University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.3
The Paschen-Back limit and the normal Zeeman pattern
Put hydrogen in a field of a few tesla and the fine-structure states of the last unit stop being the right place to start. This lesson lets the field choose the states, adds fine structure as the small correction it has become, and shows why the emitted lines fall back to the three-line pattern Lorentz predicted without spin.
Build the model
Connect the measurement to the mechanism.
Which basis is good depends on which perturbation is larger, not on which one was met first. For n = 2 hydrogen the fine-structure interval is 10.95 GHz, and μB B equals it at 0.78 T; a few tesla higher the field term (μB B/ħ)(Lz + 2Sz) dominates. Because Lz and Sz commute with the Coulomb Hamiltonian, the uncoupled states |n l mₗ mₛ⟩ are exact eigenstates of H₀ + HZ, with E = Eₙ + μB B(mₗ + 2mₛ): an evenly spaced ladder whose middle rungs, for n = 2, are doubly degenerate.
Fine structure is now the small perturbation, and it may be added as a diagonal element because L² and Jz, which commute with both terms, give the two members of every degenerate pair different labels. The diagonal of L⋅S is ħ²mₗmₛ, so the shifts are of order α² × 13.6 eV and independent of B. Electric-dipole light cannot touch spin, so each line joins states of equal mₛ and the spin term cancels from its photon energy: the lines sit at ν₀ and ν₀ ± μB B/h, the normal triplet, with a fixed 7.30 GHz fine-structure doublet inside each σ component of Lyman-α.
The picture fails at both ends. At low field the off-diagonal spin-orbit element ζ/√2 enters at second order as ζ²/(2μB B) until the full n = 2 matrix is needed; at very high field the diamagnetic term (e²B²/8m)(x² + y²), dropped here, grows as B² and roughly n⁴ and mixes n near 10⁴ T.
- Simple definition
- The Paschen-Back limit is the regime in which an atom's Zeeman energy μB B far exceeds its fine-structure splitting, so the good states are |n l mₗ mₛ⟩ with energies Eₙ + μB B(mₗ + 2mₛ), and fine structure is a small diagonal correction to them.
- Example
- Hydrogen 2p at B = 6.0 T: μB B = 347 μeV against a 45.3 μeV fine-structure interval, a ratio of 7.7, so the six 2p states sit on rungs mₗ + 2mₛ = 2, 1, 0, 0, −1, −2 spaced 347 μeV apart, each moved by a fine-structure shift between −11 and −42 μeV.
Exact for H₀ + HZ, not perturbative: Lz and Sz commute with the Coulomb Hamiltonian, so the linear field term never mixes n or l.
μB = eħ/2m = 5.788 × 10⁻⁵ eV T⁻¹; B in T; Eₙ = −13.6 eV/n²; mₗ and mₛ are pure numbers
The raising-lowering pair moves mₗ + 2mₛ by ±1, so it joins rungs μB B apart and acts only at second order.
L⋅S = Lz Sz + ½(L₊S₋ + L₋S₊); L and S in J s, so ħ² carries the units of J² s²
Relativistic kinetic term plus the spin-orbit diagonal. It does not depend on B, so it offsets each rung without changing its slope.
l ≥ 1; α = 1/137.036, dimensionless; 13.6 eV stands for the Rydberg energy 13.6057 eV, giving 13.6 eV α²/n³ = 90.565 μeV for n = 2; for l = 0 the Darwin term makes the ratio 1
The spin term 2mₛμB B is the same at both ends of every line and cancels, taking the anomalous spin moment with it.
ν in Hz, h = 6.626 × 10⁻³⁴ J s; μB/h = 13.996 GHz T⁻¹; ν₀ is the field-free line
Sets the low-field edge: once it rivals the diagonal shifts, the n = 2 matrix must be diagonalised in blocks of fixed mⱼ.
ζ = 13.6 eV α²/[n³l(l + ½)(l + 1)], 30.19 μeV for 2p; an energy squared over an energy; for 2p, ΔE⁽²⁾/h = 1.90 GHz T ÷ B
Commutes with Lz but not with H₀, so unlike the linear term it mixes n; it nears the 1.89 eV gap from n = 2 to n = 3 only near 10⁴ T.
e in C, B in T, m in kg, x and y in m; for 1s its mean is e²a²B²/4m = 1.23 × 10⁻¹⁰ eV at 1 T
The larger perturbation chooses the basis
Hydrogen in a field has three pieces: H₀, the Zeeman term HZ = (μB B/ħ)(Lz + 2Sz), and the fine structure Hfs. Perturbation theory takes the larger correction first, and the field decides which one that is. For n = 2 the interval 2p₃/₂ − 2p₁/₂ is 45.3 μeV, or 10.95 GHz, and μB B = 5.788 × 10⁻⁵ eV T⁻¹ × B matches it at 0.78 T. At 6 T the field term is 7.7 times larger, so HZ goes first. It needs no approximation: Lz and Sz commute with the spin-free central Hamiltonian, so the uncoupled states |n l mₗ mₛ⟩ are exact eigenstates of H₀ + HZ with E = Eₙ + μB B(mₗ + 2mₛ), derived here in one line. For n = 2 that puts eight states on five rungs, mₗ + 2mₛ = 2, 1, 0, −1 and −2, holding 1, 2, 2, 2 and 1 states, spaced 347 μeV apart at 6 T.
Why fine structure may go on the diagonal
Three rungs are doubly degenerate: 2s ↑ with 2p 0 ↑ at +1, 2p −1 ↑ with 2p +1 ↓ at 0, and 2s ↓ with 2p 0 ↓ at −1, where the number is mₗ and the arrow is mₛ. Degenerate perturbation theory says the first-order shifts are the eigenvalues of Hfs inside each pair; they are simply the diagonal elements when an operator commuting with H₀, HZ and Hfs gives the two members different eigenvalues, because Hfs then has no element between them. Two operators do it. L² separates the s–p pairs, since Hfs never mixes l. Jz = Lz + Sz separates the middle pair, whose members have mⱼ = −½ and +½. Lz and Sz alone would not do, because L⋅S commutes with neither. The couplings that survive, such as ⟨2p 0 ↑|Hfs|2p +1 ↓⟩ with both states at mⱼ = +½, join rungs μB B apart and enter only at second order, smaller than the first-order shifts by roughly ζ/(2μB B).
The diagonal elements, term by term
Unit 8 derived the relativistic kinetic correction, which depends on n and l only: (13.6 eV α²/n³)[3/(4n) − 1/(l + ½)]. Unit 8 also derived the spin-orbit term, (e²/8πε₀m²c²r³)L⋅S with the Thomas factor ½ included. Write L⋅S = LzSz + ½(L₊S₋ + L₋S₊); the second pair changes mₗ and mₛ, so the diagonal element is ħ²mₗmₛ. With ⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³], quoted from the table of hydrogen expectation values, and a = ħ/(mcα), the spin-orbit shift is (13.6 eV α²/n³)mₗmₛ/[l(l + ½)(l + 1)]. Writing 1/(l + ½) as l(l + 1)/[l(l + ½)(l + 1)] merges the two into the bracket on the card. For s states spin-orbit vanishes, and the Darwin term, quoted from the Dirac equation, adds 13.6 eV α²/n³ and turns the ratio into 1. For n = 2, with 13.6 eV standing for the Rydberg energy 13.6057 eV, the unit is 90.565 μeV: the 2p shifts are −11.32, −26.41 and −41.51 μeV for mₗmₛ = +½, 0 and −½, and 2s sits at −56.60 μeV. Every one is of order α² times 13.6 eV.
Three checks on the strong-field shifts
First, the stretched state mₗ = l, mₛ = +½ is the coupled state |j = l + ½, mⱼ = j⟩ at every field. The bracket gives [l(l + 1) − l/2]/[l(l + ½)(l + 1)] = 1/(l + 1) = 1/(j + ½), the weak-field value, and its Zeeman energy μB B(l + 1) equals gJ mⱼ μB B with gJ = (2l + 2)/(2l + 1). Second, |mₗ = 0, ↑⟩ = √(2/3)|3/2, 1/2⟩ − √(1/3)|1/2, 1/2⟩, so its diagonal element must be the weighted mean (2/3)(−11.32) + (1/3)(−56.60) = −26.41 μeV, and it is. Third, a trace does not depend on the basis: the six 2p shifts sum to −158.49 μeV here, as do four 2p₃/₂ and two 2p₁/₂ states. Run back to B = 0, the strong-field lines give three 2p energies instead of the true two, but the right centre of gravity. Against measurement: the 2p interval is 10.949 GHz here and 10.969 GHz measured. Scaling spin-orbit by g − 1 = 1.00232, a QED value of relative order α/π quoted here, and by the recoil factor 1 − m/mₚ gives 10.969 GHz.
Dipole light cannot see the spin
An electric-dipole line joins states through er, which acts on position and leaves spin alone, so ⟨mₛ′|mₛ⟩ forces Δmₛ = 0. From [Lz, z] = 0 and [Lz, x ± iy] = ±ħ(x ± iy), the orbital label changes by Δmₗ = 0 or ±1. Upper and lower states therefore carry the same 2mₛμB B, and it cancels from the photon energy: hν = hν₀ + μB B Δmₗ. Three lines spaced by μB B/h = eB/4πm, 14.00 GHz per tesla, the Larmor frequency of Lorentz's classical electron: the normal triplet, although every level is split by spin. The cancellation is exact, so the anomalous spin moment, g = 2.00232 rather than 2, drops out as well. For Lyman-α at 6.0 T the σ components sit ±84.0 GHz, or ∓4.14 pm, from the centre. Fine structure leaves a trace: the two spin states feeding each σ component have mₗmₛ = +½ and −½, so each σ line is a doublet 30.19 μeV, or 7.30 GHz, wide, centred on ±μB B/h and fixed as B changes, while the π line stays single at first order. Line strengths and polarisations wait for Unit 13.
Where the strong-field picture stops
Two neglected terms bound the regime. From below, the off-diagonal spin-orbit element ζ/√2, with ζ = 30.19 μeV for 2p, pushes each member of a coupled pair away from its partner by ζ²/(2μB B): 0.32 GHz at 6 T, about 4% of the doublet splitting, and 0.63 GHz at 3 T. Near 1 T the expansion in ζ/μB B fails and the n = 2 matrix must be diagonalised in blocks of fixed mⱼ, the next lesson. From above, the A² part of minimal coupling, (e²B²/8m)(x² + y²), grows as B² and, through ⟨x² + y²⟩, roughly as n⁴. With ⟨r²⟩ quoted from the hydrogen table, 1s gives e²a²B²/4m, a fraction B/(2B₀) of μB B with B₀ = ħ/ea² = 2.35 × 10⁵ T, and 2s has ⟨x² + y²⟩ = 28a² against 2a². It commutes with Lz but not with H₀, so it mixes n, and its shifts reach the 1.89 eV gap from n = 2 to n = 3 between about 1.5 × 10⁴ T for 3s and 3 × 10⁴ T for 2s. The linear ladders alone would cross, without interacting, near 6.5 × 10³ T. Magnetic white dwarfs, with fields up to about 10⁵ T, reach past this edge.
Change one variable at a time
Make the relationship visible.
Set k to 0 for the bare normal triplet, then raise it to 1: each σ line splits into a 7.30 GHz doublet whose centre stays put. Now lower B to 3 T and watch the doublets crowd the centre while the neglected second-order shift climbs.
TRIPLET SPACING84.0 GHz
SPACING IN WAVELENGTH4.14 pm
FINE STRUCTURE / ZEEMAN0.130
NEGLECTED 2ND-ORDER SHIFT0.32 GHz
Live interpretationTRIPLET SPACING: 84.0 GHz. SPACING IN WAVELENGTH: 4.14 pm. FINE STRUCTURE / ZEEMAN: 0.130. NEGLECTED 2ND-ORDER SHIFT: 0.32 GHz
Catch the common trap
Explain before calculating.
In the strong-field limit, the states |2p, mₗ = +1, mₛ = −½⟩ and |2p, mₗ = −1, mₛ = +½⟩ share the same Zeeman energy. Why may fine structure be added to each as its own diagonal element, without diagonalising Hfs inside the pair?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen discharge runs in the bore of a 6.0 T magnet. Decide whether the n = 2 level is in the strong-field regime, then predict the frequency and wavelength offsets of the Lyman-α components, taking the field-free line at 121.567 nm.
- Compare energies: μB B = 5.788 × 10⁻⁵ eV T⁻¹ × 6.0 T = 3.47 × 10⁻⁴ eV, against the 2p fine-structure interval of 4.53 × 10⁻⁵ eV. The ratio is 7.7, so the field term goes first and |n l mₗ mₛ⟩ are the good states.
- A dipole line keeps mₛ and changes mₗ by 0 or ±1, so hν = hν₀ + μB B Δmₗ; the spin term 2mₛμB B is common to both ends and cancels.
- Frequency offset: Δν = μB B/h = 13.996 GHz T⁻¹ × 6.0 T = 84.0 GHz.
- Wavelength offset: |Δλ| = λ²Δν/c = (121.567 × 10⁻⁹ m)² × 8.40 × 10¹⁰ Hz ÷ 2.998 × 10⁸ m s⁻¹ = 4.14 × 10⁻¹² m, the higher-frequency component having the shorter wavelength.
- Check: 84.0 GHz is 3.41 × 10⁻⁵ of the 2.466 × 10¹⁵ Hz line frequency, and 4.14 pm is the same fraction of 121.567 nm, as a small shift must be.
AnswerStrong field, ratio 7.7. Three components at ν₀ and ν₀ ± 84.0 GHz, that is 121.567 nm and 121.567 nm ∓ 4.14 pm.
MediumFor hydrogen 2p at B = 6.0 T, find all six strong-field energies relative to E₂, in μeV and including fine structure. Check them with the trace, and say which pair is still degenerate and why that does no harm.
- Zeeman rungs: μB B = 5.78838 × 10⁻⁵ eV T⁻¹ × 6.0 T = 347.303 μeV, so mₗ + 2mₛ = 2, 1, 0, 0, −1 and −2 give 694.606, 347.303, 0, 0, −347.303 and −694.606 μeV.
- Fine structure, with 13.6057 eV × α²/8 = 90.565 μeV and l(l + ½)(l + 1) = 3: mₗmₛ = +½ gives 90.565 × (3/8 − 1.5/3) = −11.321 μeV; mₗmₛ = 0 gives 90.565 × (3/8 − 2/3) = −26.415 μeV; mₗmₛ = −½ gives 90.565 × (3/8 − 2.5/3) = −41.509 μeV.
- Add them: (+1, ↑) 683.29; (0, ↑) 320.89; (−1, ↑) and (+1, ↓) both −41.51; (0, ↓) −373.72; (−1, ↓) −705.93, all in μeV.
- Trace: the Zeeman parts cancel in pairs, and the fine-structure parts sum to 2(−11.321 − 26.415 − 41.509) = −158.49 μeV. In the coupled basis, 4 × (−11.321) + 2 × (−56.603) = −158.49 μeV. They agree, as a trace must.
- The pair (−1, ↑) and (+1, ↓) stays degenerate, but its members have mⱼ = −½ and +½. Hfs commutes with Jz, so no element joins them, and the diagonal values are the correct first-order energies.
Answer683.29, 320.89, −41.51 (twice), −373.72 and −705.93 μeV relative to E₂; the fine-structure parts sum to −158.49 μeV in either basis.
HardA 30 T magnet is to be calibrated from the separation of the two σ-component centres of Lyman-α, read as 2μB B/h. Size three effects the strong-field formula leaves out: second-order spin-orbit, the diamagnetic term, and the proton's share of the orbital motion, which makes the orbital g-factor gL = 1 − m/mₚ (quoted). Which one limits the calibration?
- The separation: 2μB B/h = 2 × 13.996 GHz T⁻¹ × 30 T = 839.8 GHz, with μB B = 1736.5 μeV.
- Second order: V = ζ/√2 = 21.346 μeV joins (+1, ↓) to (0, ↑) one rung above and (−1, ↑) to (0, ↓) one rung below. Each is pushed away by V²/(μB B) = (21.346 μeV)² ÷ 1736.5 μeV = 0.2624 μeV, or 63.4 MHz. (+1, ↓) moves down and (−1, ↑) up, so each σ centre moves 31.7 MHz inward and the separation shrinks by 63.4 MHz.
- Diamagnetic: e²a²/8m = 6.157 × 10⁻¹¹ eV T⁻², and, with ⟨r²⟩ = 30a² for 2p quoted from the table, ⟨x² + y²⟩ = ⟨r²⟩⟨sin²θ⟩ = 30a² × 4/5 = 24a² for mₗ = ±1. Both σ upper states rise by 6.157 × 10⁻¹¹ × 900 × 24 eV = 1.33 μeV, so the separation is unchanged; only the σ midpoint moves 161 MHz from π, where mₗ = 0 has 12a².
- Orbital g-factor: with m/mₚ = 5.446 × 10⁻⁴, gL μB B Δmₗ replaces μB B Δmₗ and the separation shrinks by 5.446 × 10⁻⁴ × 839.8 GHz = 457 MHz.
- Compare: 457 MHz from gL, 63 MHz from second order, nothing from the diamagnetic term, and nothing from the anomalous spin moment, which cancels because Δmₛ = 0.
AnswerThe gL correction dominates: −457 MHz, 5.4 × 10⁻⁴ of the 839.8 GHz separation. Second-order spin-orbit adds −63 MHz (7.6 × 10⁻⁵); the diamagnetic term only moves the σ midpoint 161 MHz from π.