University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.8
Optical theorem at one loop: the φ⁴ bubble
Unitarity says the one-loop φ⁴ amplitude owes an exact, finite imaginary part, and this topic collects the debt. One contour integral leaves a single energy denominator whose iε becomes a delta function, no renormalisation is needed, and the answer matches the tree amplitude squared down to the last factor of ½.
Build the model
Connect the measurement to the mechanism.
At order λ² the forward φφ → φφ amplitude must acquire an imaginary part: topic 5.07's unitarity relation sets 2 Im M equal to ½∫dΠ₂|Mₜᵣₑₑ|², which is of order λ² and positive above threshold, yet the tree amplitude −λ is real. The only diagram that can supply it is the one-loop s-channel bubble, and power counting says that bubble diverges logarithmically. The resolution is this topic's argument.
Integrate the loop energy l⁰ by residues in the centre-of-mass frame and exactly one energy denominator survives that can vanish, √s − 2Eₗ + iε. The iε splits it into a principal value, which carries the whole divergence into the real part, and −iπδ(√s − 2Eₗ), which forces both internal particles onto their mass shell, back to back with momentum p = √(s/4 − m²). An integral over a sphere of radius p cannot diverge, so the imaginary part is finite and exact, (λ²/32π)√(1 − 4m²/s), derived here, and twice it equals λ² over two-body phase space with the ½ for identical particles: unitarity checked at one loop.
Below s = 4m² no real pair exists and the imaginary part is zero, and the forward t- and u-channel bubbles, whose invariants sit below 4m², stay real. Cutkosky's rule extends the cut to any diagram, stated without proof. What the cut cannot give is the real part's divergent constant, which needs Unit 6's regularisation and renormalisation.
- Simple definition
- The imaginary part of the φ⁴ bubble is the piece of the one-loop amplitude in which both internal particles are on shell; it is finite, vanishes below s = 4m², and unitarity fixes it at half of λ² integrated over two-body phase space.
- Example
- With m = 0.50 GeV and √s = 2.00 GeV, each internal particle carries p = √(1.00 − 0.25) GeV = 0.866 GeV, so the dimensionless factor √(1 − 4m²/s) = 2p/√s = 0.866 and Im Mₛ = (λ²/32π)(0.866) = 8.6 × 10⁻³ λ², under 1% of the tree amplitude's magnitude when λ = 1.
Four powers of l above and four below: logarithmically divergent by power counting. This is the integral whose imaginary part the topic extracts.
λ dimensionless; m, l and P = p₁ + p₂ in GeV with P² = s; the ½ is the symmetry factor of the two identical internal lines
Only the first denominator can vanish, and only when √s ≥ 2m; the second is real and nonzero for every |l|.
Eₗ = √(|l|² + m²) in GeV; closing below picks the poles l⁰ = Eₗ − iε and l⁰ = √s + Eₗ − iε. Derived.
The principal value is real and carries the divergence into Re Mₛ; the delta function puts both internal lines on shell and builds Im Mₛ.
PV the principal value; ε → 0⁺ taken after the |l| integral; both sides in GeV⁻¹. Quoted: the Sokhotski–Plemelj identity.
Finite although the loop is not; it opens in proportion to p at threshold and saturates at λ²/(32π) = 9.95 × 10⁻³ λ² once m²/s → 0.
s = Ecm² in GeV²; the same as λ²p/(16π√s), with p = √(s/4 − m²) the on-shell loop momentum. Derived.
Twice the bubble's Im Mₛ meets it exactly: topic 5.07's theorem at order λ², equivalently Im M = 2Ecm p σₜₒₜ with σₜₒₜ = λ²/(32πs).
½ = 1/2! for the identical φφ pair; ∫dΠ₂ = p/(4π√s) from topic 5.02; |Mₜᵣₑₑ|² = λ²; all dimensionless. Derived.
Cutting the bubble's two lines gives (−iλ)²(½)(2π)²∫d⁴l δδ/(2π)⁴ = −λ²p/(8π√s) = −2 Im Mₛ: the residue calculation in one line.
k the cut line's four-momentum in GeV, kept apart from the on-shell momentum magnitude p; the cuts sum to Disc(iM) = −2 Im M, as Disc M = M(s + iε) − M(s − iε) = 2i Im M
Why the bubble must have an imaginary part
Work in natural units, ħ = c = 1, with the metric (+, −, −, −) and L = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴. At tree level φφ → φφ has M = −λ, which is real. Topic 5.07's unitarity relation, 2 Im M(forward) = Σf ∫dΠf |M(i → f)|², has a right-hand side of order λ² that is strictly positive above threshold, so the amplitude needs an imaginary part at order λ² that the tree cannot supply. At one loop topic 3.05's rules give three bubbles, one in each of s, t and u. The s-channel one, with P = p₁ + p₂ flowing through it, is iMₛ = (λ²/2)∫d⁴l/(2π)⁴ 1/[(l² − m² + iε)((P − l)² − m² + iε)], the ½ being the symmetry factor of its two identical internal lines. Count powers: d⁴l above, l⁴ below, so the integral diverges logarithmically. Unitarity demands a finite number at order λ² from a divergent integral, and the rest of the topic shows how it gets one.
The loop energy by residues
Go to the centre-of-mass frame, P = (√s, 0), and integrate l⁰ first, as topic 1.06 did for the propagator. With Eₗ = √(|l|² + m²), the first propagator has poles at l⁰ = Eₗ − iε and −Eₗ + iε, the second at l⁰ = √s + Eₗ − iε and √s − Eₗ + iε: two just below the real axis, two just above. Close the contour in the lower half-plane, which costs −2πi. The pole at l⁰ = Eₗ − iε has residue 1/(2Eₗ) × 1/[√s(√s − 2Eₗ + iε)], the iε arriving because the second propagator is evaluated at a complex l⁰; the pole at √s + Eₗ − iε gives 1/(2Eₗ) × 1/[√s(√s + 2Eₗ)]. Hence Mₛ = −(λ²/2)∫d³l/(2π)³ [1/(√s − 2Eₗ + iε) + 1/(√s + 2Eₗ)]/(2Eₗ√s). Every factor is real except one denominator, which vanishes when √s = 2Eₗ: there the upper pole √s − Eₗ + iε sits directly above the lower pole Eₗ − iε and pinches the contour between them.
The iε turns a denominator into a delta function
As a distribution, 1/(x + iε) tends to PV(1/x) − iπδ(x) as ε → 0⁺, and because the limit is taken after the |l| integral the δ keeps a finite weight. Apply it with x = √s − 2Eₗ. The principal value leaves a real integrand: at large |l| the bracket falls as −√s/(2|l|²) while d³l/(2Eₗ√s) grows as |l| d|l|, so Re Mₛ gains λ²/(16π²|l|) per unit |l| and grows as (λ²/16π²) ln Λ with a cutoff Λ. The whole divergence sits in the real part. The δ part gives Im Mₛ = (πλ²/2)∫d³l/(2π)³ δ(√s − 2Eₗ)/(2Eₗ√s). Read what it enforces: l⁰ = Eₗ from the residue, and (P − l)⁰ = √s − Eₗ = Eₗ with |P − l| = |l|, so l² = (P − l)² = m². Both internal lines are on shell, real particles flying back to back with |l| = p = √(s/4 − m²). The imaginary part of the loop is the amplitude to make a real intermediate pair.
Evaluating the cut: finite, and zero below threshold
In spherical coordinates the δ needs its Jacobian: d(2Eₗ)/d|l| = 2|l|/Eₗ, so δ(√s − 2Eₗ) = (Eₗ/2p)δ(|l| − p). Then ∫d³l/(2π)³ δ(√s − 2Eₗ)/(2Eₗ√s) = (4π/8π³)p²(Eₗ/2p)/(2Eₗ√s) = p/(8π²√s), and Im Mₛ = λ²p/(16π√s) = (λ²/32π)√(1 − 4m²/s), using 2p/√s = √(1 − 4m²/s). Derived. Nothing needs cutting off: the δ lives on a sphere of radius p, which no ultraviolet divergence can reach. Dimensions: λ is dimensionless in four dimensions and d³l δ/(Eₗ√s) is GeV³ × GeV⁻¹ × GeV⁻², so Im Mₛ is a pure number, as a 2 → 2 amplitude must be. Limits: below √s = 2m, √s − 2Eₗ ≤ √s − 2m is negative for every |l| and Im Mₛ = 0; just above threshold it opens in proportion to p, the s-wave onset of phase space; for m²/s → 0 it saturates at λ²/(32π) = 9.95 × 10⁻³ λ². With m = 0.50 GeV it is already half that by √s = 4m/√3 = 1.15 GeV.
Unitarity checked at order λ²
Now the right-hand side. The only final state at order λ² is φφ, with |Mₜᵣₑₑ|² = λ², two-body phase space ∫dΠ₂ = p/(4π√s) from topic 5.02, and a factor ½ because the final particles are identical and integrating over every angle counts each configuration twice. So ½∫dΠ₂|Mₜᵣₑₑ|² = λ²p/(8π√s), exactly twice Im Mₛ. In topic 5.07's form, with pcm = p because the beams and the loop share the mass m: σₜₒₜ = λ²/(32πs) from topic 5.04, and 2Ecm p σₜₒₜ = λ²p/(16π√s) = Im Mₛ. The t- and u-channel bubbles must add nothing, and they do not. Each is the same function of its own invariant q², so any frame will do: for q² ≤ 0 take q⁰ = 0 (q = 0 itself when q² = 0), where the two lower-half-plane residues combine into −1/[2E₁E₂(E₁ + E₂)], E₁ and E₂ the energies of the two internal lines, which never vanishes; for q² between 0 and 4m² the centre-of-mass argument applies. Going forward, t = 0 and u = 4m² − s ≤ 0. The ½ for identical particles, mirrored by the loop's symmetry factor, is the factor most often lost.
Cutkosky's rule, and what the cut leaves open
What the residues did is general. Cutkosky's rule, stated here and not proved: cut a diagram in every way that separates initial from final particles through lines that can go on shell together, replace each cut propagator i/(k² − m² + iε), k the line's four-momentum, by 2πδ(k² − m²)θ(k⁰), and sum; the result is Disc(iM) = −2 Im M. For the bubble, (−iλ)²(½)(2π)²∫d⁴l δδ/(2π)⁴ = −λ²p/(8π√s), the residue calculation in one line; at one loop each cut is a product of tree amplitudes integrated over phase space. The cut even fixes the logarithm. At s ≫ m² the bubble behaves as (λ²/32π²) ln(Λ²/(−s − iε)), quoted from analyticity: its real part falls as −(λ²/32π²) ln s, and its imaginary part is π × λ²/(32π²) = λ²/(32π), the plateau just derived. What no cut fixes is the constant beside the logarithm, where the divergence lands; Unit 6 absorbs it into λ with a counterterm. No experiment measures a φ⁴ coupling, but Unit 13 meets the same cut as pair creation in the photon's vacuum polarisation, where measured rates enter.
Change one variable at a time
Make the relationship visible.
Shrink ε: the spike narrows and grows while its area closes in on the Im Mₛ ÷ λ² readout. Then drag √s below 2m: p² turns negative, the spike sinks to a low hump that vanishes as ε → 0, and the marker drops onto the axis.
ON-SHELL MOMENTUM p0.866 GeV
Im Mₛ ÷ λ²8.61 × 10⁻³
2√s p σₜₒₜ ÷ λ²8.61 × 10⁻³
σₜₒₜ ÷ λ²0.97 μb
Live interpretationON-SHELL MOMENTUM p: 0.866 GeV. Im Mₛ ÷ λ²: 8.61 × 10⁻³. 2√s p σₜₒₜ ÷ λ²: 8.61 × 10⁻³. σₜₒₜ ÷ λ²: 0.97 μb
Catch the common trap
Explain before calculating.
In φ⁴ theory with m = 0.50 GeV, what is the imaginary part of the forward φφ → φφ amplitude at order λ² when √s = 1.25 GeV?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTake φ⁴ theory with m = 0.40 GeV and λ = 1.2 at √s = 1.00 GeV. Find the on-shell momentum p of the cut and Im Mₛ from the s-channel bubble, then check it against 2Ecm p σₜₒₜ using σₜₒₜ = λ²/(32πs) from topic 5.04, and give σₜₒₜ in microbarns.
- Threshold first: 2m = 0.80 GeV lies below √s = 1.00 GeV, so the cut is open. On shell each internal particle has Eₗ = √s/2 = 0.500 GeV and p = √(0.500² − 0.400²) GeV = 0.300 GeV.
- The kinematic factor is √(1 − 4m²/s) = √(1 − 0.64) = 0.600, which equals 2p/√s = 0.600, as it must.
- Im Mₛ = (λ²/32π) × 0.600 = (1.44/100.53) × 0.600 = 1.432 × 10⁻² × 0.600 = 8.59 × 10⁻³, a pure number as a 2 → 2 amplitude should be.
- Cross-section: σₜₒₜ = λ²/(32πs) = 1.44/(100.53 × 1.00 GeV²) = 1.432 × 10⁻² GeV⁻², so 2Ecm p σₜₒₜ = 2 × 1.00 GeV × 0.300 GeV × 1.432 × 10⁻² GeV⁻² = 8.59 × 10⁻³, equal to Im Mₛ.
- Restore units with (ħc)² = 0.3894 GeV² mb: σₜₒₜ = 1.432 × 10⁻² × 0.3894 mb = 5.58 × 10⁻³ mb = 5.58 μb. Against the tree amplitude's magnitude λ = 1.2 the imaginary part is 0.72%, so perturbation theory is in good shape.
Answerp = 0.300 GeV and Im Mₛ = 8.59 × 10⁻³ (dimensionless); σₜₒₜ = 5.58 μb, and 2Ecm p σₜₒₜ = 8.59 × 10⁻³ reproduces Im Mₛ.
MediumIn φ⁴ theory with m = 0.30 GeV and λ = 0.80, explain why the bubble, as a function of its invariant s, is real at √s = 0.50 GeV, a value it can take inside a larger diagram. Then at √s = 0.65 GeV carry its cut through by hand: the on-shell momentum, the Jacobian of δ(√s − 2Eₗ), the radial integral and Im Mₛ. Finally show that the forward t- and u-channel bubbles add nothing there.
- At √s = 0.50 GeV every internal line has Eₗ ≥ m, so √s − 2Eₗ ≤ 0.50 − 0.60 = −0.10 GeV for every |l|. The δ has no zero to sit on and Im Mₛ = 0 exactly: no real φφ pair has an invariant mass of 0.50 GeV.
- At √s = 0.65 GeV the δ fires at Eₗ = 0.325 GeV, so p = √(0.325² − 0.300²) GeV = √0.015625 GeV = 0.125 GeV.
- Jacobian: d(2Eₗ)/d|l| = 2p/Eₗ = 0.250/0.325 = 0.769 at |l| = p, so δ(√s − 2Eₗ) = δ(|l| − p)/0.769.
- Radial integral: ∫d³l/(2π)³ δ(√s − 2Eₗ)/(2Eₗ√s) = (1/2π²) × p² × (1/0.769)/(2Eₗ√s) = p/(8π²√s) = 0.125/(78.96 × 0.65) = 2.436 × 10⁻³.
- Im Mₛ = (πλ²/2) × 2.436 × 10⁻³ = 1.005 × 2.436 × 10⁻³ = 2.45 × 10⁻³. The closed form agrees: (0.64/32π) × 2p/√s = 6.37 × 10⁻³ × 0.385 = 2.45 × 10⁻³.
- Going forward, t = 0 and u = 4m² − s = 0.36 − 0.4225 = −0.0625 GeV², both below 4m² = 0.36 GeV², so neither bubble can put a pair on shell and both are real: Im M(forward) = Im Mₛ.
AnswerReal at 0.50 GeV; at 0.65 GeV, p = 0.125 GeV and Im M(forward) = Im Mₛ = 2.45 × 10⁻³, already 38% of its high-energy value λ²/(32π) = 6.37 × 10⁻³ only 50 MeV above threshold.
HardAdd a second real scalar χ of mass m′ = 0.60 GeV to φ⁴ theory (m = 0.20 GeV, λ = 1.5) through the interaction −(g/4)φ²χ² with g = 2.0. At √s = 2.0 GeV find Im M for forward φφ → φφ at second order in the couplings, then verify the optical theorem by computing σ(φφ → φφ) and σ(φφ → χχ) in microbarns.
- Feynman rule: −(g/4)φ²χ² gives the vertex −i(g/4)(2!)(2!) = −ig. A χ loop between two such vertices has symmetry factor ½, like the φ bubble, so it is the same integral with λ → g and m → m′: its imaginary part is (g²/32π)√(1 − 4m′²/s) above √s = 2m′ = 1.2 GeV, and its forward t and u versions are real.
- At s = 4.00 GeV²: √(1 − 4m²/s) = √0.96 = 0.9798 and √(1 − 4m′²/s) = √0.64 = 0.800. So Im M = (2.25 × 0.9798 + 4.00 × 0.800)/100.53 = 2.193 × 10⁻² + 3.183 × 10⁻² = 5.376 × 10⁻².
- Momenta: p = √(1.00 − 0.04) GeV = 0.9798 GeV for the φφ pair and p′ = √(1.00 − 0.36) GeV = 0.800 GeV for χχ. Elastic: σ(φφ → φφ) = λ²/(32πs) = 2.25/(100.53 × 4.00) GeV⁻² = 5.595 × 10⁻³ GeV⁻² = 2.18 μb.
- Inelastic, with tree amplitude −g and ½ for the identical χχ: σ = ½ × g²/(4p√s) × p′/(4π√s) = g²p′/(32πs p) = (4.00 × 0.800)/(100.53 × 4.00 × 0.9798) GeV⁻² = 8.122 × 10⁻³ GeV⁻² = 3.16 μb.
- Optical theorem: σₜₒₜ = (5.595 + 8.122) × 10⁻³ GeV⁻² = 1.3717 × 10⁻² GeV⁻², so 2Ecm p σₜₒₜ = 2 × 2.00 × 0.9798 × 1.3717 × 10⁻² = 5.376 × 10⁻², equal to Im M. Channel by channel, 2Ecm p σ(φφ) = 2.193 × 10⁻² and 2Ecm p σ(χχ) = 3.183 × 10⁻², each matching its own cut: the χχ channel is UAQM-02.09's inelastic channel, now read off a loop.
- Below √s = 1.2 GeV the χχ cut closes: at √s = 1.0 GeV only the φ bubble contributes, Im M = (2.25/32π)√0.84 = 2.05 × 10⁻², and σ(φφ → χχ) = 0.
AnswerIm M = 5.38 × 10⁻², of which 2.19 × 10⁻² comes from the φφ cut and 3.18 × 10⁻² from the χχ cut; σ(φφ → φφ) = 2.18 μb and σ(φφ → χχ) = 3.16 μb, so σₜₒₜ = 5.34 μb and 2Ecm p σₜₒₜ = 5.38 × 10⁻² = Im M.