University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.4

The imaginary part of vacuum polarisation is pair creation

Here a loop stops being an algebraic correction and becomes something a detector counts. Watch the iε give Π̂₂ an imaginary part above q² = 4m², identify it with the rate of e⁺e⁻ → μ⁺μ⁻, then run the argument backwards, so that measured rates, R(s) for hadrons, can be fed into the photon propagator.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Π̂₂(q²) is an analytic function, not just a correction factor, and its imaginary part is the most physical thing about it. The Feynman propagators' iε survives in the Feynman-parameter logarithm as m² − x(1 − x)q² − iε. Below q² = 4m², every spacelike q² included, that argument is positive for all x and Π̂₂ is real.

Above it a window of x opens where the argument is negative, the logarithm acquires −iπ, and Im Π̂₂ = −(α/3)β(1 + 2m²/q²), with β the speed of each lepton in the pair's rest frame. Unitarity reads that number as a rate: topic 5.07's optical theorem turns −(4πα/s) Im Π̂₂ into σ(e⁺e⁻ → f f̄), and for a muon loop the result is topic 9.06's tree-level cross-section, mass factor and all. Analyticity then runs the argument backwards. Π̂₂ has no singularity off the cut, so Cauchy's theorem rebuilds the real part from the imaginary part, with one subtraction at q² = 0 that is exactly the definition of α.

The relation outlives the formula behind it: insert a measured R(s) in place of β(3 − β²)/2 and hadrons, which no perturbative loop describes at low energy, enter the photon propagator as data. The one-loop imaginary part itself fails close to threshold, where Coulomb attraction between the slow leptons changes the rate at relative order πα/(2β), and every further channel, a heavier lepton or a hadronic state, opens its own cut with its own threshold and cusp.

Simple definition: The imaginary part of the renormalised photon self-energy Π̂₂(q²) is the piece that exists only above q² = 4m², where a virtual photon is heavy enough to become a real lepton pair; the optical theorem makes it the rate for producing that pair.
Example: At q² = 9m², β = 0.745 and Im Π̂₂ = −2.22 × 10⁻³ for any lepton. For a muon loop that is √s = 317 MeV, and −(4πα/s) Im Π̂₂ = 788 nb, the cross-section σ(e⁺e⁻ → μ⁺μ⁻) there, 91% of the point-like 864 nb.

Imaginary part above thresholdIm Π̂₂(q² + iε) = −(α/3)√(1 − 4m²/q²)(1 + 2m²/q²) for q² above 4m², and 0 below it

Derived from ln(Δ − iε) = ln|Δ| − iπ where Δ is negative. It tends to −α/3 = −2.43 × 10⁻³ far above threshold and vanishes like β at it.

Π̂₂ dimensionless; q² and m² in GeV²; α = 1/137.036; real at every spacelike q²

The window where Δ is negativeΔ(x) = m² − x(1 − x)q² is negative for x between (1 − β)/2 and (1 + β)/2

Derived. Over this window ∫ x(1 − x) dx = β(3 − β²)/12, and −2α times it is Im Π̂₂. The window shuts at threshold, which is why the rate starts from zero.

x the Feynman parameter, 0 to 1; β = √(1 − 4m²/q²), the lepton speed in the pair rest frame, in units of c

Optical theorem for the bubbleσ(e⁺e⁻ → f f̄) = −(4πα/s) Im Π̂₂^(f)(s)

Derived from the forward amplitude −e²Π̂₂(s) and Im ℳ = sσ. It reproduces topic 9.06's (4πα²/3s) × β(3 − β²)/2 term by term: checked.

s in GeV²; σ in GeV⁻², times 0.3894 mb GeV² for barns; electrons massless; f a unit-charge lepton

Once-subtracted dispersion relationΠ̂₂(q²) = (q²/π) ∫ ds Im Π̂₂(s)/[s(s − q² − iε)], s from 4m² to ∞

Derived from Cauchy's theorem for Π̂₂(z)/z. Without the subtraction the integral diverges like ∫ds/s, because Im Π̂₂ tends to −α/3.

s and q² in GeV²; the subtraction uses Π̂₂(0) = 0, the definition of α at q² = 0

The slope at q² = 0, from the cutdΠ̂₂/dq² at q² = 0: (1/π) ∫ ds Im Π̂₂(s)/s² = −α/(15πm²)

Derived, and equal to the −αq²/(15πm²) topic 13.02 found by expanding the logarithm: low-q² physics is made by pairs produced near threshold.

m in GeV, slope in GeV⁻²; 46% of the integral comes from √s/m below 3, and 95% from below 10

Hadrons enter through measured R(s)Im Π̂_had(s) = −(α/3) R(s), with R(s) = σ(e⁺e⁻ → hadrons)/(4πα²/3s)

Derived by the optical theorem with any final state. In the dispersion relation it makes the hadronic Π̂ an integral over data, with no quark loop needed.

R dimensionless; 4πα²/3s = 86.85 nb ÷ (s in GeV²); free quarks would give R = 3Σq_f²

The iε decides where the logarithm is complex

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 13.03 left Π̂₂(q²) = −(2α/π)∫₀¹ dx x(1 − x) ln[m²/(m² − x(1 − x)q² − iε)], the −iε being the Feynman propagators' m² − iε carried through the Feynman parameter. Since x(1 − x) ≤ 1/4, Δ = m² − x(1 − x)q² is positive at every x when q² is below 4m²: Π̂₂ is real in the whole spacelike region and up to threshold. Above 4m², Δ is negative on the window between x₋ and x₊, x± = (1 ± β)/2 with β = √(1 − 4m²/q²), and there ln(Δ − iε) = ln|Δ| − iπ. So Im Π̂₂ = −2α times ∫ x(1 − x) dx over the window. With x = 1/2 + y the integrand is 1/4 − y² for |y| below β/2, which integrates to β(3 − β²)/12, and since 3 − β² = 2(1 + 2m²/q²), Im Π̂₂(q² + iε) = −(α/3)β(1 + 2m²/q²), derived. At q² = 9m², β = 0.745 and Im Π̂₂ = −2.216 × 10⁻³, which a direct numerical integration of the logarithm reproduces. The sign is not a convention: the opposite iε would make the next section's cross-section negative.

Unitarity reads the imaginary part as a rate

Topic 5.07's optical theorem, Im ℳ(s, 0) = 2E_cm p_cm σₜₒₜ, counts what a forward amplitude can produce, final state by final state. Take e⁺e⁻ → e⁺e⁻ forward through one s-channel photon, electrons massless, with the bubble of a heavier lepton f on the photon. The resummed propagator multiplies the tree by 1/(1 − Π̂), and the spin average of the two currents, (1/4)Σ J^μ J_μ† = −s, makes the forward amplitude ℳ = −e²/(1 − Π̂(s)); at order e⁴ its imaginary part is −e² Im Π̂₂^(f)(s), derived. That bubble is the only order-e⁴ diagram with an f f̄ intermediate state, so with 2E_cm p_cm = s the theorem gives σ(e⁺e⁻ → f f̄) = −(4πα/s) Im Π̂₂^(f)(s) = (4πα²/3s)β(1 + 2m²/s). Topic 9.06 squared the tree diagram and found 4πα²/(3s) × β(3 − β²)/2, whose high-energy form PETRA tested, and (3 − β²)/2 = 1 + 2m²/s: the loop's cut and the tree's rate agree, checked. The −iπ of the logarithm was the pair being made on shell.

Threshold, high energy and the spin of the pair

Π̂₂ is a pure number, so −(4πα/s) Im Π̂₂ has dimension GeV⁻², an area, 0.3894 mb per GeV⁻². Just above threshold β(1 + 2m²/s) ≈ 3β/2: the rate rises linearly in β, the mark of an s wave, since a 1⁻⁻ photon makes the pair in ³S₁ with no centrifugal barrier. A spin-0 loop, quoted from scalar QED, gives −Im Π̂ = (α/12)β³ instead, a p wave, which is the point-like π⁺π⁻ rate R = β³/4: the threshold power measures the spin of what is made. Far above threshold β(1 + 2m²/s) = 1 − 6m⁴/s² + …, so −Im Π̂₂ → α/3 and σ → 86.85 nb ÷ (s in GeV²), with no m²/s correction at all. At q² = 0 there is no imaginary part: a real photon cannot decay into a pair. The one-loop rate fails near threshold, where Coulomb attraction of the slow pair multiplies it by the Sommerfeld factor (πα/β)/(1 − e^(−πα/β)), quoted, about 1 + πα/(2β). BESIII's τ-mass scan fitted exactly this threshold, Coulomb and radiative factors included, and found m_τ = 1776.9 MeV, quoted.

Analytic off the cut, so Cauchy's theorem applies

Continue q² to a complex z. The logarithm ln[1 − x(1 − x)z/m²] is singular only where its argument is real and not positive, which happens only for real z ≥ 4m², so Π̂₂(z) is analytic in the plane cut along the real axis from 4m² to ∞. It is real below threshold, so Π̂₂(z*) = Π̂₂(z)*, and the jump across the cut is Π̂₂(s + iε) − Π̂₂(s − iε) = 2i Im Π̂₂(s + iε). Apply Cauchy's theorem to Π̂₂(z)/z, which is regular at z = 0 because Π̂₂(0) = 0, on a contour that hugs the cut and closes on a large circle. Π̂₂ grows only like ln|z|, so the circle contributes nothing, and Π̂₂(q²) = (q²/π)∫ ds Im Π̂₂(s)/[s(s − q² − iε)], s from 4m² to ∞, derived. Without the division by z the integral would be ∫ ds Im Π̂₂(s)/(s − q²), which diverges logarithmically because Im Π̂₂ → −α/3: topic 13.02's pole in another form. Dividing by z is subtracting at q² = 0, and Π̂₂(0) = 0 is topic 13.03's definition of α. One subtraction, one counterterm, and no regulator anywhere.

Rebuilding the real part, three ways

Each limit of the dispersion relation reproduces something already derived. Small q²: with 1/(s − q²) ≈ 1/s, the slope at q² = 0 is (1/π)∫ ds Im Π̂₂/s² = −α/(15πm²), topic 13.02's result from the Feynman-parameter logarithm. Large spacelike q² = −Q²: the integrand per unit ln s is (α/3π)β(1 + 2m²/s)Q²/(s + Q²), zero at s = 4m², close to α/3π from a few times 4m² up to s ≈ Q² and small beyond, so Π̂₂(−Q²) ≈ (α/3π)[ln(Q²/4m²) − 0.280] = (α/3π)[ln(Q²/m²) − 5/3]. The logarithm is the length of that window in ln s, and 5/3 = ln 4 + 0.280: the late start at 4m², plus ∫[1 − β(1 + 2m²/s)] d ln s = 0.280 for the slow opening of the channel. At Q/m = 10 the integral gives 2.319 × 10⁻³, equal to the closed form, while the logarithm alone gives 2.275 × 10⁻³. Timelike, above threshold, the principal value gives Re Π̂₂ = (α/3π)[−5/3 − 4m²/s + β(1 + 2m²/s) ln((1 + β)/(1 − β))], which reaches −8α/(9π) = −2.06 × 10⁻³ at threshold with a square-root cusp, computed: a channel opening leaves a cusp in the real part.

Hadrons enter as measured R(s)

No perturbative quark loop describes what a virtual photon of 1 GeV makes: pion pairs and the ρ, ω and φ resonances. The dispersion relation never needs the loop, only Im Π̂, and the optical-theorem argument with any final state gives Im Π̂_had(s) = −(α/3)R(s), derived, with R(s) = σ(e⁺e⁻ → hadrons)/(4πα²/3s). So Π̂_had(q²) = −(αq²/3π)∫ ds R(s)/[s(s − q²)] from the hadronic threshold up, with R from experiment. Free quarks would give R = 3Σq_f²: 2 below charm, 10/3 from charm up to the open-bottom threshold at 10.56 GeV, 11/3 above it; the measured values are about 2.2 at 2 to 3 GeV and 3.56 at 10.5 GeV, just below open bottom, quoted, the excess over 2 and 10/3 being roughly QCD's factor 1 + αₛ/π. The kernel decides which data matter. At small |q²| it weights by 1/s², so the region below 1 GeV dominates, and the hadronic part of the muon's g − 2, about 7 × 10⁻⁸ (quoted), is computed from e⁺e⁻ data with a kernel of the same kind. At Q = M_Z every e-fold of s up to M_Z² counts about equally, and topic 13.05 turns that integral into 1/α(M_Z²).

Change one variable at a time

Make the relationship visible.

Interactive model

Spacelike Q: log₁₀(Q/m)2.0

Dot position: log₁₀(√s/m)0.50

Raise Q/m from 1 to 3000: the dashed window stretches right and its area grows by one unit per e-fold of Q², so Π̂₂(−Q²) meets the large-Q² form once Q/m passes about 10. Drag the dot below 2 and both −Im Π̂₂ and the μ⁺μ⁻ cross-section fall to exactly zero.

−Im Π̂₂ AT THE DOT2.261 × 10⁻³

σ(e⁺e⁻ → μ⁺μ⁻) AT THE DOT723.171 nb

Π̂₂(−Q²) FROM THE CUT5.841 × 10⁻³

LARGE-Q² LOG FORM5.841 × 10⁻³

Live interpretation−Im Π̂₂ AT THE DOT: 2.261 × 10⁻³. σ(e⁺e⁻ → μ⁺μ⁻) AT THE DOT: 723.171 nb. Π̂₂(−Q²) FROM THE CUT: 5.841 × 10⁻³. LARGE-Q² LOG FORM: 5.841 × 10⁻³

Catch the common trap

Explain before calculating.

Deep-inelastic electron-proton scattering exchanges a photon with q² = −(100 GeV)², so |q²| is about 10¹⁰ times 4mₑ². At one loop, what is the imaginary part of the electron loop's Π̂₂ at this q²?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA collider runs at √s = 4.00 GeV, above the τ-pair threshold. Find Im Π̂₂ for the τ loop (m_τ = 1.77686 GeV) and for the muon loop at this energy, then use the optical theorem to obtain σ(e⁺e⁻ → τ⁺τ⁻) in nanobarns and compare it with the point-like 4πα²/(3s).

s = 16.00 GeV² and 4m_τ² = 12.629 GeV², so the τ cut is open. β = √(1 − 12.629/16.00) = √0.21069 = 0.4590, and 1 + 2m_τ²/s = 1 + 6.3145/16.00 = 1.3947.
Im Π̂₂ = −(α/3)β(1 + 2m²/s) = −(2.4325 × 10⁻³)(0.4590)(1.3947) = −1.557 × 10⁻³ for the τ loop.
For the muon, m_μ²/s = 6.98 × 10⁻⁴ and β(1 + 2m²/s) = 1 − 6(m²/s)² = 0.999997, so Im Π̂₂ = −2.432 × 10⁻³ = −α/3: this far above threshold the mass has dropped out to order m⁴/s².
Optical theorem: σ = −(4πα/s) Im Π̂₂ = 0.091701 × 1.5572 × 10⁻³/(16.00 GeV²) = 8.925 × 10⁻⁶ GeV⁻², and with 0.3894 mb GeV² this is 3.475 × 10⁻⁶ mb = 3.475 nb.
The point-like value is 86.85 nb/16.00 = 5.428 nb, so τ pairs come at 0.640 of it, which is β(3 − β²)/2 = 0.4590 × 2.7893/2 = 0.6402. Checks: σ is an area; it vanishes linearly in β at threshold and tends to the point-like value far above it.

AnswerIm Π̂₂ = −1.56 × 10⁻³ for the τ loop and −2.43 × 10⁻³ = −α/3 for the muon loop; σ(e⁺e⁻ → τ⁺τ⁻) = 3.48 nb, 0.640 of the point-like 5.43 nb.

MediumExpand the once-subtracted dispersion relation about q² = 0 to find the slope dΠ̂₂/dq² at q² = 0 for a lepton of mass m, and compare it with topic 13.02. Then find what fraction of the slope comes from pairs with √s/m below 3 and below 10, and compare with the share of Π̂₂(−Q²) at Q/m = 100 that comes from √s/m below 10.

For |q²| well below 4m², 1/(s − q²) = 1/s + q²/s² + …, so Π̂₂(q²) = (q²/π)∫ ds Im Π̂₂(s)/s² + O(q⁴). The slope is (1/π)∫ ds Im Π̂₂(s)/s² = −(α/3π)∫ ds β(1 + 2m²/s)/s², s from 4m² to ∞.
Put u = 4m²/s, so ds/s² = −du/(4m²), β = √(1 − u) and 1 + 2m²/s = 1 + u/2. The integral becomes (1/4m²)∫₀¹ √(1 − u)(1 + u/2) du = (1/4m²)(2/3 + 2/15) = 1/(5m²), using ∫₀¹ u√(1 − u) du = 4/15.
So the slope is −α/(15πm²) and Π̂₂ ≈ −αq²/(15πm²), exactly topic 13.02's result from expanding the Feynman-parameter logarithm. Checks: it has dimension GeV⁻², and it vanishes as m → ∞, a heavy loop decoupling. For the electron it is −5.93 × 10⁻⁴ MeV⁻².
Cutting the u integral at u = 4/9 (√s/m = 3) and at u = 0.04 (√s/m = 10) and integrating numerically, the region below √s/m = 3 supplies 46.0% of the slope and the region below 10 supplies 95.0%.
At Q/m = 100 the integrand per unit ln s is (α/3π)β(1 + 2m²/s)Q²/(s + Q²). The whole integral is 7.544 in units of α/3π, so Π̂₂ = 5.841 × 10⁻³; √s/m below 10 supplies 2.929 of it, 38.8%, and √s/m below 3 only 7.6%.
The 1/s² weight of the slope favours the threshold, while at large Q² every e-fold of s up to Q² counts about equally: low-q² quantities are set by low-energy pairs, large-Q² ones by the whole spectrum.

AnswerSlope −α/(15πm²), topic 13.02's result. 46% of it comes from √s/m below 3 and 95% from below 10, against 7.6% and 38.8% of Π̂₂(−Q²) = 5.84 × 10⁻³ at Q/m = 100.

HardA narrow vector meson V of mass M and electronic width Γₑₑ adds ∫σ(e⁺e⁻ → V → anything) ds = 12π²Γₑₑ/M, quoted from the spin-1 Breit-Wigner formula. Find its Im Π̂ and, by the dispersion relation, its Π̂(q²). For the J/ψ (M = 3.0969 GeV, Γₑₑ = 5.53 keV), find the shift of 1/α at spacelike Q = M_Z = 91.19 GeV, and Π̂_V at timelike √q² = 3.000 GeV and 3.200 GeV.

R(s) = 3sσ/(4πα²). Away from its peak a resonance this narrow acts as a δ-function: R(s) = (3M²/4πα²)(12π²Γₑₑ/M)δ(s − M²) = (9πMΓₑₑ/α²)δ(s − M²), so Im Π̂_V(s) = −(α/3)R(s) = −(3πMΓₑₑ/α)δ(s − M²).
The dispersion relation does the integral at once: Π̂_V(q²) = (q²/π)(−3πMΓₑₑ/α)/[M²(M² − q²)] = −(3Γₑₑ/αM) q²/(M² − q²), derived. Checks: Γₑₑ/M is dimensionless, Π̂_V(0) = 0 as the definition of α requires, and for |q²| ≪ M² it falls as q²/M², a heavy state decoupling.
Numerically 3Γₑₑ/(αM) = 3 × 5.53 × 10⁻⁶ GeV × 137.036/3.0969 GeV = 7.341 × 10⁻⁴.
At q² = −M_Z² = −8315.6 GeV², Π̂_V = 7.341 × 10⁻⁴ × 8315.6/(9.591 + 8315.6) = 7.333 × 10⁻⁴. Since 1/α_eff = (1 − Π̂)/α, the J/ψ lowers 1/α(M_Z²) by 7.333 × 10⁻⁴ × 137.036 = 0.100, about 2.7% of the 3.78 by which hadrons must take the three-lepton 132.73 to the measured 128.95 (topic 13.05).
At √q² = 3.000 GeV, 97 MeV below the peak and about a thousand total widths (Γ = 92.6 keV) away: Π̂_V = −7.341 × 10⁻⁴ × 9.000/0.591 = −0.0112. At 3.200 GeV: Π̂_V = +7.341 × 10⁻⁴ × 10.240/0.649 = +0.0116.
The real part changes sign through the resonance and near it is fifteen times its value at M_Z: the absorptive peak in R has a dispersive partner that lowers α_eff just below the J/ψ and raises it just above. Within a few widths of the peak the δ-function must give way to the full Breit-Wigner shape.

AnswerIm Π̂_V = −(3πMΓₑₑ/α)δ(s − M²) and Π̂_V(q²) = −(3Γₑₑ/αM)q²/(M² − q²), with 3Γₑₑ/αM = 7.34 × 10⁻⁴. The J/ψ lowers 1/α(M_Z²) by 0.100; Π̂_V = −0.0112 at 3.000 GeV and +0.0116 at 3.200 GeV.