β, with τ = a/v and ΔE = 3R for He⁺

Derived. Sudden when the β electron outruns the orbital speed αc = 2.19 × 10⁶ m s⁻¹: at the measured mean 5.7 keV, β = 0.148 and the ratio is 0.074.

v = βc the β-electron speed; α = 1/137.04; R = 13.606 eV; a = 5.29 × 10⁻¹¹ m; the ratio is dimensionless

The state cannot move in no time

Quantum Mechanics I gives iħ ∂|ψ⟩/∂t = H(t)|ψ⟩, first order in time. Integrate it across a switch running from t = 0 to τ: |ψ(τ)⟩ − |ψ(0)⟩ = −(i/ħ)∫H(t)|ψ(t)⟩ dt over that interval. If H stays finite on the state, the right side shrinks with τ, and in the limit the state leaves the switch exactly as it entered: |ψ(0⁺)⟩ = |ψ(0⁻)⟩, derived. What the switch changes is the set of stationary states. Afterwards the system evolves under H_new, whose eigenstates |φₙ⟩ are complete, so the kept state is Σ cₙ|φₙ⟩ with cₙ = ⟨φₙ|ψ⟩, each component then turning its phase at E′ₙ/ħ. An energy measurement finds E′ₙ with probability Pₙ = |cₙ|², and Σ Pₙ = ⟨ψ|ψ⟩ = 1. Finiteness matters. An impulsive term −xJδ(t), with J an impulse in N s, is infinite for an instant and integrates to a finite unitary, exp(iJx/ħ): the momentum jumps by J though no time passes. The sudden rule covers Hamiltonians that are large, not infinite.

How short is short: the ħ/ΔE clock

Expand U(τ) to second order and write H̄ = (1/τ)∫H dt for the Hamiltonian averaged over the switch. The probability of finding the system anywhere but in its initial state is then 1 − |⟨ψ|U|ψ⟩|² ≈ (τ/ħ)²(⟨H̄²⟩ − ⟨H̄⟩²), derived here: the first-order term is a pure phase, and the second-order terms cancel down to the variance. It is dimensionless, as a probability must be. A constant added to H changes nothing, since it only rephases the state; what moves population is the spread ΔH̄, so the condition is τΔH̄ ≪ ħ, in practice τ ≪ ħ/ΔE with ΔE the spacing of the levels the overlaps populate. That is the adiabatic clock of the previous lesson read from the other side, and between the two limits neither holds, which is the Landau-Zener problem that follows. For tritium, let the charge the electron feels ramp from 1 to 2 over τ: then H̄ = H_old − e²/(8πε₀r), the 1s variance of 1/r gives ΔH̄ = e²/(8πε₀a) = 13.6 eV, and with τ = 1.2 × 10⁻¹⁸ s the error is (τΔH̄/ħ)² = 6 × 10⁻⁴.

A well that doubles in width

Take the ground state of an infinite well on 0 ≤ x ≤ L, ψ₁ = √(2/L) sin(πx/L), and move the right wall to 2L faster than that clock. The new eigenstates are φₙ = √(1/L) sin(nπx/2L) on 0 ≤ x ≤ 2L, and ψ₁ vanishes beyond L, so cₙ = (√2/L)∫ sin(πx/L) sin(nπx/2L) dx from 0 to L. With u = πx/L this is (√2/π) sin(nπ/2)/(1 − n²/4), and for n = 2, where that form is 0/0, the integral is π/2 directly. Hence P₁ = 32/(9π²) = 0.360, P₂ = 1/2, and for odd n, Pₙ = 32/(π²(n² − 4)²): P₃ = 0.130, P₅ = 0.0074. Every other even level gets nothing, and the reason is worth seeing: on 0 ≤ x ≤ L, φ₂ₖ = √(1/L) sin(kπx/L) is the old kth state over √2, so φ₂ is ψ₁/√2 there while φ₄, φ₆ and the rest are old excited states, orthogonal to ψ₁. Checks: with no widening the same integral returns P₁ = 1, P₁ + P₂ + P₃ = 0.990, the full sum is 1 to six figures numerically, and neither the mass nor L appears, because only the shapes overlap.

Energy is fixed on average, not level by level

The new levels are E′ₙ = n²E₁/4, so the particle can now be found below its old energy (E′₁ = E₁/4, probability 0.360), exactly at it (E′₂ = E₁, 1/2) or above it (E′₃ = 9E₁/4, 0.130). Their weighted mean is ⟨ψ₁|H_old|ψ₁⟩ + ⟨ψ₁|ΔV|ψ₁⟩, and here ΔV, the removal of the wall between L and 2L, acts only where ψ₁ = 0, so ⟨H⟩ = E₁ exactly; the sum of Pₙn²/4 does reach 1, slowly. Slow widening, quoted from the previous lesson, would leave the particle in n = 1 at E₁/4, the retreating wall having taken work from it. The sudden wall does no work, yet it does not conserve energy either: the energy afterwards is spread, and spread very far. Pₙ falls as n⁻⁴ while E′ₙ² grows as n⁴, so ⟨H²⟩ diverges: an instantaneous jump of an infinite wall puts a little probability into arbitrarily high levels. No real wall is that fast. Levels whose spacing exceeds ħ/τ are not set by overlaps but barely excited, which trims this tail and leaves the low levels, which hold the probability, as computed.

Tritium: a nucleus that changes its charge

Tritium decays as ³H → ³He⁺ + e⁻ + ν̄, and its 1s electron suddenly orbits a charge 2e. The switch lasts as long as the β electron takes to leave the atom, since until then its charge screens the extra proton. The measured spectrum ends at 18.6 keV and averages 5.7 keV: β = 0.148, v = 4.44 × 10⁷ m s⁻¹, τ = a/v = 1.2 × 10⁻¹⁸ s. The He⁺ levels at stake are 3R = 40.8 eV apart, matching the measured He II line at 30.38 nm, so ħ/ΔE = 1.6 × 10⁻¹⁷ s. In α, τΔE/ħ = (3/2)α/β = 0.074: the switch is sudden because the β electron outruns the orbital speed αc. The overlap, derived: with ψ₁ₛ = (Z³/πa³)¹⁄² e^(−Zr/a), ⟨1s, Z′|1s, Z⟩ = (4/a³)(ZZ′)³⁄²∫r² e^(−σr/a) dr from 0 to ∞, with σ = Z + Z′, which is 8(ZZ′)³⁄²/σ³, equal to 1 when Z′ = Z. For Z = 1 and Z′ = 2 it is 16√2/27, so P₁ₛ = 512/729 = 0.702. The Rydberg energy, the electron mass and α have cancelled: the probability is of order α⁰, a pure number set by the charge ratio.

Where the other 30% goes, and what is left out

Projecting the same state on He⁺ 2s, ψ₂ₛ = (Z′³/8πa³)¹⁄²(1 − Z′r/2a)e^{(−Z′r/2a)}, gives the overlap 32√2(ZZ′)³⁄²(Z − Z′)/(2Z + Z′)⁴ = −1/2 by the same two integrals, so P₂ₛ = 1/4, derived. Only s states are reached, since the kept state has l = 0 and the new H is spherical. The general result, Pₙₛ = 2⁹n⁵(n − 2)^{(2n − 4)}/(n + 2)^{(2n + 4)}, quoted from the Laguerre-polynomial integral and checked here numerically, gives P₃ₛ = 0.0127, and all bound states from 3s up hold 0.0214; completeness leaves 0.0263, so 2.6% of decays ionise the new ion. Three idealisations remain. Recoil: the ³He nucleus moves off at up to 1.5 × 10⁴ m s⁻¹, giving the electron a relative wavenumber k with ka = 0.0068, which lowers P₁ₛ by about 2 parts in 10⁵, estimated here from the same overlap with exp(ik⋅r) inserted. Slow β electrons: the ramp model's error passes 1% below 340 eV, and about 3% of decays, a fraction quoted from the allowed β spectrum with its Coulomb factor rather than derived here, fall below it. And molecules: neutrino-mass experiments use T₂, whose final states spread over molecular levels, a sudden-approximation calculation beyond this one.

Change one variable at a time

Make the relationship visible.

Interactive model

New width L′/L2.0

New level shown, n1

At L′/L = 2 step n from 1 to 6: n = 2 takes exactly 1/2, because φ₂ has the old ground state's shape there, and n = 4 and 6 get nothing. Back at 1 a single bar holds everything; at 4 the level n = 4, sitting at the old energy, takes only 1/4.

OVERLAP cₙ0.600

PROBABILITY Pₙ0.360

NEW LEVEL E′ₙ / E₁0.250

SLOW WIDENING E / E₁0.250

Live interpretationOVERLAP cₙ: 0.600. PROBABILITY Pₙ: 0.360. NEW LEVEL E′ₙ / E₁: 0.250. SLOW WIDENING E / E₁: 0.250

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading of a sudden change is that the particle lands in one definite new level: the level with its old quantum number, as if n were carried across, or the level with its old energy, as if energy were conserved. The doubled well sets a trap for the second, because E′₂ = 4E₁/4 = E₁ exactly, so energy conservation seems to demand n = 2 with certainty. It gets P₂ = 1/2. Both readings fail for one reason: the state is kept, and the old ground state is not an eigenstate of the new Hamiltonian, so it is a superposition, and a measurement finds each level with its squared overlap. Energy is not conserved because H depended on time; only the mean is pinned, here at E₁, by the work done on the frozen state. Keeping the quantum number is the adiabatic result, and it needs the opposite, slow, limit.

In tritium β decay the 1s electron is treated as seeing its nuclear charge jump suddenly from 1 to 2. Which statement correctly justifies the sudden approximation?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron sits in the ground state of a model quantum well 10 nm wide with infinitely high walls. A voltage step moves the right wall out to 30 nm. Treating the move as sudden, find the probabilities of the new n = 1, 2 and 3 levels, name the levels it can never reach, and estimate how fast the move must be for the treatment to hold.

Overlap the old state with φₙ = √(2/3L) sin(nπx/3L), L = 10 nm, over the old well only. With u = πx/L, cₙ = (2/(π√3))∫ sin u sin(ν/3) du from 0 to π, which is (2/(π√3)) sin(nπ/3)/(1 − n²/9) for n ≠ 3; for n = 3 the integral is π/2 and c₃ = 1/√3.
n = 1: sin(π/3) = √3/2 and 1 − 1/9 = 8/9, so c₁ = 9/(8π) and P₁ = 81/(64π²) = 0.128. n = 2: sin(2π/3) = √3/2 and 1 − 4/9 = 5/9, so c₂ = 9/(5π) and P₂ = 81/(25π²) = 0.328. n = 3: P₃ = 1/3 = 0.333.
Never reached: sin(nπ/3) = 0 for n = 6, 9, 12 and so on, because on the old well φ₃ₖ is proportional to the old kth state, orthogonal to ψ₁ unless k = 1.
Timing: E₁ = h²/(8mL²) = 3.760 meV, and E′ₙ = n²E₁/9 gives 0.418, 1.671 and 3.760 meV. These three levels hold 0.790 of the probability across E′₃ − E′₁ = 3.34 meV, so the move must take much less than ħ/(3.34 meV) = 0.20 ps.

AnswerP₁ = 0.128, P₂ = 0.328, P₃ = 0.333, together 0.790; n = 6, 9, 12, … are never reached. The wall must move in well under 0.2 ps, and n = 3, sitting exactly at the old energy of 3.76 meV, takes only a third.

MediumA hydrogen-like carbon-14 ion, ¹⁴C⁵⁺ with its electron in 1s, is stored in a ring and β-decays to ¹⁴N⁶⁺. The measured β spectrum of carbon-14 averages 49.5 keV. Check that the change is sudden, find the probabilities that the electron ends in the 1s and 2s states of the nitrogen ion, and compare the disturbance with tritium's.

Timing: the 1s electron lives near a/Z = 8.8 × 10⁻¹² m. At 49.5 keV the β electron has γ = 1.0969 and β = 0.411, so v = 1.23 × 10⁸ m s⁻¹ and τ = 7.2 × 10⁻²⁰ s. The N⁶⁺ 1s to 2s spacing is (3/4)(49)(13.606 eV) = 500 eV, so ħ/ΔE = 1.32 × 10⁻¹⁸ s and τΔE/ħ = 0.054: sudden.
1s: P₁ₛ = [4ZZ′/(Z + Z′)²]³ with Z = 6 and Z′ = 7 gives (168/169)³ = 0.9824.
2s: the overlap derived for tritium, 32√2(ZZ′)³⁄²(Z − Z′)/(2Z + Z′)⁴, gives 32√2 × 272.2 × (−1)/19⁴ = −0.0945, so P₂ₛ = 0.0089.
Compare: the electron leaves 1s with probability 1 − 0.9824 = 0.0176, against 0.298 in tritium, just under the large-Z estimate 3/(2Z + 1)² = 0.0178; about half of it goes to 2s here, against 84% in tritium.

AnswerSudden, with τΔE/ħ = 0.054. P₁ₛ = 0.982 and P₂ₛ = 0.0089, so only 1.8% of decays leave the electron excited or free, against 29.8% for tritium: the probabilities depend only on Z′/Z, and 7/6 is far closer to 1 than 2/1.

HardShow that when a hydrogenic 1s electron's nuclear charge changes suddenly from Z to Z′, its mean energy just afterwards exceeds the new ground-state energy by (Z′ − Z)²R, whatever Z is. Then, for tritium, compare that mean with what the bound He⁺ states can supply, given P₁ₛ = 512/729, P₂ₛ = 1/4 and states from 3s up holding 0.0214 with an energy contribution of −0.097 eV, and find the probability and mean energy of ionisation.

The kept state is 1s of charge Z, with ⟨H_old⟩ = −Z²R and, from Quantum Mechanics I, ⟨1/r⟩ = Z/a. The switch adds ΔV = −(Z′ − Z)e²/(4πε₀r), whose mean is −(Z′ − Z)Z e²/(4πε₀a) = −2Z(Z′ − Z)R, since e²/(4πε₀a) = 2R.
So ⟨H_new⟩ = −Z²R − 2Z(Z′ − Z)R = −(2ZZ′ − Z²)R, and subtracting E′₁ = −Z′²R leaves (Z′² − 2ZZ′ + Z²)R = (Z′ − Z)²R. Checks: an energy of order α²mc², zero when Z′ = Z, and one Rydberg for every β⁻ or β⁺ decay, 13.6 eV for ¹⁴C⁵⁺ as for tritium.
Tritium: ⟨H_new⟩ = −R − 2R = −3R = −40.82 eV, above the He⁺ ground state at −4R = −54.42 eV, the measured ionisation energy of He⁺.
Bound states: (512/729)(−54.42 eV) = −38.22 eV, (1/4)(−13.61 eV) = −3.40 eV, and −0.097 eV from 3s up, total −41.72 eV. That is below the mean, so the bound states alone cannot supply −40.82 eV and the continuum must carry positive energy.
Completeness: the continuum holds 1 − 0.7023 − 0.2500 − 0.0214 = 0.0263 of the probability and −40.82 − (−41.72) = 0.90 eV of the mean energy, so an ionised electron leaves with about 0.90/0.0263 = 34 eV on average.

Answer⟨H_new⟩ − E′₁ = (Z′ − Z)²R, one Rydberg for any β decay. In tritium the mean, −40.8 eV, lies 0.90 eV above what the bound states supply, so 2.6% of decays ionise the ion and the freed electron carries about 34 eV on average.