University Physics IV · Angular Momentum and Spin · 9.7
The Spin g Factor & Magnetic Resonance
Orbital motion fixes the exchange rate between angular momentum and magnetic moment; spin breaks it by a factor of two. This lesson turns that anomaly into an instrument — a level splitting, a precession frequency, and a resonance line whose slope hands you g to four figures.
Build the model
Connect the measurement to the mechanism.
The classical link between magnetic moment and angular momentum is fixed by geometry: a charge −e circulating with angular momentum L carries μ = −(e/2mₑ)L, so orbital motion has gₗ = 1 and one ħ of angular momentum buys exactly one Bohr magneton. Spin refuses that rate. Stern–Gerlach deflects silver atoms as though each carried ±μB, yet the spin projection is only ±ħ/2, so the spin moment runs at twice the orbital rate: μₛ = −gₛ μB S/ħ with gₛ ≈ 2.
The factor is not a patch on a broken model — Dirac's relativistic equation returns exactly 2 for a spin-½ point charge with nothing inserted by hand, and the leftover 0.00232 is a QED radiative correction whose first term is α/2π. What makes the topic practical is that gₛ then becomes measurable to many digits. A static field B₀ splits the two spin states by gₛ μB B₀ and sets the moment precessing at ω = γB₀ with γ = gₛ μB/ħ, which is 28.0 GHz per tesla; a weak transverse field oscillating at that frequency drives transitions, and absorption peaks when hf = g μB B₀.
Plot resonance frequency against field, take the slope, and you have g. The cost is that the number belongs to that electron in that place: bind it in a radical and DPPH reads 2.0036, put it in a transition-metal ion and g turns anisotropic and can leave the neighbourhood of 2 altogether.
- Simple definition
- The spin g factor gₛ is the dimensionless ratio between an electron's spin magnetic moment and μB S/ħ; magnetic resonance is the measurement that fixes it, by finding where a photon's energy hf equals the Zeeman gap gₛ μB B₀.
- Example
- At B₀ = 0.340 T the free-electron gap is gₛ μB B₀ = 2.00232 × 57.884 µeV T⁻¹ × 0.340 T = 39.4 µeV, absorbed at f = 9.53 GHz — the X-band microwave frequency an ESR spectrometer sits at.
Fixes the exchange rate between angular momentum and moment. Orbital motion uses the identical form with gₗ = 1 exactly.
S in J s; μB = eħ/2mₑ = 9.2740×10⁻²⁴ J T⁻¹ = 57.884 µeV T⁻¹; gₛ dimensionless
Why a spin-½ beam splits by one Bohr magneton in Stern–Gerlach even though Sz is only ħ/2 — the 2 in gₛ cancels the ½.
for mₛ = ±½; the leading minus sign is the electron's negative charge, so μ opposes S
mₛ = +½ is the upper level: spin along B₀ puts the moment against it. The gap is 39.4 µeV at 0.340 T.
B₀ in T, E in eV with μB = 5.7884×10⁻⁵ eV T⁻¹, so gₛμB = 115.90 µeV T⁻¹
Exact, not an analogy: H linear in S makes Ehrenfest's equation close on ⟨S⟩ with no ħ left in it.
γ = 1.7609×10¹¹ rad s⁻¹ T⁻¹ for a free electron, so γ/2π = 28.025 GHz T⁻¹
The whole experiment. A fitted intercept catches a field-calibration offset that a single (f, B₀) pair would hide inside g.
f in GHz, B₀ in T; 13.9962 GHz T⁻¹ is μB/h. Fit a slope, never one point.
Dirac's 2 is the tree-level answer; every digit past it is a radiative correction, right here to 0.15%.
α = 1/137.036, so Schwinger's one-loop α/2π = 0.00116141 against the measured 0.00115965
The classical exchange rate, and where spin breaks it
Send a charge −e round a loop of radius r with period T. The current is −e/T and the moment is μ = IA = −eπr²/T, while the angular momentum is L = mₑvr = 2πmₑr²/T. Divide, and the geometry cancels: μ/L = −e/2mₑ. Promote it to operators and orbital motion gives μₗ = −(e/2mₑ)L = −μB L/ħ, which is gₗ = 1 by the definition of the Bohr magneton μB = eħ/2mₑ = 5.7884×10⁻⁵ eV T⁻¹. Now take the Stern–Gerlach result at face value. Silver's beam splits into two, and the deflection measures a moment projection of one Bohr magneton. But the two states carry Sz = ±ħ/2, half the orbital unit. So the spin moment runs at twice the classical rate, and the only honest response is to put the factor in explicitly: μₛ = −gₛ μB S/ħ, with gₛ ≈ 2 read off the data. Nothing in the loop argument predicts that, and nothing in it forbids it — μ/L is a property of how charge and mass are distributed through a body, and spin is not a distribution of anything.
Two levels, and which one sits higher
In a uniform field B₀ẑ the energy is H = −μ⋅B₀ = +gₛ μB B₀ Sz/ħ, so E(mₛ) = gₛ μB B₀ mₛ and the gap is ΔE = gₛ μB B₀. The sign matters and it is routinely got backwards: because the electron's charge is negative, μ points against S, so mₛ = +½ — spin along the field — is the upper state, and mₛ = −½ is the ground state. Numerically gₛμB = 115.90 µeV T⁻¹, so at 0.340 T the gap is 39.4 µeV. Set that against room temperature: kT at 295 K is 25.4 meV, 645 times larger. Boltzmann then gives a fractional excess in the lower state of only ΔE/2kT = 7.8×10⁻⁴, about one spin in 1290. Magnetic resonance is a bulk technique by necessity — the signal is the small imbalance across a very large number of spins, never the flipping of one.
Precession is exact here, not a classical analogy
Ehrenfest's theorem gives d⟨S⟩/dt = (i/ħ)⟨[H, S]⟩. With H = −μ⋅B₀ linear in S, and the spin commutators [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ, the algebra closes on itself: every ħ cancels and what is left is d⟨S⟩/dt = ⟨μ⟩ × B₀, the same torque equation a classical magnetised top obeys. So the expectation value of the spin precesses about B₀ at a fixed angle, at the Larmor rate ωL = γB₀ with γ = gₛ μB/ħ = 1.7609×10¹¹ rad s⁻¹ T⁻¹. This is not an approximation and not a picture — it is exact for a uniform field, and it is the one place in spin physics where a classical vector is safe. In ordinary frequency the rate is γ/2π = 28.025 GHz per tesla, so at 0.340 T the moment turns once every 105 ps. What the picture still does not license is definite transverse components: ⟨Sₓ⟩ and ⟨Sy⟩ trace a circle, but Sₓ and Sy have no simultaneous values.
What makes the absorption sharp
Add a small field B₁ rotating in the xy plane at frequency f, and move to a frame turning with it. In that frame the static field shrinks to B₀ − 2πf/γ along z while B₁ stands still along x. On resonance, where 2πf = γB₀, the z part vanishes and the only field left is B₁ — so the spin precesses about B₁ and flips completely, however weak B₁ is. Off resonance the effective field tilts back towards z, and the largest flip probability falls to ω₁²/(ω₁² + Δω²), with ω₁ = γB₁ and Δω the detuning. Detune by much more than γB₁ and the transition dies. That is the whole reason the line is narrow: with B₁ = 0.10 mT, ω₁/2π is 2.8 MHz, three parts in 10⁴ of a 9.5 GHz carrier. In practice the carrier is pinned by a resonant cavity and B₀ is swept through the line, because sweeping a magnet is far easier than sweeping a cavity.
Reading g off a slope, and the digits past 2
One (f, B₀) pair gives g = hf/(μB B₀), and it swallows every calibration error in the magnet. Take several fields instead and fit f against B₀: the slope is gμB/h, and a non-zero intercept exposes a residual field offset rather than hiding it inside g. A slope of 28.025 GHz T⁻¹ divided by μB/h = 13.9962 GHz T⁻¹ returns g = 2.00232. Dirac's equation, applied to a spin-½ point particle in an electromagnetic field, produces exactly 2 with nothing inserted by hand — the strongest reason to stop calling gₛ = 2 an empirical fudge. The measured free-electron value is 2.00231930436, and the anomaly aₑ = (g − 2)/2 = 0.00115965 is the QED correction from the electron emitting and reabsorbing virtual photons. Schwinger's one-loop term α/2π = 0.00116141 reproduces it to 0.15%, and the full calculation against the full measurement is the sharpest test any physical theory has passed.
When g is not 2: bound electrons and nuclei
gₛ = 2.0023 belongs to a free electron. Bind it and orbital angular momentum leaks back in through spin–orbit mixing with excited states, shifting g by roughly the ratio of the spin–orbit coupling to the orbital level spacing. DPPH's unpaired electron sits at g = 2.0036, 0.064% above free — small, because the radical's orbital moment is well quenched — while a transition-metal ion can run from about 1.4 to 6 and depend on which crystal axis lies along B₀. A measured g is therefore a probe of the electron's surroundings, not a universal constant. Nothing in the resonance argument was electronic, either. A proton has spin ½ with μ = gₚ μN I/ħ, where μN = μB mₑ/mₚ = 3.1525×10⁻⁸ eV T⁻¹ and gₚ = 5.5857 — measured, not 2, because the proton is composite. That gives γₚ/2π = 42.577 MHz T⁻¹, so in a 1.5 T scanner protons resonate at 63.9 MHz while electrons in the same magnet need 42.0 GHz, a factor of 658.
Change one variable at a time
Make the relationship visible.
Leave f at 9.5 GHz and sweep B₀ to 0.34 T: the gap arrow and the photon bar come level, and the readout puts resonance at 0.339 T. Now drag g down to 1.00, the orbital value, and the resonance field doubles to 0.679 T — the fan's slope is g, which is what the experiment measures.
ZEEMAN GAP g µB B39.36 µeV
GAP AS FREQUENCY9.517 GHz
SLOPE g µB / h27.992 GHz/T
RESONANCE FIELD0.3394 T
Live interpretationZEEMAN GAP g µB B: 39.36 µeV. GAP AS FREQUENCY: 9.517 GHz. SLOPE g µB / h: 27.992 GHz/T. RESONANCE FIELD: 0.3394 T
Catch the common trap
Explain before calculating.
An ESR spectrometer holds its microwave frequency fixed at 9.500 GHz and finds a free-electron line (g = 2.00232) at B₀ = 0.33898 T. A DPPH sample, whose unpaired electron has g = 2.0036, is then measured on the same instrument. Where does its line sit?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn X-band ESR spectrometer holds its microwave frequency at 9.388 GHz. At what static field does a free electron, gₛ = 2.00232, absorb, and how large is the gap in µeV?
- Resonance means the photon energy equals the Zeeman gap: hf = gₛ μB B₀, so B₀ = hf/(gₛ μB).
- Photon energy: h = 4.135668×10⁻¹⁵ eV s, so 1 GHz is worth 4.135668 µeV, and hf = 4.135668 × 9.388 = 38.83 µeV.
- Gap per tesla: gₛ μB = 2.00232 × 57.8838 µeV T⁻¹ = 115.90 µeV T⁻¹.
- B₀ = 38.83 / 115.90 = 0.3350 T. Cross-check with the slope gₛμB/h = 28.025 GHz T⁻¹: 9.388 / 28.025 = 0.3350 T.
AnswerB₀ = 0.3350 T, at a gap of 38.83 µeV. The quick route is the same arithmetic: divide the frequency by the free-electron slope of 28.025 GHz T⁻¹.
MediumA DPPH sample (g = 2.0036) sits in a static field of 0.6500 T at 295 K. Find its Larmor frequency and precession period, then the fractional excess of spins in the lower state at thermal equilibrium.
- Slope for this sample: gμB/h = 2.0036 × 57.8838 / 4.135668 = 28.043 GHz T⁻¹, slightly above the free-electron 28.025.
- Larmor frequency: fL = 28.043 GHz T⁻¹ × 0.6500 T = 18.228 GHz.
- Period: T = 1/fL = 1/(1.8228×10¹⁰ Hz) = 5.486×10⁻¹¹ s = 54.9 ps.
- Gap: ΔE = h fL = 4.135668 µeV GHz⁻¹ × 18.228 GHz = 75.38 µeV.
- Thermal scale: kT = 8.6173×10⁻⁵ eV K⁻¹ × 295 K = 25.42 meV = 25 421 µeV, so ΔE/kT = 75.38/25 421 = 2.965×10⁻³.
- Population difference: (N₋ − N₊)/(N₋ + N₊) = tanh(ΔE/2kT) ≈ ΔE/2kT = 1.48×10⁻³, one spin in 674.
AnswerfL = 18.23 GHz, a period of 54.9 ps, and a lower-state excess of 1.48×10⁻³ — about one spin in 674. The signal is that imbalance across a huge number of spins, never a single flip.
HardA variable-frequency ESR bridge on DPPH records resonance at (9.412 GHz, 0.33564 T) and (14.118 GHz, 0.50346 T). Each field is known to ±0.00020 T and the frequency uncertainty is negligible. Find g from the slope with its uncertainty, and say whether this run can separate DPPH's 2.0036 from the free-electron 2.00232.
- Both points lie on f = (gμB/h)B₀ + c. Use the slope, not either point alone: a slope cancels any constant field offset in the magnet, which a single (f, B₀) pair would absorb into g.
- Slope = (14.118 − 9.412) GHz / (0.50346 − 0.33564) T = 4.706 / 0.16782 = 28.042 GHz T⁻¹.
- Divide by μB/h = 13.9962 GHz T⁻¹: g = 28.042 / 13.9962 = 2.0035.
- Only ΔB carries uncertainty, and it is a difference, so u(ΔB) = √2 × 0.00020 T = 0.00028 T — a relative 0.169%. Hence ug = 0.00169 × 2.0035 = 0.0034.
- Compare the candidates: 2.0036 − 2.00232 = 0.00128, which is 0.38 u. They sit less than half an uncertainty apart, so this run cannot choose between them.
- To separate them at 3u needs ug ≤ 0.00043, eight times smaller: fields calibrated to about 25 µT, or the same 0.20 mT calibration spread over a field span eight times wider, near 1.3 T.
Answerg = 2.0035 ± 0.0034 (k = 1). The value sits on DPPH's 2.0036, but the 0.00128 gap to the free-electron 2.00232 is only 0.4 u — this run cannot resolve the two. It needs the field to ~25 µT, or a far longer lever arm.