University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.8
Hydrogen's vacuum-polarisation shift: about −27 MHz for 2s
This is where the polarised vacuum first meets a measured number. Feed topic 13.07's contact term to first-order perturbation theory, find hydrogen's 2s level 27.1 MHz lower and 2p untouched, then account for why the effect shares the order of Bethe's self-energy yet is 38 times smaller and opposite in sign.
Build the model
Connect the measurement to the mechanism.
Vacuum polarisation reaches hydrogen through a potential the atom cannot resolve. Topic 13.07 found that the Uehling correction strengthens the Coulomb attraction only within 1/(2m) = 193 fm, the reach of the virtual pairs, while the wavefunction varies over the Bohr radius, 274 times farther out. Across the cloud an s-state density changes by under 1 per cent, so the correction acts as its volume integral times a delta function, −(4α(Zα)/(15m²))δ³(r), and first-order perturbation theory asks the atom for one number, |ψ(0)|².
With |ψₙₛ(0)|² = (Zαm)³/(πn³) the shift is −4α(Zα)⁴m/(15πn³), derived: exactly −1/5 of the unit 4α(Zα)⁴m/(3πn³) in which Bethe's self-energy is ln(m/K₀) ≈ 7.7. Hydrogen's 2s falls by 27.1 MHz against Bethe's rise of about 1040 MHz, and every l ≥ 1 level, vanishing at the nucleus, stays put. The two share the order α(Zα)⁴m because both are short-range corrections sampled through |ψ(0)|²; they differ in size because the self-energy gathers photons from K₀ up to m, a wide band that makes a logarithm, while the pair cloud has one scale; and in sign because the polarised vacuum sharpens the attraction where the self-energy smears it.
The measured 1057.8 MHz needs the term: without it one-loop QED misses by 21.5 MHz. The contact result stops at relative order Zα, where the density's slope across the cloud trims the shift by 0.9 per cent, and at two loops.
- Simple definition
- The vacuum-polarisation shift of an atomic level is the first-order energy change caused by the Uehling correction to the nuclear Coulomb potential; for hydrogen-like s states it is −4α(Zα)⁴m/(15πn³), and for l ≥ 1 it vanishes at this order.
- Example
- For hydrogen's 2s level, α⁵m/(30π) = 1.12 × 10⁻⁷ eV, so δE/h = −27.1 MHz, one fifth of the 135.6 MHz unit; the 1s level falls by 217 MHz and the 2p level not at all.
From topic 13.07: the excess's volume integral, fixed by the slope of Π̂₂ at q² = 0, where Π̂₂(q²) ≈ −αq²/(15πm²). Valid while the wavefunction is flat across 1/(2m) = 193 fm.
α = 1/137.036; Z the nuclear charge number; m = 0.511 MeV, the loop's mass; δ³(r) in fm⁻³; δV an energy, in MeV
The delta function picks out the density at r = 0, so the whole atom enters through one number. First order is ample: a second insertion costs a further factor of order α(Zα) ≈ 5 × 10⁻⁵, set by momenta up to m rather than by the binding energy.
|ψₙₗ(0)|² the non-relativistic density at the nucleus, in fm⁻³; δE in eV; zero for every l ≥ 1
Falls as 1/n³ and exists for s states only, which is why 1s moves 8 times as far as 2s and 2p does not move.
n the principal quantum number; Zαm = 1/a = 3.73 keV for hydrogen, a = 52 918 fm; a nucleus of finite mass puts the reduced mass here
Exactly −1/5 of Bethe's unit 4α(Zα)⁴m/(3πn³), 135.6 MHz for n = 2: the order of the self-energy, the opposite sign, and no logarithm.
an energy, in eV; hydrogen 2s: −1.12 × 10⁻⁷ eV, or δE/h = −27.1 MHz with h = 4.136 × 10⁻¹⁵ eV s
−2.6 per cent for hydrogen 2s. K₀ grows as Z², so the logarithm shrinks and the polarisation share rises along the sequence, to −3.2 per cent in He⁺.
K₀ the mean excitation energy, 16.64 Ry = 226 eV for 2s (Bethe used 17.8 Ry); ln(m/K₀) = 7.72, or 7.65
Derived in the Hard example: for s levels the polarised vacuum acts as a nucleus of negative mean-square radius, 196 times the proton's in magnitude.
matched to the finite-size shift (2πZα/3)⟨r²⟩|ψ(0)|²; with 1/m = 386.2 fm it is −139 fm², against the proton's +0.707 fm²
A delta function the atom cannot resolve
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout; an energy becomes a frequency through h = 4.136 × 10⁻¹⁵ eV s. Topic 13.07's Uehling potential exceeds Coulomb's only within the reach of the virtual pairs: the excess dies as e(−2mr), and 1/(2m) = 193 fm. Hydrogen's wavefunction changes on the Bohr radius a = 1/(Zαm) = 52 918 fm, and 2ma = 2/(Zα) = 274. Every s state has the same cusp at the nucleus, |ψ(r)|² ≈ |ψ(0)|²(1 − 2Zαmr), so across the cloud its density falls by only 2 × 193/52 918 = 0.73 per cent. To that accuracy only the integrated strength of the excess matters, and it may be replaced by a delta function of the same volume integral, −4α(Zα)/(15m²), quoted from topic 13.07. A check in momentum space: the corrected exchange is −4πZα/(|q|²(1 − Π̂₂)), and since Π̂₂(−|q|²) ≈ α|q|²/(15πm²) for |q| ≪ m, its excess over Coulomb is the constant −(4πZα/|q|²)(α|q|²/(15πm²)) = −4α(Zα)/(15m²). Dimensions: 1/m² times δ³(r) is energy⁻² times energy³, an energy.
First order: one number from the atom
With δV = −(4α(Zα)/(15m²))δ³(r), the first-order shift is ⟨nl|δV|nl⟩ = −(4α(Zα)/(15m²))|ψₙₗ(0)|², derived. Quantum Mechanics II supplies |ψₙₛ(0)|² = (Zαm)³/(πn³), quoted, so δEₙₛ = −4α(Zα)⁴m/(15πn³), derived. For hydrogen's 2s that is −α⁵m/(30π): with α⁵ = 2.069 × 10⁻¹¹ and m = 510 999 eV it is −1.122 × 10⁻⁷ eV, and dividing by h gives −27.13 MHz. A proton of finite mass puts the reduced mass into |ψ(0)|² but not into the loop's 1/m², multiplying by (μ/m)³ = 0.99837 and giving −27.08 MHz; both round to −27.1 MHz. First order is ample, though not because the shift is 3.3 × 10⁻⁸ of the 3.40 eV binding energy: for a potential this short-ranged the intermediate states that matter carry momenta up to m, so a second insertion costs a further factor of order α(Zα) ≈ 5 × 10⁻⁵ and enters only at order α²(Zα)⁵m. The 1s level falls 8 times as far, by 217.0 MHz, and 3s by 8/27 of the 2s value, 8.04 MHz.
Nothing for 2p at this order
A state of angular momentum l vanishes as rl at the nucleus, so every l ≥ 1 level has ψ(0) = 0 and no contact shift. Averaged over mₗ, the 2p density is (Zαmr)²/12 times the 2s density at the origin, which at r = 193 fm is 1.1 × 10⁻⁶: the cloud sits where a p electron almost never is. The full Uehling potential does reach 2p, at order α(Zα)⁶m, through the next term in Π̂₂'s expansion, a ∇²δ³(r) weighted by the curvature of |ψ|² at the origin; a non-relativistic evaluation gives −77 Hz, computed, which relativistic corrections of the same order modify. That is three parts per million of the 2s shift. So at the order of this lesson vacuum polarisation lowers 2S₁/₂ and leaves 2P₁/₂ in place, and the 2S₁/₂ − 2P₁/₂ interval, the Lamb shift, shrinks by 27.1 MHz. Every n behaves alike: only s levels move, by amounts in the ratio 1/n³.
Bethe's order, without Bethe's logarithm
UAQM-13.07 wrote Bethe's self-energy of an s level as 4α(Zα)⁴m/(3πn³) times ln(m/K₀): a unit of 135.6 MHz for n = 2 times a logarithm of 7.65 with Bethe's K₀ = 17.8 Ry, or 7.72 with the modern 16.64 Ry. In that unit the polarisation shift is exactly −1/5, derived, so for 2s it is −27.1 MHz against about +1040 MHz, −2.6 per cent. The order is shared because both are order-α corrections confined to short distances and sampled through |ψ(0)|²: in Welton's picture the self-energy smears the electron over a mean-square displacement of order (α/m²) ln(m/K₀), while the pair cloud acts as a nuclear mean-square radius of order −α/m², with no logarithm. The sizes differ because the self-energy collects virtual photons from K₀ ≈ 0.23 keV up to m with equal weight per logarithmic interval, a range of 2257 that builds a logarithm. The pair spectrum starts at 2m and has no second scale, so the contact strength is a pure number, the 1/30 of ∫₀¹ x²(1 − x)² dx that sets Π̂₂'s slope.
Lower, because the attraction strengthens
The sign follows from the running charge. α = 1/137.036 is the charge seen from far away, where the vacuum's screening is complete; inside 1/(2m) the electron sees part of the unscreened charge, so the attraction there exceeds Coulomb's with that α, δV is negative, and any state with density at the nucleus is pulled down. Bethe's self-energy works the other way: the jittering electron averages the potential over a small sphere, and because ∇²(−Zα/r) = 4πZαδ³(r) is positive, the averaged attraction is weaker and s levels rise. Limits check the result. As n grows the shift dies as 1/n³. A heavier loop decouples: a muon loop gives (mₑ/mμ)² = 2.3 × 10⁻⁵ of the electron loop's shift, −0.63 kHz for 2s, computed. And along the hydrogen-like sequence the shift grows as Z⁴, one Z from the nuclear charge and three from |ψ(0)|², reaching −434 MHz for the 2s level of He⁺, computed.
What the measurement demands, and where the formula stops
UAQM-13.09 assembled the one-loop 2S₁/₂ − 2P₁/₂ interval from 1015.6 MHz of self-energy, +67.8 MHz from g − 2, −4.1 MHz from 2P₁/₂'s own logarithm and −27.1 MHz from this lesson, quoted. Without the polarisation term the sum is 1079.3 MHz, 21.5 MHz above the measured 1057.8 MHz; with it, 1052.2 MHz, 5.6 MHz below. The term cuts the discrepancy almost fourfold, and measurements good to a fraction of a megahertz by the early 1950s made it the first observed effect of vacuum polarisation, quoted. The remaining 5.6 MHz belongs to higher orders, left to topic 15.09, and two of them refine this lesson's term. The density's slope across the cloud gives a correction of relative order Zα: the full Uehling potential against the non-relativistic 2s wavefunction gives −26.89 MHz, 0.89 per cent smaller, computed, matching the known +(5/48)α(Zα)⁵m/n³, quoted. Two-loop polarisation adds about −0.24 MHz, quoted.
Change one variable at a time
Make the relationship visible.
Start on 2s at Z = 1: the curve is flat across the shaded cloud, so only |ψ(0)|² counts and the shift is −27.1 MHz. Switch to 2p: inside the cloud its density is a millionth of that, so it does not move. Then raise Z: a/Z closes in on the cloud while the shift grows as Z⁴.
SHIFT δE/h-27.1 MHz
BETHE'S UNIT 4α(Zα)⁴m/(3πn³)135.6 MHz
a/Z OVER 1/(2m)274.1
DENSITY AT 193 fm99.2726 % of |ψₙₛ(0)|²
Live interpretationSHIFT δE/h: −27.1 MHz. BETHE'S UNIT 4α(Zα)⁴m/(3πn³): 135.6 MHz. a/Z OVER 1/(2m): 274.1. DENSITY AT 193 fm: 99.2726 % of |ψₙₛ(0)|²
Catch the common trap
Explain before calculating.
A hypothetical hydrogen-like atom binds a particle of charge −e and mass 2mₑ to an infinitely heavy proton; the vacuum loop is still electrons. Given hydrogen's 2s shift of −27.1 MHz, what does the contact formula give for this atom's 2s level?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUsing δEₙₛ = −4α(Zα)⁴m/(15πn³), find hydrogen's vacuum-polarisation shifts of 1s, 2s and 3s in eV and MHz for an infinitely heavy nucleus, then correct 2s for the proton's recoil, mₚ/mₑ = 1836.15. Take α = 1/137.036, m = 510 999 eV and h = 4.135 67 × 10⁻¹⁵ eV s.
- For Z = 1 the prefactor is 4α⁵m/(15π). α⁵ = 2.0693 × 10⁻¹¹, so α⁵m = 1.0574 × 10⁻⁵ eV and 4α⁵m/(15π) = 8.9756 × 10⁻⁷ eV.
- 1s, n = 1: δE = −8.976 × 10⁻⁷ eV, and δE/h = (−8.976 × 10⁻⁷ eV)/(4.135 67 × 10⁻¹⁵ eV s) = −217.0 MHz.
- 2s divides by n³ = 8: −1.122 × 10⁻⁷ eV, or −27.13 MHz. 3s divides by 27: −3.324 × 10⁻⁸ eV, or −8.04 MHz.
- Recoil: the reduced mass μ = m/(1 + m/mₚ) enters only |ψ(0)|², as μ³, while the loop's 1/m² keeps the electron mass. (μ/m)³ = (1 + 1/1836.15)⁻³ = 0.99837, so 2s becomes −27.08 MHz.
- Checks: α is dimensionless, so δE carries the dimension of m, an energy; the shifts stand as 1 : 1/8 : 1/27; and a second insertion of the potential would cost a further factor of order α(Zα) ≈ 5 × 10⁻⁵, so first order suffices.
Answer1s: −8.98 × 10⁻⁷ eV, −217.0 MHz; 2s: −1.12 × 10⁻⁷ eV, −27.1 MHz (−27.08 MHz with recoil); 3s: −3.32 × 10⁻⁸ eV, −8.04 MHz.
MediumBethe's self-energy of an s level is (4α(Zα)⁴m/(3πn³)) ln(m/K₀), with K₀ = 16.64 Z² Ry for 2s and Ry = 13.606 eV. For the 2s levels of hydrogen and of He⁺ (Z = 2), find Bethe's unit, the self-energy and the polarisation shift in MHz, ignoring recoil, and their ratio. Then compare the He⁺ polarisation shift with the measured He⁺ 2S₁/₂ − 2P₁/₂ interval, 14.04 GHz.
- Hydrogen: the unit for n = 2 is 4α⁵m/(24π) = 5.610 × 10⁻⁷ eV, or 135.64 MHz. K₀ = 16.64 × 13.606 eV = 226.4 eV, m/K₀ = 2257 and ln(m/K₀) = 7.722, so the self-energy is +1047.4 MHz.
- The polarisation shift is −1/5 of the unit, −27.13 MHz, and the ratio is −0.2/7.722 = −2.59 per cent.
- He⁺: the unit scales as Z⁴ = 16, to 2170.3 MHz, and the polarisation shift to −434.1 MHz. K₀ scales as Z², so ln(m/K₀) = 7.722 − ln 4 = 6.336 and the self-energy is 2170.3 × 6.336 = 13 750 MHz.
- The ratio for He⁺ is −0.2/6.336 = −3.16 per cent. The polarisation share grows with Z because its coefficient is a fixed −1/5, while Bethe's logarithm shrinks as K₀, climbing as Z², closes on m.
- Measured: 434 MHz is 3.1 per cent of the 14.04 GHz interval, quoted, so He⁺ needs the polarisation term just as hydrogen does. Check: both shifts scale as Z⁴, three powers from |ψ(0)|² and one from the nuclear charge; only the logarithm carries extra Z dependence.
AnswerH 2s: unit 135.6 MHz, self-energy +1047 MHz, polarisation −27.1 MHz, ratio −2.6 per cent. He⁺ 2s: unit 2170 MHz, self-energy +13 750 MHz, polarisation −434 MHz, ratio −3.2 per cent, and 3.1 per cent of the measured 14.04 GHz.
HardA nucleus of mean-square charge radius ⟨r²⟩ multiplies the Coulomb exchange −4πZα/|q|² by 1 − |q|²⟨r²⟩/6 and so shifts s levels by (2πZα/3)⟨r²⟩|ψ(0)|². Find the ⟨r²⟩ that mimics the polarisation contact term, evaluate it in fm², compare it with the proton's rₚ = 0.841 fm, find the proton-size shift of hydrogen 2s in kHz, and say how well the polarisation term must be known to extract rₚ to 1 per cent.
- Both effects are q² terms in the exchange: the polarised vacuum multiplies −4πZα/|q|² by 1/(1 − Π̂₂) ≈ 1 + α|q|²/(15πm²), the nucleus by 1 − |q|²⟨r²⟩/6. Matching gives ⟨r²⟩VP = −6α/(15πm²) = −2α/(5πm²), derived; check: (2πZα/3)(−2α/(5πm²)) = −4α(Zα)/(15m²), this lesson's coefficient.
- 1/m = ħc/mc² = 197.327 MeV fm / 0.510 999 MeV = 386.16 fm, so ⟨r²⟩VP = −9.2913 × 10⁻⁴ × 149 120 fm² = −138.6 fm², or −(11.77 fm)².
- The proton has rₚ² = 0.841² = 0.7073 fm², so the polarisation shift is −138.6/0.7073 = −196 times the proton-size shift, for every s level.
- For 2s, (2πα/3)rₚ²(αm)³/(8π) = α⁴m(m rₚ)²/12. With m rₚ = 0.841/386.16 = 2.178 × 10⁻³ and α⁴m = 1.4490 × 10⁻³ eV, the shift is +5.727 × 10⁻¹⁰ eV, or +138.5 kHz, and indeed 27.13 MHz/196 = 138 kHz.
- rₚ to 1 per cent needs rₚ² to 2 per cent, 2.77 kHz of the size term, which is 1.0 × 10⁻⁴ of the 27.1 MHz polarisation term: its 0.9 per cent Zα correction and its two-loop part must both be in hand.
- Checks: α/m² has dimension energy⁻², a length squared, as ⟨r²⟩ must; the ratio −196 is independent of n and of the orbiting mass while the contact form holds, which in muonic hydrogen it does not (topic 13.09).
Answer⟨r²⟩VP = −2α/(5πm²) = −138.6 fm² = −(11.8 fm)², −196 times the proton's 0.707 fm²; the proton-size 2s shift is +138 kHz, so the polarisation term must be known to about 1 part in 10⁴ to give rₚ to 1 per cent.