University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.1
The Zeeman Hamiltonian, and what counts as a weak field
Before a single Zeeman shift can be computed a basis has to be chosen, and the magnet cannot choose it for you. This lesson writes down the magnetic Hamiltonian hydrogen actually needs, sizes every term it throws away, and turns “weak field” into a number: μB B over the splitting it competes with.
Build the model
Connect the measurement to the mechanism.
Put hydrogen in a uniform field B along z and Unit 3's minimal coupling, with A = ½B × r, and its spin moment add three terms: the orbital (μB B/ħ)Lz, the spin (gₛ μB B/ħ)Sz and the diamagnetic (e²B²/8mₑ)(x² + y²). The model keeps the first two with gₛ = 2, quoted from the Dirac equation, so H'Z = (μB B/ħ)(Lz + 2Sz), and drops the rest: the diamagnetic term is 3 × 10⁻⁵ of μB B for 2s at 1 T, the proton's moment is 1.5 × 10⁻³ of the electron's, and gₛ − 2 = 0.0023 decides nothing. What H'Z cannot settle alone is its basis, because fine structure is already present.
Spin–orbit coupling contains L⋅S, which commutes with J² but not with Lz or Sz; H'Z contains Lz + 2Sz = Jz + Sz, which commutes with Lz and Sz but not with J². No basis diagonalises both, so the bigger term chooses, and the measure is x = μB B/ΔEfs: for n = 2, ΔEfs = 45.3 μeV and x = 1 at B* = 0.78 T. Weak and strong mean x ≪ 1 and x ≫ 1, a verdict on the level, not on the magnet.
What both terms respect, L² and Jz, survives at every field, so no block is ever larger than 2 × 2. The model ends where its dropped terms stop being small: near 10 mT hyperfine structure couples in the proton, and at high n the diamagnetic term, growing as n⁴B², overtakes the fine structure and breaks l itself, at 0.55 T for n = 10.
- Simple definition
- The Zeeman Hamiltonian H'Z = (μB B/ħ)(Lz + 2Sz) is the energy of hydrogen's orbital and spin moments in a uniform field B along z; the field is weak or strong for a level as μB B is much smaller or much larger than the fine-structure splitting it competes with.
- Example
- μB B/h is 14.0 GHz per tesla and hydrogen's 2P₃/₂–2P₁/₂ interval is 10.95 GHz, so the crossover is B* = 0.78 T: a 0.05 T magnet is weak for n = 2 (x = 0.064) and a 5 T magnet is strong (x = 6.4).
Unit 3's orbital and spin terms, with gₛ = 2 quoted from the Dirac equation. One μB B is 57.9 μeV, or 14.0 GHz, per tesla: the yardstick for every comparison here.
μB = eħ/2mₑ = 5.788 × 10⁻⁵ eV T⁻¹ (9.274 × 10⁻²⁴ J T⁻¹); B in T; Lz and Sz in J s
Derived here. For 2s, ⟨x² + y²⟩ = 28a² and the ratio is 7B/Bₐₜ = 3.0 × 10⁻⁵ at 1 T. It grows as B and as n⁴, which is what finally breaks the model.
Hdia = (e²B²/8mₑ)(x² + y²); Bₐₜ = ħ/(ea²) = 2.35 × 10⁵ T; a = 5.29 × 10⁻¹¹ m; the ratio is dimensionless
Nonzero, so no basis diagonalises both terms, while Jz = Lz + Sz commutes with L⋅S. Had gₛ been 1, the field term would be Jz and every field would be weak.
L and S in J s, the commutator in J³ s³; Lz alone gives +iħ(L × S)z and Sz gives −iħ(L × S)z
0.78 T for 2p, 0.23 T for 3p, 0.077 T for 3d, and Z⁴ times more for a hydrogen-like ion. In atomic units of field the crossover is of order α², far below 1.
ΔEfs = α²(13.6 eV)/[n³ l(l + 1)], Unit 8's j = l ± ½ interval; α ≈ 1/137; α²Bₐₜ = 12.5 T
The linear weak-field shifts then err by fractions x/3 for 2P₃/₂ and 2x/3 for 2P₁/₂, so per-cent accuracy needs x ≈ 0.015, about 0.01 T, not merely a field below 0.78 T.
ε and x dimensionless; valid for x ≪ 1; from the element (√2/3)μB B joining 2P₃/₂ and 2P₁/₂ at equal mⱼ
Below it the proton spin couples in and F and mF, not mⱼ, are the good labels; above it j breaks. It is under two decades wide, and the 0.01 T that per-cent accuracy asks for lies below its floor.
ΔEhfs/h = 24 to 178 MHz for the n = 2 levels, quoted; ΔEfs/h = 10.95 GHz; so 13 mT ≪ B ≪ 0.78 T
What the field adds, term by term
Quantum Mechanics I's p²/2mₑ + V(r) becomes, with Unit 3's minimal coupling and spin moment for charge −e, (p + eA)²/2mₑ + V(r) + (gₛ μB/ħ)S⋅B. For a uniform B along z take A = ½B × r, which has ∇⋅A = 0, so p⋅A = A⋅p and the cross term is (e/mₑ)A⋅p = (e/2mₑ)B⋅(r × p) = (μB B/ħ)Lz, with μB = eħ/2mₑ. The square of A gives (e²/2mₑ)|½B × r|² = (e²B²/8mₑ)(x² + y²). So three terms appear: the orbital (μB B/ħ)Lz and the diamagnetic (e²B²/8mₑ)(x² + y²), both derived here from the expansion, and the spin (gₛ μB B/ħ)Sz, carried over from Unit 3's spin moment. The value gₛ = 2 is quoted from the Dirac equation, and the measured 2.00232, whose excess α/π comes from quantum electrodynamics, is quoted too. The working model is H'Z = (μB B/ħ)(Lz + 2Sz). Dimensions: μB B is in J T⁻¹ × T = J and (Lz + 2Sz)/ħ is a pure number. Its scale, μB = 5.788 × 10⁻⁵ eV T⁻¹, is 57.9 μeV or 14.0 GHz per tesla.
Sizing what the model throws away
Each dropped term is judged by its ratio to μB B. The diamagnetic one gives (eB/4ħ)⟨x² + y²⟩ = (B/Bₐₜ)⟨x² + y²⟩/(4a²), with Bₐₜ = ħ/(ea²) = 2.35 × 10⁵ T. For 2s, Unit 2's R₂₀ gives ⟨r²⟩ = 42a², and spherical symmetry gives ⟨x² + y²⟩ = (2/3)⟨r²⟩ = 28a², so the ratio is 7B/Bₐₜ = 3.0 × 10⁻⁵ at 1 T, a shift of 1.7 neV or 0.42 MHz. It is quadratic in B and grows with ⟨r²⟩ ∝ n⁴, negligible for n = 2 in any laboratory magnet and not for Rydberg levels. The proton's moment, gₚ(e/2mₚ)Sₚ with gₚ = 5.586 quoted from experiment, is (gₚ/2)(mₑ/mₚ) = 1.5 × 10⁻³ of the electron's; its direct Zeeman energy is dropped, and its coupling to the electron, hyperfine structure, waits for the hyperfine topics. Replacing gₛ = 2 by 2.00232 changes the spin term by 0.12 per cent and no comparison below. Order check: μB Bₐₜ = ħ²/2mₑ a² = 13.6 eV, so at 1 T the field perturbs the Bohr energies at the 4 × 10⁻⁶ level.
The operators both terms respect
Inside one n level the good basis must diagonalise Hfs and H'Z together. H₀ and every piece of Hfs, the p⁴ correction, the Darwin term and f(r)L⋅S, are unchanged when orbit and spin rotate together, so they commute with J² and Jz, and because p² = pᵣ² + L²/r² they commute with L². H'Z commutes with L², S², Lz, Sz and Jz, and even with H₀, so it never mixes different n. The two disagree about the rest: [L⋅S, Lz] = iħ(L × S)z and [L⋅S, Sz] = −iħ(L × S)z, so [L⋅S, Lz + 2Sz] = −iħ(L × S)z ≠ 0. The spin's factor 2 does the damage: with gₛ = 1, H'Z would be (μB B/ħ)Jz, commute with L⋅S, and leave every field weak. What survives is L², S² and Jz, so l and mⱼ are good in every regime and the Hamiltonian splits into blocks of fixed l and mⱼ. For spin ½ a block holds at most |mₗ = mⱼ − ½, ↑⟩ and |mₗ = mⱼ + ½, ↓⟩, so the intermediate field never needs more than a 2 × 2 diagonalisation: n = 2 gives four 1 × 1 and two 2 × 2 blocks.
Weak, strong, and how well either holds
The bigger perturbation is diagonalised first, measured by x = μB B/ΔEfs with Unit 8's ΔEfs = α²(13.6 eV)/[n³l(l + 1)]: 45.3 μeV or 10.95 GHz for 2p, against a measured 10.969 GHz. For x ≪ 1, J² stays good and H'Z acts inside each (2j + 1)-fold level, already diagonal there because it commutes with Jz; the next lesson evaluates it. For x ≫ 1, Lz and Sz are good and fine structure is the correction. For n = 2, x = 1 at B* = 0.78 T. How far below counts as weak follows from the mⱼ = +½ pair. Unit 5 gives |3/2, ½⟩ = √(2/3)|0↑⟩ + √(1/3)|1↓⟩ and |½, ½⟩ = −√(1/3)|0↑⟩ + √(2/3)|1↓⟩, and (Lz + 2Sz)/ħ is 1 on |0↑⟩ and 0 on |1↓⟩, so the off-diagonal element is −(√2/3)μB B. It admixes the other j with amplitude (√2/3)x and shifts each level by (2/9)x²ΔEfs, so the linear shifts err by fractions x/3 and 2x/3: per-cent accuracy needs B ≈ 0.01 T. But μB B meets the measured n = 2 hyperfine splittings, 24 to 178 MHz, at 2 to 13 mT, so there the proton is no longer negligible. For 2p the weak-field formulas hold to about ten per cent near 0.05 T, and no field makes them good to one per cent.
The Lamb shift is not the competitor
n = 2 has a second small splitting, the measured Lamb shift of 1057.8 MHz, 4.37 μeV, lifting 2S₁/₂ above 2P₁/₂, and μB B passes it at 0.076 T. It does not compete for the basis. H'Z commutes with L² and, built from the axial vectors L and S, is even under parity, so it has no matrix element between 2s and 2p; the Lamb shift only fixes where 2s and 2p levels cross as the field grows, which the intermediate-field lesson plots. The competitor is always a splitting between states the perturbation connects. For 2s, with l = 0, only j = ½ exists, the coupled and uncoupled bases coincide, and E = E₂s ± μB B at every field, derived in one line from 2Sz/ħ = ±1. The 1s ground state is the same, so in this model it has no Paschen–Back transition; its only competitor is the 1420 MHz hyperfine splitting. An electric field is the opposite case: eFz is odd under parity, joins 2s to 2p₀, and meets the Lamb shift head on, as the Stark lessons show.
The same magnet, different verdicts
Because B* = α²Bₐₜ/[n³l(l + 1)] with α²Bₐₜ = 12.5 T, the verdict belongs to the level. In a 5.0 T solenoid μB B/h = 70.0 GHz. Hydrogen 2p has x = 6.4 and is Paschen–Back; 3p, B* = 0.23 T, and 3d, B* = 0.077 T, are deeper still. He⁺ n = 2, fine structure ×Z⁴, has B* = 12.5 T and x = 0.40, between the limits. Sodium 3p, with a measured 515.5 GHz D₁–D₂ interval, has x = 0.14 and is weak, which is why sodium's D lines keep the anomalous pattern at 5 T while hydrogen's n = 2 is already Paschen–Back. The order-in-α check reads the same way: in atomic units the n = 2 crossover is α²/16, so an argument that compares B only with Bₐₜ = 2.35 × 10⁵ T calls every laboratory field weak, and is wrong for hydrogen above 1 T. At the other end the diamagnetic term grows as n⁴B² while ΔEfs falls as n⁻³: for n = 10 p states with mₗ = ±1 they are equal at 0.55 T, and above that l itself is lost, as in Rydberg spectra in laboratory fields and hydrogen lines from magnetic white dwarfs at 10² to 10⁵ T.
Change one variable at a time
Make the relationship visible.
At 1 T with n = 2, x = 1.28: neither limit. Drop B to 0.05 T for x = 0.064, weak enough for j and mⱼ to be good. Then set n = 10 at 1 T: x ≈ 160, deep Paschen–Back, yet the dashed diamagnetic curve has passed the fine-structure level at 0.55 T, so the term the model drops now breaks l.
FIELD B1.000 T
ZEEMAN SCALE μB B57.88 μeV
x = μB B / ΔEfs1.278
DIAMAGNETIC = ΔEfs AT175.06 T
Live interpretationFIELD B: 1.000 T. ZEEMAN SCALE μB B: 57.88 μeV. x = μB B / ΔEfs: 1.278. DIAMAGNETIC = ΔEfs AT: 175.06 T
Catch the common trap
Explain before calculating.
Hydrogen in its n = 3, l = 2 level sits in a 0.50 T field. Unit 8 gives the 3D₅/₂–3D₃/₂ interval as 4.47 μeV. Which labels, besides n, describe the good states to a good approximation?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 5.0 T superconducting solenoid holds a hydrogen discharge and a sodium lamp. Hydrogen's 2P₃/₂–2P₁/₂ interval is 10.95 GHz (Unit 8) and sodium's 3P₃/₂–3P₁/₂ interval is a measured 515.5 GHz. With μB/h = 13.996 GHz T⁻¹, decide the regime of each level and name its good quantum numbers.
- Zeeman scale: μB B/h = 13.996 GHz T⁻¹ × 5.0 T = 70.0 GHz, which is μB B = 5.788 × 10⁻⁵ eV T⁻¹ × 5.0 T = 289 μeV.
- Hydrogen 2p: x = 70.0/10.95 = 6.4. The field term is the larger perturbation, so l, mₗ and mₛ are good: the Paschen–Back regime. Its crossover is B* = 10.95/13.996 = 0.78 T.
- Sodium 3p: x = 70.0/515.5 = 0.136. Fine structure dominates, so l, j and mⱼ are good: the weak-field regime, with its crossover at B* = 515.5/13.996 = 36.8 T.
- In both, l and mⱼ are good whatever the field, because fine structure and H'Z both commute with L² and Jz; the regime only decides whether j or the pair mₗ, mₛ joins them.
AnswerHydrogen 2p: x = 6.4, strong field, labels l, mₗ, mₛ. Sodium 3p: x = 0.14, weak field, labels l, j, mⱼ. One 5.0 T magnet, opposite regimes.
MediumUsing Unit 2's R₂₀(r) = (2a³)(−1/2)(1 − r/2a)e(−r/2a), find ⟨r²⟩ for 2s, then the diamagnetic shift (e²B²/8mₑ)⟨x² + y²⟩ of 2s at 5.0 T. Compare it with μB B and with the 45.3 μeV fine-structure interval, and find the field at which it would reach 1 per cent of that interval.
- ⟨r²⟩ = ∫ r⁴R₂₀² dr from 0 to ∞ = (a²/2)∫ u⁴(1 − u/2)²e(−u) du with u = r/a. Expand to u⁴ − u⁵ + u⁶/4 and use ∫uⁿe(−u) du = n!: (a²/2)(24 − 120 + 180) = 42a².
- 2s is spherically symmetric, so ⟨x²⟩ = ⟨y²⟩ = ⟨z²⟩ = ⟨r²⟩/3 and ⟨x² + y²⟩ = 28a².
- Prefactor: e²a²/8mₑ = (1.602 × 10⁻¹⁹ C)²(5.292 × 10⁻¹¹ m)²/(8 × 9.109 × 10⁻³¹ kg) = 9.86 × 10⁻³⁰ J T⁻² = 6.16 × 10⁻¹¹ eV T⁻². At 5.0 T the shift is 28 × 6.16 × 10⁻¹¹ eV T⁻² × 25 T² = 4.31 × 10⁻⁸ eV = 43.1 neV, or 10.4 MHz.
- Against μB B = 289 μeV the ratio is 1.49 × 10⁻⁴, matching 7B/Bₐₜ = 35/(2.35 × 10⁵). Against the fine structure it is 43.1 neV/45.3 μeV = 9.5 × 10⁻⁴. Both 2s sublevels move by the same amount, so the pair shifts without splitting.
- The shift is 1.72 × 10⁻⁹ eV T⁻² × B², so it reaches 1 per cent of 45.3 μeV, 0.453 μeV, at B = √[(0.453 × 10⁻⁶)/(1.72 × 10⁻⁹)] T = 16 T.
Answer⟨r²⟩ = 42a²; the 2s diamagnetic shift at 5.0 T is 43 neV (10 MHz), 1.5 × 10⁻⁴ of μB B and 9.5 × 10⁻⁴ of the fine structure, reaching 1 per cent of the fine structure only near 16 T.
HardHydrogen excited to n = 10, l = 1, mₗ = +1, mₛ = +½ sits in a 1.00 T field. Take ⟨r²⟩ = (n²a²/2)[5n² + 1 − 3l(l + 1)], quoted from standard tables, and ⟨sin²θ⟩ = 4/5 for Y₁(±1). Compute ΔEfs, μB B and the diamagnetic shift, rank them, say which of this lesson's claims fails, and find the field at which the diamagnetic shift equals ΔEfs.
- Fine structure: ΔEfs = α²(13.606 eV)/[n³l(l + 1)] = 7.245 × 10⁻⁴ eV/2000 = 3.62 × 10⁻⁷ eV = 0.362 μeV, or 87.6 MHz. Zeeman: μB B = 57.9 μeV, so x = 160, deep Paschen–Back, with B* = 12.5 T/2000 = 6.3 mT.
- Diamagnetic: ⟨r²⟩ = (100a²/2)(500 + 1 − 6) = 24 750a², so ⟨x² + y²⟩ = ⟨r²⟩⟨sin²θ⟩ = 19 800a². With e²a²/8mₑ = 6.157 × 10⁻¹¹ eV T⁻², the shift is 19 800 × 6.157 × 10⁻¹¹ eV = 1.22 μeV at 1.00 T.
- Rank: μB B = 57.9 μeV, diamagnetic 1.22 μeV, ΔEfs = 0.362 μeV. The dropped term is still only 2.1 per cent of μB B, but it is 3.4 times the fine structure.
- In the strong field, 10p and 10f with equal mₗ and mₛ are degenerate under H₀ + H'Z, split only by fine structure, 0.16 μeV here (quoted). Hdia is even and commutes with Lz, so it joins l = 1 to l = 3; a numerical integral, quoted not derived, gives −0.31 μeV for that element, about twice the gap, so l mixes at tens of per cent. The claim that l is good in every regime fails; mₗ and mₛ hold.
- Crossover of the dropped term: 1.219 × 10⁻⁶ eV T⁻² × B² = 3.62 × 10⁻⁷ eV at B = √0.297 T = 0.545 T. Check: for n = 2 the same condition gives 175 T, and 175/0.545 = 321 = (10/2)3.5 × √(0.99/0.75), the n⁷B² scaling with its (n² − 1) factor.
AnswerΔEfs = 0.362 μeV (87.6 MHz), μB B = 57.9 μeV, diamagnetic shift 1.22 μeV: x = 160, so mₗ and mₛ are good, but the dropped term is 3.4 times the fine structure and mixes 10p with 10f, so l is not. They are equal at 0.55 T.