University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.4
F₂ in a Coulomb field is the Pauli term, and Uehling joins
Unit 14 derived a = α/2π for a free electron, and Advanced QM moved hydrogen's levels with an a it had to borrow. This lesson joins the two: F₂ inside an atom is Pauli's term, its 2S and 2P shares follow and are checked, it needs none of the matching F₁ will, and the Uehling term goes into the same ledger.
Build the model
Connect the measurement to the mechanism.
A nucleus at rest enters the one-loop vertex as an external field that supplies momentum but no energy, so a bound electron samples F₁ and F₂ only at spacelike q² = −|q|², and an atom of size 1/(Zαm) transfers |q| of order Zαm. Topic 14.03's integral makes F₂ vary only on the scale m: F₂(−|q|²) = (α/2π)[1 − |q|²/(6m²) + …], constant to nine parts per million at the momenta that set hydrogen's size. A constant coefficient of ū′(iσ(μν)qν/2m)u is exactly the Feynman rule of Pauli's term with μ′ = QF₂(0)/2m for the electron's charge Q = −e, so the anomalous moment UAQM-06.04 added by hand is the vertex correction at low momentum, with a = α/2π now derived.
In the proton's purely electric field the term acts through its electric half, which multiplies Dirac's Darwin and spin-orbit terms by 2a: 2S₁/₂ rises by 3/8 and 2P₁/₂ falls by 1/8 of the unit Cₙ, 50.9 MHz and 17.0 MHz at n = 2. F₂(0) has no infrared logarithm, since a photon mass changes it only by the power πμγ/m, so binding cannot alter it at this order and no matching is needed. Topic 13.08's Uehling term, −Cₙ/5 on s states, joins, and the high-energy ns bracket becomes ln(m/μγ) − 1/5, with μγ left in F₁ alone.
The constant coupling stops where the s-state weight's tail reaches |q| ≈ m, a fraction 8Zα/π that belongs to order α(Zα)⁵, and it fails outright in heavy ions.
- Simple definition
- Over the momentum transfers an atom supplies, the one-loop magnetic form factor F₂(q²) equals its zero-momentum value, so a bound electron feels it as Pauli's term with a = F₂(0) = α/2π, adding 2a times Dirac's Darwin and spin-orbit energies.
- Example
- At n = 2 in hydrogen, Pauli's term with a = α/2π raises 2S₁/₂ by (3/8) × 135.64 = 50.87 MHz and lowers 2P₁/₂ by (1/8) × 135.64 = 16.96 MHz: 67.82 MHz, 6.4% of the measured 1057.8 MHz Lamb shift.
Topic 14.03's integral expanded to first order in |q|²/m². The atom never probes F₂ far from q² = 0, so there its coupling is one constant number.
|q| and m in MeV, F₂ dimensionless, q⁰ = 0 for a nucleus at rest. At |q| = Zαm = 3.729 keV (Z = 1) the bracket is 1 − 8.9 × 10⁻⁶.
The anomalous moment UAQM-06.04 added by hand is the low-momentum vertex. Its operator has mass dimension 5, so no free renormalisable coupling exists: QED must predict it.
Q = −e = −0.3028, the electron's charge, and q = p′ − p in MeV; μ′ in MeV⁻¹; a dimensionless. The right side is the vertex of −(μ′/2)ψ̄σ(μν)ψF(μν).
A power of μγ where F₁ has ln μγ. Soft photons barely feed the magnetic coupling, so binding, which reshapes only them, cannot move F₂ at this order: no matching.
z is the photon's Feynman parameter and μγ the photon mass of Unit 14, in MeV. At μγ = 0 the integrand is 2z and F₂(0) = α/2π.
2S₁/₂: +3/8, +50.87 MHz. 2P₁/₂: −1/8, −16.96 MHz. 2P₃/₂: +1/16, +8.48 MHz. These are UAQM-13.09's +51 and −17 MHz, now resting on a derived a.
Cₙ = 4α(Zα)⁴m/(3πn³), C₂/h = 135.64 MHz; a = α/2π. From UAQM-06.04's reduction, which multiplies Darwin and spin-orbit by 1 + 2a.
Derived in topic 13.08 and quoted. Vacuum polarisation corrects the photon line, not Γμ, yet on s states it enters beside F₁'s slope as −1/5 of the unit.
|ψₙₛ(0)|² = (Zαm)³/(πn³) in MeV³. 2S₁/₂ falls 27.13 MHz; p states, whose density vanishes at the origin, do not move.
F₂'s +3/8 cancels the constant beside F₁'s logarithm on s states. The photon mass, inside F₁ alone, is the last regulator, and topic 15.07 removes it.
Terms in order: F₁'s slope (topic 15.03), F₂, Uehling; μγ in MeV. On nP₁/₂ only F₂ acts at this order, giving −Cₙ/8.
A static nucleus probes the vertex at spacelike q
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, with frequencies restored through h = 4.136 × 10⁻¹⁵ eV s. Topic 15.03 put the nucleus in as an external field: a static potential supplies momentum q = p′ − p but no energy, so q⁰ = 0 and the vertex is needed only at q² = −|q|² ≤ 0. Topic 14.01's decomposition, quoted, is Γμ = γμF₁(q²) + (iσ(μν)qν/2m)F₂(q²). How large is |q|? A wave function of size 1/(Zαm) transfers momenta of order Zαm, 3.729 keV in hydrogen and 137 times below m. Topic 14.03 left F₂ = (α/2π)∫₀¹ dξ m²/(m² − ξ(1 − ξ)q²); at q² = −|q|², expanding with ∫ξ(1 − ξ)dξ = 1/6 gives F₂ = (α/2π)[1 − |q|²/(6m²) + …], derived, a correction of (Zα)²/6 = 8.9 × 10⁻⁶ at |q| = Zαm. Limits check: |q| → 0 returns Schwinger's α/2π, and far above m the exact integral falls as (α/π)(m²/|q|²)ln(|q|²/m²), the moment smeared over the electron's Compton wavelength.
The F₂ vertex is the Feynman rule of Pauli's term
Take the electron's charge as Q = −e, keeping q for the momentum transfer, with e = √(4πα) = 0.3028, so minimal coupling −Qψ̄γμψAμ gives the vertex −iQū′γμuAμ(q) and the corrected vertex −iQū′ΓμuAμ(q). Now put UAQM-06.04's Pauli term, −(μ′/2)ψ̄σ(μν)ψF(μν), in the same field. With Aμ(x) ∝ e(−iq⋅x), F(μν) becomes −i(qμAν − qνAμ), and relabelling the antisymmetric indices gives σ(μν)F(μν) = 2iσ(μν)qνAμ, so the term's vertex is −iμ′ū′(iσ(μν)qν)uAμ(q). That is the F₂ piece of the corrected vertex exactly when μ′ = QF₂/2m, derived, and since F₂ is constant across the atom, a = F₂(0) = α/2π. Dimensions agree: ψ̄σψF has mass dimension 3 + 2 = 5, so μ′ must be an inverse mass, as Qa/2m is. The same count explains why the moment is a prediction: QED has no renormalisable slot for it, so the loop must deliver it finite. Restoring units, aμB = 6.723 × 10⁻⁸ eV T⁻¹ against the measured aₑμB = 6.713 × 10⁻⁸ eV T⁻¹: 0.15% high, the α² term of topic 14.04.
In an electric field the moment acts through its odd half
A Coulomb field has no B, so the half of Pauli's term that makes g = 2(1 + a) has nothing to act on. UAQM-06.04's reduction, quoted, shows the other half, iμ′βα⋅E, meeting α⋅p and leaving ΔHₐ = (aπZα/m²)δ³(r) + (aZα/(2m²r³))σ⋅L for V = −Zα/r: exactly 2a times Dirac's Darwin and spin-orbit terms, so the totals carry 1 + 2a, not the 1 + a of the moment. The shares then follow from hydrogen's expectation values, quoted. On ns states |ψ(0)|² = (Zαm)³/(πn³) gives a(Zα)⁴m/n³, which divided by Cₙ = 4α(Zα)⁴m/(3πn³) with a = α/2π is 3/8: 2S₁/₂ rises 50.87 MHz. On 2P₁/₂, ⟨r⁻³⟩ = (Zαm)³/24 and ⟨σ⋅L⟩ = −2 give −a(Zα)⁴m/24, which is −1/8 of C₂: −16.96 MHz. Limits: a → 0 restores Dirac's degenerate pair, and both shares scale as Z⁴/n³. These are UAQM-13.09's +51 and −17 MHz, checked rather than recomputed; what has changed is their footing, since a is now topic 14.03's derived number, not an input.
Why F₂, unlike F₁, needs no matching
F₁'s slope carries ln(m/μγ): it counts photons at every energy down to the regulator, including the soft ones whose behaviour binding changes, which is why topics 15.05 to 15.07 must replace its infrared end. F₂ is different, and the Feynman-parameter integral shows why. At q² = 0, with z the photon's parameter, the integral over x and y gives a factor 1 − z and the integrand is 2z(1 − z)/(1 − z)², so F₂(0) = (α/2π)∫₀¹ 2z dz = α/2π, with nothing singular at the soft corner z → 1. Keep the photon mass and the denominator gains zμγ²; the corner then gives F₂(0) = (α/2π)(1 − πμγ/m), derived, a power where F₁ has a logarithm. Treat μγ as a crude model of binding, which removes photons softer than about (Zα)²m = 27 eV: the 2s share moves by π(Zα)² × 50.87 MHz = 8.5 kHz, an order-α(Zα)⁶ effect, while F₁ would lose a whole logarithm. So the free electron's F₂ is the bound electron's at this order, with no cutoff to cancel.
The Uehling term joins, and one regulator is left
Vacuum polarisation is not part of Γμ: it corrects the photon line between nucleus and electron. In a static field it strengthens the Coulomb potential by the fraction (α/15π)|q|²/m² at small |q|, which in position space is the contact term −(4α(Zα)/(15m²))δ³(r), and topic 13.08 derived its shift, quoted here: −4α(Zα)⁴m/(15πn³) = −Cₙ/5, so 2S₁/₂ falls 27.13 MHz and p states do not move. Now collect what is in hand for ns: [ln(m/μγ) − 3/8] from F₁'s slope (topic 15.03), +3/8 from F₂ and −1/5 from Uehling, so ΔEhigh(nS₁/₂) = Cₙ[ln(m/μγ) − 1/5], F₂'s Darwin share cancelling the constant beside F₁'s logarithm. On 2P₁/₂ only F₂ acts, −C₂/8, so the interval's high-energy bracket is ln(m/μγ) − 3/40. The F₂ and Uehling pieces add (3/8 + 1/8 − 1/5)C₂ = (3/10)C₂ = 40.69 MHz to the Lamb interval. Every high-energy piece is now in hand, and the only regulator left is μγ, inside F₁ alone.
Where a constant coupling stops: the s-state tail
The estimate |q| ≈ Zαm describes the bulk of the wave function, not its edge. An s state has a cusp, ψ′(0) = −Zαmψ(0), and the Fourier transform of a cusp falls as |q|⁻⁴: the density's form factor tends to 16πZαm|ψ(0)|²/|q|⁴ for every n, derived. So momentum transfers above |q| = m carry a share 8Zα/π of the contact weight |ψ(0)|², 1.9% in hydrogen. There F₂ is not constant and the non-relativistic reduction fails as well, so the constant Pauli term is exact at order α(Zα)⁴ and incomplete at relative order Zα, which is order α(Zα)⁵m. Applied to the full Uehling form factor, the same tail gives +(5/48)α(Zα)⁵m/n³, +0.24 MHz on 2s, the next-order term topic 15.09 quotes, reproduced here numerically. For hydrogen-like uranium, Zα = 0.671, F₂ at |q| = Zαm is already 0.931 of F₂(0) and 90% of the non-relativistic contact weight lies above m: no constant coupling survives, and the bound propagator of the Furry picture is needed.
Change one variable at a time
Make the relationship visible.
Hold Z = 1 and slide the probe: by |q| = 0.1 m the dashed curve holds 82% of the weight while F₂ has dropped 0.17%, and the 1.9% beyond m is the order-α(Zα)⁵ tail. Raise Z towards 92 and the weight marches into the fall of F₂, where no constant coupling fits.
F₂ DEFICIT AT |q| = Zαm8.9 ppm
1s WEIGHT ABOVE |q| = m1.86 %
F₂/F₂(0) AT PROBE0.9983
1s WEIGHT BELOW PROBE81.7 %
Live interpretationF₂ DEFICIT AT |q| = Zαm: 8.9 ppm. 1s WEIGHT ABOVE |q| = m: 1.86 %. F₂/F₂(0) AT PROBE: 0.9983. 1s WEIGHT BELOW PROBE: 81.7 %
Catch the common trap
Explain before calculating.
Topic 14.03 derives a = α/2π. Using the n = 2 unit 135.64 MHz, how much does the F₂ part of the one-loop vertex add to hydrogen's 2S₁/₂ − 2P₁/₂ interval at order α(Zα)⁴, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTopic 14.03 gives F₂ at spacelike q² = −|q|² as (α/2π) times ∫₀¹ dξ/(1 + ξ(1 − ξ)t), with t = |q|²/m² and closed form 4 ln[(√(t + 4) + √t)/2]/√(t(t + 4)). At the typical atomic momentum transfer |q| = Zαm, find F₂/F₂(0) for hydrogen and for hydrogen-like uranium (Z = 92), and say whether Pauli's constant coupling is justified in each.
- Hydrogen: |q| = αm = 0.51100 MeV ÷ 137.036 = 3.729 keV, so t = α² = 5.325 × 10⁻⁵.
- Expand: ∫ξ(1 − ξ)dξ = 1/6, so F₂/F₂(0) = 1 − t/6 = 1 − 8.88 × 10⁻⁶ = 0.9999911. The next term, +t²/30 from ∫ξ²(1 − ξ)²dξ = 1/30, is 9.5 × 10⁻¹¹.
- Uranium: Zα = 92/137.036 = 0.6714 and t = 0.4507. The series gives 1 − 0.0751 = 0.9249, but t is no longer small, so use the closed form.
- √(t + 4) = 2.1097 and √t = 0.6714, so ln[(2.1097 + 0.6714)/2] = ln 1.3905 = 0.32967; √(t(t + 4)) = √2.0060 = 1.4163; the ratio is 4 × 0.32967 ÷ 1.4163 = 0.9311.
- Verdict: in hydrogen F₂ is one number across the atom, so a = F₂(0) is Pauli's coupling; in uranium it is 6.9% low already at the typical momentum, and even the truncated series misses by 0.006, so no constant coupling describes the ion.
AnswerHydrogen: F₂/F₂(0) = 0.9999911, constant to 9 parts per million. Uranium: 0.931 (series 0.925), 7% below F₂(0), where a constant Pauli coupling fails.
MediumUsing C₂/h = 135.64 MHz, a = α/2π and topic 13.08's Uehling term, find what F₂ and vacuum polarisation do to hydrogen's 2S₁/₂, 2P₁/₂ and 2P₃/₂ levels, their net contribution to the 2S₁/₂ − 2P₁/₂ interval, and the 2S₁/₂ bracket once topic 15.03's F₁ term C₂[ln(m/μγ) − 3/8] is added.
- F₂ shares from (3/8)C₂cₗⱼ/(2l + 1): 2S₁/₂ has l = 0 and c = 1, so +3/8 of the unit, +50.87 MHz; 2P₁/₂ has c = −1 over 3, so −1/8, −16.96 MHz; 2P₃/₂ has c = ½ over 3, so +1/16, +8.48 MHz.
- Uehling: −C₂/5 = −27.13 MHz on 2S₁/₂, and nothing on the p levels, whose |ψ(0)|² vanishes.
- Interval: 2S₁/₂ moves 50.87 − 27.13 = +23.74 MHz and 2P₁/₂ moves −16.96 MHz, so the interval grows by (3/8 + 1/8 − 1/5)C₂ = (3/10)C₂ = 40.69 MHz; the rounded shares, 23.74 + 16.96, give 40.70 MHz.
- Bracket: [ln(m/μγ) − 3/8] + 3/8 − 1/5 = ln(m/μγ) − 1/5 for 2S₁/₂, and with 2P₁/₂'s −1/8 the interval's bracket is ln(m/μγ) − 3/40. Only F₁ brings in μγ.
- Check: the 2P fine structure gains (1/16 + 1/8)C₂ = 25.43 MHz, and 2a times Dirac's α⁴m/32 = 10949.28 MHz is also 25.43 MHz, as a spin-orbit coefficient raised by 1 + 2a requires.
Answer2S₁/₂: +23.74 MHz (F₂ +50.87, Uehling −27.13); 2P₁/₂: −16.96 MHz; 2P₃/₂: +8.48 MHz. Net +40.69 MHz on the Lamb interval; the 2S₁/₂ bracket is ln(m/μγ) − 1/5.
HardHydrogen's measured 2S₁/₂ → 2P₃/₂ interval, 9911.200 MHz, plus the Lamb interval, 1057.845 MHz, puts 2P₃/₂ 10969.045 MHz above 2P₁/₂. Derive F₂'s shares of the two 2P levels from the Pauli term's spin-orbit part, explain why no other order-α(Zα)⁴ piece enters this interval, and compare Dirac's interval with the reduced mass, plus F₂, against the measurement.
- The quoted reduction gives the anomalous spin-orbit term (aZα/(2m²r³))σ⋅L. For n = 2, l = 1, ⟨r⁻³⟩ = (Zαm)³/(n³l(l + ½)(l + 1)) = (Zαm)³/24, and ⟨σ⋅L⟩ = j(j + 1) − l(l + 1) − 3/4 is −2 for j = ½ and +1 for j = 3/2.
- Shares: 2P₁/₂ gets −a(Zα)⁴m/24 and 2P₃/₂ gets +a(Zα)⁴m/48. Dividing by C₂ = α(Zα)⁴m/(6π) with a = α/2π gives −1/8 and +1/16, so F₂ opens the interval by (3/16)C₂ = 25.43 MHz.
- Nothing else at this order depends on j: F₁'s slope and the Uehling term are contact terms, zero on p states, and Bethe's low-energy logarithm depends on l but not on spin, so it moves both 2P levels alike. The interval tests F₂ alone, with no photon mass and no cutoff anywhere.
- Dirac: α⁴mc²/h = 350.377 GHz, so the leading interval (Zα)⁴m/32 is 10949.28 MHz; the reduced mass multiplies it by 1/(1 + 1/1836.15) = 0.999456, giving 10943.324 MHz, and the exact Dirac formula adds the order-(Zα)⁶ term (5/256)(Zα)⁶m, 0.364 MHz after the same factor, for 10943.688 MHz.
- Total: 10943.688 + 25.433 = 10969.121 MHz against 10969.045 MHz measured. Dirac alone falls 25.357 MHz short; with F₂ the residue is 0.076 MHz, 7 parts per million. Replacing α/2π by the measured aₑ lowers the share by 0.04 MHz, so what remains is the size of the terms this order drops.
AnswerF₂ lowers 2P₁/₂ by 16.955 MHz and raises 2P₃/₂ by 8.478 MHz, opening the interval by 25.433 MHz. Dirac with the reduced mass gives 10943.688 MHz; adding F₂ gives 10969.121 MHz against 10969.045 MHz measured, a 0.076 MHz residue.