University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.3
Form factors in a Coulomb field: the s-state contact term
This is where Unit 14's vertex correction first touches an atom. The static nucleus supplies keV momentum transfers, F₁'s slope turns them into a δ-function at the nucleus, and the s levels rise while the p levels stay put, with a photon mass still in the answer, kept there on purpose until binding can remove it.
Build the model
Connect the measurement to the mechanism.
Hydrogen's electron never leaves the Coulomb field, so the one-loop corrections of Units 13 and 14 reach it only through the nucleus. A static source conserves energy but absorbs any three-momentum, so it meets the electron at spacelike q² = −|q|², and the bound wavefunction feeds it momentum transfers of order Zαm, 3.7 keV in hydrogen against m = 511 keV. That smallness is the method: expand F₁, F₂ and Π̂₂ to first order in |q|²/m² and every correction becomes a short-range potential.
At spacelike q the positive slope of F₁ in q² pulls F₁ below 1, so it removes a constant from the attractive −Ze²/|q|²; a constant in momentum space is δ³(r) in position space, a repulsion whose real range is ħ/mc = 386 fm, a point to an atom 137 times larger. Only a wavefunction alive at the nucleus feels it: the ns levels rise by (4α(Zα)⁴m/(3πn³))[ln(m/μγ) − 3/8], 135.6 MHz per unit of the bracket for 2s, and the p levels stay where Dirac put them. The model stops at its own logarithm.
The photon mass μγ is not a property of the world; it is the price of putting a free, on-shell electron inside the loop. That is right for virtual photons far above atomic energies and wrong below them, where the Furry picture would keep the Coulomb field in every internal propagator, and the next topics replace μγ with binding rather than with a guess.
- Simple definition
- The high-energy contact potential is the δ³(r) term that the slope of F₁ adds to the Coulomb potential of a static nucleus; it shifts only states whose wavefunction is non-zero at the nucleus, which are the s states.
- Example
- With the photon mass set at 200 eV the bracket ln(m/μγ) − 3/8 is 7.47, so hydrogen's 2s rises by 135.6 MHz × 7.47 = 1013 MHz, while 2P₁/₂, which vanishes at the nucleus, does not move at all.
Derived: the source's 2πδ(E′ − E) fixes energy only. Each one-loop factor multiplies the tree amplitude ū(p′)γ⁰u(p) Ze²/|q|², whose Born reading is Ṽ(q) = −Ze²/|q|².
q = p′ − p is absorbed by the nucleus, so q⁰ = 0; F₁, F₂ and Π̂₂ are dimensionless; p, q and m = 0.511 MeV in MeV
Why slopes suffice: each further power of |q|²/m² costs (Zα)², so the |q|⁴ terms of the form factors reach α(Zα)⁶m, beyond this section's order.
αm = 3.73 keV, m = 511 keV, n the principal quantum number; for hydrogen's 2s, |q|²/m² ≈ 1.3 × 10⁻⁵
Quoted from topic 15.02 with q² = −|q|² put in. F₁ falls below 1 as |q| grows: the dressed electron couples more weakly to the nucleus at short distance.
μγ is the photon mass regulating the infrared, μγ ≪ m, in MeV; F₁(0) = 1 is the charge's renormalisation condition
Derived: the slope's |q|² cancels the photon's 1/|q|², and a constant in momentum space is a delta function in position space. A positive bracket makes it repulsive, at the nucleus only.
δ³(r) in fm⁻³; restoring ħ and c, 1/m² becomes ħc(ħ/mc)², with ħc = 197.33 MeV fm and ħ/mc = 386.16 fm
Derived at first order in δV. A level with l ≥ 1 vanishes at the nucleus and gets nothing; a finite range would give it a share (Zα)² smaller, of order α(Zα)⁶m.
From |ψₙₛ(0)|² = (Zαm)³/(πn³); the unit Cₙ is h × 135.64 MHz for Z = 1, n = 2, and h × 1085.15 MHz for 1s
Derived by the same two lines. The proton's own slope enters the same way: its measured radius, 0.841 fm, is quoted, and the contact term it makes lifts hydrogen's 2s by only +0.14 MHz.
⟨r²⟩ in fm²; F₁'s slope is the case ⟨r²⟩ = (2α/πm²)[ln(m/μγ) − 3/8] of topic 15.02, about (72 fm)² at μγ = 200 eV
A static nucleus is a vertex that conserves energy only
Natural units ħ = c = 1, the metric (+, −, −, −) and Heaviside-Lorentz charges with e² = 4πα hold throughout. Hold the nucleus fixed and its field is Aμ = (Φ, 0) with Φ(r) = Ze/(4πr). Its four-dimensional Fourier transform is 2πδ(q⁰) Ze/|q|²: the delta function conserves the electron's energy, while the three-momentum q = p′ − p is swallowed by a source too heavy to recoil. Every transfer it supplies therefore has q⁰ = 0 and q² = −|q|², spacelike. For the electron, of charge −e, the tree amplitude is ℳ = ū(p′)γ⁰u(p) Ze²/|q|², and for a slow electron ū(p′)γ⁰u(p) ≈ 2m. The Born amplitude of one relativistically normalised electron is ℳ = −2mṼ(q), so Ṽ(q) = −Ze²/|q|², the transform of −Zα/r (derived). That step never needed a free final state; it is a statement about one vertex, which is why the same vertex serves a bound electron.
Atoms sample keV momentum transfers, so only slopes matter
A bound state is a superposition of momenta spread over about Zαm/n: 3.73 keV in hydrogen's 1s and 1.86 keV in 2s, beside m = 511 keV. The transfers that keep it bound are of the same size, so the form factors are needed only near q² = 0, where |q|²/m² ≈ (Zα)²/n² = 1.3 × 10⁻⁵ for 2s. At one loop γ⁰ becomes γ⁰F₁(−|q|²) + (iσ(0ν)qν/2m)F₂(−|q|²), and the Coulomb propagator gains 1/(1 − Π̂₂(−|q|²)). Renormalisation fixes F₁(0) = 1 and Π̂₂(0) = 0, so what survives is their slopes, together with F₂(0) = α/2π from Unit 14. Order counting decides the rest: a correction of relative size α|q|²/m² to a potential worth (Zα)²m moves levels by α(Zα)⁴m, and the |q|⁴ terms arrive only at α(Zα)⁶m. This lesson follows F₁; F₂'s Pauli term and Π̂₂'s Uehling term enter the same way in the next topic.
F₁'s slope becomes a δ-function at the nucleus
Put the slope of topic 15.02 into the tree potential with q² = −|q|²: F₁(−|q|²) = 1 − (α/3π)(|q|²/m²)[ln(m/μγ) − 3/8]. Then Ṽ(q) = −(Ze²/|q|²)F₁ = −Ze²/|q|² + (Ze²α/3πm²)[ln(m/μγ) − 3/8]. The photon's 1/|q|² has cancelled against the slope's |q|², leaving a constant, and a constant in momentum space transforms into δ³(r). With Ze² = 4πZα the new term is δV = (4α(Zα)/3m²)[ln(m/μγ) − 3/8] δ³(r) (derived). The sign reads physically: at spacelike q the dressed vertex couples more weakly than the bare charge as |q| grows, so near the nucleus the electron feels less attraction than −Zα/r promises. The term is repulsive whenever the bracket is positive, that is for μγ below 0.687m. Its real range is set by loop momenta of order m, ħ/mc = 386 fm, 137 times smaller than the Bohr radius, so the atom sees a point.
Only states alive at the nucleus feel it
At first order ΔE = ⟨ψ|δV|ψ⟩ = (4α(Zα)/3m²)[ln(m/μγ) − 3/8]|ψ(0)|². The centrifugal barrier makes ψ ∝ rl near the origin, so |ψₙₛ(0)|² = (Zαm)³/(πn³) for l = 0 and zero for every l ≥ 1 (quoted from Schrödinger hydrogen). Hence ΔEₙₛ = (4α(Zα)⁴m/(3πn³))[ln(m/μγ) − 3/8], and p, d and higher levels do not move at this order (derived). The prefactor is the section's unit Cₙ: for Z = 1 and n = 2 it is α⁵m/6π = 5.610 × 10⁻⁷ eV, or 135.6 MHz; for 1s it is 1085 MHz and for 3s 40.2 MHz, the 1/n³ of the chance of sitting at the nucleus. The figure shows why p drops out: across the 386 fm where the contact term acts, the 2s density is flat while the 2p density is down by (Zα)²/12 ≈ 4.4 × 10⁻⁶. Dirac's 2P₁/₂ has an s-wave small component, but with weight (Zα)², so its contact shift is of order α(Zα)⁶m.
Checks: dimensions, limits and the proton's radius
Dimensions: δ³(r) has mass dimension 3 and 1/m² has −2, so δV has dimension 1, an energy, as does ΔEₙₛ ∝ α(Zα)⁴m; restoring constants, 4α(Zα)/3m² becomes (4α(Zα)/3)ħc(ħ/mc)², an energy times a volume. Limits: as α → 0 the term vanishes and Dirac's degenerate 2S₁/₂ and 2P₁/₂ return; as n grows it falls as 1/n³. A measurable cousin: any form factor F(−|q|²) = 1 − |q|²⟨r²⟩/6 gives δV = (2π/3)(Zα)⟨r²⟩δ³(r) by the same two lines, F₁'s slope being the case ⟨r²⟩ = (2α/πm²)[ln(m/μγ) − 3/8] of topic 15.02. With the proton's measured radius, 0.841 fm, quoted, it gives the standard leading finite-size shift of hydrogen's 2s, +0.14 MHz, some 7300 times less than the vertex term at μγ = 200 eV. The vertex term itself meets experiment only once μγ is gone: topic 15.08 sets the assembled splitting, about 1052 MHz, against the measured 1057.8 MHz.
The photon mass is the price of a free electron in the loop
The answer still holds ln(m/μγ), and μγ is not a property of the world: halve it and 2s rises by another 135.6 MHz × ln 2 = 94 MHz. The dependence reports the approximation this lesson made. F₁ was computed for a free electron on its mass shell and is applied here to one bound by energies of order (Zα)²m. For virtual photons with energy k ≫ (Zα)²m the binding is a negligible correction inside the loop and the free vertex is right; for k ≲ (Zα)²m it is not, and that soft region is exactly where ln(m/μγ) is generated. The exact framework is the Furry picture: bound-state QED in which every internal electron propagator is the Dirac propagator in the Coulomb field, to all orders in Zα. Expanded in Coulomb exchanges inside the loop, its zeroth term is this lesson's free vertex; further exchanges cost powers of Zα for hard photons, but for soft ones every exchange counts, which is why Bethe's non-relativistic sum takes over there. Topics 15.05 to 15.07 cut the soft region off and remove μγ.
Change one variable at a time
Make the relationship visible.
Raise Z: the ħ/mc line walks into the atom, the 2s density there sags and the 2p density there climbs from 4.5 ppm to 8.5% of it, so the point contact fails. Lower μγ: the 2S bar grows one unit per e-fold, past the dashed line, where μγ falls below atomic binding and the free loop goes wrong.
BRACKET ln(m/μγ) − 3/810.47
2S SHIFT ÷ Z⁴1420 MHz
PHOTON MASS μγ10.000 eV
2p ÷ 2s DENSITY AT ħ/mc4.5 ppm
Live interpretationBRACKET ln(m/μγ) − 3/8: 10.47. 2S SHIFT ÷ Z⁴: 1420 MHz. PHOTON MASS μγ: 10.000 eV. 2p ÷ 2s DENSITY AT ħ/mc: 4.5 ppm
Catch the common trap
Explain before calculating.
An electron scattering off hydrogen's static proton has its vertex dressed by F₁'s one-loop slope, with the bracket ln(m/μγ) − 3/8 positive. The resulting potential is then used inside the atom. What does this term do to the n = 2 levels at order α(Zα)⁴m?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyHold the photon mass at μγ = 100 eV. Find the F₁ contact shifts of hydrogen's 1s, 2s, 3s and 2p levels in MHz, and what this term does to the 1S to 2S interval. Which of these results would survive a different choice of μγ?
- The bracket: ln(5.110 × 10⁵ eV ÷ 100 eV) − 3/8 = 8.539 − 0.375 = 8.164.
- The units Cₙ = 4α(Zα)⁴m/(3πn³) with Z = 1: C₁ = 1085.15 MHz, C₂ = C₁/8 = 135.64 MHz, C₃ = C₁/27 = 40.19 MHz.
- Shifts: 1s rises 1085.15 × 8.164 = 8859 MHz, 2s rises 1107 MHz and 3s rises 328.1 MHz. The 2p wavefunction vanishes at the nucleus, so 2p gets 0 at this order.
- The 1S to 2S interval is E(2s) − E(1s). Both rise, but 1s rises eight times more, so the interval shrinks by 8859 − 1107 = 7752 MHz, seven-eighths of the 1s shift.
- Change μγ and every number above moves, by Cₙ per e-fold. What survives is the pattern this term imposes: 1 : 1/8 : 1/27 for 1s, 2s and 3s, nothing for 2p, and the interval losing 7/8 of the 1s shift.
Answer1s +8.86 GHz, 2s +1.11 GHz, 3s +0.328 GHz, 2p 0, and the 1S to 2S interval shrinks by 7.75 GHz. Only the ratios 1 : 1/8 : 1/27, the zero for 2p and the 7/8 are independent of μγ.
MediumSet the photon mass at μγ = 200 eV. Write hydrogen's F₁ contact term as C δ³(r) with C in MeV fm³, find the 2s shift from |ψ₂ₛ(0)|² in fm⁻³ and convert it to MHz. Then picture C spread evenly over a sphere of radius ħ/mc: how high is that repulsive step, and how often is the 2s electron inside it?
- The bracket: ln(5.110 × 10⁵ eV ÷ 200 eV) − 3/8 = 7.846 − 0.375 = 7.471.
- Restore ħ and c: C = (4α(Zα)/3) ħc (ħ/mc)² × 7.471. With Z = 1, (4/3)α² = 7.100 × 10⁻⁵, ħc = 197.33 MeV fm and (ħ/mc)² = (386.16 fm)² = 1.4912 × 10⁵ fm², so C = 1.561 × 10⁴ MeV fm³.
- At the origin: a₀ = (ħ/mc)/α = 5.2918 × 10⁴ fm, so |ψ₂ₛ(0)|² = 1/(8πa₀³) = 2.685 × 10⁻¹⁶ fm⁻³.
- Shift: ΔE = C|ψ₂ₛ(0)|² = 4.191 × 10⁻¹² MeV = 4.191 × 10⁻⁶ eV. Dividing by h = 4.1357 × 10⁻¹⁵ eV s gives 1.013 × 10⁹ Hz, matching C₂ × 7.471 = 135.64 MHz × 7.471 = 1013 MHz.
- Picture: the sphere holds (4π/3)(386.16 fm)³ = 2.412 × 10⁸ fm³, so the step is C ÷ volume = 6.47 × 10⁻⁵ MeV = 64.7 eV, and the 2s electron is inside with probability |ψ₂ₛ(0)|² × volume = 6.48 × 10⁻⁸. The product, 4.19 × 10⁻⁶ eV, is the same shift: any short-range shape with the same integral gives it, because |ψ₂ₛ|² is flat across 386 fm.
AnswerC ≈ 1.56 × 10⁴ MeV fm³ and ΔE(2s) ≈ 4.19 × 10⁻⁶ eV = 1013 MHz, the same as a 65 eV repulsive step over the ħ/mc sphere that the 2s electron occupies 6.5 × 10⁻⁸ of the time.
HardMuonic hydrogen replaces the electron by a muon, mμ = 105.66 MeV, reduced mass mᵣ = 94.96 MeV. For 2s, compare two contact terms: the lepton's own F₁ term, with the leading-log bracket ln(m_ℓ/(α²mᵣ)) − 3/8, and the proton's finite size, rₚ = 0.841 fm. Find their ratio in ordinary and in muonic hydrogen, and both 2s shifts in meV for muonic hydrogen. Why does the ordering flip?
- Both are contact terms, so each shift is a coefficient times the same |ψ₂ₛ(0)|², which cancels in the ratio. The F₁ coefficient carries the lepton's own mass, (4α²/3m_ℓ²) times the bracket, because the loop is the lepton's vertex; the size term is (2π/3)α rₚ². Their ratio, size over F₁, is π(m_ℓ rₚ)²/(2α × bracket).
- Brackets, with μγ set at the binding scale α²mᵣ as a leading-log estimate: electron, α²mᵣ = 27.2 eV and ln(mₑ/27.2 eV) = 9.841, bracket 9.466; muon, α²mᵣ = 5.06 keV and ln(105.66 MeV ÷ 5.06 keV) = 9.947, bracket 9.572.
- Sizes in Compton units, m_ℓ rₚ = rₚ ÷ (ħ/m_ℓc): electron 0.841 fm ÷ 386.16 fm = 2.178 × 10⁻³; muon 0.841 fm ÷ 1.8676 fm = 0.4503.
- Ratios: electron π(2.178 × 10⁻³)²/(2α × 9.466) = 1.08 × 10⁻⁴; muon π(0.4503)²/(2α × 9.572) = 4.56. The ratio grows as m_ℓ², by 206.8² = 4.28 × 10⁴, trimmed only by the log.
- Muonic hydrogen: |ψ₂ₛ(0)|² = (αmᵣ)³/(8π) = 1.723 × 10⁻⁹ fm⁻³. F₁ term: (4α²/3)ħc(ħ/mμc)² × 9.572 = 0.468 MeV fm³, giving 0.806 meV. Size term: (2π/3)α ħc rₚ² = 2.133 MeV fm³, giving 3.68 meV.
- The flip: the F₁ coefficient falls as 1/m_ℓ² while the size term does not, and both share the mᵣ³ of the wavefunction, so a heavier lepton hands the lead to the proton's size, which is why muonic hydrogen measures rₚ. The electron-loop Uehling term that dominates it is no contact term there: ħ/mₑ c = 386 fm exceeds its Bohr radius, 285 fm.
AnswerSize ÷ F₁ contact term = 1.1 × 10⁻⁴ in hydrogen but 4.6 in muonic hydrogen, where the 2s shifts are about 0.81 meV from the muon's vertex and 3.68 meV from the proton's 0.841 fm radius.