University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.5

Where first order fails against the exact Rabi solution

Perturbation theory is usually taken on trust, because nobody hands you the exact answer to grade it against. Here someone does. Unit 3 solved the driven two-level system exactly, so put the first-order amplitude for the same drive beside it and read off, as numbers, how long and how far off resonance first order can be believed.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

First-order time-dependent perturbation theory freezes the amplitudes inside the coefficient integral at their starting values, cₐ = 1 and c_b = 0. For a rotating drive with ⟨b|H′|a⟩ = (ħω₁/2)e^(−iωt), that single integral gives P₁ = (ω₁²/δ²) sin²(δt/2). Unit 3 solved the same drive exactly, P = (ω₁²/Ω²) sin²(Ωt/2) with Ω = √(ω₁² + δ²), so for once the approximation can be graded.

Expanded in the coupling ω₁ = |V_ab|/ħ, the exact result begins with P₁: agreement at order ω₁² is guaranteed, and because the second-order amplitude into |b⟩ vanishes for two levels with no diagonal matrix elements, the first error is of order ω₁⁴. Freezing cₐ hides two approximations, and they fail separately. Ignoring the depletion of |a⟩ lets the resonant P₁ = (ω₁t/2)² grow without bound, past 1 at ω₁t = 2, while the true probability turns over at ω₁t = π.

Ignoring the shift the drive produces in the levels leaves first order oscillating at |δ| instead of Ω, so off resonance, with the probability bounded and small, its phase still slips by ω₁²t/4|δ|. Together they give the working criterion ω₁²t min(t, 1/|δ|) ≪ 1: a condition on elapsed time as well as on the coupling, derived here from the comparison rather than assumed.

Simple definition: For a driven two-level system the first-order probability P₁ = (ω₁²/δ²) sin²(δt/2) is the lowest-order term of Rabi's exact result, and it can be trusted only while ω₁²t min(t, 1/|δ|) ≪ 1.
Example: With ω₁ = 1.00 × 10⁶ rad s⁻¹ on resonance, first order gives P₁ = 0.250 at t = 1.00 μs against the exact sin²(0.500) = 0.230, 9% high; at 2.00 μs it gives 1.00 while the exact value is 0.708.

Exact benchmark, quoted from UQMII-03.04P(t) = (ω₁²/Ω²) sin²(Ωt/2), with Ω = √(ω₁² + δ²)

Quoted, not re-solved: exact for a rotating drive with no diagonal matrix elements. It is the answer key against which every first-order number here is graded.

ω₁ = |V_ab|/ħ and δ = ω − ω₀ in rad s⁻¹; t in s; P dimensionless

First-order amplitude, derivedc_b⁽¹⁾(t) = −i e^(−iδt/2) (ω₁/δ) sin(δt/2)

One integral with |a⟩ frozen. It omits both the depletion of |a⟩ and the drive's shift of the levels, and each omission returns below as a separate failure.

from −(i/ħ)∫₀ᵗ ⟨b|H′|a⟩ e^(iω₀t′) dt′ with cₐ = 1 and ⟨b|H′|a⟩ = (ħω₁/2)e^(−iωt); dimensionless

First-order probability, derivedP₁(t) = (ω₁²/δ²) sin²(δt/2)

Its ω₁ → 0 limit is the benchmark's, since Ω → |δ|. The resonant parabola passing 1 is the depletion error: first order still holds |cₐ|² = 1.

ceiling ω₁²/δ², above 1 when |δ| < ω₁; on resonance P₁ = (ω₁t/2)², which passes 1 at ω₁t = 2

Short-time error, derivedP₁/P = 1 + ω₁²t²/12 + …, for Ωt ≪ 1

First order always errs high near the start: 1% at ω₁t = 0.35, and on resonance 10% at ω₁t = 1.06 once the sines are solved exactly rather than expanded.

ω₁t is the Rabi angle in rad; the detuning cancels because Ω² − δ² = ω₁²

Frequency error off resonance, derivedΩ − |δ| ≈ ω₁²/(2|δ|), for ω₁ ≪ |δ|

First order oscillates at |δ|, so its phase slips by ω₁²t/4|δ|: a bounded probability with a growing error, half a cycle out of step, peaks on troughs, once the slip reaches π/2 at t ≈ 2π|δ|/ω₁².

rad s⁻¹; the widening of the rotating-frame splitting by second-order level shifts ±ħω₁²/4δ

Validity criterion, derivedκ = ω₁²t min(t, 1/|δ|) ≪ 1

On resonance it demands t ≪ 1/ω₁, that is |V_ab|t/ħ ≪ 1; off resonance t ≪ |δ|/ω₁². It limits the time, not only the size of the coupling.

κ dimensionless; κ/12 is the short-time relative error, κ/4 the phase slip in rad once |δ|t ≫ 1

Quote the benchmark, and set up the same drive

Take two levels |a⟩ and |b⟩ with ω₀ = (E_b − Eₐ)/ħ, and the drive Unit 3 solved: a field rotating at ω whose only matrix element is ⟨b|H′|a⟩ = (ħω₁/2)e^(−iωt), with nothing on the diagonal. For a spin, ω₁ = |γ|B₁. For the linear drive V cos ωt of the previous topic, the co-rotating half has exactly this form with ħω₁ = |V_ab|, so ω₁ = |V_ab|/ħ in both languages. With δ = ω − ω₀, quote UQMII-03.04 in one line: P(t) = (ω₁²/Ω²) sin²(Ωt/2), Ω = √(ω₁² + δ²), exact for this drive because the rotating frame makes the Hamiltonian static. It is not re-derived here and serves only as the answer key. Everything else comes from this unit's exact coefficient equations, iħ dc_b/dt = ⟨b|H′|a⟩e^(iω₀t) cₐ and iħ dcₐ/dt = ⟨a|H′|b⟩e^(−iω₀t) c_b, which hold before any approximation is made.

Derive the first-order amplitude

First order puts the starting values cₐ = 1 and c_b = 0 into the right-hand side and integrates once: c_b⁽¹⁾(t) = −(i/ħ)∫₀ᵗ ⟨b|H′(t′)|a⟩e^(iω₀t′) dt′ = −(iω₁/2)∫₀ᵗ e^(−iδt′) dt′. The integral is (1 − e^(−iδt))/(iδ) = e^(−iδt/2) × 2 sin(δt/2)/δ, so c_b⁽¹⁾ = −i e^(−iδt/2)(ω₁/δ) sin(δt/2) and P₁ = (ω₁²/δ²) sin²(δt/2). Dimensions: ω₁/δ is a ratio of angular frequencies and δt an angle, so P₁ is a pure number. Limit: as δ → 0, sin(δt/2)/δ → t/2 and P₁ → (ω₁t/2)², growing as t² because every instant of a resonant drive adds amplitude in phase. Freezing cₐ = 1 made two approximations at once. It ignored the depletion of |a⟩, whose amplitude must fall as |b⟩ fills, and it ignored the shift the drive itself produces in the two levels. Each leaves its own signature in the comparison.

Where they agree: the leading order in ω₁

Expand the benchmark in the coupling at fixed t and δ ≠ 0. Ω = |δ|(1 + ω₁²/2δ² + …), so ω₁²/Ω² = (ω₁²/δ²)(1 − ω₁²/δ² + …) and sin²(Ωt/2) = sin²(δt/2) + O(ω₁²). The leading term is P₁ exactly: agreement at order ω₁² is built in. The first error is of order ω₁⁴, not ω₁³, because the second-order Dyson amplitude into |b⟩ needs ⟨b|H′|b⟩ or ⟨a|H′|a⟩, both zero, so the correction to c_b starts at third order. At short times, Ωt ≪ 1, expand both sines: P = (ω₁t/2)²(1 − Ω²t²/12) and P₁ = (ω₁t/2)²(1 − δ²t²/12), so P₁/P = 1 + ω₁²t²/12. The detuning cancels because Ω² − δ² = ω₁². First order errs high, by an amount set by the Rabi angle ω₁t alone: 1% at ω₁t = 0.35, and roughly 10% near ω₁t = 1.1.

On resonance, first order breaks the norm

At δ = 0 the benchmark is P = sin²(ω₁t/2) and first order is its small-angle form, (ω₁t/2)². Solving x²/sin²x = 1.10 numerically, with x = ω₁t/2, puts the 10% error at ω₁t = 1.06. P₁ reaches 1 at ω₁t = 2 and keeps climbing, while the exact P turns over at the π pulse, ω₁t = π, and returns to zero at 2π, where first order claims π² = 9.9. A probability above 1 is the depletion error made visible: first order still holds |cₐ|² = 1, so its total probability is 1 + (ω₁t/2)². The condition for first order on resonance is therefore ω₁t ≪ 1, that is |V_ab|t/ħ ≪ 1, derived from the comparison rather than assumed. The measured check is routine: NMR spectrometers calibrate B₁ by locating the null of the nutation at ω₁t = 2π, the 360° pulse. That null exists only because the true probability oscillates; a parabola has none.

Off resonance, bounded but out of step

For |δ| ≫ ω₁ both results stay small, P₁ ≤ ω₁²/δ² and P ≤ ω₁²/Ω², and the first-order ceiling is high by the fraction ω₁²/δ², 4% at δ = 5ω₁. A small probability does not make first order right, though, because the two oscillate at different frequencies: |δ| and Ω = |δ| + ω₁²/2|δ| + …. The excess is physics first order cannot see. In the rotating frame the Hamiltonian is (ħ/2)(ω₁σₓ − δσ_z); second-order time-independent perturbation theory from Unit 6 moves its levels ∓ħδ/2 by ∓ħω₁²/4δ, widening the splitting by ħω₁²/2|δ|. That is the drive-induced level shift, and it appears only at second order in the energy. Hence the phase slip (Ω − |δ|)t/2 ≈ ω₁²t/4|δ| grows without limit. At δ = 5ω₁ it reaches 1 rad at ω₁t = 20 and π/2 at ω₁t = 32, where the two probabilities, oscillating as cos δt and cos Ωt, are half a cycle apart and first-order peaks fall on exact troughs.

The criterion, and what it leaves out

The two failures combine into one number, κ = ω₁²t min(t, 1/|δ|). When |δ|t ≲ 1 the error is the Rabi-angle term, P₁/P − 1 ≈ κ/12; when |δ|t ≫ 1 it is the phase slip, κ/4 rad, which caps the error near κ/4 of the ceiling. First order is trustworthy while κ ≪ 1: on resonance until t ≈ 1/ω₁, at detuning |δ| until t ≈ |δ|/ω₁², longer by the factor |δ|/ω₁. No power of α appears, since this is an expansion in ω₁t and ω₁/δ; the fine-structure constant enters only through the coupling, ħω₁ = gμ_B B₁ for an electron spin, with g = 2 quoted from relativistic quantum mechanics and its α/2π correction quoted from QED. One boundary remains. Unit 3's field rotates, so the Rabi formula needed no rotating-wave approximation. A linear drive adds a counter-rotating term to c_b⁽¹⁾, at most ω₁/(ω₀ + ω) in size, which the benchmark omits; that comparison is left to the lab, where the full equations are integrated numerically.

Change one variable at a time

Make the relationship visible.

Interactive model

Detuning δ/ω₁0.00

Rabi angle ω₁t1.0 rad

On resonance, raise ω₁t: the dashed parabola hugs the solid curve below ω₁t ≈ 1, crosses the ceiling P = 1 at ω₁t = 2 and leaves the chart while the exact P turns over at π. Then set δ = 3ω₁: both stay under their ceilings, but the dashed curve slips about a radian out of step by 4π.

EXACT P0.230

FIRST ORDER P₁0.250

PHASE SLIP (Ω − |δ|)t/20.50 rad

CRITERION κ1.00

Live interpretationEXACT P: 0.230. FIRST ORDER P₁: 0.250. PHASE SLIP (Ω − |δ|)t/2: 0.50 rad. CRITERION κ: 1.00

Catch the common trap

Explain before calculating.

Misconception check

The instinct carried over from time-independent perturbation theory is that a perturbation is weak when its matrix element is small beside the level spacing, |V_ab| ≪ ħω₀, and that first order is then good at every time. Neither the exact nor the first-order probability contains ω₀ at all, only ω₁ and δ, so that comparison cannot be the criterion. An ESR drive at 9.6 GHz with ω₁/ω₀ = 10⁻⁴ is weak by that standard, yet on resonance its first-order probability passes 1 after 2/ω₁ = 332 ns. The small quantity is |V_ab|t/ħ on resonance and ω₁²t/|δ| off it: time enters, because a weak drive acting coherently for long enough does everything a strong one does. The companion error, that a small probability guarantees an accurate one, fails off resonance, where P₁ never exceeds ω₁²/δ² but its phase keeps slipping.

A two-level system starting in |a⟩ is driven at detuning δ = 10ω₁. First order puts the ceiling at ω₁²/δ² = 0.0100, and the exact Rabi ceiling is ω₁²/Ω² = 0.0099. For which times does the first-order P(t) track the exact one?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA proton spin (γ/2π = 42.577 MHz T⁻¹, measured) starts in its lower state and is driven exactly on resonance by a field B₁ = 0.100 mT rotating with its precession. Compare the first-order and exact flip probabilities after 20.0 μs and after 60.0 μs, and find when the first-order value reaches 1.

Coupling: ω₁ = γB₁ = 2π × 42.577 × 10⁶ × 1.00 × 10⁻⁴ = 2.675 × 10⁴ rad s⁻¹. Units: rad s⁻¹ T⁻¹ times T.
At δ = 0 the derived first-order result is P₁ = (ω₁t/2)², and the quoted exact result is P = sin²(ω₁t/2).
At 20.0 μs: ω₁t = 0.535, so P₁ = (0.2675)² = 0.0716 and P = sin²(0.2675) = 0.0699. First order is 2.4% high, as the short-time estimate ω₁²t²/12 = 0.024 predicts.
At 60.0 μs: ω₁t = 1.605, so P₁ = (0.8026)² = 0.644 but P = sin²(0.8026) = 0.517. First order is now 25% high.
P₁ = 1 when ω₁t = 2, at t = 2/ω₁ = 74.8 μs, where the exact P = sin²(1) = 0.708. The true probability reaches 1 only at the π pulse, π/ω₁ = 117 μs.

AnswerP₁ = 0.0716 against P = 0.0699 at 20.0 μs; P₁ = 0.644 against P = 0.517 at 60.0 μs; P₁ reaches 1 at 74.8 μs, where the exact P is 0.708.

MediumAn ESR nutation measurement on a radical with the free-electron value |γ|/2π = 28.025 GHz T⁻¹ (measured) drives the spin on resonance with a rotating field B₁. The flip probability first reaches 0.500 after 25.0 ns. Infer B₁ from the exact solution and from first order, find the longest pulse first order describes to within 10%, and compare the first-order error with the order-α correction inside γ.

Exact: sin²(ω₁t/2) = 0.500 first at ω₁t/2 = π/4, so ω₁ = π/(2 × 25.0 ns) = 6.283 × 10⁷ rad s⁻¹ and B₁ = ω₁/|γ| = 6.283 × 10⁷ ÷ (2π × 2.8025 × 10¹⁰) T = 0.357 mT.
First order: (ω₁t/2)² = 0.500 gives ω₁t/2 = 0.7071, so ω₁ = 5.657 × 10⁷ rad s⁻¹ and B₁ = 0.321 mT, low by the factor 2√2/π = 0.900.
Why 10%: with the true ω₁, first order predicts (π/4)² = 0.617 at 25.0 ns instead of 0.500, 23% high, and because P₁ ∝ ω₁² a fit absorbs that as the factor √1.234 = 1.11 in ω₁.
The 10% window: P₁/P = x²/sin²x with x = ω₁t/2 equals 1.10 at x = 0.532, solved numerically (the estimate 1 + ω₁²t²/12 gives 0.548). So t = 2 × 0.532/(6.283 × 10⁷ rad s⁻¹) = 16.9 ns, and the 25.0 ns point lies outside it.
Order in α: |γ| = gμ_B/ħ, with g = 2 quoted from Dirac's relativistic theory and g/2 − 1 = α/2π = 0.00116 quoted from leading-order QED (measured 0.0011597). Taking g = 2 would move B₁ by 0.116%, about 86 times less than first order's 10%.

AnswerB₁ = 0.357 mT from the exact solution and 0.321 mT from first order, 10% low; first order holds to 10% only up to 16.9 ns; the QED correction to g moves B₁ by 0.12%.

HardA proton spin is driven with ω₁/2π = 2.00 kHz at a detuning δ/2π = 12.0 kHz. Compare the two ceilings, find when first order has slipped half a cycle out of step, compare the two probabilities after a 1.00 ms pulse, and find how far off resonance the drive must sit for first order to hold its phase to 0.10 rad over that pulse.

Ratios first: δ/ω₁ = 6, so the first-order ceiling is ω₁²/δ² = 1/36 = 0.0278 and the exact one is ω₁²/Ω² = 1/37 = 0.0270. They differ by only 2.8%, the fraction ω₁²/δ².
Frequencies: ω₁ = 1.2566 × 10⁴ rad s⁻¹ and δ = 7.5398 × 10⁴ rad s⁻¹, so Ω = √(ω₁² + δ²) = 7.6438 × 10⁴ rad s⁻¹ and Ω − δ = 1.040 × 10³ rad s⁻¹, against the derived estimate ω₁²/2δ = 1.047 × 10³ rad s⁻¹.
The phases δt/2 and Ωt/2 part by π/2 at t = π/(Ω − δ) = 3.02 ms, so the probabilities, oscillating as cos δt and cos Ωt, are half a cycle apart: first-order peaks sit on exact troughs, although both probabilities stay below 0.03.
At 1.00 ms: δt/2 = 12π exactly, so P₁ = 0, while Ωt/2 = 38.219 rad = 12π + 0.520 rad and P = (1/37) sin²(0.520) = 0.00667, a quarter of the exact ceiling that first order misses entirely. Check: κ = ω₁²t/δ = 2.09, and κ/4 = 0.52 rad is that slip.
For a 0.10 rad slip at 1.00 ms, ω₁²t/4δ = 0.10 needs δ = ω₁²t/0.40 = 3.95 × 10⁵ rad s⁻¹, so δ/2π = 62.8 kHz, or 31.4ω₁; the exact slip condition agrees to three figures. The ceiling error there is ω₁²/δ² = 0.10%, so over a long pulse the phase, not the ceiling, sets the detuning.

AnswerCeilings 0.0278 (first order) and 0.0270 (exact); a half-cycle slip at 3.02 ms; at 1.00 ms P₁ = 0 while P = 0.00667; holding the slip to 0.10 rad needs δ/2π ≥ 62.8 kHz, about 31ω₁.