University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.6

Bethe's low-energy part with a photon cutoff K

Topic 15.05 handed the soft photons back to the atom. Here we take Bethe's answer for them off the shelf and give it a new job: we check that its ln K carries exactly the coefficient the vertex's ln(1/μ_γ) carried, which is what lets the cutoff cancel, and we collect the small, cutoff-free push it gives 2P.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Below some photon energy K a bound electron cannot be treated as free, and the photons down there belong to the atom. Their effect is the one Advanced QM computed, quoted here rather than redone: for transverse photons below K, in Coulomb gauge and the dipole approximation, after Bethe's single mass subtraction, an ns level rises by (4α(Zα)⁴m/(3πn³)) ln(K/K₀), where K₀, a weighted log-average of the atom's excitation energies, is 226 eV for 2s and 269 eV for 1s, deep in the continuum. In this course the formula changes job.

K is no longer pushed up to m; it is a fence inside the window (Zα)²m ≪ K ≪ m, the formula is trusted only below it, and its ln K must be cancelled by the free vertex above it. That works only if the two descriptions agree on the coefficient, and they do: Bethe's coefficient is (2α/(3πm²)) times a sum over states equal to ½⟨∇²V⟩, and the F₁ slope multiplies ∇²V too, so both are (α/(3πm²))⟨∇²V⟩. Because the Coulomb potential is harmonic everywhere except at the nucleus, the logarithm lands only on s states; 2p keeps no ln K but a finite remainder, +4.07 MHz, from its own Bethe logarithm.

The description stops at both ends: near m the electron recoils relativistically and the dipole form fails, and as K comes down towards K₀ the logarithm no longer dominates what the formula leaves out.

Simple definition: Bethe's low-energy part is the level shift a bound electron gets from transverse photons below a cutoff K, treated non-relativistically in the dipole approximation; on s states it grows as ln K, at a rate set by the Laplacian of the potential where the electron sits.
Example: For hydrogen 2s with K = 10.0 keV, ΔE_low = 135.6 MHz × ln(10 000 eV/226.4 eV) = 135.6 MHz × 3.788 = 514 MHz; doubling K adds 94.0 MHz, while 2p sits at +4.07 MHz whatever K is.

Bethe's low part, quoted from UAQM-13.04 to 13.07ΔE_low(n) = (2α/(3πm²)) Σₘ |⟨m|p|n⟩|² (Eₘ − Eₙ) ln(K/|Eₘ − Eₙ|)

The whole atomic spectrum enters through two numbers: the weighted sum Σ|pₘₙ|²(Eₘ − Eₙ) and the log-average of |Eₘ − Eₙ| that it weights.

K the photon-energy cutoff and Eₘ, Eₙ atomic levels, in eV; p in eV; transverse photons, dipole form, mass subtracted once

The ns shift with Bethe's logarithmΔE_low(ns) = (4α(Zα)⁴m/(3πn³)) ln(K/K₀)

A straight line in ln K that rises 94.0 MHz per doubling of K on 2s. That K-dependence is bookkeeping, and the high-energy part must cancel it.

ln(K₀/Ry) = 2.81 for 2s and 2.98 for 1s, quoted, so K₀ = 226 eV and 269 eV; the prefactor is 135.6 MHz for n = 2

The sum rule that fixes the coefficientΣₘ |⟨m|p|n⟩|² (Eₘ − Eₙ) = ½⟨n|∇²V|n⟩ = 2πZα|ψₙ(0)|²

Shown here in one line. Coulomb's V is harmonic everywhere except at the nucleus, so ln K reaches only the states that touch the nucleus: the s states.

From ½⟨n|[p, [H, p]]|n⟩ with [H, p] = i∇V; in eV³; |ψₙₛ(0)|² = (Zαm)³/(πn³), and zero for l ≥ 1

The vertex's contact term with the Laplacian showingδV(r) = (α/(3πm²)) [ln(m/μ_γ) − 3/8] ∇²V(r)

Its coefficient of ln(1/μ_γ), (α/(3πm²))⟨∇²V⟩, is Bethe's coefficient of ln K: the condition for a total that does not depend on K.

Derived here from topic 15.02's slope of F₁ with q² = −|q|² in a static field; μ_γ the photon mass in eV; δV in eV

2p: no ln K, a finite shiftΔE_low(2p) = −(4α(Zα)⁴m/(24π)) ln(K₀(2p)/Ry) = +4.07 MHz

Computed here. The dipole coupling is spin-blind, so 2P₁/₂ and 2P₃/₂ both rise 4.07 MHz: 0.4 per cent of the 1057.8 MHz splitting, kept because the target is a few MHz.

ln(K₀(2p)/Ry) = −0.0300, quoted; the weights sum to zero, so neither K nor the reference energy Ry enters

The quoted result, and the conditions it carries

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout; an energy becomes a frequency through h = 4.136 × 10⁻¹⁵ eV s. Advanced QM (UAQM-13.04 to 13.07) coupled a non-relativistic electron in V = −Zα/r to the transverse photons of Coulomb gauge in the dipole approximation, kept photons below K, and removed the free electron's linear divergence once as a mass shift. What survives, quoted, is ΔE_low(n) = (2α/(3πm²)) Σₘ |⟨m|p|n⟩|²(Eₘ − Eₙ) ln(K/|Eₘ − Eₙ|). Defining ln K₀ as the average of ln|Eₘ − Eₙ| with those weights turns it, for ns, into (4α(Zα)⁴m/(3πn³)) ln(K/K₀), whose prefactor is 5.610 × 10⁻⁷ eV, or 135.6 MHz, for n = 2 and 1085 MHz for n = 1. Each condition matters below. Only transverse photons appear, because the Coulomb photon is already inside V. Only photons below K appear, so the formula is one half of a calculation, never the whole of it.

K₀ sits deep in the continuum, and K must sit above it

The quoted ln(K₀/Ry) = 2.81 for 2s (2.8118 to four places) and 2.98 for 1s put K₀ at 16.64 Ry = 226.4 eV and 19.77 Ry = 269.0 eV: 67 times the 2s binding energy of 3.40 eV, and 20 times the 1s binding of 13.6 eV. A weighted average of ln|Eₘ − Eₙ| cannot land there unless much of the weight lies far above ionisation, so the logarithm is decided by continuum states, the electron briefly knocked free. That fixes what large K means. The form ln(K/|Eₘ − Eₙ|) replaced the exact ln((K + Eₘ − Eₙ)/(Eₘ − Eₙ)), and what it drops, from continuum states near K, fades only once K is well above K₀. Hydrogen has room: from K₀ = 226 eV to m = 511 keV is 3.35 decades. For a hydrogen-like ion K₀ grows as Z² while m does not, and the room closes at Z ≈ 48, where the non-relativistic sum would be averaging over excitations heavier than the electron.

Why the coefficient of ln K is the Laplacian of V

The coefficient of ln K is (2α/(3πm²)) Σₘ |pₘₙ|²(Eₘ − Eₙ), and the sum is an expectation value in disguise, as UAQM-13.05 showed. Since ⟨m|[H, p]|n⟩ = (Eₘ − Eₙ)⟨m|p|n⟩, the sum equals ½⟨n|[p, [H, p]]|n⟩, and with [H, p] = [V, p] = i∇V and [pⱼ, i∂ⱼV] = ∂ⱼ∂ⱼV it is ½⟨n|∇²V|n⟩. For the Coulomb potential ∇²(−Zα/r) = 4πZα δ³(r): V is harmonic everywhere except at the nucleus, so the sum is 2πZα|ψₙ(0)|². For ns, |ψₙₛ(0)|² = (Zαm)³/(πn³) gives ⟨∇²V⟩ = 4(Zα)⁴m³/n³, which is 1.892 × 10⁸ eV³ for hydrogen 2s, and the coefficient (α/(3πm²))⟨∇²V⟩ = 4α(Zα)⁴m/(3πn³) = 5.610 × 10⁻⁷ eV. The 135.6 MHz unit is the curvature of the potential where the electron is. Dimensions agree, α/m² times eV³ being eV, and with V switched off the coefficient vanishes, as it must once the free electron's self-energy has gone into m.

The vertex's soft logarithm has the same coefficient

Topic 15.02's slope, F₁(q²) = 1 + (α/(3π))(q²/m²)[ln(m/μ_γ) − 3/8], acts here on a static field, where q⁰ = 0 and q² = −|q|². The charge coupling becomes F₁(−|q|²)Ṽ(q), F₂'s spin term being topic 15.04's business, so δṼ(q) = (α/(3πm²))[ln(m/μ_γ) − 3/8] × (−|q|²Ṽ(q)), and multiplying by −|q|² is ∇² in position space: δV(r) = (α/(3πm²))[ln(m/μ_γ) − 3/8]∇²V(r). That is topic 15.03's contact term, derived again so the Laplacian shows, and its coefficient of ln(1/μ_γ) is (α/(3πm²))⟨∇²V⟩, exactly Bethe's coefficient of ln K. It had to be. The coefficient of ln K is what each factor e of photon energy adds at the top of Bethe's range; the coefficient of ln(1/μ_γ) is what each factor e adds at the bottom of the vertex's range. In the window both describe the same photons, too soft to make the electron relativistic and too hard to feel its binding, and both see one thing: the electron's response to the kicks the potential gives it.

What the equal coefficients buy: a total free of K

Call the common coefficient C = (α/(3πm²))⟨∇²V⟩. In the free-vertex calculation the photons between μ_γ and K, which it misdescribes below (Zα)²m and which Bethe's formula takes over, carry C ln(K/μ_γ) plus a constant, since every factor e of soft photons carries C. Take them out and the vertex leaves C[ln(m/K) + c]; Bethe puts back C ln(K/K₀); the s-level total is C[ln(m/K₀) + c] for any K, and the photon mass and the cutoff have gone together. Topic 15.07 computes c with gauge and regulator stated, because a constant beside a logarithm is not protected the way its coefficient is: Bethe's side uses Coulomb gauge and a sharp cutoff, the vertex Feynman gauge and a photon mass. Had the coefficients differed by δC, the total would drift by δC per factor e of K across a window ln(1/α²) = 9.84 e-folds wide, and no K could be defended. The natural place for K is near Zαm = 3.73 keV, the geometric mean of (Zα)²m and m, where K/m and (Zα)²m/K are both Zα ≈ 1/137.

2p: no logarithm, but a finite 4 MHz

For l ≥ 1, ψ(0) = 0, so ⟨∇²V⟩ = 0 and Σₘ |pₘₙ|²(Eₘ − Eₙ) vanishes: in 2p the upward weights, to 3d, the higher s and d states and the continuum, exactly balance the downward weight of 2p → 1s. The ln K term is zero at every K, as it must be, since the vertex's contact term gives p states nothing either. What remains, −(2α/(3πm²)) Σₘ |pₘₙ|²(Eₘ − Eₙ) ln|Eₘ − Eₙ|, is finite and free of K. Quoted in the s-state normalisation as ln(K₀(2p)/Ry) = −0.0300, it gives ΔE_low(2p) = −135.6 MHz × (−0.0300) = +4.07 MHz, computed. Because the weights sum to zero, any reference energy in place of Ry gives the same number, so the 2p K₀ is a label, not an energy. The same downward term carries the 2p decay, Γ/2π = 99.7 MHz (lifetime 1.60 ns, quoted), about 25 times the shift: the 4.07 MHz moves the centre of a broad line, and the order-α(Zα)⁴ assembly, aiming at a few MHz, must keep it.

Change one variable at a time

Make the relationship visible.

Interactive model

Cutoff K, as log₁₀(K/1 eV)3.60

Nuclear charge Z1

At Z = 1, slide K through the band: 2s and 1s climb 94 and 752 MHz per doubling of K while 8 × 2s − 1s holds at 187.0 MHz and 2p stays flat: that K-dependence is what the vertex must cancel. Then raise Z: K₀ and the band's lower edge move right as Z², and m ÷ K₀ drops below 1 at Z = 48.

2s LOW PART ÷ Z⁴389 MHz

1s LOW PART ÷ Z⁴2924 MHz

(8 × 2s − 1s) ÷ Z⁴187.0 MHz

ROOM ABOVE K₀: m ÷ K₀(2s)2257.16

Live interpretation2s LOW PART ÷ Z⁴: 389 MHz. 1s LOW PART ÷ Z⁴: 2924 MHz. (8 × 2s − 1s) ÷ Z⁴: 187.0 MHz. ROOM ABOVE K₀: m ÷ K₀(2s): 2257.16

Catch the common trap

Explain before calculating.

Misconception check

The tempting move is to take the 1047 MHz that Advanced QM obtained with K = m and add the vertex's contact term on top. That counts every photon below m twice, once as the atom's and once as the free electron's, and it leaves ln(1/μ_γ) in an energy level, so the answer diverges as the fictitious photon mass is removed. In this course Bethe's formula is not a prediction with an uncertain cutoff; it is the half of the calculation valid below K, its ln K is bookkeeping, and the bookkeeping balances only because the vertex's soft logarithm has the same coefficient, (α/(3πm²))⟨∇²V⟩. The opposite slip is to read 'no ln K on p states' as 'no shift on p states'. 2p keeps a finite +4.07 MHz from its Bethe logarithm, small beside 1057.8 MHz but comparable to the 6 MHz that this order leaves unexplained.

In hydrogen, Bethe's low-energy part is evaluated with K = 5.0 keV and again with K = 10.0 keV, both inside the window (Zα)²m ≪ K ≪ m. How do the 2s and 2P₁/₂ low-energy parts change between the two evaluations?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyPut the fence at K = Zαm = 3.73 keV, the geometric mean of the window's ends for hydrogen. Using ln(K₀/Ry) = 2.8118 for 2s and −0.0300 for 2p with Ry = 13.606 eV, find K₀ for 2s and compare it with the 2s binding energy, then find the low-energy parts of 2s and 2p in MHz.

K₀(2s) = 13.606 eV × e^2.8118 = 13.606 eV × 16.64 = 226.4 eV, which is 226.4/3.401 = 66.6 times the 2s binding energy of Ry/4 = 3.401 eV: the weight lies in the continuum.
K = Zαm = 510 999 eV/137.036 = 3729 eV, so ln(K/K₀) = ln(3729/226.4) = ln 16.47 = 2.802.
ΔE_low(2s) = 135.6 MHz × 2.802 = 380 MHz, which is 380.0 × 10⁶ s⁻¹ × 4.136 × 10⁻¹⁵ eV s = 1.57 × 10⁻⁶ eV.
ΔE_low(2p) = −135.6 MHz × (−0.0300) = +4.07 MHz, with no K anywhere in it.
Check: at this K the low parts alone separate 2s from 2p by 376 MHz, but that number moves 94.0 MHz with every doubling of K, so it is not yet a prediction.

AnswerK₀(2s) = 226 eV, 66.6 times the 2s binding energy; at K = 3.73 keV, ΔE_low(2s) = 380 MHz (1.57 × 10⁻⁶ eV) and ΔE_low(2p) = +4.07 MHz at any K.

MediumFor hydrogen, form D = 8ΔE_low(2s) − ΔE_low(1s) from Bethe's low parts, with ln(K₀/Ry) = 2.8118 for 2s and 2.9841 for 1s. Show that D does not depend on K, evaluate it, and explain why the vertex's contact term drops out of D as well. Then evaluate the two low parts separately at K = 8.0 keV to see what has cancelled.

ΔE_low(ns) = (4α(Zα)⁴m/(3πn³)) ln(K/K₀(ns)), so 8ΔE_low(2s) = (4α(Zα)⁴m/(3π)) ln(K/K₀(2s)): the 8 = 2³ undoes the 1/n³, and both terms carry C₁ = 4α(Zα)⁴m/(3π) = 1085.2 MHz.
D = C₁[ln(K/K₀(2s)) − ln(K/K₀(1s))] = C₁ ln(K₀(1s)/K₀(2s)) = 1085.2 MHz × (2.9841 − 2.8118) = 1085.2 MHz × 0.1723 = 187.0 MHz. K has cancelled.
The vertex's contact term (α/(3πm²))[ln(m/μ_γ) − 3/8]⟨∇²V⟩ is proportional to |ψₙₛ(0)|², so it scales as 1/n³ and 8 × (2s) − (1s) removes it exactly, μ_γ included; so does every other contact term, the Uehling term and the nuclear size among them.
At K = 8.0 keV: ln(8000/226.39) = 3.5649 gives ΔE_low(2s) = 135.644 MHz × 3.5649 = 483.56 MHz, and ln(8000/268.97) = 3.3926 gives ΔE_low(1s) = 1085.15 MHz × 3.3926 = 3681.5 MHz; then 8 × 483.56 MHz − 3681.5 MHz = 3868.5 MHz − 3681.5 MHz = 187.0 MHz.
Compare: the full QED value of this combination, about 187.2 MHz (quoted), differs by 0.1 per cent. Spectroscopists use it because nearly everything that sits at the nucleus cancels, leaving the difference of two Bethe logarithms.

AnswerD = C₁ ln(K₀(1s)/K₀(2s)) = 187.0 MHz, independent of K and of μ_γ, out of separate low parts of 484 MHz and 3.68 GHz at K = 8.0 keV.

HardAn electron moves inside a large, uniform cloud of positive charge, so V = ½mω²r² wherever it goes, with ħω = 2.00 eV. Show that Bethe's coefficient of ln K and the vertex's coefficient of ln(1/μ_γ) still agree, find both in MHz, find K₀, and find the low-energy part at K = 2.00 keV for the ground state and for a one-quantum excited state. What happens to the spacing ħω?

∇²V = 3mω² everywhere, so ⟨∇²V⟩ = 3mω² in every state: 3 × 510 999 eV × (2.00 eV)² = 6.13 × 10⁶ eV³. No delta function, and no special role for the origin.
Ground state: on each axis p = i√(mω/2)(a† − a) reaches only the one-quantum state, with |p|² = mω/2 and Eₘ − Eₙ = ω. Summing three axes, Σ = 3(mω/2)ω = 3mω²/2 = 3.07 × 10⁶ eV³ = ½⟨∇²V⟩, the sum rule checked directly.
Bethe's coefficient (2α/(3πm²)) × 3mω²/2 = αω²/(πm). The vertex gives δV = (α/(3πm²))[ln(m/μ_γ) − 3/8] × 3mω², a constant, whose ln(1/μ_γ) coefficient is also αω²/(πm) = 7.297 × 10⁻³ × 4.00 eV²/(π × 510 999 eV) = 1.82 × 10⁻⁸ eV = 4.40 MHz.
Every |Eₘ − Eₙ| is ω, so the log-average is exact: K₀ = ħω = 2.00 eV. At K = 2.00 keV, ΔE_low = 4.40 MHz × ln(2000/2.00) = 4.40 MHz × 6.908 = 30.4 MHz.
Excited state with one quantum on x: up to two quanta, |p|² = mω and Eₘ − Eₙ = +ω; down to the ground state, |p|² = mω/2 and −ω. Net on x is ω(mω − mω/2) = mω²/2, plus mω²/2 on each other axis: 3mω²/2 again, all at |Eₘ − Eₙ| = ω, so the shift is again 30.4 MHz.
Every level moves by the same amount, so the spacing ħω gets no shift at this order. Coulomb's ∇²V is a delta function, which is why hydrogen's levels split instead.

AnswerBoth coefficients are αω²/(πm) = 4.40 MHz; K₀ = ħω = 2.00 eV exactly; at K = 2.00 keV every level rises 30.4 MHz, so the spacing ħω is unshifted at this order.