University Physics V · Atomic Spectroscopy · 13.7

The Zeeman Effect & Paschen-Back

Put an atom in a magnet and its levels fan out — but which fan you get is decided before you write anything down, by one ratio. This topic teaches you to compare μ_B B against the fine-structure interval, choose the basis that ratio makes good, and diagonalise whatever block is left over.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Minimal coupling with A = ½ B × r splits the atom-field energy into a term linear in B, (μ_B/ħ)(L + gₛ S)⋅B, and a quadratic diamagnetic term smaller by about 10⁻⁶ per tesla, which is dropped. Take B along z and rewrite the linear term as (μ_B B/ħ)[J_z + (gₛ − 1)S_z]: J_z is diagonal in the coupled basis |n l j mⱼ⟩ and returns ħmⱼ, but S_z is not, and that leftover (gₛ − 1)S_z carries the entire subject. Its diagonal part within one j is fixed by the projection theorem and gives the Landé shift g_J μ_B B mⱼ; its off-diagonal part connects j to j′ at the same mⱼ, and that part is exactly what the weak-field formula discards.

So the model comes with a bill: g_J is not a property of a level but the first-order slope of a level at B → 0, trustworthy only while μ_B B is small against the fine-structure interval Δ_fs. Beyond that you must diagonalise, which for s = ½ is one 2×2 block per mⱼ, running continuously from the Landé fan to the Paschen-Back pattern μ_B B(mₗ + gₛ mₛ) + Aħ² mₗ mₛ in the uncoupled basis. Only mⱼ survives the whole journey, because [J_z, H] = 0 at every field while [J², H] does not vanish once the field term is switched on.

Simple definition: The Zeeman effect is the splitting of atomic levels by H_Z = (μ_B/ħ)(L + gₛ S)⋅B, with the pattern set by which basis diagonalises the larger of the field and spin-orbit terms: coupled |j mⱼ⟩ in weak field, uncoupled |mₗ mₛ⟩ in strong.
Example: Sodium's ²P_{3/2} has g_J = 4/3, so at 0.60 T its mⱼ = +3/2 sublevel rises by (4/3)(3/2)(57.9 μeV T⁻¹)(0.60 T) = 69.5 μeV — a legitimate use of the formula, since μ_B B = 34.7 μeV is only 1.6% of the 2.13 meV fine-structure interval.

The Zeeman HamiltonianH_Z = (μ_B/ħ)(L + gₛ S)⋅B = (μ_B B/ħ)[J_z + (gₛ − 1)S_z]

Splitting off J_z isolates the whole difficulty in one operator, (gₛ − 1)S_z — the anomaly.

μ_B = eħ/2mₑ = 5.7884×10⁻⁵ eV T⁻¹; gₛ = 2.00232 from Dirac plus QED; B along z

Landé factor, weak-field shiftΔE = g_J μ_B B mⱼg_J = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)]

Gives 2j+1 equally spaced sublevels, and the spacing measures g_J — which is how a term assignment gets checked.

Needs μ_B B ≪ Δ_fs. ²S_½: 2 · ²P_½: 2/3 · ²P₃/2: 4/3 · any singlet: 1

The diamagnetic term you dropH_dia = (e²B²/8mₑ)(x² + y²)H_dia/Hₚₐᵣₐ ≈ eB⟨ρ²⟩/4ħ

Licenses the linear Hamiltonian below ~10⁶ T for low-n states — and forbids it for Rydberg atoms and magnetars.

≈ 1.06×10⁻⁶ per tesla for ⟨ρ²⟩ ≈ a₀²; scales as n⁴, so it catches up near n = 30 at 1 T

Intermediate field: the fixed-mⱼ block⟨l−½, mⱼ|H_Z|l+½, mⱼ⟩ = −μ_B B √[(l+½)² − mⱼ²]/(2l+1)

The element the Landé formula throws away. It vanishes at mⱼ = ±(l+½), which is why those two levels never bend.

s = ½. Diagonal spin-orbit entries +Aħ²l/2 for j = l+½ and −Aħ²(l+1)/2 for j = l−½

Paschen-Back limitE = μ_B B(mₗ + gₛ mₛ) + Aħ² mₗ mₛ, for μ_B B ≫ Δ_fs

L and S precess about B independently. Δmₛ = 0 for E1, so the spectrum collapses back to a Lorentz triplet.

Uncoupled |mₗ mₛ⟩ basis; Aħ² = 2Δ_fs/(2l+1) is the spin-orbit constant

From levels to a spectrumΔEₗᵢₙₑ = μ_B B (gᵤ mᵤ − gₗ mₗ), with Δmⱼ = 0, ±1

Six components for Na D2, four for D1, three only when gᵤ = gₗ — the 'normal' effect as the special case.

Δmⱼ = 0 is π, linear ∥ B and dark along B; Δmⱼ = ±1 is σ±, circular seen along B

Where the field term comes from, and what gets dropped

Put the atom in a uniform B and choose the symmetric gauge A = ½ B × r. Minimal coupling p → p + eA turns p²/2m into p²/2m + (e/2m)(A⋅p + p⋅A) + e²A²/2m. The cross term collapses to (e/2m)B⋅L = (μ_B/ħ)L⋅B, the spin magnetic moment adds (μ_B/ħ)gₛ S⋅B with gₛ = 2.0023 imported from the Dirac equation and QED rather than derived here, and the last term is diamagnetic: H_dia = (e²B²/8mₑ)(x² + y²). Its size relative to the linear term is eB⟨ρ²⟩/4ħ, about 1.06 × 10⁻⁶ per tesla when ⟨ρ²⟩ ≈ a₀². That is why it goes — and the same estimate says when you may not drop it. Since ⟨ρ²⟩ grows as n⁴, the two terms are comparable near n = 30 at 1 T, and in a magnetar field of 10⁸ T the diamagnetic term dominates outright and the whole level scheme has to be rebuilt.

The anomaly is gₛ − 1, and nothing else

Write L + gₛ S = J + (gₛ − 1)S, so H_Z = (μ_B B/ħ)[J_z + (gₛ − 1)S_z] for B along z. Now ask what would happen if gₛ were 1. The second term vanishes, H_Z is proportional to J_z alone, and it is already diagonal in |n l j mⱼ⟩ with shift μ_B B mⱼ for every level whatever its l, j or s — equal spacing everywhere, and, since Δmⱼ = 0, ±1, exactly three lines in any transition. That is the 'normal' Zeeman effect, and it is precisely what Lorentz's classical orbiting electron predicts. Every complication in this topic — the unequal g_J values, the six-component sodium D2 pattern, the bent intermediate-field curves, the Paschen-Back crossover — traces back to the single fact that gₛ ≈ 2 rather than 1. Singlet levels have S = 0, so (gₛ − 1)S_z has nothing to act on and they show the normal pattern; that is the only sense in which 'normal' is normal.

Weak field: the projection theorem hands you g_J

While μ_B B ≪ Δ_fs, treat H_Z as a perturbation on the fine-structure eigenstates and diagonalise it inside one j. J_z is already diagonal there. For S_z use the projection theorem — the Wigner-Eckart theorem specialised to a vector operator acting within a fixed-j subspace — which gives ⟨j mⱼ|S_z|j mⱼ⟩ = ⟨J⋅S⟩ ħ mⱼ / [ħ² j(j+1)]. With J⋅S = (J² + S² − L²)/2 = (ħ²/2)[j(j+1) + s(s+1) − l(l+1)], this returns the Landé factor g_J = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)] and the shift g_J μ_B B mⱼ. So ²S_{1/2} gets g_J = 2, ²P_{1/2} gets 2/3, ²P_{3/2} gets 4/3, and any singlet gets 1. Note what the theorem is not: it is not the classical claim that S precesses so fast about J that only its parallel component survives. The vector model reproduces the same fraction, but the real statement is an exact one about matrix elements inside one irreducible subspace.

The regime is a ratio — and the wrong one is easy to check

'Weak' means μ_B B small compared with the fine-structure interval, not compared with the transition energy. For sodium's 3p term Δ_fs = 17.20 cm⁻¹ = 2.132 meV, while μ_B B = 57.88 μeV per tesla, so the two are equal at 36.8 T: a 1 T lab magnet sits at a ratio of 0.027, comfortably weak. Hydrogen's 2p is a different story. Its interval is only 45.3 μeV (10.9 GHz), so the crossover falls at 0.783 T and a 2 T superconducting magnet has already left the Landé regime behind. The difference is spin-orbit strength, which scales roughly as Z_eff⁴/n³. Push down one more level of structure: hydrogen's 1s hyperfine interval is 5.87 μeV, so its Back-Goudsmit crossover sits near 0.05 T. That is why hyperfine Breit-Rabi diagrams bend at a few hundred gauss while fine-structure fans stay straight to several tesla.

Intermediate field: diagonalise the fixed-mⱼ block

With B along z, [J_z, H] = 0 at every field, so mⱼ stays a good quantum number throughout; [J², H] ≠ 0 once H_Z is present, so j does not. H therefore block-diagonalises by mⱼ, and for s = ½ each block is at most 2 × 2, spanning |l+½, mⱼ⟩ and |l−½, mⱼ⟩. Its diagonal entries are the spin-orbit energies Aħ²l/2 and −Aħ²(l+1)/2 plus the Landé shifts; its off-diagonal entry is ⟨l−½, mⱼ|H_Z|l+½, mⱼ⟩ = −μ_B B √[(l+½)² − mⱼ²]/(2l+1). Two consequences follow. First, the coupling leaves the trace untouched, so the two eigenvalues are pushed apart symmetrically — level repulsion, not a common drift. Second, the coupling vanishes at mⱼ = ±(l+½): the stretched states have no partner to mix with, so they are exact eigenstates and stay exactly linear in B at any field. For hydrogen 2p at 0.500 T the mⱼ = +½ eigenvalues are +37.60 and −23.75 μeV, while Landé predicts +34.40 and −20.55 — each off by 3.20 μeV, or 9.3%.

Strong field: Paschen-Back, and the triplet returning

When μ_B B ≫ Δ_fs, swap the roles: H_Z becomes the zeroth-order term and spin-orbit the perturbation. L and S then precess about B independently, mₗ and mₛ are separately good, and E = μ_B B(mₗ + gₛ mₛ) + Aħ² mₗ mₛ, because ⟨L⋅S⟩ reduces to ⟨L_z S_z⟩ = ħ² mₗ mₛ once the ladder terms average away. Now run a dipole transition through it. The photon does not couple to spin, so Δmₛ = 0 sits on top of Δmₗ = 0, ±1, the spin term cancels between upper and lower levels, and the field-linear shift is just μ_B B Δmₗ — three lines at 0 and ±μ_B B. The Lorentz triplet comes back, which is what Paschen and Back saw and why the limit carries their names. What survives of the anomaly is the residual Aħ² mₗ mₛ, splitting each outer component into a close doublet of separation Aħ² = 2Δ_fs/(2l+1): the spin-orbit constant, read off at a field that has otherwise beaten the coupling flat.

Change one variable at a time

Make the relationship visible.

Interactive model

Field μ_B B / Δ_fs2.00

Spin anomaly gₛ − 11.0

Set the spin anomaly to 0 and every level straightens into slope mⱼ with the mixing shift exactly zero — the whole anomaly is gₛ ≠ 1. Restore 1.0 and drag the field past 1 unit (0.78 T for hydrogen 2p) to watch the mⱼ = ±½ curves peel off the dashed Landé line.

FIELD μ_B B / Δ_fs2.00

EXACT UPPER mⱼ = +½2.092 Δ_fs

LANDÉ LINE, SAME LEVEL1.667 Δ_fs

j-MIXING SHIFT0.425 Δ_fs

Live interpretationFIELD μ_B B / Δ_fs: 2.00. EXACT UPPER mⱼ = +½: 2.092 Δ_fs. LANDÉ LINE, SAME LEVEL: 1.667 Δ_fs. j-MIXING SHIFT: 0.425 Δ_fs

Catch the common trap

Explain before calculating.

Misconception check

g_J gets treated as a property of the level, so g_J μ_B B mⱼ is quoted at whatever field the magnet happens to reach. It is not a property: it is the first-order slope of the level at B → 0, produced by degenerate perturbation theory inside a single j. The condition it needs is μ_B B ≪ Δ_fs, and the usual slip is to test the wrong denominator — comparing μ_B B with the optical transition energy instead of with the fine-structure interval. For sodium's D lines the photon carries 2.10 eV while Δ_fs is 2.13 meV, a factor of a thousand: the weak-field formula is safe to a few tesla, not to a few thousand. Past the crossover the off-diagonal elements between j = 3/2 and j = 1/2 at fixed mⱼ stop being negligible, the levels bend away from their straight lines, and only the stretched mⱼ = ±(l+½) states — which have no partner to mix with — remain exactly linear.

A 2p electron (l = 1, s = ½) sits in a uniform field along z, with the spin-orbit and Zeeman terms both kept exactly and no assumption made about their relative size. Which of the six sublevels have energies that are exactly linear in B at every field strength?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium's D2 line runs from ²P_{3/2} to ²S_{1/2}. Find the Landé factor of each level, list the allowed components in units of μ_B B, and evaluate the pattern in μeV at B = 0.60 T. Take gₛ = 2 exactly.

Use g_J = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)]. For ²P_{3/2}: j = 3/2, l = 1, s = ½, so g_J = 1 + (3.75 + 0.75 − 2)/(2 × 3.75) = 1 + 2.5/7.5 = 4/3.
For ²S_{1/2}: l = 0, j = ½, so g_J = 1 + (0.75 + 0.75 − 0)/(2 × 0.75) = 1 + 1 = 2. A pure-spin level simply carries gₛ.
Each line shifts by μ_B B(gᵤ mᵤ − gₗ mₗ) subject to Δmⱼ = 0, ±1. The six allowed (mᵤ, mₗ) pairs give gᵤ mᵤ − gₗ mₗ = +5/3 (½,−½), +1 (3/2,½), +1/3 (−½,−½), −1/3 (½,½), −1 (−3/2,−½) and −5/3 (−½,½).
μ_B B = (57.884 μeV T⁻¹)(0.60 T) = 34.73 μeV, so the components sit at ±11.58, ±34.73 and ±57.88 μeV, spanning (10/3)μ_B B = 115.8 μeV. Six components, not three, because gᵤ ≠ gₗ.

Answerg(²P_{3/2}) = 4/3, g(²S_{1/2}) = 2. Six components at gᵤ mᵤ − gₗ mₗ = ±1/3, ±1, ±5/3; at 0.60 T that is ±11.6, ±34.7 and ±57.9 μeV, a total span of 116 μeV (0.93 cm⁻¹).

MediumHydrogen's 2p fine-structure interval is Δ_fs = 45.3 μeV. At B = 0.500 T, find the two mⱼ = +½ eigenvalues of the 2p manifold exactly and compare them with the Landé prediction. Measure energies from the 2p centroid, where the j = 3/2 levels sit at +Δ_fs/3 and the j = 1/2 levels at −2Δ_fs/3.

Field scale first: μ_B B = (57.884 μeV T⁻¹)(0.500 T) = 28.94 μeV, so μ_B B/Δ_fs = 0.639. Neither limit applies, so neither limiting formula may be quoted.
Build the 2 × 2 in the coupled basis (|3/2, ½⟩, |1/2, ½⟩). Diagonals: Δ_fs/3 + (4/3)(½)μ_B B = 15.10 + 19.29 = 34.39 μeV, and −2Δ_fs/3 + (2/3)(½)μ_B B = −30.20 + 9.65 = −20.55 μeV. These two numbers are exactly the Landé answers.
Off-diagonal: −μ_B B√[(3/2)² − (½)²]/(2l+1) = −28.94 × √2/3 = −13.64 μeV. This is the term the Landé formula sets to zero.
Diagonalise: λ± = (34.39 − 20.55)/2 ± √([(34.39 + 20.55)/2]² + 13.64²) = 6.92 ± √(754.8 + 186.1) = 6.92 ± 30.67.
So λ+ = +37.60 μeV and λ− = −23.75 μeV, against Landé's +34.39 and −20.55. Both are displaced by 3.20 μeV in opposite directions — the trace is untouched by the off-diagonal element, so the repulsion is exactly symmetric, which is a free check on the arithmetic.

AnswerExact: +37.60 μeV and −23.75 μeV. Landé: +34.39 and −20.55 μeV, each off by 3.20 μeV — 9.3% of the upper level's energy. At μ_B B/Δ_fs = 0.64 the first-order formula is already unusable.

HardStay with hydrogen 2p (Δ_fs = 45.3 μeV) and set the field so that μ_B B = Δ_fs exactly. (a) What is B? (b) Find all six sublevel energies about the centroid. (c) Check them against the trace. (d) In the weak-field extrapolation two straight lines cross; find the field where they would meet, and the true gap there.

(a) B = Δ_fs/μ_B = 45.3 μeV ÷ 57.884 μeV T⁻¹ = 0.783 T. Write x = μ_B B = 45.3 μeV from here on.
(b) The stretched states mⱼ = ±3/2 are alone in their blocks, so their energies are bare diagonal entries: E = Δ_fs/3 ± (4/3)(3/2)x = 15.10 ± 90.60, giving +105.70 and −75.50 μeV.
(b) For mⱼ = +½ the diagonals are 15.10 + (2/3)(45.3) = 45.30 and −30.20 + (1/3)(45.3) = −15.10 μeV, with off-diagonal −x√2/3 = −21.35 μeV. λ± = 15.10 ± √(30.20² + 21.35²) = 15.10 ± 36.99, so +52.09 and −21.89 μeV.
(b) For mⱼ = −½ the diagonals are 15.10 − 30.20 = −15.10 and −30.20 − 15.10 = −45.30 μeV, same off-diagonal. λ± = −30.20 ± √(15.10² + 21.35²) = −30.20 ± 26.15, giving −4.05 and −56.35 μeV.
(c) Trace check: 105.70 + 52.09 − 4.05 − 21.89 − 56.35 − 75.50 = 0.00 μeV, as required. Tr(L⋅S) = 0 over a complete l–s manifold and Σ(mₗ + 2mₛ) = 0 as well, so no unitary change of basis can move the sum.
(d) The weak-field lines Δ_fs/3 − (2/3)x for (j = 3/2, mⱼ = −½) and −2Δ_fs/3 − (1/3)x for (j = 1/2, mⱼ = −½) meet when x/3 = Δ_fs, i.e. x = 135.9 μeV, B = 2.35 T. They share mⱼ, so they cannot cross: the true gap is 2|V| = 2x√2/3 = 128 μeV.

AnswerB = 0.783 T. The six levels lie at +105.70, +52.09, −4.05, −21.89, −56.35 and −75.50 μeV, summing to zero. The two mⱼ = −½ lines would cross at 2.35 T; sharing mⱼ, they instead repel with a 128 μeV gap.