University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.6
The fine-structure formula: energies depend on n and j only
This is where the week's three corrections stop being separate calculations and become one result. Add them state by state, watch l cancel, and you hold a formula in n and j that draws hydrogen's level diagrams in gigahertz, says which levels stay paired, and tells you what a spectrometer aimed at the red Balmer line can hope to see.
Build the model
Connect the measurement to the mechanism.
Three corrections reach hydrogen at order α⁴mc², and each on its own depends on l: the relativistic kinetic shift, derived this week and largest for s states that plunge close to the proton; spin-orbit coupling, derived with a quoted Thomas factor, which raises j = l + ½ and lowers j = l − ½; and the Darwin term, quoted from Dirac theory, which touches s states alone. All three are diagonal in the coupled states |n l j mⱼ⟩, so first-order perturbation theory simply adds them, and the sum holds a small surprise: the l that the kinetic term depends on is handed back, exactly, by spin-orbit or Darwin, leaving Eₙⱼ = −(13.6 eV/n²)[1 + (α²/n²)(n/(j + ½) − 3/4)]. Three predictions follow.
Every level moves down, by less than 2 parts in 10⁵ of its Bohr energy. Levels of equal n and j stay paired, 2S₁/₂ with 2P₁/₂ and 3P₃/₂ with 3D₃/₂. And the gaps are microwave-sized: 2P₃/₂ sits 45.3 μeV, 10.95 GHz, above the n = 2 pair, within 0.2% of the measured 2P₁/₂ to 2P₃/₂ interval of 10.969 GHz, while Balmer-α dissolves into seven components at five frequencies across 14.2 GHz, only 2.5 times the Doppler blur of a room-temperature lamp.
The formula stops where its ingredients stop. It is first order in a perturbation of relative size α², so terms of order α⁶mc² lie beyond it; it treats the proton as a fixed point charge; and it knows nothing of the quantised field, which lifts 2S₁/₂ above 2P₁/₂ by 1057.8 MHz.
- Simple definition
- The fine-structure formula gives each hydrogen level's energy through order α⁴mc², with the relativistic kinetic, spin-orbit and Darwin corrections included, as a function of the principal quantum number n and the total angular momentum j only.
- Example
- At n = 2, 2S₁/₂ and 2P₁/₂ both sit 56.6 μeV below the Bohr energy of −3.40 eV and 2P₃/₂ sits 11.3 μeV below it, a gap of 45.3 μeV or 10.95 GHz; the measured 2P₁/₂ to 2P₃/₂ interval is 10.969 GHz.
Each term alone carries l; their sum carries only j. For j = l + ½ it rests on the identity 2/(l + ½) − 1/[(l + ½)(l + 1)] = 2/(l + 1).
Eₙ = −13.6057 eV/n² and mc² = 0.511 MeV, so Eₙ²/mc² = 362.3 μeV/n⁴. The same for j = l + ½ and j = l − ½.
This is the l ≥ 1 result at j = ½, so s states join the same curve. Hydrogen 2S: −147.2 μeV kinetic and +90.6 μeV Darwin make −56.6 μeV, as for 2P₁/₂.
ED = 2n Eₙ²/mc² = mc²α⁴/(2n³), quoted from Dirac theory. Eₛₒ = 0, since L⋅S vanishes when l = 0.
Since n/(j + ½) ≥ 1 always exceeds 3/4, every level moves down, by at most 1.7 × 10⁻⁵ of its Bohr energy (the n = 2, j = ½ pair). No l appears anywhere in it.
α = 1/137.036, a pure number; n = 1, 2, 3, …; j = ½, 3/2, …, n − ½; energies in eV.
Widest next to j = ½ and falling as 1/n³, so the n = 2 gap dominates Balmer-α. The measured 2P₁/₂ to 2P₃/₂ interval, 10.969 GHz, is 0.18% larger.
mc²α⁴ = 1.449 meV. n = 2: 45.3 μeV, 10.95 GHz. n = 3: 13.4 and 4.47 μeV, or 3.24 and 1.08 GHz.
Seven components, but two pairs share both their j values, so only five frequencies appear, spread over 3.24 + 10.95 = 14.2 GHz.
Chosen by Δl = ±1 (Unit 5) and Δj = 0, ±1 (Wigner–Eckart); 3D₅/₂ → 2P₁/₂ fails, with Δj = 2.
The 14.2 GHz Hα spread is only 2.5 Doppler widths at 300 K, so gaps near 1 GHz, the 3D doublet and the Lamb shift alike, vanish inside a lamp's line.
Full width at half maximum, in Hz. Hα: ν₀ = 4.567 × 10¹⁴ Hz, Mc² = 938.8 MeV; at T = 300 K, 5.64 GHz.
Three shifts, and the one basis that adds them
Nothing here is new about any single correction; the work is adding three the week has produced. In the unit Eₙ²/mc² = mc²α⁴/(4n⁴) = 362.3 μeV/n⁴, the relativistic kinetic correction, derived earlier this week, gives Eᵣₑₗ = (Eₙ²/mc²)[3/2 − 2n/(l + ½)]. Spin-orbit coupling, derived this week with the Thomas factor ½ quoted from relativistic kinematics, gives Eₛₒ = (Eₙ²/mc²) n[j(j + 1) − l(l + 1) − 3/4]/[l(l + ½)(l + 1)] for l ≥ 1. The Darwin term, quoted from Dirac theory, gives ED = 2n Eₙ²/mc² for l = 0 and zero otherwise. First-order shifts may be added only in a basis that makes all three diagonal inside the 2n²-fold degenerate shell. The coupled states |n l j mⱼ⟩ are that basis: p⁴ and δ³(r) are scalars, commuting with L², S², J² and Jz, and L⋅S = (J² − L² − S²)/2 is diagonal by construction. Each shift is one diagonal element, and diagonal elements add.
Watching l cancel
Take j = l + ½ first. The spin-orbit bracket j(j + 1) − l(l + 1) − 3/4 collapses to l, so Eₛₒ = (Eₙ²/mc²) n/[(l + ½)(l + 1)], and the l-dependent part of Eᵣₑₗ + Eₛₒ is −n[2/(l + ½) − 1/((l + ½)(l + 1))] = −n(2l + 1)/[(l + ½)(l + 1)] = −2n/(l + 1), which is −2n/(j + ½). For j = l − ½ the bracket is −(l + 1), Eₛₒ = −(Eₙ²/mc²) n/[l(l + ½)], and the l-part is −n[2/(l + ½) + 1/(l(l + ½))] = −2n/l, again −2n/(j + ½) because now j + ½ = l. For l = 0 there is no spin-orbit shift, and the Darwin term turns 3/2 − 4n into 3/2 − 2n, the j = ½ value. In numbers at n = 2: 2S₁/₂ gets −147.2 μeV kinetic and +90.6 μeV Darwin, 2P₁/₂ gets −26.4 μeV kinetic and −30.2 μeV spin-orbit, and both total −56.6 μeV. The kinetic shifts differ by 120.8 μeV, and the other terms hand back exactly that.
Reading the formula before trusting it
Add the sum to Eₙ and use Eₙ/mc² = −α²/(2n²) to reach Eₙⱼ = −(13.6 eV/n²)[1 + (α²/n²)(n/(j + ½) − 3/4)]. Test it four ways. Dimensions: α and the bracket are pure numbers, so Eₙⱼ carries the eV of 13.6 eV. Limit: as α → 0, which is c → ∞, the Bohr energy returns. Order: the correction is of order α² ≈ 5.3 × 10⁻⁵ relative to Eₙ, α⁴mc² in absolute terms, one factor α² below Bohr as the hierarchy demands. Sign: j + ½ ≤ n makes n/(j + ½) at least 1, above 3/4, so every level is lowered, j = ½ most. That includes 2P₃/₂, whose spin-orbit shift is upward, +15.1 μeV, but whose kinetic shift, −26.4 μeV, is larger. The ground state drops 181.1 μeV, or 43.8 GHz. For a hydrogen-like ion, 13.6 eV becomes 13.6Z² eV and α becomes Zα, so the shifts grow as Z⁴. The result is first order in a perturbation of size α²Eₙ and is silent on terms of order α⁶mc²; the next topic tests it against the quoted Dirac energy.
Counting levels: n energies holding 2n² states
Since j runs from ½ to n − ½, a shell splits into exactly n energies, not the 2n − 1 it would have if every (l, j) pair had its own. Each j below the top is shared by l = j − ½ and l = j + ½ and holds 2(2j + 1) states; the top level, j = n − ½, can only have l = n − 1 and holds 2n. The total, 2n(n − 1) + 2n = 2n², accounts for every state of the Bohr shell. So n = 2 splits into two levels of four states each, 2S₁/₂ with 2P₁/₂ and then 2P₃/₂, and n = 3 into three levels holding 4, 8 and 6 states, with 3S₁/₂ paired with 3P₁/₂ and 3P₃/₂ paired with 3D₃/₂. Fine structure has lifted only part of the Coulomb degeneracy. The quoted Dirac energy, which also depends on n and j alone, keeps the pairs together at every order. The measured 1057.8 MHz that separates 2S₁/₂ from 2P₁/₂ is the Lamb shift, of order α⁵ln(1/α)mc², and it belongs to quantum electrodynamics.
The n = 2 and n = 3 diagrams to scale
Measured from each Bohr energy, n = 2 has its j = ½ pair at −56.6 μeV (−13.69 GHz) and 2P₃/₂ at −11.3 μeV (−2.74 GHz): a gap of 45.3 μeV, 10.95 GHz, a microwave of wavelength 2.74 cm. The measured 2P₁/₂ to 2P₃/₂ interval is 10.969 GHz, 0.18% more. The residue is understood but not derived here: the reduced mass lowers the prediction by about 6 MHz, and the electron's anomalous moment, gₑ = 2.00232 from quantum electrodynamics, strengthens spin-orbit coupling and adds about 25 MHz. For n = 3 the levels sit at −20.1, −6.7 and −2.2 μeV for j = ½, 3/2 and 5/2, that is −4.87, −1.62 and −0.54 GHz, with gaps of 3.24 and 1.08 GHz. Their 3 to 1 ratio comes from 1/(j + ½) − 1/(j + 3/2), which is ½ and then 1/6. The j = ½ to 3/2 gap falls as 1/n³, and 10.95 GHz × 8/27 = 3.24 GHz, so drawn to one scale the whole n = 3 structure, 4.33 GHz across, is narrower than half the n = 2 gap.
Balmer-α: seven components, five frequencies
An n = 3 → 2 photon must obey Δl = ±1, from Unit 5, and Δj = 0 or ±1, from the Wigner–Eckart theorem for the vector operator r. Among the pairs that pass Δl, only 3D₅/₂ → 2P₁/₂ fails Δj, leaving seven components. Two pairs share both their j values, 3S₁/₂ → 2P₁/₂ with 3P₁/₂ → 2S₁/₂ and 3P₃/₂ → 2S₁/₂ with 3D₃/₂ → 2P₁/₂, so five frequencies appear: −2.13, +1.12, +2.20, +8.82 and +12.06 GHz from the Bohr line, 456.7 THz with the reduced mass. Their spread, 14.19 GHz, is the n = 3 gap plus the n = 2 gap, 3 parts in 10⁵ of the line. A 300 K lamp blurs each component to 5.64 GHz, the Doppler width quoted from Unit 13. If the n = 3 sublevels fill in proportion to their degeneracy, tabulated rates put about three quarters of the light in 3D₅/₂ → 2P₃/₂ and 3D₃/₂ → 2P₁/₂, 9.87 GHz apart: the lamp shows a doublet split by 1.75 Doppler widths, and every gap near 1 GHz, the Lamb shift included, is lost inside it.
Change one variable at a time
Make the relationship visible.
At the default n = 2, j = ½ the 2S kinetic bar is 5.6 times the 2P one, yet Darwin and spin-orbit bring both columns to one dashed line. Raise n with j fixed: the bars reshape but still end together. Set n − (j + ½) to 0 for the top level, j = n − ½, which has no partner.
SHIFT Eₙⱼ − Eₙ-56.603 μeV
SHIFT AS A FREQUENCY-13.687 GHz
SHIFT ÷ |Eₙ|-16.64 ppm
KINETIC ALONE, l = j − ½-147.168 μeV
Live interpretationSHIFT Eₙⱼ − Eₙ: −56.603 μeV. SHIFT AS A FREQUENCY: −13.687 GHz. SHIFT ÷ |Eₙ|: −16.64 ppm. KINETIC ALONE, l = j − ½: −147.168 μeV
Catch the common trap
Explain before calculating.
The n = 4 shell of hydrogen holds seven (l, j) states, from 4S₁/₂ to 4F₇/₂. To first order in the relativistic kinetic, spin-orbit and Darwin corrections, how many distinct energies does it split into, and where do they lie relative to the Bohr energy E₄?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyHydrogen's 3P₃/₂ (l = 1) and 3D₃/₂ (l = 2) differ in l but share n = 3 and j = 3/2. Add the relativistic kinetic, spin-orbit and Darwin shifts of each, taking E₃²/mc² = 4.4724 μeV, and show they land on one energy. Give the common shift in μeV, in GHz, and as a fraction of |E₃| = 1.512 eV.
- Kinetic, Eᵣₑₗ = (E₃²/mc²)[3/2 − 6/(l + ½)]. For l = 1 the bracket is 1.5 − 4 = −2.5, giving −11.181 μeV; for l = 2 it is 1.5 − 2.4 = −0.9, giving −4.025 μeV. The P state is pushed down almost three times as hard.
- Spin-orbit, Eₛₒ = (E₃²/mc²) × 3[j(j + 1) − l(l + 1) − 3/4]/[l(l + ½)(l + 1)] with j(j + 1) = 15/4. For 3P₃/₂ the bracket is 15/4 − 2 − 3/4 = 1 and the denominator 1 × 1.5 × 2 = 3, so Eₛₒ = 4.4724 × 3 × 1/3 = +4.472 μeV. For 3D₃/₂ the bracket is 15/4 − 6 − 3/4 = −3 and the denominator 2 × 2.5 × 3 = 15, so Eₛₒ = 4.4724 × 3 × (−3)/15 = −2.683 μeV.
- Darwin: zero for both, since neither is an s state and neither has density at the nucleus.
- Totals: 3P₃/₂ gives −11.181 + 4.472 = −6.709 μeV and 3D₃/₂ gives −4.025 − 2.683 = −6.708 μeV, equal to rounding. The kinetic shifts differ by 7.156 μeV and the spin-orbit shifts by the same 7.156 μeV the other way.
- Check with the formula: (E₃²/mc²)[3/2 − 6/2] = 4.4724 × (−1.5) = −6.709 μeV. As a frequency, 6.709 μeV ÷ 4.1357 μeV GHz⁻¹ = 1.622 GHz. As a fraction, 6.709 × 10⁻⁶ eV ÷ 1.512 eV = 4.44 × 10⁻⁶, which matches (α²/9)(3/2 − 3/4) = 4.44 × 10⁻⁶, order α² as it should be.
AnswerBoth states shift by −6.71 μeV, −1.62 GHz as a frequency, 4.4 × 10⁻⁶ of |E₃|: 3P₃/₂ and 3D₃/₂ stay degenerate although their kinetic shifts differ by 7.16 μeV.
MediumLyman-α (n = 2 → 1) has ν₀ = 2.466 × 10¹⁵ Hz. Which fine-structure components are allowed, how far apart are they, and can a 300 K hydrogen lamp resolve them? Use the Doppler width ΔνD = ν₀√(8kT ln 2/(Mc²)) quoted from Unit 13, with Mc² = 938.8 MeV for the hydrogen atom and k = 8.617 × 10⁻⁵ eV K⁻¹.
- Lower level: n = 1 holds only 1S₁/₂, shifted by −(13.6057 eV)α²(1 − 3/4) = −181.13 μeV.
- Upper levels: 2S₁/₂ and 2P₁/₂ at −56.60 μeV, 2P₃/₂ at −11.32 μeV. Δl = ±1 forbids 2S₁/₂ → 1S₁/₂, which has Δl = 0, leaving 2P₁/₂ → 1S₁/₂ (Δj = 0) and 2P₃/₂ → 1S₁/₂ (Δj = −1): two components.
- Offsets from the Bohr frequency: (−56.60 + 181.13) μeV = 124.53 μeV, or 30.11 GHz; (−11.32 + 181.13) μeV = 169.81 μeV, or 41.06 GHz. They are 10.95 GHz apart, the n = 2 gap, because the lower level is single. The measured gap is 10.969 GHz.
- Doppler width: kT = 8.617 × 10⁻⁵ eV K⁻¹ × 300 K = 0.02585 eV, and √(8 × 0.02585 eV × 0.6931 ÷ (938.8 × 10⁶ eV)) = 1.236 × 10⁻⁵, so ΔνD = 1.236 × 10⁻⁵ × 2.466 × 10¹⁵ Hz = 30.5 GHz.
- Separation over width is 10.95/30.5 = 0.36: the doublet is buried. Since ΔνD grows as √T, the width falls to 10.95 GHz at T = 300 K × (10.95/30.47)² = 39 K.
- Contrast with Hα: the Doppler width scales with ν₀ and the fine-structure spread does not. Lyman-α has 5.4 times the frequency of Hα, so the same gas that shows Hα's 14.2 GHz pattern over a 5.64 GHz width hides Lyman-α's.
AnswerTwo components, 2P₁/₂ → 1S₁/₂ and 2P₃/₂ → 1S₁/₂, at +30.1 and +41.1 GHz from the Bohr frequency and 10.95 GHz apart. The 300 K Doppler width is 30.5 GHz, so they merge unless the gas is cooled to roughly 40 K.
HardLi²⁺ (Z = 3) is hydrogen-like. For its n = 3 shell, find the relativistic kinetic, spin-orbit and Darwin shifts of all five (l, j) states in μeV, show that they sum to a function of j alone, and give the two fine-structure gaps in GHz. Check the result by scaling from hydrogen, and estimate how much the next order in Zα could change it.
- For charge Z, Eₙ = −13.6057 eV × Z²/n². With Z = n = 3, E₃ = −13.606 eV, hydrogen's ground-state energy, so E₃²/mc² = (13.6057 eV)²/(0.511 × 10⁶ eV) = 362.26 μeV, the unit for every term below.
- Kinetic, (E₃²/mc²)[3/2 − 6/(l + ½)]: l = 0 gives 362.26 × (−10.5) = −3803.7 μeV; l = 1 gives × (−2.5) = −905.7 μeV; l = 2 gives × (−0.9) = −326.0 μeV.
- Spin-orbit, (E₃²/mc²) × 3[j(j + 1) − l(l + 1) − 3/4]/[l(l + ½)(l + 1)]: 3P₁/₂ −724.5, 3P₃/₂ +362.3, 3D₃/₂ −217.4 and 3D₅/₂ +144.9 μeV. Darwin, for 3S₁/₂ only: 2n(E₃²/mc²) = 6 × 362.26 = +2173.6 μeV.
- Sums: 3S₁/₂ −3803.7 + 2173.6 = −1630.2; 3P₁/₂ −905.7 − 724.5 = −1630.2; 3P₃/₂ −905.7 + 362.3 = −543.4; 3D₃/₂ −326.0 − 217.4 = −543.4; 3D₅/₂ −326.0 + 144.9 = −181.1 μeV. Three energies, fixed by j, matching (E₃²/mc²)[3/2 − 6/(j + ½)] = −4.5, −1.5 and −0.5 units.
- Gaps: 1086.8 μeV and 362.3 μeV; dividing by h = 4.1357 μeV GHz⁻¹ gives 262.8 GHz and 87.6 GHz, in the 3 to 1 ratio of any n = 3 shell. Scaling check: hydrogen's 3.244 and 1.081 GHz times Z⁴ = 81 give 262.8 and 87.6 GHz. And 3D₅/₂ at −181.1 μeV equals hydrogen 1S, because Z⁴/n⁴ = 1 and n/(j + ½) − 3/4 = 1/4 in both.
- Order: the j = ½ shift is 1630 μeV ÷ 13.606 eV = 1.2 × 10⁻⁴ of |E₃|, of order (Zα)²/n² with (Zα)² = 4.8 × 10⁻⁴. The next term, a further (Zα)² smaller, moves the j = ½ level by about −0.09 GHz in the quoted Dirac energy of the next topic, 0.03% of the 262.8 GHz gap. Quantum electrodynamics is outside this count: the Lamb shift, of order α(Zα)⁴mc² with a logarithm, sits below fine structure by one power of α rather than (Zα)², so it lifts 3S₁/₂ above 3P₁/₂ by far more than 0.03% of the gap.
Answer3S₁/₂ and 3P₁/₂ at −1630 μeV, 3P₃/₂ and 3D₃/₂ at −543.4 μeV, 3D₅/₂ at −181.1 μeV; gaps of 262.8 GHz and 87.6 GHz, 81 times hydrogen's. The next Dirac order in Zα moves the larger gap by about 0.03%; the Lamb shift, not included, moves 3S₁/₂ by far more.