University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.1
What order α(Zα)⁴ contains, and the unit 135.6 MHz
Before a single form factor is expanded, this week needs a list and a ruler. The list names every one-loop effect that can split 2S₁/₂ from 2P₁/₂, the states it reaches and the topic that computes it; the ruler, 4α(Zα)⁴m/(3πn³), turns each entry into a pure number you can check at a glance.
Build the model
Connect the measurement to the mechanism.
Dirac's equation for an electron in the fixed Coulomb field of a point nucleus is solved to all orders in Zα, and its levels depend on n and j alone (UAQM-07.07, quoted), so 2S₁/₂ and 2P₁/₂ coincide however far that series is carried. Whatever splits them lies outside it: loops of the quantised fields, each costing α with no Z attached, the nucleus's recoil, m/M, or its size. The loops dominate, and power counting says where they start.
The bound electron trades momentum |q| ~ Zαm with the nucleus, far below m, so the one-loop vertex and vacuum polarisation can be expanded in q²/m² ~ (Zα)². Their q⁰ terms only restate the charge, which renormalisation has already fixed, so the first observable effect is α q²/m² times the Coulomb energy (Zα)²m: order α(Zα)⁴m. Because q² cancels the photon's 1/q², the correction is a contact term, felt through |ψ(0)|² by s states alone; the anomalous moment instead rescales Dirac's spin couplings and reaches every l.
Each entry is a pure number times Cₙ = 4α(Zα)⁴m/(3πn³), 135.6 MHz at n = 2, the unit UAQM-13.07 used: −1/5 for vacuum polarisation, +3/8 for the moment on 2S, and about 7.5 for the charge self-energy, whose logarithm ln(1/(Zα)²) = 9.84 comes from soft photons the free vertex cannot treat. The counting fails as Zα grows: the first omitted term rises as Z against Cₙ and overtakes the whole leading 2S shift near Z = 24.
- Simple definition
- The order-α(Zα)⁴m Lamb shift is the part of a hydrogen-like level's energy carrying one power of α from a single loop of the quantised fields and four powers of Zα from the binding; it is the leading term of the 2S₁/₂ − 2P₁/₂ splitting.
- Example
- In hydrogen at n = 2 the unit 4α(Zα)⁴mc²/(3πn³) is 5.610 × 10⁻⁷ eV, or 135.64 MHz; vacuum polarisation's −27.13 MHz on 2S is −1/5 of it, and the measured 1057.8 MHz interval is 7.80 units.
From UAQM-07.07, exact in Zα. 2S₁/₂ and 2P₁/₂ share n = 2, j = ½, so no power of Zα can split them; loops, recoil or nuclear size must.
ħ = c = 1, metric (+, −, −, −); m = 0.511 MeV; n principal, j total angular momentum; nucleus fixed. No l appears.
F₁(0) − 1 and Π̂₂(0) vanish because the measured charge is defined at q = 0, so the first observable loop effect costs (Zα)² = 5.3 × 10⁻⁵ beyond α.
|q|: momentum traded with the nucleus, 3.73 keV/n in hydrogen; (Zα)²m = 27.2 eV, the binding scale; the loop supplies α.
First-order perturbation theory. Uehling's c = −4/15 from Unit 13 reads at once as −1/5 unit, the −27.13 MHz on 2S.
c a pure number; δ³(r) in MeV³; |ψ(0)|² = (Zαm)³/(πn³) for nS; for l ≥ 1, ψ(0) = 0 and the first-order shift vanishes.
Each unit of F₁'s slope bracket is one Cₙ, so every one-loop entry at n = 2 is a pure number times 135.64 MHz, the unit UAQM-13.07 used for Bethe's logarithm.
Restore c: 4α(Zα)⁴mc²/(3πn³h), with α⁵mc²/h = 2α³cR∞ = 2556.83 MHz; C₂ = 5.610 × 10⁻⁷ eV. Grows as Z⁴/n³.
The logarithm counts the e-folds of photon energy between the binding scale (Zα)²m and m, 9.84 of them in hydrogen, or 4.27 decades. It makes the charge part 7.49 units on 2S, 37 times vacuum polarisation.
K₀: Bethe's mean excitation energy, ln(K₀/Ry) = 2.812 for 2s and 2.984 for 1s (Advanced QM); ln(1/α²) = 9.840.
In hydrogen 0.0526 units, 7.14 MHz on 2S. Its ratio to Cₙ grows linearly with Z and reaches one unit at Z = 19: the Zα expansion has a domain.
A₅₀ = 4π(139/128 − ½ ln 2) + 5π/48: self-energy plus vacuum polarisation, s states only. Set aside for topic 15.09.
Dirac keeps the pair together, so the split starts at α
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout; a shift ΔE is quoted as the frequency ΔE/h, with h = 4.1357 × 10⁻¹⁵ eV s, so 1 MHz is 4.1357 neV. For a point nucleus held fixed, Dirac's levels are exact in Zα (UAQM-07.07, quoted), and for n = 2, j = ½ they reduce to m√((1 + √(1 − (Zα)²))/2), shared by 2S₁/₂ and 2P₁/₂: a binding of 3.4015 eV, 13.69 GHz deeper than Bohr's level, with 2P₃/₂ 10.95 GHz above. No l appears, so no power of Zα separates the pair. Three things the external-field equation omits can: loops of the quantised photon and electron fields, each costing α with no Z; the nucleus's recoil, m/M = 1/1836 for hydrogen; and its charge radius, 0.84 fm. In hydrogen α and Zα are both 1/137, but they count different physics, exchanges with the nucleus's classical field against the electron's own photons and the vacuum's pairs, and only the first grows with Z.
Counting powers: why one loop lands at α(Zα)⁴m
Draw the bound electron in the nucleus's static field, Unit 8's photon line sourced by Ze. At one loop that photon carries Unit 13's polarisation insertion, and the electron's coupling to it Unit 14's vertex correction, with the self-energy whose δ₂ cancels the vertex's ultraviolet pole. In the atom the electron trades momentum |q| ~ Zαm/n with the nucleus, 3.73 keV/n in hydrogen, so the loop functions expand in q²/m² ~ (Zα)² = 5.3 × 10⁻⁵. Their q⁰ terms vanish, F₁(0) = 1 and Π̂₂(0) = 0, since the measured charge is defined at q = 0: an α(Zα)²m shift would only change e. The first observable term is (α/π)(q²/m²) times −4πZα/|q|². The q² cancels, leaving a strength α(Zα)/m² constant in q, a δ³(r), felt through |ψ(0)|² = (Zαm)³/(πn³): α(Zα)⁴m/n³, s states only. The magnetic part, F₂(0) = α/2π, rescales Dirac's spin couplings, already (Zα)⁴m/n³, and reaches every l. Check: α(Zα)² = 3.9 × 10⁻⁷, and the measured 4.375 μeV over the 3.40 eV binding is 1.29 × 10⁻⁶, larger by 3.3: the 7.8-unit bracket times 8/(3πn) = 0.42.
The ruler: a contact term's shift, and 135.6 MHz
Make the counting exact for any contact term. First-order perturbation theory gives δV(r) = c α(Zα)δ³(r)/m² a shift c α(Zα)|ψ(0)|²/m² = c α(Zα)⁴m/(πn³) on nS and nothing on l ≥ 1, derived. The course measures in Cₙ = 4α(Zα)⁴m/(3πn³), so a contact term of strength c is worth 3c/4 units. The 4/(3π) is not arbitrary: F₁'s slope arrives as (α/3π)(q²/m²) times a bracket, and the Coulomb transform as 4πZα/|q|², so each unit of that bracket is c = 4/3, exactly one Cₙ, as topics 15.02 and 15.03 will show. Numerically α⁵mc²/h = 2α³cR∞ = 2556.83 MHz and 4/(3π × 2³) = 1/(6π), so C₂/h = 135.64 MHz, an energy of 5.610 × 10⁻⁷ eV, computed. α is pure and m an energy, so Cₙ is an energy; it vanishes as the loop coupling α → 0 with Zα held fixed, the free-field limit of the radiation, where Dirac's levels survive. It scales as Z⁴/n³: C₁/h = 1085.15 MHz, and He⁺ has C₂/h = 2170.3 MHz. Uehling's c = −4/15 reads as −1/5 unit, −27.13 MHz; the measured interval, 1057.8 MHz, is 7.80 units.
The inventory: three sources, their states and their status
Self-energy in the nuclear field, charge part. F₁'s slope carries ln(m/μγ), an infrared logarithm the free vertex cannot settle in an atom, because there a virtual photon softer than the binding meets a bound electron, not a free one (topic 15.05). So photon energies are split at K, (Zα)²m ≪ K ≪ m: above it the free vertex, derived in Unit 14 and topics 15.02 to 15.03; below it Bethe's dipole calculation, derived in UAQM-13.04 to 13.07 and quoted in 15.06; the two are joined in 15.07. On nS this gives ln(1/(Zα)²) − ln(K₀/Z²Ry) + 11/24 units, 7.49 on 2S; on 2P₁/₂ only its own Bethe logarithm, +0.030, quoted. Self-energy, magnetic part: F₂(0) = α/2π, derived in Unit 14, acting through Advanced QM Unit 6's Pauli term, +3/8 on nS and −1/8 on 2P₁/₂ (topic 15.04). Vacuum polarisation: Π̂₂, derived in Unit 13, as Uehling's contact term, −1/5 on nS and zero on 2P₁/₂. Totals: 7.662 units on 2S₁/₂ and −0.095 on 2P₁/₂, an interval of 7.757 units or 1052.2 MHz, assembled in topic 15.08.
What is set aside, and where the counting stops
Four effects are named here and accounted for in topic 15.09. Recoil: the order-α(Zα)⁴ terms scale as mᵣ³ through |ψ(0)|² or ⟨r⁻³⟩, so the reduced mass mᵣ = m/(1 + m/M) lowers them by 3(m/M) = 0.163%, about 1.7 MHz on the interval. Size: a proton of radius rₚ = 0.841 fm alters the potential only inside rₚ, lifting nS by (2/3)(Zα)⁴m³rₚ²/n³ (quoted), 1.02 × 10⁻³ units or 0.14 MHz. One more Zα: (3/4)A₅₀Zα units with A₅₀ = 9.618 for s states (quoted), 0.0526 units or 7.14 MHz. One more loop: a scale α/π = 0.0023 units, 0.3 MHz per unit coefficient. The counting assumes Zα ≪ 1, so that q²/m² is small and a window (Zα)²m ≪ K ≪ m exists, and a Bohr radius, 52 918 fm/Z, far beyond the 386 fm Compton wavelength 1/m. Against Cₙ the first omitted term grows as Z: one unit at Z = 19, the whole leading 2S shift near Z = 24. For uranium, Zα = 0.671 and (Zα)²m = 230 keV against m = 511 keV: no window is left.
Change one variable at a time
Make the relationship visible.
Start at Z = 1, n = 2: the filled dot is the 135.6 MHz unit. Drag Z: the Cₙ line stays parallel to Dirac's, because the loop's α never grows, while the open α(Zα)⁵ dot closes in and passes Cₙ at Z = 19. Change n: Cₙ and the omitted term fall as n⁻³ with |ψ(0)|², binding only as n⁻².
UNIT Cₙ135.6 MHz
Cₙ AS AN ENERGY0.561 μeV
α(Zα)⁵ TERM ÷ Cₙ, s STATES0.053
LOGARITHM ln(1/(Zα)²)9.84
Live interpretationUNIT Cₙ: 135.6 MHz. Cₙ AS AN ENERGY: 0.561 μeV. α(Zα)⁵ TERM ÷ Cₙ, s STATES: 0.053. LOGARITHM ln(1/(Zα)²): 9.84
Catch the common trap
Explain before calculating.
The unit Cₙ = 4α(Zα)⁴m/(3πn³) is 135.64 MHz for hydrogen's n = 2. Neglecting recoil, what is the unit for the n = 2 level of He⁺ (Z = 2), and why does it scale that way?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFrom α = 1/137.036, mc² = 510 998.95 eV and h = 4.135 668 × 10⁻¹⁵ eV s, compute the unit Cₙ = 4α(Zα)⁴mc²/(3πn³) for hydrogen's n = 2 as an energy and as a frequency, and give C₁/h. Then express three shifts quoted from Advanced QM in units of C₂: vacuum polarisation's −27.13 MHz on 2S, and the anomalous moment's +50.87 MHz on 2S and −16.96 MHz on 2P₁/₂.
- Powers: α⁵ = 2.069 32 × 10⁻¹¹, so α⁵mc² = 2.069 32 × 10⁻¹¹ × 510 998.95 eV = 1.057 42 × 10⁻⁵ eV.
- For Z = 1 and n = 2 the prefactor is 4/(3π × 2³) = 1/(6π) = 0.053 052, so C₂ = 1.057 42 × 10⁻⁵ eV × 0.053 052 = 5.6098 × 10⁻⁷ eV. α and 4/(3πn³) are pure numbers, so C₂ is an energy, as it must be.
- As a frequency: C₂/h = 5.6098 × 10⁻⁷ eV ÷ 4.135 668 × 10⁻¹⁵ eV s = 1.3564 × 10⁸ Hz = 135.64 MHz. The unit falls as n⁻³, so C₁/h = 2³ × 135.644 MHz = 1085.15 MHz.
- Ratios: −27.13/135.64 = −0.2000, 50.87/135.64 = 0.3750 and −16.96/135.64 = −0.1250, that is −1/5, +3/8 and −1/8 of a unit: the unit turns megahertz into the simple fractions the loop calculations produce.
AnswerC₂ = 5.610 × 10⁻⁷ eV, so C₂/h = 135.64 MHz and C₁/h = 1085.15 MHz; the three shifts are −1/5, +3/8 and −1/8 of C₂.
MediumThe one-loop interval is 1052.19 MHz and the measurement 1057.85 MHz. Before any coefficient is known, estimate by power counting the scale of each effect set aside for topic 15.09, in MHz with C₂ = 135.64 MHz: one more power of Zα, one more loop, recoil through the reduced mass (m/M = 1/1836.15), and a proton charge radius rₚ = 0.841 fm. Say what the estimates decide and what they cannot.
- One more power of Zα: Zα × C₂ = 0.007 297 × 135.64 MHz = 0.990 MHz per unit coefficient. One more loop: (α/π)C₂ = 0.002 323 × 135.64 MHz = 0.315 MHz per unit coefficient.
- Recoil: every order-α(Zα)⁴ term scales as mᵣ³ through |ψ(0)|² or ⟨r⁻³⟩, with mᵣ = m/(1 + m/M), so it shrinks by 1 − (1 + m/M)⁻³ = 0.001 632, and the interval falls by 0.001 632 × 1052.19 MHz = 1.72 MHz.
- Size: mrₚ = 0.511 MeV × 0.841 fm/197.327 MeV fm = 2.178 × 10⁻³. The quoted size shift (2/3)(Zα)⁴m³rₚ²/n³ is (π/2)(mrₚ)²/α = (π/2)(4.743 × 10⁻⁶)/0.007 297 = 1.021 × 10⁻³ units, 0.138 MHz, raising 2S and so widening the interval.
- What counting decides: size and two loops sit near or below 0.3 MHz, while recoil and the Zα term can reach the 1 MHz level and must be computed. What it cannot decide is the coefficient: for s states the Zα term carries (3/4)A₅₀ = 7.21 (quoted), making it 7.14 MHz, larger than the whole 5.66 MHz gap.
AnswerScales: about 0.99 MHz per unit coefficient for one more Zα (7.14 MHz with its quoted 7.21), 0.32 MHz for two loops, −1.72 MHz from recoil and +0.14 MHz from the proton's size; set aside means accounted for later, not negligible.
HardFor a hydrogen-like ion with recoil neglected, the order-α(Zα)⁴ 2S shift is C₂[ln(1/(Zα)²) − 2.81177 + 11/24 + 3/8 − 1/5], with ln(K₀/Z²Ry) = 2.81177 at every Z, and the first omitted term is (3/4)A₅₀Zα C₂ with A₅₀ = 9.618, both quoted. Evaluate both for hydrogen and for C⁵⁺ (Z = 6) in units and in MHz, find the Z at which they are equal, and say why for U⁹¹⁺, Zα = 0.6714, the leading formula has no authority.
- The constants combine to −2.81177 + 0.45833 + 0.375 − 0.2 = −2.17844, and (3/4)A₅₀ = 7.2135. Hydrogen: ln(1/α²) = 9.84049, so the bracket is 7.6621 units, 1039.3 MHz with C₂ = 135.64 MHz; the omitted term is 7.2135 × 0.007 297 = 0.0526 units, 7.14 MHz, or 0.69%.
- C⁵⁺: Zα = 0.043 784 and ln(1/(Zα)²) = 6.2570, so the bracket is 4.0785 units. C₂ = 6⁴ × 135.64 MHz = 175.79 GHz, so the 2S shift is 717.0 GHz; the omitted term is 0.3158 units, 55.5 GHz, now 7.7% of it.
- Equality: 9.84049 − 2 ln Z − 2.17844 = 0.052 64 Z. At Z = 24 the sides are 1.306 and 1.263; at Z = 25, 1.224 and 1.316. They cross near Z = 24.3, beyond which the correction exceeds the term it corrects.
- Uranium: (Zα)² = 0.4507, so the binding scale (Zα)²m = 230 keV sits within a factor 2.2 of m = 511 keV. The window (Zα)²m ≪ K ≪ m is gone, q²/m² is no longer small, and the bracket, 0.797 − 2.178 = −1.38, is outweighed by an omitted term of 4.84 units.
AnswerHydrogen: 7.662 units (1039.3 MHz) against 0.0526 (7.14 MHz), 0.69%; C⁵⁺: 4.079 units (717.0 GHz) against 0.316 (55.5 GHz), 7.7%; equal near Z = 24; for uranium no Zα expansion holds.