University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.4

Magnetic resonance solved exactly in the rotating frame

Every NMR magnet, ESR spectrometer and spin qubit runs on one calculation: a spin in a strong static field, nudged by a weak field that turns with it. Step into the frame that turns too and the problem becomes a static two-state system, so the flip probability, the line width and the pulse length all come out exact.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A spin in B₀ẑ precesses at ω₀ = |γ|B₀ and never changes its S_z populations, because S_z commutes with its Hamiltonian. Moving population needs a transverse field, and a weak one only accumulates an effect if it keeps pace with the precession, so the drive is a field B₁ rotating about z at ω. That makes H(t) time-dependent and puts exp(−iHt/ħ) out of reach, but H(t) is only a rotated copy of one static Hamiltonian. Write the state as exp(−iωtS_z/ħ) acting on a rotating-frame ket, and that ket obeys a Schrödinger equation with the constant Hamiltonian (ω₀ − ω)S_z + ω₁Sₓ: the frame change subtracts ω/|γ| from B₀ and freezes B₁ into a fixed transverse field.

Quantum Mechanics I's two-state propagator then applies with no approximation, and Rabi's formula follows: the flip probability oscillates at Ω = √(ω₁² + δ²) between 0 and a ceiling ω₁²/Ω². The unitary is diagonal in S_z, so the populations it predicts are the lab-frame populations. The exactness has edges. It holds for a circularly polarised drive; a linear coil adds a counter-rotating field that can be dropped only when ω₁ ≪ ω₀.

It describes one isolated spin, with no relaxation to damp the oscillation. And the γ it uses is quoted, not derived: the electron's g = 2.0023 comes from relativistic quantum mechanics plus QED.

Simple definition: Magnetic resonance is the flipping of a spin in a static field B₀ by a weak field B₁ rotating about B₀ near the Larmor frequency; in the frame that turns with B₁ the Hamiltonian is static, so the flip probability follows exactly.
Example: A proton in 1.50 T precesses at 63.87 MHz. A 10.0 μT field rotating at that rate gives ω₁ = |γ|B₁ = 2.68 × 10³ rad s⁻¹, so a pulse of π/ω₁ = 1.17 ms inverts the spin, and the resonance envelope is 2ω₁/2π = 852 Hz wide.

Driven-spin Hamiltonian in the laboratoryH(t) = ω₀S_z + ω₁(Sₓ cos ωt + S_y sin ωt)

B₁ turns with the precession. H at different times do not commute, so exp(−iHt/ħ) is not the propagator.

ω₀ = |γ|B₀ and ω₁ = |γ|B₁ in rad s⁻¹, ω the drive's angular frequency, S in J s. Written for negative γ, the electron

Rotating-frame Hamiltonian|ψ⟩ = exp(−iωtS_z/ħ)|ψ̃⟩ gives H_eff = (ω₀ − ω)S_z + ω₁Sₓ

Derived. H_eff is static, so the two-state propagator applies exactly, and the unitary is diagonal in S_z, so populations match in both frames.

ψ̃ is the rotating-frame ket. The −ωS_z comes from differentiating the unitary, a fictitious field ω/|γ| against B₀

Rabi's formulaP(t) = [ω₁²/(ω₁² + δ²)] sin²(√(ω₁² + δ²) t/2)δ = ω − ω₀

Derived, exact for a circular drive: the probability that a spin starting in |↓⟩ is found in |↑⟩, in either frame.

ω₁ and δ in rad s⁻¹, t in s, P dimensionless. Ω = √(ω₁² + δ²) is the generalised Rabi frequency

Resonance envelopePₘₐₓ(δ) = ω₁²/(ω₁² + δ²)

The width belongs to the drive, not the spin: doubling B₁ doubles the line. Off resonance the flip is capped, not slowed.

a Lorentzian in δ of height 1 and full width 2ω₁ = 2|γ|B₁ in rad s⁻¹, or 2B₁ in a field sweep

π pulset_π = π/ω₁ = π/(|γ|B₁)

A π pulse inverts the populations. Half of it, a π/2 pulse, makes the equal superposition whose precessing moment is the NMR or ESR signal.

t_π in s, on resonance. Proton with B₁ = 0.80 mT: ω₁ = 2.14 × 10⁵ rad s⁻¹ and t_π = 14.7 μs

Linear drive and the rotating-wave approximation2B₁ cos ωt x̂ = B₁(x̂ cos ωt + ŷ sin ωt) + B₁(x̂ cos ωt − ŷ sin ωt)

A linear coil drives either sign of γ, since one half always co-rotates. The other half turns at −2ω in the rotating frame and averages away.

Keep the half that co-rotates and drop the other when ω₁ ≪ ω₀. Quoted cost: the line moves up by ω₁²/4ω₀ (Bloch–Siegert)

Why the drive has to rotate

Start from the Larmor topic: in B₀ẑ alone, H₀ = −γB₀S_z, S_z is conserved, and ⟨S⟩ circles the field at ω₀ = |γ|B₀ without changing the populations of |↑⟩ and |↓⟩. Take the electron, whose γ is negative, so H₀ = +ω₀S_z and the circling is anticlockwise seen from +z. A static transverse field B₁ only tilts the precession axis by B₁/B₀, so the flip probability never exceeds about (B₁/B₀)², some 10⁻⁷ when B₁/B₀ = 3 × 10⁻⁴. A field that turns with the spin pushes it the same way on every cycle instead, so add B₁(x̂ cos ωt + ŷ sin ωt). With ω₁ = |γ|B₁ the Hamiltonian is H(t) = ω₀S_z + ω₁(Sₓ cos ωt + S_y sin ωt). It depends on time, its values at different times do not commute, and exp(−iHt/ħ) is simply wrong. For a proton, with positive γ, the free spin turns clockwise, so B₁ must too, and every probability below is unchanged.

Change to the frame that turns with B₁

The rotation operator R(t) = exp(−iωtS_z/ħ) turns Sₓ into Sₓ cos ωt + S_y sin ωt, so H(t) = R(t)(ω₀S_z + ω₁Sₓ)R(t)†: one static Hamiltonian, rotated. Define the rotating-frame ket by |ψ(t)⟩ = R(t)|ψ̃(t)⟩ and substitute into iħ∂|ψ⟩/∂t = H|ψ⟩. The product rule adds iħ(∂R/∂t)|ψ̃⟩ = ωS_zR|ψ̃⟩, which moves to the right-hand side, and multiplying by R† leaves iħ∂|ψ̃⟩/∂t = [(ω₀ − ω)S_z + ω₁Sₓ]|ψ̃⟩. The time dependence has gone. Read as a field, the rotating observer sees B_eff = (B₀ − ω/|γ|)ẑ + B₁x̂: the ω/|γ| is a fictitious field, the spin's version of the Coriolis force, and at ω = ω₀ it cancels B₀ exactly. One more fact makes this usable: R(t) is diagonal in the S_z basis, so it only multiplies |↑⟩ and |↓⟩ by phases, and the probabilities of up and down are identical in both frames.

Quantum Mechanics I finishes the job, exactly

With δ = ω − ω₀ the rotating-frame Hamiltonian is (ħ/2)(ω₁σₓ − δσ_z) = (ħΩ/2) n̂⋅σ, where Ω = √(ω₁² + δ²) and n̂ = (ω₁, 0, −δ)/Ω. Quote the two-state propagator, exp(−iΩt n̂⋅σ/2) = cos(Ωt/2) I − i sin(Ωt/2) n̂⋅σ. Its amplitude from |↓⟩ to |↑⟩ is −i(ω₁/Ω) sin(Ωt/2), so P(t) = (ω₁²/Ω²) sin²(Ωt/2), with no approximation for a circular drive. On the Bloch sphere the rotating-frame vector cones around n̂, which sits at angle θ from the z axis with sin θ = ω₁/Ω. Starting at a pole it swings out to 2θ and back, so the largest P is sin²θ. Dimensions: ω₁, δ and Ω are each |γ| times a field, in rad s⁻¹, so Ωt is a pure angle. Limit: put ω = 0, a static transverse field, and δ = −ω₀ makes the ceiling B₁²/(B₁² + B₀²), the tilt met in the first section.

Reading the line: height, width and the π pulse

The prefactor ω₁²/(ω₁² + δ²) is the ceiling on the flip at detuning δ. As a function of δ it is a Lorentzian of height 1 and full width 2ω₁, and since δ and ω₁ both carry the factor |γ|, a field sweep at fixed frequency sees a line 2B₁ wide for any spin. The drive sets the width: double B₁ and the line doubles, which is power broadening. On resonance P = sin²(ω₁t/2), so a pulse of π/ω₁ inverts the spin and one of π/2ω₁ leaves the equal superposition that NMR detects. For a proton with B₁ = 0.80 mT, ω₁ = 2.14 × 10⁵ rad s⁻¹: the π pulse lasts 14.7 μs, the π/2 pulse 7.34 μs, and the envelope is 68.1 kHz wide. A line recorded with a fixed pulse is not the envelope, though. A π pulse scanned in δ gives (ω₁²/Ω²) sin²(πΩ/2ω₁): solved numerically, its full width at half height is 1.60ω₁, and it falls to zero wherever Ω = 2ω₁, 4ω₁ and so on.

A linear coil and the rotating-wave approximation

Most coils make a linearly polarised field, 2B₁ cos ωt x̂, which is exactly B₁ turning anticlockwise plus B₁ turning clockwise. In the frame that follows the first, the second turns at −2ω, so the rotating-frame Hamiltonian keeps a term oscillating near 2ω₀. It averages almost to zero when it cycles many times in each Rabi period, so dropping it, the rotating-wave approximation, needs ω₁ ≪ ω₀, that is B₁ ≪ B₀. The leading correction, quoted from Bloch and Siegert's second-order treatment, raises the resonance by ω₁²/4ω₀. At 9.40 T with B₁ = 0.80 mT that is 0.72 Hz on a 400 MHz proton line: invisible. Push to ω₁ = 0.2ω₀ and a numerical integration of the full linear drive lands anywhere from 0.977 to 1.000 at the nominal π pulse, depending on the phase of the drive: besides the steady shift ω₁²/4ω₀ = 0.05ω₁, which alone would cost only 0.0025, the dropped half leaves a ripple at 2ω. Both errors shrink as (ω₁/ω₀)², so at ω₁ = 0.02ω₀ the worst case is already 0.9998. A linear coil also drives either sign of γ, because one half always co-rotates; a circularly polarised coil flips electrons or protons, never both.

Checks, and the benchmark for Unit 12

Let ω₁ → 0 at fixed δ: Ω → |δ| and P → (ω₁/δ)² sin²(δt/2), exactly the first-order result of Unit 12, and at short times P ≈ (ω₁t/2)², the quadratic start of any coherent drive. So the exact formula grades the perturbative one: on resonance first order predicts (ω₁t/2)², which passes 1 once t exceeds 2/ω₁, while the true P turns back at π/ω₁. The only quoted input is γ. The proton's γ/2π = 42.58 MHz T⁻¹ is measured, which is why a 1.50 T scanner transmits at 63.87 MHz. The electron's g = 2 is quoted from relativistic quantum mechanics, and the leading QED term α/2π = 0.00116 makes it 2.0023; the measured g/2 − 1 = 0.0011597 sits 1.8 × 10⁻⁶ below α/2π, the size of the order-α² term. At a fixed 9.600 GHz that order-α shift moves the resonance field by 0.40 mT, five full widths when B₁ = 0.040 mT, so ESR sees it directly. What the model omits is relaxation: real oscillations decay over a time T₂, and the line is never narrower than about 2/T₂ in rad s⁻¹.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotating field B₁50 μT

Detuning f − f₀0.0 kHz

Pulse length t0.23 ms

With zero detuning, set t on the faint π-pulse line and both dots reach 1. Detune by 2 kHz, about one Rabi frequency: the top curve cycles faster yet tops out near one half. Raise B₁ and the dashed line widens while the π pulse shortens: both belong to the drive.

RABI FREQUENCY f₁2.129 kHz

INVERSION PULSE LENGTH235 μs

FLIP CEILING1.000

FLIP AFTER THIS PULSE0.999

Live interpretationRABI FREQUENCY f₁: 2.129 kHz. INVERSION PULSE LENGTH: 235 μs. FLIP CEILING: 1.000. FLIP AFTER THIS PULSE: 0.999

Catch the common trap

Explain before calculating.

Misconception check

Students bring two pictures to this problem, a pulse that always completes the flip if it runs long enough and the damped classical oscillator whose line width is set by its own damping, and expect three things: that off resonance the spin still flips completely, only more slowly; that the resonance width is a property of the spin; and that a stronger drive only makes the flip faster. Rabi's formula contradicts all three. Detuning speeds the oscillation up, to Ω = √(ω₁² + δ²), while capping its height at ω₁²/Ω², because the rotating-frame axis tilts toward z and the Bloch vector can only cone around it. Nothing in this Hamiltonian damps anything, so only the drive can set the width: the envelope's full width is 2ω₁ = 2|γ|B₁, and turning B₁ up broadens the line in proportion. A narrow line needs a weak drive and a long pulse, and in a real sample it is finally limited by relaxation, which this model leaves out.

A spin starts in |↓⟩. A rotating field with Rabi frequency ω₁ is applied at detuning δ = ω₁, for the time π/ω₁ that would be a π pulse on resonance. What is the probability of then finding the spin in |↑⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyProtons (γ/2π = 42.58 MHz T⁻¹) sit in B₀ = 0.800 T and are driven exactly on resonance by a field B₁ = 0.250 mT rotating in their precession sense, starting from the lower state. Find the drive frequency, the Rabi frequency, the π-pulse length, and the probability that a proton has flipped after 30.0 μs.

Resonance means ω = ω₀, so the drive frequency is f₀ = (γ/2π)B₀ = 42.58 MHz T⁻¹ × 0.800 T = 34.06 MHz.
Rabi frequency: ω₁ = γB₁ = 2π × 42.58 × 10⁶ × 2.50 × 10⁻⁴ = 6.688 × 10⁴ rad s⁻¹, so ω₁/2π = 10.65 kHz. Units: (rad s⁻¹ T⁻¹)(T) = rad s⁻¹.
With δ = 0, Rabi's formula reduces to P(t) = sin²(ω₁t/2), which first reaches 1 at t_π = π/ω₁ = π/(6.688 × 10⁴ rad s⁻¹) = 4.697 × 10⁻⁵ s = 47.0 μs.
After 30.0 μs: ω₁t/2 = 6.688 × 10⁴ × 3.00 × 10⁻⁵/2 = 1.003 rad, so P = sin²(1.003) = 0.711.
Check the premises: the drive is circular, so no rotating-wave approximation was made, and ω₁/ω₀ = B₁/B₀ = 3.1 × 10⁻⁴ would make one harmless anyway.

Answerf₀ = 34.06 MHz, ω₁ = 6.69 × 10⁴ rad s⁻¹ (10.6 kHz), t_π = 47.0 μs, and P = 0.711 after 30.0 μs.

MediumAn ESR spectrometer runs at a fixed 9.600 GHz and sweeps B₀. A free electron (|γ|/2π = 28.025 GHz T⁻¹) feels a rotating B₁ = 0.040 mT. Find the resonance field. With B₀ set 0.030 mT above it, find the detuning, the largest flip probability and the time it is first reached, then give the line's full width in field.

Resonance: Bᵣₑₛ = 9.600 GHz ÷ 28.025 GHz T⁻¹ = 0.34255 T, that is 342.55 mT.
Rates: ω₁ = |γ|B₁ = 2π × 28.025 × 10⁹ × 4.0 × 10⁻⁵ = 7.043 × 10⁶ rad s⁻¹. The extra field raises ω₀ above ω, so δ = ω − ω₀ = −|γ| × 3.0 × 10⁻⁵ T = −5.283 × 10⁶ rad s⁻¹, or δ/2π = −0.841 MHz.
Generalised Rabi frequency: Ω = √(ω₁² + δ²) = 8.804 × 10⁶ rad s⁻¹. Since |δ|/ω₁ = 0.75, this is a 3-4-5 triangle and Ω = 1.25ω₁.
Ceiling: Pₘₐₓ = ω₁²/Ω² = 1/1.5625 = 0.640, first reached when Ωt/2 = π/2, at t = π/Ω = 357 ns. That is sooner than the resonant π pulse, π/ω₁ = 446 ns: detuning speeds the oscillation while capping it.
The ceiling falls to one half at |δ| = ω₁, a field offset of B₁ either side, because δ and ω₁ carry the same factor |γ|. The envelope's full width is therefore 2B₁ = 0.080 mT, whatever the spin.

AnswerBᵣₑₛ = 342.55 mT; δ = −5.28 × 10⁶ rad s⁻¹ (−0.841 MHz); Pₘₐₓ = 0.640, first reached at 357 ns; full width 0.080 mT.

HardAn electron spin in B₀ = 0.500 T (|γ|/2π = 28.025 GHz T⁻¹) starts in its lower state |↓⟩. A linearly polarised field 2B₁ cos ωt x̂ of amplitude 0.100 mT is applied 2.00 MHz below the Larmor frequency for 200 ns. Find the flip probability in the rotating-wave approximation, compare first-order perturbation theory, and justify dropping the counter-rotating half.

Split the drive: 2B₁ cos ωt x̂ is B₁ turning anticlockwise plus B₁ turning clockwise. For the electron's negative γ the anticlockwise half co-rotates, so B₁ = 0.0500 mT and ω₁ = |γ|B₁ = 2π × 28.025 × 10⁹ × 5.00 × 10⁻⁵ = 8.804 × 10⁶ rad s⁻¹.
Detuning δ = −2π × 2.00 × 10⁶ = −1.2566 × 10⁷ rad s⁻¹, so Ω = √(ω₁² + δ²) = 1.5344 × 10⁷ rad s⁻¹ and ω₁²/Ω² = 0.3293.
Rabi's formula: Ωt/2 = 1.5344 × 10⁷ × 2.00 × 10⁻⁷/2 = 1.534 rad and sin²(1.534) = 0.9987, so P = 0.3293 × 0.9987 = 0.329.
First order, the ω₁ → 0 limit: P₁ = (ω₁/δ)² sin²(δt/2) = 0.4909 × sin²(1.2566) = 0.4909 × 0.9045 = 0.444. It overshoots by 35% because ω₁t/2 = 0.88 is not small, and the true oscillation runs at Ω, not |δ|.
RWA: with ω₀ = |γ|B₀ = 8.804 × 10¹⁰ rad s⁻¹, the quoted Bloch–Siegert shift ω₁²/4ω₀ is 220 rad s⁻¹, or 35 Hz, about 6 × 10⁴ times smaller than |δ|/2π. Integrating the lab-frame equations numerically with the full linear drive gives 0.3288.
Order in α: with g = 2 in place of the QED value 2.0023, ω₀/2π would come out 16.2 MHz lower, eight times the stated detuning, so the quoted correction decides where the drive must sit.

AnswerP = 0.329 in the rotating-wave approximation, and 0.3288 with the full linear drive; first order gives 0.444, 35% too high; the dropped half shifts the line by only 35 Hz.