University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.5
Binding cuts off the soft photons: the overlap window
Topic 15.03 left the 2s shift resting on ln(m/μγ), which a massless photon makes infinite. This topic shows why the atom never meets that infinity, measures how far up in photon energy binding still matters, and finds the range, 27 eV to 511 keV in hydrogen, in which the free vertex and Bethe's calculation can be joined.
Build the model
Connect the measurement to the mechanism.
The free vertex of topic 15.03 shifts 2s by 135.6 MHz × [ln(m/μγ) − 3/8], which grows by 312 MHz for every decade the photon mass falls and diverges as μγ → 0. The cause is kinematic: next to a soft photon an on-shell electron's propagator is −1/(2p⋅k), proportional to 1/ω, so every octave of photon energy down to μγ counts the same. In scattering, Unit 14 cancelled that logarithm against real photons below the detector's resolution; a stationary level offers no such sum.
What saves the atom is that its electron is off shell: each momentum component carries energy m + Eₙ, the propagator becomes −1/(2m(ω + ε)) with ε = |Eₙ| + |p|²/(2m), and photons softer than ε are no longer answered freely. Bethe's energy denominators (UAQM-13.08) are the exact version and end the logarithm near K₀ = 226 eV for 2s. Each picture has a domain.
The free vertex holds for photons far above the off-shellness, k ≫ (Zα)²m, and the cusp at the nucleus makes its error fall only as 2√2 Zα √(m/k); Bethe's non-relativistic dipole treatment holds for k ≪ m, with errors of order k/m. In hydrogen the domains overlap from 27 eV to 511 keV, a factor 1/α² = 18 779, and the part of each error that depends on the joining point K is shared and cancels. The window spans 1/(Zα)²: a factor 2.2 for uranium, and none as Zα nears 1.
- Simple definition
- The overlap window is the range of photon-energy cutoffs K, (Zα)²m ≪ K ≪ m, hard enough that a bound electron answers virtual photons above K almost as a free one and soft enough that Bethe's non-relativistic dipole treatment still holds below it.
- Example
- In hydrogen (Zα)²m = α²m = 27.2 eV and m = 511.0 keV, a factor 1/α² = 18 779, or 4.27 decades. At K = 10 keV the 2s electron still lacks 12% of the free response, computed, while Bethe's dropped terms are of order K/m = 2%.
Derived. Since E exceeds |p| it vanishes linearly in ω with no other scale, so the soft loop region is ∫dω/ω, cut only by μγ: topic 15.02's ln(m/μγ).
p on shell, p² = m², with energy E; the photon at its pole, k² = 0, with energy ω at angle θ to p; all in eV (natural units)
Derived, dropping terms of relative order v and (Zα)². ε ≥ |Eₙ| keeps it finite as ω → 0; by the virial theorem ⟨ε⟩ = 2|Eₙ|, 6.80 eV for 2s.
|p| the three-momentum of one component of level n, Eₙ its negative non-relativistic energy, ω the photon energy; ε and ω in eV
Built on UAQM-13.08's factor k/(k + ΔEₘ), quoted. Computed for 2s: 0.74 at 27.2 eV, 0.47 at K₀ = 226 eV, 0.18 at 3.73 keV, 0.020 at 511 keV.
wₘ = |pₘₙ|²ΔEₘ are Bethe's weights, ΔEₘ = Eₘ − Eₙ; k and ΔEₘ in eV; D is a pure number, 0 for a free electron
Argued from the cusp, whose Fourier transform falls as P⁻⁴, and checked on the 2s sum: 0.0206 against 0.0197 at k = m. A square root, not (Zα)²m/k.
The same as 4√(Z²Ry/k) with Ry = 13.606 eV; k and m in eV. The weights' tail falls as ΔE(−3/2), fixed by ψ(0)
Argued here. It spans 1/(Zα)²: 18 779 for Z = 1, 188 for Z = 10, 2.2 for Z = 92. The free side drops D(K), Bethe's side terms of order K/m.
K the photon energy at which the two descriptions are joined, in eV. Hydrogen: 27.2 eV to 511.0 keV
Derived. Bethe's large-K form drops L(K) and the free form above K overcounts by the same L(K), so it cancels. For 2s, L = 0.41 at 3.73 keV: 55 MHz.
L(K) = Σₘ wₘ ln(1 + ΔEₘ/K) ÷ Σₘ wₘ, a pure number; K₀ is Bethe's mean excitation energy, 226 eV for 2s
Why ln(m/μγ) has no floor
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 15.03 derived an ns shift of (4α(Zα)⁴m/(3πn³))[ln(m/μγ) − 3/8], whose unit is 135.6 MHz for 2s. With μγ = 1 eV that is 1732 MHz and with 1 meV 2669 MHz, while the measured 2S₁/₂–2P₁/₂ splitting is 1057.8 MHz, quoted; every factor of ten in μγ adds 135.6 × ln 10 = 312 MHz. The logarithm is made by soft virtual photons. Next to one the electron propagator is 1/((p − k)² − m²) = −1/(2p⋅k), and on shell p⋅k = ω(E − |p| cos θ) is proportional to ω with no other scale. Two such propagators and the photon's d³k/(2ω) give ∫dω/ω: each decade of photon energy between μγ and m adds the same 312 MHz. In Unit 14 this infrared logarithm was harmless because a cross-section is inclusive: elastic scattering was summed with emission of photons below the resolution ΔE, and ΔE replaced μγ. A bound level is not a sum over final states, so something else must stop the count.
A bound electron is off shell by atomic energies
In level n each momentum component, of three-momentum |p|, carries energy m + Eₙ, not √(m² + |p|²), so its four-momentum has p² − m² ≈ 2mEₙ − |p|² = −2mε with ε = |Eₙ| + |p|²/(2m), and next to the photon (p − k)² − m² ≈ −2m(ω + ε), derived up to terms of relative order v and (Zα)². Dimensions: ε and ω are both energies. Limits: ε → 0 returns the free 1/ω, while for ω ≪ ε the propagator freezes at −1/(2mε) and soft photons stop counting. The virial theorem gives ⟨|p|²⟩/(2m) = |Eₙ|, so ⟨ε⟩ = 2|Eₙ| = (Zα)²m/n²: 27.2 eV for 1s, 6.80 eV for 2s. The exact version replaces one ε by the spectrum: each intermediate state contributes k/(k + Eₘ − Eₙ) per unit ln k (UAQM-13.08, quoted), 1 far above its transition and k/ΔE below it. With Bethe's weights the logarithm ends at K₀ = 226 eV for 2s, quoted, 33 times the mean off-shellness, because the weights favour the cusp at r = 0. Binding, not a detector, replaces μγ, and topic 15.07 will find the replacement worth 2K₀e(−5/6) ≈ 197 eV.
How slowly binding lets go
Let D(k) be the share of the free response a bound electron still lacks at photon energy k: ΔEₘ/(k + ΔEₘ) averaged with Bethe's weights wₘ = |pₘₙ|²ΔEₘ. Summing 2s's bound and continuum states for this lesson gives D = 0.74 at 27.2 eV, 0.47 at K₀, 0.29 at 1 keV, 0.18 at Zαm = 3.73 keV, 0.12 at 10 keV, 0.043 at 100 keV and 0.020 at 511 keV, computed. That is a square root, not (Zα)²m/k. States far above the level are reached only through the cusp of ψ at r = 0, whose Fourier transform falls as P⁻⁴, so the weight per unit ΔE falls as ΔE(−3/2) and D(k) → 4√(Z²Ry/k) = 2√2 Zα √(m/k), the power argued and the 4 taken from hydrogen's high-energy oscillator strengths; at k = m that is 0.0206 against the summed 0.0197. Dimensions: Z²Ry/k and m/k are ratios of energies. Limits: D → 1 as k → 0, where the electron is frozen, and D → 0 as k grows. Physically, a photon of energy k probes momenta P ≈ √(2mk), and the share of ψ(0) carried by momenta above P is of order Zαm/P.
Where Bethe's picture runs out
Bethe's part is non-relativistic and dipole: it keeps |p|²/(2m) for the electron, drops the photon's momentum from e(ik⋅r) and ignores recoil. At photon energy k the states that matter have kinetic energies of order k, so momenta P ≈ √(2mk). Three corrections follow, each of relative order k/m, argued: recoil, k²/(2m) beside k; the relativistic correction to the kinetic energy, a fraction P²/(4m²) = k/(2m) of it; and the photon's momentum beside P, whose square is (k/P)² = k/(2m), the first power averaging away over photon directions. The tempting dipole test ka₀ ≪ 1 would end Bethe's part at Zαm = 3.73 keV, but it uses the wrong length: the states carrying the weight live within 1/P of the nucleus, not across the Bohr orbit, so the test is k ≪ m. At 100 keV the dropped terms are of order K/m = 0.2. Near m all three fail together, with pair creation and spin left out as well, and only Unit 14's relativistic vertex is valid there.
The window, and what each side drops inside it
The free vertex needs K ≫ (Zα)²m and Bethe needs K ≪ m; for Z = 1 they overlap from 27.2 eV to 511.0 keV, a factor 1/α² = 18 779. The free side drops D(K) and Bethe's side terms of order K/m, and no K makes both tiny: 12% and 2% at 10 keV, 10% and 3% at 15 keV, 4.3% and 20% at 100 keV. Joining works anyway, because what must agree in the window is the form, not the value. Per state, Bethe's exact low part is ln(1 + K/ΔE) = ln(K/ΔE) + ln(1 + ΔE/K), while the free form overcounts the photons above K by ∫(1/k − 1/(k + ΔE)) dk = ln(1 + ΔE/K) for an upper limit far above ΔE. Summed, Bethe's large-K form ln(K/K₀) drops a leftover L(K) that the free form adds back: 0.41 of the unit at 3.73 keV, 55 MHz on 2s, cancelled exactly, derived. What the forms leave out is the leftover at the top, L(m) = 0.040 or 5.5 MHz: binding that lasts to k ≈ m, where neither side holds, of relative order Zα, the size of the α(Zα)⁵ term topic 15.09 quotes.
Heavier nuclei close the window
In a hydrogen-like ion every atomic energy scales as Z², so the lower edge (Zα)²m rises while m stays put, and the window spans 1/(Zα)²: 188 for neon, Z = 10, 11.7 for zirconium, Z = 40, and 2.2 for uranium, Z = 92, where (Zα)²m = 230 keV. The deficit scales the same way, DZ(k) = DH(k/Z²), so binding lasts to higher photon energies: at k = m the 2s electron still lacks 2.0% of the free response for Z = 1, 16% for Z = 10, 42% for Z = 40 and 64% for Z = 92, computed. Bethe's logarithm loses 2 ln Z with it, as UAQM-13.08 found. The window's width in e-folds, ln(1/(Zα)²) = 9.84 for hydrogen, is the large logarithm that lets the α(Zα)⁴ ln term dominate; as Zα nears 1 it is no larger than the constants beside it. Then no K exists at which a free description and a non-relativistic one both hold, the expansion in Zα that organises this unit has nothing to expand in, and heavy ions need the bound-state calculation to all orders in Zα, quoted, where this course stops.
Change one variable at a time
Make the relationship visible.
Slide K up from 10 eV: the missing share falls only about as 1/√K while K/m climbs, so no K makes both tiny. Then raise Z: the window's left edge rises as Z², the curve moves with it, and at Z = 92 a photon of energy m still finds 64% of the response missing.
2s RESPONSE MISSING AT K11.9 %
BETHE'S TERMS, ORDER K/m2.0 %
STILL MISSING AT k = m2.0 %
WINDOW WIDTH 1/(Zα)²18779
Live interpretation2s RESPONSE MISSING AT K: 11.9 %. BETHE'S TERMS, ORDER K/m: 2.0 %. STILL MISSING AT k = m: 2.0 %. WINDOW WIDTH 1/(Zα)²: 18779
Catch the common trap
Explain before calculating.
Hydrogen's 2s self-energy is to be split at K = 30 keV: the free vertex for virtual photons above K, Bethe's non-relativistic dipole result below. Roughly how large are the corrections each description drops at that K?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen 2s electron has energy E₂ = −3.401 eV. Use the virial theorem to find its mean off-shellness ε = |E₂| + ⟨|p|²⟩/(2m), and the photon energy at which the single-ε propagator ratio ω/(ω + ε) reaches 0.90. Compare it with α²m, and with the 15.1 keV at which the full 2s sum first reaches 90% of the free response.
- For a Coulomb potential the virial theorem gives ⟨|p|²⟩/(2m) = −E₂ = 3.401 eV, so ⟨ε⟩ = 3.401 eV + 3.401 eV = 6.802 eV, which is α²m/n² = 27.21 eV/4, as it should be.
- The bound propagator over the free one at photon energy ω is (1/(2m(ω + ε)))/(1/(2mω)) = ω/(ω + ε). Setting it to 0.90 gives ω = 9ε = 61.2 eV.
- Against α²m = 27.21 eV this is 2.25 times the window's lower edge: a single off-shellness would make the free vertex good to 10% almost at once.
- The real spectrum disagrees. The summed 2s deficit falls to 0.10 only at 15.1 keV, computed, 247 times higher, because the cusp spreads Bethe's weight to transition energies far above ⟨ε⟩.
Answer⟨ε⟩ = 6.80 eV, and ω/(ω + ε) = 0.90 at 61 eV, about 2.2 α²m. The full 2s sum needs 15 keV for the same 90%, about 250 times higher.
MediumUse the large-k law D(k) ≈ 4√(Z²Ry/k), with Ry = 13.606 eV, to estimate the photon energy above which a 2s electron lacks less than 10% of the free response, and less than 1%, in hydrogen and in hydrogen-like carbon, Z = 6. Say whether each lies inside the window, and compare with the summed hydrogen value D(21.8 keV) = 0.085.
- Invert the law: k = 16Z²Ry/D². For D = 0.10, k = 1600 Z²Ry; for D = 0.01, k = 160 000 Z²Ry.
- Hydrogen: 1600 × 13.606 eV = 21.8 keV, well inside 27.2 eV to 511 keV. 160 000 × 13.606 eV = 2.18 MeV, four times m, so the bound electron never comes within 1% of free while Bethe's picture still holds.
- Carbon: multiply by Z² = 36. The 10% point moves to 784 keV and the 1% point to 78.4 MeV, both above m = 511 keV, although the window itself, 0.980 keV to 511 keV, still spans a factor 522.
- Check against the sum: at 21.8 keV hydrogen's summed D is 0.085, not 0.10, because the law is only the leading term and overstates D by about 17% here; the summed 10% point is 15.1 keV. The Z² scaling is exact, DZ(k) = DH(k/Z²).
AnswerHydrogen: under 10% above about 22 keV (15 keV summed), inside the window; under 1% only above 2.2 MeV, outside it. Carbon: 784 keV and 78 MeV, both beyond m = 511 keV.
HardReplace 2s's spectrum by one transition at ΔE = K₀ = 226.4 eV carrying all of Bethe's weight. Join at K = Zαm = 3.729 keV, with the free form running up to m = 511.0 keV. In units of 135.6 MHz, find Bethe's exact low part, its large-K form, the free form's overcount above K and what is left at m. Then repeat the comparison with the summed 2s values L(3.729 keV) = 0.408 and L(m) = 0.040.
- Exact low part: ln(1 + K/ΔE) = ln(1 + 3729/226.4) = ln 17.4708 = 2.86053. Large-K form: ln(K/ΔE) = ln 16.4708 = 2.80159. The form drops L(K) = ln(1 + ΔE/K) = 0.05894.
- Above K the exact integrand per unit ln k is k/(k + ΔE), giving ln((m + ΔE)/(K + ΔE)) = ln(511 226/3955.4) = 4.86173, while the free form gives ln(m/K) = ln(511 000/3729) = 4.92023: an overcount of 0.05850.
- Forms used: 2.80159 + 4.92023 = 7.72182 = ln(m/ΔE). Exact: 2.86053 + 4.86173 = 7.72226 = ln(1 + m/ΔE). They differ by ln(1 + ΔE/m) = 0.00044, or 0.06 MHz, so the 0.0589 leftover, 8.0 MHz, has cancelled.
- The real 2s leftover at the same K is 0.408, or 55 MHz, seven times the toy's, because the cusp puts weight at transition energies far above K₀. It cancels in exactly the same way.
- What survives is the leftover at the top: 0.040, or 5.5 MHz, against the toy's 0.00044. That is binding lasting up to k ≈ m, where neither description holds, of relative order Zα, like the α(Zα)⁵ term.
AnswerToy: a leftover of 0.059 (8.0 MHz) cancels, leaving 0.0004 (0.06 MHz). Real 2s: a leftover of 0.41 (55 MHz) cancels, leaving 0.040 (5.5 MHz), of relative order Zα.