University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.3
Larmor precession and the electron and proton rates
The first thing a spin Hamiltonian does is make the spin turn. Three commutators turn H = −γS⋅B into a gyroscope equation, the Bloch sphere of Quantum Mechanics I starts to rotate, and two quoted numbers, 28.0 GHz and 42.6 MHz per tesla, set the clock for every ESR and NMR experiment that follows.
Build the model
Connect the measurement to the mechanism.
Give a spin the Hamiltonian H = −γS⋅B of the previous topic, with B uniform, and the Heisenberg equation needs nothing but the angular-momentum commutators to return dS/dt = γS × B: the spin operator turns about the field at ω₀ = |γ|B₀, while its component along B commutes with H and is frozen. Because that equation is linear in S, the expectation value obeys it exactly, so ⟨S⟩ sweeps a cone of fixed half-angle like a classical gyroscope, with no correction in ħ for any state or any field strength. The Schrödinger picture tells the same story in another language: |↑⟩ and |↓⟩ are stationary states split by ħω₀, so their populations never change while their relative phase advances at ω₀, and that relative phase is the azimuth on the Bloch sphere.
What the derivation cannot supply is γ. It enters through quoted g-factors: for the electron, 2 from the Dirac equation raised by a fraction close to α/2π by quantum electrodynamics; for the proton, a measured 5.586. They give 28.025 GHz T⁻¹ and 42.577 MHz T⁻¹, in opposite senses.
The picture of one isolated spin in a uniform classical field stops working when the field varies across the particle, so spin couples to motion; when the spin trades energy with its surroundings, so populations relax and phases dephase; and inside atoms and molecules, whose electrons change both the field the spin feels and the moment that turns.
- Simple definition
- Larmor precession is the rotation of a spin's expectation value ⟨S⟩ about a uniform magnetic field at the angular frequency ω₀ = |γ|B₀, with its component along the field, and so its tilt from the field, held fixed.
- Example
- A proton in the 3.00 T magnet of a clinical MRI scanner precesses at 42.577 MHz T⁻¹ × 3.00 T = 127.7 MHz, one turn every 7.83 ns; an electron in the same field turns 658 times faster, at 84.07 GHz, and in the opposite sense.
Derived from [Sᵢ, Sⱼ] = iħεᵢⱼₖSₖ alone. The right side is linear in S, so ⟨S⟩ obeys the same equation with no correction in ħ.
S in J s, γ = gq/2m in s⁻¹ T⁻¹ with g quoted, B in T; both sides in J. B may vary in time but not in space.
The sense follows from the sign of γ alone: counterclockwise seen from the tip of B for the electron, clockwise for the proton.
Ω in rad s⁻¹, |Ω| = ω₀ = |γ|B₀. A negative γ (electron) puts Ω along the field; a positive γ (proton) puts it against.
θ and ⟨Sz⟩ never change. Only the azimuth turns, and its rate does not depend on the tilt.
θ, φ₀ in rad, set by the initial state; B₀ in T, t in s. A mixed state traces the same cone with a shorter vector.
Populations stay cos²(θ/2) and sin²(θ/2); the relative phase is the Bloch azimuth, so precession is the Bohr frequency ΔE/ħ seen geometrically.
|↑⟩, |↓⟩ are Sz eigenstates with energies ∓γħB₀/2 in J. The overall phase has no effect on any single-spin measurement.
The ratio 658.2 is the ratio of the two moments. With gₑ = 2 exactly the electron would give 27.992 GHz T⁻¹, 0.116% low: the order-α term.
f₀ in Hz. Quoted: electron 28.025 GHz T⁻¹ (gₑ = 2.00232), proton 42.577 MHz T⁻¹ (gₚ = 5.5857).
The velocity turns at ωc, the spin at (g/2)ωc. The electron's spin gains half a turn on its velocity every 1/(2aₑ) = 431 orbits, whatever the field.
All in rad s⁻¹; aₑ = 1.1597 × 10⁻³ (quoted). Velocity and spin turn in the same sense; non-relativistic speeds.
Three commutators give the equation of motion
Take the Hamiltonian of the previous topic as given: H = −μ⋅B = −γS⋅B, with γ = gq/2m the ratio of moment to spin and g quoted, not derived. Put B = B₀ẑ, uniform and static, so H = −γB₀Sz. The Heisenberg equation dA/dt = (i/ħ)[H, A] then needs only [Sz, Sₓ] = iħSy and [Sz, Sy] = −iħSₓ: dSₓ/dt = (i/ħ)(−γB₀)(iħSy) = γB₀Sy, dSy/dt = −γB₀Sₓ, and dSz/dt = 0 because Sz commutes with H. For a uniform field in any direction the same algebra, with [Sᵢ, Sⱼ] = iħεᵢⱼₖSₖ, collects into dS/dt = γS × B, derived here with no approximation. The dimensions check: γB₀ is s⁻¹ T⁻¹ × T = s⁻¹, a rate, and both sides carry J. The field may even change in time, provided it stays uniform, and the equation keeps its form; that is the door the next topic opens for magnetic resonance.
The solution is a cone of fixed angle
Differentiate once more: d²Sₓ/dt² = γB₀ dSy/dt = −(γB₀)²Sₓ, an oscillator equation. Keeping ω = γB₀ signed, Sₓ(t) = Sₓ cos ωt + Sy sin ωt and Sy(t) = Sy cos ωt − Sₓ sin ωt, operator identities valid in every state. Take expectation values in the state whose Bloch vector starts at polar angle θ and azimuth φ₀: ⟨S⟩ = (ħ/2)(sin θ cos φ, sin θ sin φ, cos θ) with φ = φ₀ − γB₀t. The tilt θ, ⟨Sz⟩ and the length of ⟨S⟩ never change; only the azimuth turns, at ω₀ = |γ|B₀ whatever the tilt. The energy −γB₀⟨Sz⟩ is fixed too, as it must be for a time-independent H. At 1.00 T the electron's ⟨S⟩ goes round once every 35.7 ps and the proton's once every 23.5 ns. Two limits check the picture: at θ = 0 or π the state is an eigenstate of H and the cone collapses onto the axis, and as B₀ → 0 the period grows without bound.
Linear means no quantum correction
Ehrenfest's theorem returns a classical trajectory exactly only when the operator equation is linear, because only then is ⟨f(S)⟩ = f(⟨S⟩). Here it is: with B a c-number, γS × B depends linearly on S, so ⟨S⟩ obeys dS/dt = γS × B exactly, in any state, pure or mixed, at any field strength. The classical gyroscope is the quantum answer for the mean, with no correction in ħ. The quantum content sits elsewhere. ⟨S⟩ has length ħ/2 in a pure state while √⟨S²⟩ = (√3/2)ħ, so the spin is not an arrow of either length; any component still measures ±ħ/2; and ΔSₓΔSy ≥ (ħ/2)|⟨Sz⟩| holds throughout. In a pure state the turning direction n̂(t) is the axis along which S is certain to give +ħ/2. Linearity fails once B depends on position: ⟨S × B(r)⟩ is not ⟨S⟩ × B(⟨r⟩), and spin becomes correlated with motion, which is the Stern–Gerlach story.
The same motion as a relative phase
The Schrödinger picture must agree, and it shows where the rate comes from. |↑⟩ and |↓⟩ are eigenstates of H = −γB₀Sz with energies ∓γħB₀/2, split by ħω₀. Start from cos(θ/2)|↑⟩ + e(iφ₀) sin(θ/2)|↓⟩ and let each coefficient pick up e(−iEt/ħ). The magnitudes, and with them the populations, never move; only the relative phase advances, as φ₀ − γB₀t. On the Bloch sphere of Quantum Mechanics I that relative phase is the azimuth, so Larmor precession is the two-level Bohr frequency ΔE/ħ drawn as a rotation. For the electron at 1.00 T, ΔE = 115.9 μeV = h × 28.02 GHz. One detail has no classical partner: after one full turn, t = 2π/ω₀, the factored-out phase e(iγB₀t/2) has become −1, so the spinor returns as −|χ⟩. No single-spin measurement sees that sign, but neutron interferometers did in 1975: precession through 2π in one arm reversed the interference and 4π restored it.
Rates for the electron and the proton
The derivation fixes ω₀ = |γ|B₀ but not γ, which enters through quoted g-factors. For the electron, g = 2 comes from the Dirac equation of relativistic quantum mechanics, and the extra 0.00232 from quantum electrodynamics, whose first term is α/π; with gₑ = 2.00232, |γₑ|/2π = 28.025 GHz T⁻¹. Setting g = 2 would give 27.992 GHz T⁻¹, 0.116% low: an order-α correction, far larger than ESR's precision. For the proton, gₚ = 5.586 is a measured number, nowhere near the 2 a structureless Dirac particle would have, because the proton is made of quarks; γₚ/2π = 42.577 MHz T⁻¹. The rates differ by 658.2, the ratio of the two moments. The senses differ too: μₑ points against Sₑ, so γₑ is negative and Ω = −γB lies along B, counterclockwise seen from its tip, while the proton turns clockwise. Measured anchors: X-band ESR at 9.50 GHz needs 0.339 T, a 1.5 T MRI scanner runs at 63.9 MHz, and Earth's 50 μT turns protons at 2.13 kHz.
Checks, and where the isolated spin stops
Every rate above passes three checks. Dimensions: γ in s⁻¹ T⁻¹ times B in T is a rate. Limits: no field, no turning, and a spin along ±z is stationary. Order in α: the electron's rate departs from its g = 2 value at first order in α, by a fraction 1.16 × 10⁻³. A sharper check comes from the electron's own orbit. In a uniform field its velocity turns at ωc = eB₀/m and its spin at (gₑ/2)ωc in the same sense, so the spin gains on the velocity at aₑωc with aₑ = (gₑ − 2)/2; Penning-trap measurements of that slip now fix gₑ to about one part in 10¹³. The model itself is one spin in a uniform classical field with nothing else present. It stops working when the spin trades energy with its surroundings, so populations relax and phases dephase; when neighbouring electrons shield the field, as the 26 ppm shift for protons in water shows; and inside atoms, where the moment that precesses is set by the total angular momentum and a Landé factor.
Change one variable at a time
Make the relationship visible.
Put the electron in 1.5 mT and the proton in 1.0 T: the rates nearly match, 42.0 and 42.6 MHz, though the fields differ 667-fold, and the two ⟨Sy⟩ curves become near mirror images because the spins turn opposite ways. Set either field to zero and its arrow stops.
ELECTRON f₀28.0 MHz
PROTON f₀42.6 MHz
ELECTRON TURNS (+ = CCW)0.14 turns
PROTON TURNS (+ = CCW)-0.21 turns
Live interpretationELECTRON f₀: 28.0 MHz. PROTON f₀: 42.6 MHz. ELECTRON TURNS (+ = CCW): 0.14 turns. PROTON TURNS (+ = CCW): −0.21 turns
Catch the common trap
Explain before calculating.
A proton spin sits in a uniform field B₀ẑ whose strength makes its Larmor frequency f₀ = 100 MHz. At t = 0 it is prepared with ⟨S⟩ along +x. At t = 2.50 ns, Sy is measured. What is found?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA proton-precession magnetometer on a field survey records a free-precession signal at 2105.3 Hz from the protons in its liquid sample. Treating them as free protons, find the magnetic field, then the rate and sense at which a free electron would precess in the same field.
- The signal frequency is the Larmor frequency f₀ = |γₚ|B₀/2π, with the quoted γₚ/2π = 42.577 MHz T⁻¹. It is quoted, not derived, because gₚ is a measured number.
- B₀ = f₀/(γₚ/2π) = 2105.3 Hz ÷ 4.2577 × 10⁷ Hz T⁻¹ = 4.9447 × 10⁻⁵ T = 49.45 μT. Units: Hz ÷ (Hz T⁻¹) = T.
- Protons in a liquid are shielded by about 26 ppm, 0.0013 μT here, below the last digit kept, so the free-proton value is safe.
- Electron: f = 28.025 GHz T⁻¹ × 4.9447 × 10⁻⁵ T = 1.386 MHz, 658 times the proton's rate. Its γ is negative, so it turns counterclockwise seen from the tip of B, opposite to the proton.
- Check: 49.45 μT lies inside Earth's range of about 25 to 65 μT, and the proton period of 0.475 ms is slow enough to count with ordinary electronics, part of why these instruments use protons.
AnswerB₀ = 49.45 μT; a free electron would precess at 1.386 MHz, counterclockwise seen from the tip of B, against the proton's clockwise turning.
MediumAn electron in B₀ = 0.250 T along +z is prepared at t = 0 in |χ(0)⟩ = (√3/2)|↑⟩ + (1/2)|↓⟩. Write |χ(t)⟩, find the probability that a measurement of Sₓ at t = 50.0 ps gives +ħ/2, and check the result against the Heisenberg-picture cone.
- With γₑ negative, H = −γₑB₀Sz = +|γₑ|B₀Sz, so E_↑ = +ħω₀/2 and E_↓ = −ħω₀/2, where ω₀ = |γₑ|B₀ = 2π × 28.025 GHz T⁻¹ × 0.250 T = 4.402 × 10¹⁰ rad s⁻¹, f₀ = 7.006 GHz.
- Each coefficient gains e(−iEt/ħ): |χ(t)⟩ = e(−iω₀t/2)[(√3/2)|↑⟩ + (1/2)e(iω₀t)|↓⟩]. The populations 3/4 and 1/4 stay fixed; the relative phase φ = ω₀t grows, counterclockwise seen from +z.
- At t = 50.0 ps, φ = 4.402 × 10¹⁰ rad s⁻¹ × 5.00 × 10⁻¹¹ s = 2.201 rad = 126.1°.
- With |+x⟩ = (|↑⟩ + |↓⟩)/√2, P₊ = ½|√3/2 + (1/2)e(iφ)|² = ½[1 + (√3/2) cos φ] = ½[1 + 0.8660 × (−0.5894)] = 0.2448.
- Heisenberg check: cos θ = 3/4 − 1/4 = 1/2, so θ = 60° and ⟨Sₓ⟩ = (ħ/2) sin 60° cos φ = −0.510(ħ/2), giving P₊ = [1 + ⟨Sₓ⟩/(ħ/2)]/2 = 0.245 ✓. ⟨Sz⟩ = ħ/4 has not moved.
Answer|χ(t)⟩ = e(−iω₀t/2)[(√3/2)|↑⟩ + (1/2)e(iω₀t)|↓⟩] with ω₀ = 4.402 × 10¹⁰ rad s⁻¹; P(Sₓ = +ħ/2) = 0.245 at 50.0 ps, the azimuth having reached 126°.
HardAn electron circles at non-relativistic speed in a uniform 1.50 T field, with its spin expectation value lying in the orbital plane and pointing along its velocity at t = 0. Using the quoted gₑ = 2.0023193, find how long the spin takes to point against the velocity and how many orbits that is, and say what gₑ = 2 and Schwinger's aₑ = α/2π would each predict for that count.
- Velocity: the Lorentz force m dv/dt = qv × B turns v about B at ωc = eB₀/m = 1.75882 × 10¹¹ C kg⁻¹ × 1.50 T = 2.63823 × 10¹¹ rad s⁻¹, fc = 41.9887 GHz. For q = −e the rotation vector −(q/m)B lies along B, so the turn is counterclockwise seen from its tip.
- Spin: Ω = −γₑB with γₑ = −gₑe/2m also lies along B, so the spin turns the same way at ωₛ = (gₑ/2)ωc: fₛ = 1.0011597 × 41.9887 GHz = 42.0374 GHz, the same as 28.02495 GHz T⁻¹ × 1.50 T.
- The angle between spin and velocity grows at ωₐ = ωₛ − ωc = aₑωc, with aₑ = (gₑ − 2)/2 = 1.1597 × 10⁻³: fₐ = aₑfc = 1.1597 × 10⁻³ × 41.9887 GHz = 48.69 MHz, the 0.0487 GHz gap between the two rates above.
- Antiparallel after half a slip turn: t = 1/(2fₐ) = 1/(2 × 4.869 × 10⁷ Hz) = 10.27 ns. Orbits in that time: fc t = fc/(2aₑfc) = 1/(2aₑ) = 431.2, independent of both B₀ and the speed.
- Limits: with gₑ = 2 exactly, as the Dirac equation gives, ωₐ = 0 and the spin stays locked to the velocity forever. Schwinger's first QED term, aₑ ≈ α/2π = 1.1614 × 10⁻³, quoted, gives 1/(2aₑ) = 430.5 orbits, 0.15% fewer; the measured aₑ carries the higher orders in α.
Answerfₐ = 48.69 MHz, so the spin points against the velocity after 10.27 ns, 431 orbits, a count that is the same in any field. gₑ = 2 gives no slip at all; α/2π alone gives 430.5 orbits.