University Physics IV · Angular Momentum and Spin · 9.4
Orbital Moments & the Normal Zeeman Effect
Until you switch on a field, mₗ is pure bookkeeping — 2l+1 states with nothing to tell them apart. Here the label earns an energy: the orbiting electron is a current loop, the loop is a magnet, and B fans every level into rungs μB B apart. Read the fan with the selection rules and a spectral line hands you e/mₑ.
Build the model
Connect the measurement to the mechanism.
The model is one classical line of magnetostatics promoted intact. An electron circling a nucleus is a current loop, and computing its moment gives μ = −(e/2mₑ)L: radius and speed cancel, leaving pure charge-to-mass bookkeeping. Because μ is proportional to L as an operator, quantisation is inherited for free — gₗ = 1 exactly — and the quantised projection Lz = mₗℏ becomes a quantised moment μz = −mₗ μB.
A field B then adds −μ⋅B to the energy, fanning each level into 2l+1 rungs spaced μB B, and the previously idle label mₗ becomes a measurable energy. The spectrum, though, shows fewer lines than the rungs suggest: every level's ladder has the same spacing, so the nine allowed d → p transitions land on exactly three frequencies, ν₀ and ν₀ ± μB B/h. That coincidence is the model's cost as well as its content.
It holds only when the whole moment is orbital — S = 0, singlet terms — because spin carries g ≈ 2 and wrecks the equal spacing; hence hydrogen and the alkalis, with their lone unpaired electron, are "anomalous". And Lorentz's classical electron theory predicts the same triplet with the same spacing, so the normal Zeeman effect measured e/mₑ a year before Thomson without proving quantisation at all.
- Simple definition
- An electron's orbital motion gives it a magnetic moment μ = −(e/2mₑ)L, so a field B shifts each sublevel by mₗ μB B and splits every level into 2l+1 equally spaced rungs — the normal Zeeman effect is the three-line spectrum those rungs produce in singlet atoms.
- Example
- At B = 1.0 T each rung sits μB B = 5.79 × 10⁻⁵ eV from its neighbour, so a d level (l = 2) spans 4 × 5.79 × 10⁻⁵ = 2.3 × 10⁻⁴ eV — about one part in 10⁴ of the 2 eV optical photon that reveals it.
One classical line that survives quantisation exactly — μ and L are proportional operators, so quantising L quantises μ with it.
e = 1.602 × 10⁻¹⁹ C, mₑ = 9.109 × 10⁻³¹ kg; μ in J T⁻¹ when L is in J s. The minus sign: the electron's charge is negative.
Every splitting in this unit is a dimensionless g times μB B: learn this one number and you can size them all.
The moment carried by one ℏ of orbital angular momentum — the per-tesla energy unit of atomic magnetism.
Turns the idle mₗ degeneracy into an equally spaced ladder: the field picks the z axis and breaks the symmetry that protected it.
B along z, in tesla; mₗ = −l … +l gives 2l+1 rungs spaced μB B — 57.9 µeV at 1.0 T.
Collapses the many allowed transitions to three frequencies, because every jump shifts by Δmₗ times the same μB B.
Δmₗ = 0 is the π line, polarised ∥ B and absent along B; Δmₗ = ±1 are the σ lines, circular about B.
What the spectrometer must resolve — and the slope from which Zeeman read e/mₑ a year before Thomson.
μB/h = 13.996 GHz T⁻¹; at 500 nm and 1.0 T, Δλ ≈ 11.7 pm — parts in 10⁵.
An orbit is a current loop, and a loop is a magnet
Let an electron of charge −e circle at speed v on radius r. It passes any point once per period T = 2πr/v, so the loop carries current I = −e/T = −ev/(2πr), and a loop's moment is current times area: μ = IA = −(ev/2πr)(πr²) = −evr/2. Multiply and divide by the mass and the orbit's geometry disappears: μ = −(e/2mₑ)(mₑvr) = −(e/2mₑ)L. Radius and speed have cancelled; nothing survives but the charge-to-mass ratio, which is why the same relation holds for any orbit once L is read as the angular momentum. The minus sign is physical — a negative charge circulating one way is a current circulating the other — so μ points opposite L. And because the result says one operator is a constant multiple of another, it passes into quantum mechanics unchanged: no correction factor appears, gₗ = 1 exactly, and every eigenstate of Lz is automatically an eigenstate of μz.
The field turns mₗ into an energy
Project onto the field direction: μz = −(e/2mₑ)Lz = −mₗ μB, one Bohr magneton per unit of mₗ. A moment in a field carries energy U = −μ⋅B, so with B along z each sublevel shifts by ΔE = mₗ μB B: the 2l+1 values of mₗ, degenerate in zero field, spread into a ladder of rungs μB B apart. The degeneracy was never an accident — with no field there is no preferred axis, and states differing only in orientation must share an energy. The field supplies the axis, and the symmetry argument dies with it. Scale matters: at 1.0 T the spacing is 5.79 × 10⁻⁵ eV, about 450 times smaller than room-temperature kT = 0.0259 eV, so the rungs stay essentially equally populated, and about 3 × 10⁻⁵ of a 2 eV optical transition, so the splitting is a fine perturbation on the line, not a rearrangement of the atom.
Count distinct energies, not transitions
Take a singlet d → p line. The upper level fans into five rungs, the lower into three, and the dipole rules Δl = ±1, Δmₗ = 0, ±1 allow nine distinct transitions. The photon energy is hν = hν₀ + (mₗ′ − mₗ)μB B, because both ladders — whatever their n and l — share the identical spacing μB B. So the frequency depends on the difference Δmₗ alone, and nine transitions deliver exactly three frequencies: ν₀, and ν₀ ± μB B/h. Nothing here is approximate; it is a strict coincidence enforced by gₗ = 1 being the same number in every orbital level. That is also its fragility. Give the two levels different g factors — say 1.2 and 1.0 — and hν picks up g′m′ − gm, which takes up to nine distinct values: the triplet fans apart. The number of Zeeman lines counts the distinct values of the energy shift, never the sublevels and never the arrows between them.
Polarisation labels each line with its Δmₗ
The three lines are distinguishable by more than frequency. A Δmₗ = 0 transition leaves the z angular momentum unchanged: its charge oscillation runs along B, and an oscillating dipole radiates strongest broadside and not at all along its own axis. So the π line is linearly polarised parallel to B, brightest viewed transverse to the field, and absent when you sight along it. A Δmₗ = ±1 transition removes ±ℏ of z angular momentum, carried off by charge circulating in the plane perpendicular to B: viewed along the field these σ lines are circularly polarised with opposite senses; viewed transverse they appear linearly polarised perpendicular to B. Geometry is therefore a prediction: transverse observation shows three lines, longitudinal shows only the two σ components. Both checks pass, and the sense of the circular polarisation seen along B told Zeeman the radiating charge is negative — a sign determined before the electron was isolated.
Size the experiment before trusting the story
Put numbers on cadmium's 643.85 nm singlet line at 1.0 T. The spacing is Δν = μB B/h = 14.0 GHz against an optical frequency ν = c/λ = 4.66 × 10¹⁴ Hz — a shift of three parts in 10⁵. In wavelength, Δλ = λ²Δν/c ≈ 19 pm, demanding resolving power λ/Δλ ≈ 3 × 10⁴: a large grating or a Fabry–Pérot étalon, and hopeless for a prism spectroscope. Run the logic backwards and the experiment measures a fundamental constant: the observed Δν per tesla is μB/h, and since μB = eℏ/2mₑ, the charge-to-mass ratio follows as e/mₑ = 4π(μB/h) = 4π × 1.40 × 10¹⁰ ≈ 1.76 × 10¹¹ C kg⁻¹. That is how a wavelength shift of a few parts per hundred thousand delivered e/mₑ in 1896, a year ahead of Thomson's cathode-ray deflections.
Why the 'normal' pattern is the exception
Every step above assumed the moment is purely orbital, g = 1. Spin breaks that: the intrinsic moment carries gₛ ≈ 2.002, so any term with S ≠ 0 has a Landé factor gJ depending on l, s and j, and the upper and lower ladders of a transition no longer share a spacing. The three-line coincidence collapses and the pattern fans out — sodium's D₁ line shows four components, D₂ six. Only singlet terms, where paired spins cancel the spin moment — bright lines of cadmium, zinc, mercury — show the clean triplet. The name is a historical accident: Lorentz's classical theory predicted exactly this triplet with exactly the μB B/h spacing, so it was 'normal' and everything else 'anomalous'. The joke is on the labels. Hydrogen, the simplest atom, is anomalous, because one electron cannot pair its spin; and the anomalous majority is the pattern classical physics can never produce — the one whose explanation had to wait for spin.
Change one variable at a time
Make the relationship visible.
Take B to zero and the whole pattern closes onto the single field-free line; back at 1 T, drag the upper g from 1 to 2 and watch three lines become nine — the spacing coincidence, not the selection rules, is what makes the normal triplet a triplet.
π–σ SPACING μB B/h14.0 GHz
SUBLEVEL SHIFT μB B57.9 µeV
Δλ AT 500 nm11.7 pm
OUTERMOST σ OFFSET14.0 GHz
Live interpretationπ–σ SPACING μB B/h: 14.0 GHz. SUBLEVEL SHIFT μB B: 57.9 µeV. Δλ AT 500 nm: 11.7 pm. OUTERMOST σ OFFSET: 14.0 GHz
Catch the common trap
Explain before calculating.
A singlet line from an l = 3 → l = 2 transition is viewed transverse to a 2.0 T field. How many Zeeman components appear, and how far apart are they?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFrom e = 1.602 × 10⁻¹⁹ C, ℏ = 1.055 × 10⁻³⁴ J s and mₑ = 9.109 × 10⁻³¹ kg, evaluate the Bohr magneton in J T⁻¹ and eV T⁻¹, then find the spacing and total span of a p level (l = 1) in a 0.50 T field.
- μB = eℏ/2mₑ = (1.602 × 10⁻¹⁹ × 1.055 × 10⁻³⁴)/(2 × 9.109 × 10⁻³¹) = 1.690 × 10⁻⁵³ / 1.822 × 10⁻³⁰ = 9.28 × 10⁻²⁴ J T⁻¹.
- Divide by e to convert the unit: μB = 9.28 × 10⁻²⁴ / 1.602 × 10⁻¹⁹ = 5.79 × 10⁻⁵ eV T⁻¹.
- A p level splits into 2l + 1 = 3 sublevels, mₗ = −1, 0, +1, with adjacent spacing μB B = 5.79 × 10⁻⁵ × 0.50 = 2.9 × 10⁻⁵ eV.
- The span runs from mₗ = −1 to mₗ = +1: 2μB B = 5.8 × 10⁻⁵ eV — about 450 times smaller than room-temperature kT = 0.0259 eV, so all three rungs stay almost equally populated.
AnswerμB = 9.28 × 10⁻²⁴ J T⁻¹ = 5.79 × 10⁻⁵ eV T⁻¹; the p level's rungs sit 2.9 × 10⁻⁵ eV apart and span 5.8 × 10⁻⁵ eV at 0.50 T.
MediumCadmium's red singlet line (¹D₂ → ¹P₁) sits at λ = 643.85 nm. In a 1.20 T field, viewed transverse to B: how many lines appear, what are the frequency and wavelength splittings between adjacent components, and what resolving power must the spectrometer beat?
- Both levels are singlets (S = 0, g = 1), so all nine allowed transitions (Δmₗ = 0, ±1 between five and three rungs) collapse onto three frequencies: ν₀ and ν₀ ± μB B/h.
- Adjacent spacing in frequency: Δν = μB B/h = 13.996 GHz T⁻¹ × 1.20 T = 16.8 GHz.
- Convert to wavelength: |Δλ| = λ²Δν/c = (643.85 × 10⁻⁹)² × 1.68 × 10¹⁰ / (2.998 × 10⁸) = 2.32 × 10⁻¹¹ m = 23.2 pm.
- Resolving power: R = λ/Δλ = 643.85 nm / 0.0232 nm ≈ 2.8 × 10⁴ — Fabry–Pérot or large-grating territory; a pocket spectroscope at R ≈ 10³ sees a single unsplit line.
AnswerThree lines, adjacent components 16.8 GHz (23.2 pm) apart, needing R ≈ 2.8 × 10⁴ to resolve.
HardIn 1896 the splitting of a spectral line in a field gave the first measurement of e/mₑ. A normal triplet at λ = 500.0 nm shows its outer components displaced by Δλ = 11.7 pm each side of centre at B = 1.00 T. Extract e/mₑ and compare with the accepted 1.759 × 10¹¹ C kg⁻¹.
- Convert the wavelength displacement to frequency: |Δν| = cΔλ/λ² = (2.998 × 10⁸ × 11.7 × 10⁻¹²)/(500.0 × 10⁻⁹)² = 3.508 × 10⁻³ / 2.500 × 10⁻¹³ = 1.403 × 10¹⁰ Hz.
- Each σ component is displaced by one quantum of ladder spacing, hΔν = μB B, so μB = hΔν/B = 6.626 × 10⁻³⁴ × 1.403 × 10¹⁰ / 1.00 = 9.30 × 10⁻²⁴ J T⁻¹.
- μB = eℏ/2mₑ rearranges to e/mₑ = 2μB/ℏ = 2 × 9.30 × 10⁻²⁴ / 1.055 × 10⁻³⁴ = 1.76 × 10¹¹ C kg⁻¹.
- Compare: 1.763 versus 1.759 — agreement to about 0.3%, extracted from a wavelength shift of only 23 parts per million. The sense of circular polarisation of the σ lines viewed along B fixed the sign of the orbiting charge as negative.
Answere/mₑ ≈ 1.76 × 10¹¹ C kg⁻¹, within about 0.3% of the accepted value — e/mₑ from light alone, a year before Thomson.