University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.2

The weak-field Zeeman effect and the Landé g-factor

This is the calculation that makes the anomalous Zeeman effect ordinary. Project the atom's magnetic moment onto J, get one number gⱼ per level, and every line of the pattern follows from l and j, provided you first check that the field is weak beside the fine structure it competes with.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A weak magnetic field meets a hydrogen atom whose fine structure has already chosen the basis. Each level |n, l, j, mⱼ⟩ is a (2j + 1)-fold block, the field's first-order effect is the average of (μ_B B/ħ)(L_z + 2S_z) inside it, and that average would be μ_B B mⱼ if the magnetic moment were parallel to J. It is not, because spin carries twice the moment per unit of angular momentum that orbital motion does.

Inside one j-multiplet only the moment's projection on J survives, and the projection theorem turns that into a single number per level, the Landé factor gⱼ, so the sublevels split evenly by gⱼμ_B B. Because gⱼ depends on l and j alone, two quantum numbers predict the whole pattern: the ten components of hydrogen's Lyman-α, the identical pattern of sodium's D lines, and, when S = 0 leaves the doubling nothing to act on, the plain triplet of a singlet line such as cadmium's. The model is first-order perturbation theory inside one fine-structure level, so it holds only while μ_B B is small beside the fine-structure interval, which it equals at 0.78 T for hydrogen's 2P pair but at 37 T for sodium's 3P pair, and only above the field at which hyperfine coupling makes F, not mⱼ, the good quantum number.

Simple definition: In a field weak beside the fine structure, a level of angular momentum j splits into 2j + 1 evenly spaced sublevels at E_Z = gⱼμ_B B mⱼ, where the Landé factor gⱼ measures how much of the moment L + 2S lies along J.
Example: Hydrogen's 2P₃/₂ has gⱼ = 4/3, so in B = 0.10 T its four sublevels sit at mⱼ × 1.87 GHz from the zero-field level, using μ_B/h = 13.996 GHz T⁻¹; 2P₁/₂, with gⱼ = 2/3, splits by only 0.93 GHz.

The moment projected on J (derived)gⱼ = 1 + ⟨J⋅S⟩/[j(j + 1)ħ²]

Inside one level the part of S perpendicular to J averages away; what is left adds to J and sets the slope of every sublevel.

From L + 2S = J + S and Unit 5's projection theorem ⟨S⟩ = ⟨S⋅J⟩⟨J⟩/[j(j + 1)ħ²]; a ratio of ħ² to ħ², so dimensionless.

Landé factor for one electron (derived)gⱼ = 1 + [j(j + 1) − l(l + 1) + ¾]/[2j(j + 1)]

Hydrogen n = 2: 2 for 2S₁/₂, 2/3 for 2P₁/₂, 4/3 for 2P₃/₂. No n and no radial integral enter, so sodium's 3S and 3P share these values.

From J⋅S = (J² + S² − L²)/2 with s = ½, gₗ = 1 and gₛ = 2 quoted; j = l ± ½ gives (2l + 2)/(2l + 1) or 2l/(2l + 1).

Weak-field sublevels (derived)E = Eₙⱼ + gⱼμ_B B mⱼ

2j + 1 evenly spaced sublevels gⱼμ_B B apart, centred on Eₙⱼ: 1.87 GHz apart for 2P₃/₂ at 0.10 T.

μ_B = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹ = 5.788 × 10⁻⁵ eV T⁻¹, and μ_B/h = 13.996 GHz T⁻¹ (CODATA); B in T.

Zeeman components of a line (derived)hΔν = μ_B B(g′m′ − g m), with m′ − m = 0 or ±1

Lyman-α: four lines at ±2/3 and ±4/3 from 2P₁/₂, six at ±1/3, ±1 and ±5/3 from 2P₃/₂, in units of μ_B B/h. None stays at zero shift.

Primes mark the upper level; Δν in Hz from the zero-field line; the Δmⱼ rule is Unit 5's.

Singlet levels, S = 0 (derived)hΔν = μ_B B Δm, Δm = 0 or ±1

The normal triplet: however many sublevels the two levels have, the nine allowed transitions of ¹D₂ → ¹P₁ land on three frequencies.

S = 0 makes the moment parallel to J, so g = 1 on every level. Cd 643.847 nm at 1.00 T: σ lines ±14.0 GHz, ±0.0194 nm.

First correction beyond weak field (derived)E⁽²⁾ = ±(2/9)(μ_B B)²/ΔE_fs, for 2P with |mⱼ| = ½

Beside 2P₁/₂'s first-order (1/3)μ_B B the error is (2/3)μ_B B/ΔE_fs: 10% at 0.12 T, far below the 0.78 T where μ_B B equals ΔE_fs.

Plus for 2P₃/₂, minus for 2P₁/₂; ΔE_fs = h × 10.97 GHz = 45.4 μeV, measured. mⱼ = ±3/2 get no correction.

Choose the basis first: why first order is already diagonal

From Unit 3 and the previous topic, in one line: a field B along z adds H′_Z = (μ_B B/ħ)(L_z + 2S_z), with μ_B = eħ/2mₑ = 5.788 × 10⁻⁵ eV T⁻¹. When μ_B B is small beside the fine-structure interval, 45.4 μeV for hydrogen's 2P pair, which μ_B B matches only at 0.78 T, H′_Z acts on each fine-structure level separately, and each level is a (2j + 1)-fold degenerate block. Degenerate perturbation theory asks for H′_Z inside that block, and here the answer costs nothing. Write L_z + 2S_z = J_z + S_z. J_z is diagonal in |l, ½, j, mⱼ⟩, and S_z commutes with J_z, so ⟨j mⱼ′|S_z|j mⱼ⟩ vanishes unless mⱼ′ = mⱼ. The block is already diagonal: inside it L² and J² are fixed and J_z commutes with H′_Z, so no two states of the block are connected, Unit 5's coupled states are the good states, and the shifts are the diagonal elements ⟨L_z + 2S_z⟩. What H′_Z fails to commute with is J², and its elements between j = l + ½ and j = l − ½ at equal mⱼ are what the last section sizes.

Project the moment onto J

In a coupled state that is not stretched, the diagonal element ⟨S_z⟩ is neither ½ħ nor −½ħ, because the spin is shared between two product states. Unit 5's projection theorem, quoted, handles it in one line: inside one j-multiplet a vector operator V has ⟨V⟩ = ⟨V⋅J⟩⟨J⟩/[j(j + 1)ħ²]. With V = S and J⋅S = (J² + S² − L²)/2, ⟨J_z + S_z⟩ = mⱼħ(1 + [j(j + 1) + ¾ − l(l + 1)]/[2j(j + 1)]) = gⱼmⱼħ, so E_Z = gⱼμ_B B mⱼ, derived. Dimensions check: μ_B B is J T⁻¹ × T = J, and gⱼ is a ratio of angular momenta. For one electron, j = l ± ½ reduces gⱼ to (2l + 2)/(2l + 1) or 2l/(2l + 1), and both tend to 1 at large l, where the orbital moment dominates. The vector model shows why only the projection counts: μ = −(μ_B/ħ)(L + 2S) tilts away from J, and L and S precess about J at the fine-structure rate, 10.97 GHz for 2P, about six times faster than J precesses about B, 1.87 GHz for 2P₃/₂ at 0.10 T, so only the component along J survives the average, and the survival improves as the ratio grows. gⱼ carries no n, no radial integral and no α: it is geometry.

Check the projection by brute force

A derived formula deserves an independent check, and Unit 5's coefficients provide one. Writing product states as |mₗ, mₛ⟩, the 2P states with mⱼ = ½ are |3/2, ½⟩ = √(2/3)|0, ↑⟩ + √(1/3)|1, ↓⟩ and |½, ½⟩ = −√(1/3)|0, ↑⟩ + √(2/3)|1, ↓⟩. L_z + 2S_z has eigenvalue ħ on |0, ↑⟩ and 0 on |1, ↓⟩, so its expectation value is a probability-weighted average: 2/3 for j = 3/2 and 1/3 for j = ½, in units of ħ. Divided by mⱼ = ½, these are 4/3 and 2/3. The stretched state |3/2, 3/2⟩ = |1, ↑⟩ gives 1 + 1 = 2 = (4/3)(3/2) with no averaging at all, and 2S₁/₂ = |0, ↑⟩ gives 1 = 2 × ½, the l = 0 limit in which gⱼ = gₛ. A second check is a trace: at fixed mⱼ the sum of gⱼmⱼ over the two j-levels equals the sum of mₗ + 2mₛ over the two product states, 2/3 + 1/3 = 1 + 0, because a change of basis cannot change a trace. That sum is what the intermediate-field topic conserves as the field grows.

From sublevels to lines: the anomalous pattern of Lyman-α

A line from an upper level, primed, to a lower one shifts by hΔν = μ_B B(g′m′ − g m), and Unit 5's Δmⱼ = 0, ±1 decides which pairs appear. For 2P₁/₂ → 1S₁/₂, g′ = 2/3 and g = 2 give four components at ±2/3 and ±4/3 in units of μ_B B/h. For 2P₃/₂ → 1S₁/₂, g′ = 4/3 puts the upper sublevels at ±2 and ±2/3, and six components survive at ±1/3, ±1 and ±5/3; the pairs 3/2 → −½ and −3/2 → ½ would need Δmⱼ = ±2. That is ten lines where the orbital picture predicts two triplets, and none sits at the unshifted frequency: the anomalous Zeeman effect. At 0.10 T the unit is 1.40 GHz, so each pattern spans under 5 GHz, while the thermal Doppler width of Lyman-α in a room-temperature gas is near 30 GHz, a kinetic-theory estimate, so in a room-temperature gas the pattern blurs into one line: hydrogen's Lyman-α pattern is computed here rather than read off a spectrum. Sodium's D lines show the same four-and-six pattern, since gⱼ never saw n or the radial function, and a sodium lamp at 1 T resolves them.

The singlet limit: why the normal triplet is the special case

Set S = 0 and the moment is −(μ_B/ħ)L = −(μ_B/ħ)J, exactly parallel to J; projection changes nothing, and g = 1 for every level whatever its L. Every line then shifts by μ_B BΔm/h, and Δm = 0, ±1 gives three evenly spaced components however many sublevels the levels have. Cadmium's red line at 643.847 nm, whose ¹D₂ → ¹P₁ assignment is quoted from spectroscopic tables, is the standard case: at 1.00 T its σ components sit ±14.0 GHz, or ±0.0194 nm, from the central π line, and nine allowed transitions land on those three frequencies. Read backwards, the splitting measures μ_B/h = e/(4πmₑ): a measured 14.0 GHz at 1.00 T gives e/mₑ = 1.76 × 10¹¹ C kg⁻¹, Zeeman's route to the electron's charge-to-mass ratio, against the modern 1.7588 × 10¹¹ C kg⁻¹. History called this the normal effect and everything else anomalous, but the logic runs the other way: the triplet is what the projection gives when the spin's double weighting has nothing to act on, and it returns in any atom once a strong field decouples the spin, the next topic's case.

Where the weak-field result stops

Three things end it. From above, H′_Z joins 2P₁/₂ and 2P₃/₂ at equal mⱼ = ±½ through an element of size (√2/3)μ_B B, computed from the states of the brute-force check, and Unit 6's second-order formula pushes 2P₁/₂ down and 2P₃/₂ up by (2/9)(μ_B B)²/ΔE_fs. Against 2P₁/₂'s first-order (1/3)μ_B B the error is (2/3)μ_B B/ΔE_fs, 10% at 0.12 T, well below the 0.78 T where μ_B B equals the interval. The stretched mⱼ = ±3/2 states have no partner and stay linear at every field, as does 2S₁/₂, since the field cannot change l. For n = 2, Unit 8's ΔE_fs = α²Ry/16 = 45.3 μeV, against 45.4 μeV measured, so the whole weak-field window is an α² statement. From below, hyperfine coupling ties J to the proton spin: μ_B B matches 2P₃/₂'s 23.7 MHz interval near 2 mT, and 2μ_B B the 1420 MHz ground-state splitting near 0.05 T; beneath such fields F, not mⱼ, is good. And gⱼ itself moves at order α: with gₛ = 2.00232 from QED, quoted, 2P₃/₂ has gⱼ = 1.33411, and the measured 1S₁/₂ value lies about 17.7 parts per million below gₛ, matching the α²/3 = 17.8 ppm binding correction, also quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

Magnetic field B0.10 T

Spin g-factor gₛ in the Zeeman term2.0

Lower gₛ from 2 to 1: the ten lines close onto two normal triplets and the error falls to zero, because L + S = J makes the weak-field result exact. At gₛ = 2, raise B to 0.26 T, where the two patterns touch; by then the error readout has passed 20%.

gⱼ OF 2P₁/₂0.667

gⱼ OF 2P₃/₂1.333

ZEEMAN UNIT μ_B B/h1.40 GHz

2P₁/₂ SECOND-ORDER ERROR8.5 %

Live interpretationgⱼ OF 2P₁/₂: 0.667. gⱼ OF 2P₃/₂: 1.333. ZEEMAN UNIT μ_B B/h: 1.40 GHz. 2P₁/₂ SECOND-ORDER ERROR: 8.5 %

Catch the common trap

Explain before calculating.

Misconception check

The commonest slip carries the strong-field formula into the weak field: take 2P₁/₂, mⱼ = +½, read off its larger component |mₗ = 1, ↓⟩, and report a shift μ_B B(mₗ + 2mₛ) = 0. In a weak field mₗ and mₛ are not good quantum numbers. The state is the superposition √(2/3)|1, ↓⟩ − √(1/3)|0, ↑⟩, and first-order theory asks for the expectation value of L_z + 2S_z in it, which weights each component by its probability: (2/3)(0) + (1/3)(1) = 1/3, the gⱼmⱼ = (2/3)(½) of the Landé formula, a shift of 0.47 GHz at 0.10 T rather than none. Behind the slip sits a picture of the moment as parallel to J, or as belonging to one product state. It is neither: spin counts double, so L + 2S tilts away from J, and only its projection on J survives inside a level.

Hydrogen's 2P₃/₂ → 1S₁/₂ line is observed in a field weak beside the fine structure but strong beside the hyperfine structure. In units of μ_B B/h, where do its Zeeman components lie?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen's n = 3 shell contains 3D₅/₂ and 3D₃/₂, 1.08 GHz apart. In B = 5.00 mT, find gⱼ for each level from the Landé formula, check the 3D₅/₂ value on its stretched state, and give each level's sublevel spacing in MHz.

Weak-field check: μ_B B/h = 13.996 GHz T⁻¹ × 0.00500 T = 69.98 MHz, 6.5% of the 1.08 GHz interval, so the coupled states |3, 2, j, mⱼ⟩ are good.
3D₅/₂ (l = 2, j = 5/2): gⱼ = 1 + [35/4 − 6 + 3/4]/(2 × 35/4) = 1 + 3.5/17.5 = 6/5. 3D₃/₂ (j = 3/2): gⱼ = 1 + [15/4 − 6 + 3/4]/(2 × 15/4) = 1 − 1.5/7.5 = 4/5.
Check: |5/2, 5/2⟩ = |mₗ = 2, ↑⟩ is a single product state, so its shift is μ_B B(2 + 1) = 3μ_B B exactly, and gⱼmⱼ = (6/5)(5/2) = 3 agrees. The j = l + ½ shortcut (2l + 2)/(2l + 1) = 6/5 agrees too.
Spacings: (6/5) × 69.98 MHz = 84.0 MHz between the six 3D₅/₂ sublevels, and (4/5) × 69.98 MHz = 56.0 MHz between the four 3D₃/₂ sublevels. The fans span 420 MHz and 168 MHz, so they never meet across the 1.08 GHz gap.

Answergⱼ = 6/5 for 3D₅/₂ and 4/5 for 3D₃/₂; sublevel spacings 84.0 MHz and 56.0 MHz at 5.00 mT.

MediumA sodium lamp sits in B = 1.20 T. Sodium's 3S₁/₂, 3P₁/₂ and 3P₃/₂ have the same l and j as hydrogen's 1S₁/₂, 2P₁/₂ and 2P₃/₂, and its D lines, D2 at 588.995 nm and D1 at 589.592 nm, are 515.5 GHz apart. Predict every Zeeman component of both lines in GHz, find the full width of the D2 pattern in pm, and check that the field is weak.

gⱼ depends only on l, s and j, so 3S₁/₂, 3P₁/₂ and 3P₃/₂ carry 2, 2/3 and 4/3, exactly as in hydrogen.
Zeeman unit: μ_B B/h = 13.996 GHz T⁻¹ × 1.20 T = 16.80 GHz. Weak-field check: 16.80/515.5 = 0.033, and the largest second-order shift, (2/9)(16.80 GHz)²/515.5 GHz = 0.12 GHz, is about 2% of the smallest component offset.
D1, 3P₁/₂ → 3S₁/₂: (2/3)m′ − 2m with m′ − m = 0 or ±1 gives ±2/3 and ±4/3 units, that is ±11.2 GHz for Δmⱼ = 0 and ±22.4 GHz for Δmⱼ = ±1.
D2, 3P₃/₂ → 3S₁/₂: (4/3)m′ − 2m gives ±1/3 unit for Δmⱼ = 0 and ±1 and ±5/3 units for Δmⱼ = ±1, that is ±5.60, ±16.8 and ±28.0 GHz.
D2 spans 10/3 units = 56.0 GHz. Δλ = λ²Δν/c = (588.995 × 10⁻⁹ m)² × 56.0 × 10⁹ Hz ÷ 2.998 × 10⁸ m s⁻¹ = 6.48 × 10⁻¹¹ m = 64.8 pm, about a ninth of the 597 pm D1 to D2 gap, so the two patterns stay apart.
Neighbouring components are at least 11.2 GHz apart, several times the lamp's Doppler width of about 1.6 GHz and the 1.3 GHz hyperfine substructure each component carries, so the anomalous pattern is resolved.

AnswerD1: ±11.2 and ±22.4 GHz; D2: ±5.60, ±16.8 and ±28.0 GHz. The D2 pattern is 56.0 GHz, or 64.8 pm, wide, and μ_B B is 3.3% of the fine-structure interval.

HardIn hydrogen's n = 2 shell the Lamb shift puts 2S₁/₂ 1057.8 MHz above 2P₁/₂ (measured, quoted), and 2P₃/₂ lies 10.97 GHz above 2P₁/₂. Using weak-field energies, find the field at which 2S₁/₂, mⱼ = −½ meets 2P₁/₂, mⱼ = +½; then correct it with second-order theory and decide whether the weak-field answer can be trusted.

Measure energies from zero-field 2P₁/₂ and write x = μ_B B and L = h × 1057.8 MHz. With gⱼ = 2 and 2/3: E(2S₁/₂, −½) = L − x and E(2P₁/₂, +½) = x/3.
Setting them equal gives L = 4x/3, so x/h = 0.75 × 1057.8 MHz = 793.4 MHz and B = 793.4 MHz ÷ 13.996 GHz T⁻¹ = 0.0567 T.
Size the error: x/ΔE_fs = 793.4/10 969 = 0.072. The (√2/3)x element to 2P₃/₂, mⱼ = +½ pushes 2P₁/₂, +½ down by (2/9)(793.4 MHz)²/10 969 MHz = 12.8 MHz. 2S₁/₂ has no n = 2 partner of equal l, so it stays linear.
Corrected condition: L = 4x/3 − (2/9)x²/ΔE_fs. Iterating x = 0.75(L + (2/9)x²/ΔE_fs) gives 802.9 MHz and then converges on 803.2 MHz, so B = 0.0574 T, 1.2% above the first-order value.
The field cannot couple the two states, since it conserves l, so they cross rather than repel. gₛ = 2.00232 from QED raises gⱼ(2S₁/₂) + gⱼ(2P₁/₂) from 8/3 to 2.66821, lowering B by only 0.06%. At the next crossing, 2S₁/₂, −½ with 2P₁/₂, −½, the second-order correction grows to 5%, moving it from 0.113 T to 0.119 T.
Hyperfine structure, ignored so far, matters more. At this field the proton spin is largely decoupled, so 2S₁/₂ and 2P₁/₂ shift by a quarter of their measured 177.6 MHz and 59.2 MHz hyperfine splittings, ±44.4 MHz and ±14.8 MHz, with signs set by the proton's m_I. For equal m_I that moves x by ±(3/4)(44.4 + 14.8) MHz = ±44.4 MHz, or B by ±3.2 mT; for unequal m_I, by ±(3/4)(44.4 − 14.8) MHz = ±22.2 MHz, or ±1.6 mT. One crossing becomes four, spread over 5.5% either side, more than four times the weak-field error.

AnswerB ≈ 0.057 T: 0.0567 T from the weak-field gⱼ values, 0.0574 T after the second-order correction. That 1.2% is smaller than the ±3.2 mT, or 5.5%, spread of the four hyperfine crossings, so hyperfine structure, not the weak-field approximation, first limits the prediction.