University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.7
The Uehling potential: vacuum polarisation in position space
Atoms measure vacuum polarisation in position space: a bound electron samples a potential, not a momentum transfer. This topic turns the finite Π̂₂ into Uehling's potential and teaches you to read its three regimes, so that before you compute any level shift you already know which one the orbit actually feels.
Build the model
Connect the measurement to the mechanism.
Topic 13.04 wrote the renormalised loop as a once-subtracted dispersion integral over the pair spectrum, and in the static limit that integral is already a statement about forces. A pair of invariant mass M ≥ 2m enters the corrected photon exchange as a propagator 1/(|q|² + M²), whose Fourier transform is a Yukawa potential e(−Mr)/r, so the potential of a fixed point charge Ze is Coulomb's law with the Thomson-defined α plus a superposition of Yukawas weighted by the same spectrum that counts real pairs. The subtraction at q² = 0 is what keeps the far field exactly −Zα/r: the measured charge is the fully screened one.
A probe well inside 1/m = 386 fm sits within part of the cloud, sees less screening and feels a stronger attraction, larger by (2α/(3π))[ln(1/(mr)) − γE − 5/6], which is the running coupling read in position space. Beyond 1/(2m) = 193 fm only the lightest pairs reach and the excess dies as e(−2mr), and an orbit 274 times wider than that feels nothing but its integrated weight, a contact term. The model stops where its assumptions do.
It is one electron loop, first order in α and in the nuclear field, so two-loop terms and higher orders in Zα, large for heavy nuclei, are missing; the nucleus is a static point, which is wrong inside a few fermi; and muon and hadron loops add clouds of their own, of range 0.93 fm and about 0.7 fm, that hydrogen barely sees.
- Simple definition
- The Uehling potential is the static potential of a point charge Ze with the one-loop electron vacuum polarisation included: Coulomb's law with the measured α, plus a sum of Yukawa potentials, one for each virtual-pair mass, felt only within about 193 fm.
- Example
- At r = 1.0 fm from a point nucleus, mr = 1/386.16 and the excess is (2α/(3π))(ln 386.16 − 1.4105) = 0.70%, so the attraction there corresponds to 1/α ≈ 136.08; at r = 1000 fm the excess is down to 0.86 parts per million.
Derived from Π̂₂'s dispersion relation. Coulomb's 1/|q|² cancels against the subtraction, so every term is a massive exchange and none can change the far field.
ρ(s) = (1 + 2m²/s)√(1 − 4m²/s) is dimensionless; s = M² in MeV²; |q| in MeV; δṼ in MeV⁻²
Derived by a contour integral over |q|. Each slice of the spectrum is a short-range force: 193 fm for the lightest pair, 19.3 fm for a 10.2 MeV pair.
M = √s ≥ 2m is the pair's invariant mass in MeV; r in fm through ħc = 197.327 MeV fm; the range is 1/M
Derived: Coulomb's law with the Thomson-defined charge plus every pair-mass Yukawa. The excess over Coulomb is positive at every r, so the attraction is always stronger.
u = M/(2m) labels the pair mass; the integral is called I(mr); Z is the nuclear charge number; 1/m = 386.16 fm
Derived. It is the running charge in position space, equal to the momentum-space αeff at Q = e(−γE)/r = 0.561/r, and it rises 0.357% per decade of falling r.
γE = 0.5772 is Euler's constant; within about 1% of the exact I for r ≤ 5 fm, 13% low by 30 fm
Derived from the threshold of the spectrum. Only the lightest pairs reach this far, and their √(u − 1) phase space costs the power (mr)(−3/2).
1/(2m) = 193 fm is the decay length; the exponent is exact, the prefactor only asymptotic: 2.5 times high at mr = 1, 9% at mr = 20
Derived from the slope of Π̂₂ at q² = 0, or equally from ∫ x I(x) dx = 1/10. It carries hydrogen's whole vacuum-polarisation shift at leading order.
Restoring units, the coefficient is −418 Z MeV fm³; valid when the wavefunction varies on scales far beyond 193 fm
The static limit turns the dispersion integral into forces
Work in natural units ħ = c = 1 with the metric (+, −, −, −). For a nucleus held fixed, the exchanged photon carries no energy, so q² = −|q|² and the corrected exchange is Ṽ(q) = −(4πZα/|q|²)[1 + Π̂₂(−|q|²)], with α the Thomson-scattering charge of topic 13.03. Topic 13.04's once-subtracted dispersion relation writes Π̂₂(−|q|²) = (α/(3π)) ∫ (ds/s) ρ(s) |q|²/(s + |q|²) over s ≥ 4m², with ρ(s) = (1 + 2m²/s)√(1 − 4m²/s). The |q|² that the subtraction leaves in the numerator cancels Coulomb's 1/|q|², so the correction is δṼ = −(4Zα²/3) ∫ (ds/s) ρ(s)/(|q|² + s): a superposition of massive propagators and nothing else. That cancellation is the renormalisation condition at work. A piece of the loop proportional to 1/|q|² would have changed the coefficient of 1/r at large distance; Π̂₂(0) = 0 says that from far away the nucleus has exactly the measured charge. This step is derived.
Each pair mass is a Yukawa, and the spectrum weights them
Fourier transforming one propagator gives ∫ d³q/(2π)³ e(iq⋅r) 4πα/(|q|² + M²) = α e(−Mr)/r, so a pair of invariant mass M = √s contributes a Yukawa potential of range 1/M. Writing M = 2mu with u ≥ 1 turns ds/s into 2 du/u and the spectrum into w(u) = (1 + 1/(2u²))√(u² − 1)/u², and the sum assembles into V(r) = −(Zα/r)[1 + (2α/(3π)) I(mr)] with I(mr) = ∫₁^∞ du e(−2mru) w(u). That is Uehling's 1935 potential, derived. The weight is familiar: ρ = β(3 − β²)/2 with β = √(1 − 1/u²) is the threshold factor of e⁺e⁻ → μ⁺μ⁻ with the muon mass replaced by m, so the Yukawas are weighted by the rate at which a timelike photon makes real pairs, as the optical theorem promised. w vanishes at threshold like √(u − 1), peaks at 0.64 near u = 1.30, a 1.33 MeV pair, and falls as 1/u for heavy pairs; the lightest pair reaches 193 fm, a 10.2 MeV pair (u = 10) only 19.3 fm.
Inside 1/m: the running charge, read in position space
For mr ≪ 1 the exponential cuts the integral off near u ≈ 1/(2mr), far out where w(u) ≈ 1/u, so I grows logarithmically. Split w = 1/u + (w − 1/u). The first piece gives ∫₁^∞ e(−2mru) du/u ≈ −γE − ln(2mr); the second converges with the exponential set to 1, to ∫₁^∞ (w − 1/u) du = ln 2 − 5/6, elementary once u = cosh t. The ln 2 cancels, leaving I(mr) ≈ ln(1/(mr)) − γE − 5/6, derived. In momentum space topic 13.05 found αeff = α[1 + (2α/(3π))(ln(Q/m) − 5/6)] at large Q, quoted here, and the two agree term by term if Q = e(−γE)/r = 0.561/r. The sign is the physics. Thomson scattering defines α as the charge seen from outside the whole cloud; a probe at r = 1.0 fm is inside most of it, sees less screening, and feels an attraction 0.70% stronger than −Zα/r, as if 1/α were 136.08. The excess grows by (2α/(3π)) ln 10 = 0.357% per decade, so even 1% needs r = 0.15 fm.
Beyond 1/(2m): the cloud ends exponentially
For mr ≫ 1 only the lightest pairs contribute, so expand about threshold, u = 1 + τ: w ≈ (3/2)√(2τ), and ∫₀^∞ √τ e(−2mrτ) dτ = (√π/2)(2mr)(−3/2) gives I(mr) ≈ (3√π/8) e(−2mr)/(mr)³⁄², derived. The excess over Coulomb is then (α/(4√π)) e(−2mr)/(mr)³⁄². The decay length is 1/(2m) = 193 fm, not 1/m, because the lightest state the photon can fluctuate into is a pair of mass 2m, and the threshold factor √(u − 1) costs the extra (mr)(−3/2). The prefactor is only asymptotic, 2.5 times too large at mr = 1 and 9% at mr = 20, but the exponent is exact. Computed values of the full I show the edge: the excess is 0.028% at 193 fm, 56 ppm at 386 fm and 0.86 ppm at 1000 fm, and at hydrogen's Bohr radius a₀ = 1/(mα) = 52 918 fm, where 2mr = 2/α = 274, it is about 10⁻¹²⁵.
At atomic distances only the integrated weight survives
A wavefunction that varies over a₀ cannot resolve a correction confined within a few hundred fermi. Across that region ψ(r) ≈ ψ(0), so δV acts as C δ³(r) with C = ∫ d³r δV = δṼ(q → 0). In momentum space C is set by the slope: Π̂₂(−|q|²) ≈ (α/(15π))|q|²/m² near zero, so δṼ(0) = −4πZα × α/(15πm²) = −4α(Zα)/(15m²), derived. Position space agrees: ∫ d³r δV = −Zα (2α/(3π))(4π/m²) ∫₀^∞ x I(x) dx, and since ∫ x e(−2xu) dx = 1/(4u²) the moment is ∫₁^∞ w(u)/(4u²) du = 1/10, giving the same 4/15. Restoring units, C = −418 Z MeV fm³. Each Yukawa enters C with weight 1/M², so C is dominated by pairs near threshold, the opposite of the short-distance log. Topic 13.08 multiplies C by |ψ₂ₛ(0)|²; here the point is validity: the cloud's weighted mean radius, 25π/(128m) = 237 fm, is 0.45% of a₀.
Checks: dimensions, limits and a measurement
Dimensions: I is a pure number and α/r has the dimension of mass, as a potential must in natural units; the contact coefficient α²/m² times δ³(r), of dimension mass³, is again a mass. Limits: α → 0 returns Coulomb; r → ∞ returns −Zα/r exactly, because the charge was defined there; and m → ∞ shrinks the cloud to nothing and the contact weight as 1/m², which is why the muon loop, with range 1/(2mμ) = 0.93 fm, adds only (mₑ/mμ)² = 2.3 × 10⁻⁵ of the electron's weight in hydrogen. Measurement: hydrogen's 2s–2p splitting, measured at 1057.8 MHz, cannot be matched without the roughly −27 MHz this term produces (quoted here, computed in topic 13.08), and in muonic hydrogen, whose Bohr radius of 285 fm is only 1.5 times 193 fm, vacuum polarisation dominates the Lamb shift and only the full potential, not its contact limit, gets it right (topic 13.09).
Change one variable at a time
Make the relationship visible.
Lower the heaviest pair from 12.6 GeV: inside r ≈ 1/M the solid curve leaves the dashed log and flattens, because the short-distance rise is built from heavy pairs; below 1 MeV no pair exists and Coulomb is exact. Then push r past 193 fm and watch the excess vanish.
LOG FORM AT r0.454 %
EQUIVALENT Q = 0.561/r22.1 MeV
2mr, CLOUD EDGE AT 10.026
RANGE 1/M OF HEAVIEST PAIR0.016 fm
Live interpretationLOG FORM AT r: 0.454 %. EQUIVALENT Q = 0.561/r: 22.1 MeV. 2mr, CLOUD EDGE AT 1: 0.026. RANGE 1/M OF HEAVIEST PAIR: 0.016 fm
Catch the common trap
Explain before calculating.
Hydrogen's 2s wavefunction varies over a₀ = 52 918 fm, while the Uehling excess lives within a few hundred fermi of the proton. A student must predict the sign and the form of the vacuum-polarisation shift of 2s. Which reasoning is right?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTreat a nucleus as a static point charge. Using the short-distance form of Uehling's potential, find by what percentage the attraction at r = 5.0 fm exceeds Coulomb's Zα/r, check that the form applies, and find the momentum transfer at which the momentum-space running coupling is enhanced by the same amount.
- Restore units for the length scale: 1/m = ħc/(mc²) = 197.327 MeV fm ÷ 0.51100 MeV = 386.16 fm, so mr = 5.0/386.16 = 0.01295. That is far below 1, so the probe sits deep inside the pair cloud and the logarithmic form applies.
- The bracket: ln(1/(mr)) = ln 77.23 = 4.347, and γE + 5/6 = 0.5772 + 0.8333 = 1.4105, so ln(1/(mr)) − γE − 5/6 = 2.936.
- Multiply by 2α/(3π) = 2/(3π × 137.036) = 1.5485 × 10⁻³: the excess is 4.55 × 10⁻³, or 0.455%. A numerical evaluation of the full I(mr) gives 2.967 rather than 2.936, so the exact excess is 0.459%; the log form is 1% low here, and its error grows roughly in proportion to r.
- The matching momentum is Q = e(−γE)/r = 0.5615 × 197.327 MeV fm ÷ 5.0 fm = 22.2 MeV. Check: (2α/(3π))(ln(Q/m) − 5/6) = 1.5485 × 10⁻³ × (ln 43.36 − 0.8333) = 1.5485 × 10⁻³ × 2.936 = 4.55 × 10⁻³, the same number.
- Dimensions: mr and Q/m are pure numbers, as the arguments of logarithms must be. The charge felt at 5.0 fm corresponds to 1/α = 137.036/1.00459 = 136.41.
AnswerAbout 0.46%: 0.455% from the log form, 0.459% exact, the same enhancement as the momentum-space running at Q ≈ 22 MeV; the attraction there corresponds to 1/α ≈ 136.4.
MediumAt leading order hydrogen feels the Uehling potential only through its contact weight, which is proportional to J = ∫₁^∞ du (1 + 1/(2u²))√(u² − 1)/u⁴, where u = M/(2m) labels the pair mass. Evaluate J exactly, find the fraction of it supplied by pairs lighter than 4m = 2.04 MeV, and compare with the share of the excess at r = 0.1 fm that those pairs supply.
- Why u⁻⁴: a Yukawa of mass M integrates to ∫ d³r e(−Mr)/r = 4π/M², so each slice of w(u) = (1 + 1/(2u²))√(u² − 1)/u² enters the contact weight with an extra 1/u². Heavy pairs are suppressed as 1/M², unlike the short-distance log, which counts them equally per e-fold of mass.
- Substitute u = sec θ, so √(u² − 1) = tan θ and du = sec θ tan θ dθ. The integrand becomes (sin²θ cos θ + ½ sin²θ cos³θ) dθ, with θ running from 0 to π/2.
- Integrate: ∫ sin²θ cos θ dθ = sin³θ/3 and ∫ sin²θ cos³θ dθ = sin³θ/3 − sin⁵θ/5. At θ = π/2, J = 1/3 + ½(1/3 − 1/5) = 1/3 + 1/15 = 2/5. Then ∫ x I(x) dx = J/4 = 1/10, which is what makes the contact coefficient 4/15.
- Pairs lighter than 4m have u ≤ 2, so θ ≤ π/3 and sin θ = √3/2, giving sin³θ = 0.6495 and sin⁵θ = 0.4871. The partial integral is 0.6495/3 + ½(0.6495/3 − 0.4871/5) = 0.2165 + 0.0595 = 0.2760.
- Fraction: 0.2760/0.4000 = 0.690. By contrast, a numerical evaluation of I at r = 0.1 fm finds that the same pairs supply only 8.2% of the excess there; the rest comes from pairs up to about 1/r ≈ 2 GeV.
AnswerJ = 2/5 exactly, and 69% of it comes from pairs below 4m ≈ 2.04 MeV, against 8% of the excess at 0.1 fm: hydrogen's shift is a near-threshold effect, the running charge a high-mass one.
HardThe contact form treats |ψ(r)|² as constant across the pair cloud. For any s state the cusp condition gives |ψ(r)|² ≈ |ψ(0)|²(1 − 2Zr/a) near the nucleus, with a = 1/(mᵣ α) and mᵣ the reduced mass. Find the fractional correction this slope makes to the contact estimate of an s-level shift in hydrogen, and the nuclear charge Z at which it reaches 10% for an electronic hydrogen-like ion.
- The shift is ΔE = ∫ d³r δV |ψ|² = −Zα (2α/(3π)) 4π ∫₀^∞ r dr I(mr) |ψ(r)|². With x = mr the leading term carries ∫ x I(x) dx = 1/10, and the slope adds −(2Z/(am)) ∫ x² I(x) dx.
- Second moment: ∫₀^∞ x² e(−2xu) dx = 1/(4u³), so ∫ x² I dx = ¼ ∫₁^∞ w(u)/u³ du. With u = sec θ this is ¼ ∫ (sin²θ cos²θ + ½ sin²θ cos⁴θ) dθ over 0 to π/2, which is ¼(π/16 + π/64) = 5π/256.
- Ratio of correction to leading term: −(2Z/(am)) × (5π/256)/(1/10) = −(2Z/(am)) × 25π/128. The cloud's weighted mean radius is 25π/(128m) = 0.6136 × 386.16 fm = 237 fm, and Z/(am) = Zα mᵣ/m.
- Hydrogen: mᵣ/m = 1/(1 + 0.51100/938.272) = 0.99946, so the fraction is −(25π/64) × α × 0.99946 = −1.2272 × 7.2974 × 10⁻³ × 0.99946 = −8.95 × 10⁻³. The contact form overstates the size of the shift by 0.90%.
- Setting (25π/64)Zα = 0.10, with mᵣ ≈ m for a heavy nucleus, gives Z = 0.10/(1.2272 × 7.2974 × 10⁻³) = 11.2. Check: the correction times the leading −(4/(15π))α(Zα)⁴m/n³ is +(5/48)α(Zα)⁵m/n³, the published coefficient at this order (quoted).
AnswerThe fraction is −(25π/64)Zα(mᵣ/m): −0.90% for hydrogen, reaching 10% near Z = 11, beyond which the full Uehling potential, not its contact limit, must be used.