University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.7

Unitarity of S and the optical theorem

This is the one result in the unit that no Feynman rule can break, because it is only the conservation of probability. It ties the forward elastic amplitude to the total cross-section into everything, and from here on it is the check every loop must pass, the reason a real tree is never complete, and the ceiling on any partial wave.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Probability is conserved, and in field theory that is S†S = 1. Writing S = 1 + iT turns it into −i(T − T†) = T†T: the anti-Hermitian part of T, linear on the left, equals something quadratic and positive on the right. Take the forward matrix element in a two-particle state a, insert a complete set of out states between T† and T, strip the common (2π)⁴δ⁴, and the left side becomes 2 Im ℳ of the forward elastic amplitude while the right side becomes Σ_f ∫dΠ_f |ℳ(a → f)|², which topic 5.01's flux turns into 4E_cm p_cm σₜₒₜ.

So Im ℳ(s, 0) = 2E_cm p_cm σₜₒₜ: the amplitude for carrying straight on knows the probability of everything else, because whatever leaves the beam is removed by forward interference between the scattered and incident waves. With f = ℳ/(8π√s) it is quantum mechanics' σₜₒₜ = (4π/k) Im f(0), now for any final state, particle creation included. The same identity organises perturbation theory, since a real tree forces the loops one order later to supply an imaginary part equal to the trees squared, and it confines each partial wave to the unitarity circle while only elastic scattering is possible, and to its interior once inelastic channels open.

It rests on topic 4.09's asymptotic completeness, assumed rather than proved; it fixes only the imaginary part; and a massless exchange, as in Coulomb scattering, makes both sides infinite.

Simple definition: The optical theorem says that, because S is unitary, the imaginary part of the forward elastic amplitude is fixed by the total cross-section into every final state: Im ℳ(s, 0) = 2E_cm p_cm σₜₒₜ.
Example: Two spinless particles of mass 0.90 GeV colliding at E_cm = 3.0 GeV have p_cm = 1.2 GeV. If σₜₒₜ = 25 mb = 64.2 GeV⁻², then Im ℳ(s, 0) = 2 × 3.0 × 1.2 × 64.2 = 462, however the 25 mb divides between elastic and inelastic channels.

Unitarity of S written for T−i(T − T†) = T†T

Derived from S†S = 1. Linear in T on the left, quadratic and positive on the right, so T's anti-Hermitian part at order g⁴ is fixed by T at order g².

S = 1 + iT on Fock space; ⟨b|iT|a⟩ = (2π)⁴δ⁴(pₐ − p_b) iℳ(a → b), with ℳ of mass dimension 4 − n for n external legs

Forward amplitude against every final state2 Im ℳ(a → a) = Σ_f ∫dΠ_f |ℳ(a → f)|²

Derived by inserting a complete set of out states and stripping the common (2π)⁴δ⁴. Every term is dimensionless, and the sum needs topic 4.09's asymptotic completeness.

a two particles in; dΠ_f topic 5.02's phase space, 1/n! for n identical; ℳ(a → f) of mass dimension 2 − n_f, dΠ_f of 2n_f − 4

The optical theoremIm ℳ(s, 0) = 2E_cm p_cm σₜₒₜ

Derived with topic 5.01's flux 4E_cm p_cm. Never negative, and it counts every channel, particle creation included; at high energy Im ℳ(s, 0) = sσₜₒₜ.

E_cm, p_cm in GeV; σₜₒₜ in GeV⁻² (mb ÷ 0.3894); ℳ dimensionless; 2E_cm p_cm = 2√((k₁⋅k₂)² − m₁²m₂²), which tends to s

The quantum-mechanical formσₜₒₜ = (4π/p_cm) Im f(0)

UQMII-15.05's partial-wave theorem, derived there for elastic scattering by a potential, here derived for any final state, including ones a potential cannot produce.

f = ℳ/(8π√s) in GeV⁻¹ (× 0.1973 for fm), so |f|² = dσ/dΩ for elastic scattering; p_cm = k in GeV

One partial wave, elastic and inelasticIm aₗ = |aₗ|² + (1 − ηₗ²)/4

Shown by projecting onto Pₗ. aₗ rides |aₗ − i/2| = 1/2 while only elastic scattering is open and lies inside above, so |aₗ| ≤ 1 and |Re aₗ| ≤ 1/2.

aₗ = (ηₗ e^(2iδₗ) − 1)/(2i), dimensionless; δₗ the phase shift; 0 ≤ ηₗ ≤ 1, with ηₗ = 1 below inelastic thresholds

The forward elastic peakdσₑₗ/dt(t = 0) = (1 + ρ²) σₜₒₜ²/(16π)

Derived from dσ/dt = |ℳ|²/(64π s p_cm²) for any masses. Read backwards, the forward elastic rate measures σₜₒₜ without the luminosity.

ρ = Re ℳ/Im ℳ at t = 0, dimensionless; σₜₒₜ in GeV⁻² gives GeV⁻⁴, and × 0.3894 mb GeV² gives mb GeV⁻²

Unitarity is linear on one side, quadratic on the other

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 4.01 showed that S is unitary when the in states and the out states are each complete, and wrote S = 1 + iT so that the 1 carries the particles that miss each other. Substitute: S†S = (1 − iT†)(1 + iT) = 1 + i(T − T†) + T†T, so unitarity is −i(T − T†) = T†T, derived. Two features carry the rest of the topic. The left side is linear in T and the right side quadratic, so in an expansion in a coupling g the anti-Hermitian part of T at one order is fixed by T itself at half that order. And the right side is a positive operator, ⟨a|T†T|a⟩ = ‖T|a⟩‖² ≥ 0, so the anti-Hermitian part of T has a definite sign in every state. Nothing here used a Lagrangian, a Feynman rule or perturbation theory: it is probability conservation for an operator that creates and destroys particles.

Sandwich between two particles, insert everything

Between any two states ⟨b|iT|a⟩ = (2π)⁴δ⁴(pₐ − p_b) iℳ(a → b), and so ⟨b|T†|a⟩ = (2π)⁴δ⁴(pₐ − p_b) ℳ*(b → a). Between T† and T insert the identity as a sum over out states, 1 = Σ_f (1/S_f) ∫∏ⱼ d³qⱼ/((2π)³2Eⱼ) |f⟩⟨f|, with S_f = n! for n identical particles so that each state is counted once. Each of ⟨b|T†|f⟩ and ⟨f|T|a⟩ brings a delta function: one becomes the common (2π)⁴δ⁴(pₐ − p_b), the other completes topic 5.02's dΠ_f. Strip the common one: −i[ℳ(a → b) − ℳ*(b → a)] = Σ_f ∫dΠ_f ℳ*(b → f) ℳ(a → f), derived, and at b = a it is 2 Im ℳ(a → a) = Σ_f ∫dΠ_f |ℳ(a → f)|². The step that carries the physics is completeness. The out states must span the space, topic 4.09's asymptotic completeness, assumed here, so every bound state and multiparticle state the pair can reach belongs in the sum, while an unstable particle appears only through its decay products.

The flux factor turns the sum into σₜₒₜ

Topic 5.01 gave the cross-section into f as σ_f = (1/F) ∫dΠ_f |ℳ(a → f)|², with flux F = 4E₁E₂|v₁ − v₂|. In the centre-of-mass frame the speeds are p_cm/E₁ and p_cm/E₂ in opposite directions, so F = 4p_cm(E₁ + E₂) = 4E_cm p_cm, or invariantly 4√((k₁⋅k₂)² − m₁²m₂²). Summed over f the right side is F σₜₒₜ, and writing ℳ(s, t) with t = 0 forward, Im ℳ(s, 0) = 2E_cm p_cm σₜₒₜ, derived. Dimensions: a 2 → 2 amplitude has mass dimension 4 − 4 = 0, and E_cm p_cm σₜₒₜ is GeV × GeV × GeV⁻². High energy: the masses drop out, 2E_cm p_cm → s, and Im ℳ(s, 0) → sσₜₒₜ. Sign: σₜₒₜ ≥ 0, so the forward amplitude's imaginary part is never negative, whatever the interaction. Scale: for proton-proton scattering at √s = 13 TeV, TOTEM measured σₜₒₜ = 110.6 mb, quoted, which is 284.0 GeV⁻², so Im ℳ(s, 0) = 1.690 × 10⁸ GeV² × 284.0 GeV⁻² = 4.80 × 10¹⁰.

Quantum mechanics' theorem, now with particle creation

For elastic scattering in the centre-of-mass frame, topic 5.04 found dσ/dΩ = |ℳ|²/(64π²s), so f(θ) = ℳ/(8π√s) has |f|² = dσ/dΩ, and with k = p_cm the theorem reads σₜₒₜ = (4π/k) Im f(0). That is the optical theorem UQMII-15.05 and UAQM-02.09 reached through partial waves, quoted. What is new is its reach: the derivation never asked what the final states were, so σₜₒₜ now includes channels that create particles, which no potential can describe; at 13 TeV about 79.5 mb of the 110.6 mb is inelastic, quoted. The forward peak shows the theorem at work. Since dσ/dt = |ℳ|²/(64π s p_cm²) and |ℳ(s, 0)|² = (1 + ρ²)(Im ℳ)², with ρ = Re ℳ/Im ℳ at t = 0, the elastic peak at t = 0 is (1 + ρ²)σₜₒₜ²/(16π), derived for any masses. With ρ ≈ 0.10, quoted, the 13 TeV σₜₒₜ predicts 631 mb GeV⁻². TOTEM runs the relation backwards, dividing the forward elastic rate by the total event rate, and so measures σₜₒₜ without knowing the luminosity.

A real tree, and an imaginary part one order later

Expand ℳ = ℳ⁽²⁾ + ℳ⁽⁴⁾ + …, with ℳ⁽ⁿ⁾ of order gⁿ in a coupling g. Tree amplitudes are real in the physical region: a t-channel propagator 1/(t − m²) has t ≤ 0 and never reaches its pole, so its iε does nothing, and an s-channel pole sits at s = m², the resonance topic 5.09 treats. So at order g² the left side of 2 Im ℳ(a → a) = Σ_f ∫dΠ_f |ℳ(a → f)|² vanishes, while the right side starts at |g²|² = g⁴. The theorem holds order by order only if the g⁴ loops supply 2 Im ℳ⁽⁴⁾(a → a) = Σ_f ∫dΠ_f |ℳ⁽²⁾(a → f)|², the trees squared, shown. For ψψ' → ψψ' in topic 5.06's model the tree g²/(m² − t) is real, ψψ' cannot annihilate, and below √s = M + M' + m no φ can be made, so the box diagram must carry Im ℳ(s, 0) = 2E_cm p_cm σₑₗ, with σₑₗ = g⁴/(16π s m²(m² + 4p_cm²)) from the tree. An imaginary part switches on only where a channel opens, which is why loop amplitudes have branch points at thresholds; topic 5.08 checks all of this on the φ⁴ bubble.

Partial waves: on the circle, then inside it

For distinguishable spinless particles expand ℳ(s, θ) = (8π√s/p_cm) Σₗ (2l + 1) aₗ(s) Pₗ(cos θ), so that aₗ is the dimensionless k fₗ of quantum mechanics. Projecting the off-forward identity onto Pₗ, shown in outline, gives Im aₗ = |aₗ|² + (1 − ηₗ²)/4, where aₗ = (ηₗ e^(2iδₗ) − 1)/(2i) and the last term is that wave's inelastic cross-section times p_cm²/(4π(2l + 1)). Below the first inelastic threshold ηₗ = 1, and aₗ = e^(iδₗ) sin δₗ rides the circle |aₗ − i/2| = 1/2; above it 0 ≤ ηₗ ≤ 1 and aₗ lies inside, on the circle of radius ηₗ/2. Three bounds follow: |aₗ| ≤ 1, |Re aₗ| ≤ 1/2, and σₗ ≤ 4π(2l + 1)/p_cm², the limit topic 5.09's peak reaches. They also say where perturbation theory fails. A real tree aₗ lies on the real axis, which touches the circle only at 0, and once |aₗ| nears 1/2 the missing |aₗ|² is already half the tree, so the loops are no longer small. φ⁴'s s wave, with the identical pair's 1/2, is a₀ = −λβ/(32π), with β = 2p_cm/√s the speed of each particle, so at high energy, where β → 1, the loops stop being corrections near λ = 16π ≈ 50.

Change one variable at a time

Make the relationship visible.

Interactive model

Phase shift δₗ60 °

Inelasticity ηₗ0.60

Set η = 1 and sweep δ: the point never leaves the outer circle and the open, inelastic part of the bar vanishes. Then set δ = 0 and lower η: aₗ climbs the imaginary axis as pure absorption, and at η = 0 the elastic share is exactly one half, the black disk.

TOTAL Im aₗ0.650

ELASTIC |aₗ|²0.490

INELASTIC (1 − η²)/40.160

ELASTIC SHARE |aₗ|²/Im aₗ0.754

Live interpretationTOTAL Im aₗ: 0.650. ELASTIC |aₗ|²: 0.490. INELASTIC (1 − η²)/4: 0.160. ELASTIC SHARE |aₗ|²/Im aₗ: 0.754

Catch the common trap

Explain before calculating.

Misconception check

The forward amplitude is elastic, a → a, so it is tempting to read the theorem as Im ℳ(s, 0) = 2E_cm p_cm σₑₗ. The derivation forbids it: the right side came from inserting a complete set of states, so every channel the pair can reach appears, particle creation included. Physically, the forward direction is where the scattered wave interferes with the unscattered one, and anything that removes flux from the beam, by deflecting it or by turning it into new particles, must cast a shadow there. A black disk shows it most sharply: it absorbs everything that hits it, yet its shadow is itself diffraction, so σₑₗ = σᵢₙₑₗ and elastic scattering is only half of σₜₒₜ. At 13 TeV, σₑₗ = 31.0 mb against σₜₒₜ = 110.6 mb, quoted, so the elastic-only reading understates Im ℳ(s, 0) by a factor of 3.6; at order g⁴ it would drop the squared trees of every inelastic channel.

Two different spinless particles collide with E_cm = 2.0 GeV and p_cm = 0.60 GeV. Elastic scattering has σₑₗ = 1.5 mb, and the one inelastic channel that is open adds 2.5 mb. What value does unitarity require for Im ℳ(s, 0), the imaginary part of the forward elastic amplitude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo different spinless particles, of masses 0.50 GeV and 1.0 GeV, collide with p_cm = 0.30 GeV, and the total cross-section into all final states is 12 mb. Find E_cm, the imaginary part of the forward amplitude and Im f(0) in femtometres, and check the result against the invariant flux and the quantum-mechanical form of the theorem.

Energies: E₁ = √(0.50² + 0.30²) GeV = 0.5831 GeV and E₂ = √(1.0² + 0.30²) GeV = 1.0440 GeV, so E_cm = 1.6271 GeV.
Units: σₜₒₜ = 12 mb ÷ 0.3894 mb GeV² = 30.82 GeV⁻².
Theorem: Im ℳ(s, 0) = 2E_cm p_cm σₜₒₜ = 2 × 1.6271 GeV × 0.30 GeV × 30.82 GeV⁻² = 30.09, a pure number, as a 2 → 2 amplitude must be.
Invariant flux: k₁⋅k₂ = E₁E₂ + p_cm² = 0.6988 GeV², so 2√((k₁⋅k₂)² − m₁²m₂²) = 2√(0.4883 − 0.2500) GeV² = 0.9763 GeV², the same as 2E_cm p_cm = 2 × 1.6271 × 0.30 GeV² = 0.9763 GeV².
Quantum-mechanical form: Im f(0) = Im ℳ/(8πE_cm) = 30.09/40.89 GeV⁻¹ = 0.7357 GeV⁻¹ = 0.1452 fm, and (4π/p_cm) Im f(0) = 4π × 0.7357/0.30 GeV⁻² = 30.82 GeV⁻² = 12 mb, as it must.

AnswerE_cm ≈ 1.627 GeV, Im ℳ(s, 0) ≈ 30.1 and Im f(0) ≈ 0.145 fm, however the 12 mb is shared between elastic and inelastic channels.

MediumIn the scalar Yukawa model with a second species ψ' (topic 5.06), take M = 1.0 GeV, M' = 2.0 GeV, a φ of mass m = 0.14 GeV and g = 2.0 GeV, and collide ψψ' at p_cm = 0.10 GeV. The tree amplitude g²/(m² − t) is real. Find the imaginary part of the forward amplitude that unitarity requires at order g⁴, compare it with the tree, and say which diagram must supply it.

Kinematics: E_ψ = √(1.0² + 0.10²) GeV = 1.0050 GeV and E_ψ' = √(2.0² + 0.10²) GeV = 2.0025 GeV, so √s = 3.0075 GeV and s = 9.045 GeV². Making a φ needs √s ≥ M + M' + m = 3.14 GeV, and ψψ' cannot annihilate, so elastic scattering is the only open channel.
Tree: forward, at t = 0, ℳ(s, 0) = g²/m² = 4.0/0.0196 = 204.1, real and dimensionless. At order g² the left side of 2 Im ℳ(a → a) = Σ_f ∫dΠ_f |ℳ(a → f)|² is zero, while the right side is already of order g⁴.
Elastic cross-section from the tree, with t = −2p_cm²(1 − cos θ): σₑₗ = ∫dΩ |ℳ|²/(64π²s) = g⁴/(16π s m²(m² + 4p_cm²)) = 16/(16π × 9.045 × 0.0196 × 0.0596) GeV⁻² = 30.13 GeV⁻² = 11.73 mb.
Unitarity at order g⁴: Im ℳ(s, 0) = 2E_cm p_cm σₑₗ = 2 × 3.0075 × 0.10 × 30.13 = 18.12, which is 18.12/204.1 = 8.9% of the tree. It must come from the box diagram, the only g⁴ graph that can be cut through an on-shell ψψ' pair; the crossed box and the vertex and self-energy corrections are real here, because none of their cuts opens in this region.
Checks: g²/m² and E_cm p_cm σₑₗ are both dimensionless. In quantum-mechanical form Im f(0) = Im ℳ/(8π√s) = 0.2397 GeV⁻¹ = 0.0473 fm, equal to kσₑₗ/(4π) = 0.10 × 30.13/(4π) GeV⁻¹: the second Born term balancing the first one squared (UQMII-15, quoted). The ratio g²p_cm/(8π√s(m² + 4p_cm²)) grows as g², a direct measure of how perturbative the coupling is.

AnswerUnitarity requires Im ℳ(s, 0) ≈ 18.1 at order g⁴, about 9% of the real tree value 204, from σₑₗ ≈ 11.7 mb; only the box diagram can supply it.

HardFor pp scattering at √s = 13 TeV, TOTEM measured σₜₒₜ = 110.6 mb and σₑₗ = 31.0 mb, with ρ = Re ℳ/Im ℳ ≈ 0.10 at t = 0, all quoted. Ignoring spin and the proton mass, find Im ℳ(s, 0) and the forward elastic peak dσₑₗ/dt at t = 0 in mb GeV⁻². Then model the proton as a grey disk of N partial waves, each with the same purely absorptive aₗ = i(1 − η)/2, and find η, the radius R = N/p_cm in fm, and N.

Natural units: σₜₒₜ = 110.6 ÷ 0.3894 GeV⁻² = 284.0 GeV⁻² and s = (13 000 GeV)² = 1.690 × 10⁸ GeV². With mₚ²/s = 5 × 10⁻⁹ the masses drop out, 2E_cm p_cm = s, and Im ℳ(s, 0) = sσₜₒₜ = 4.80 × 10¹⁰, dimensionless.
Forward peak: |ℳ(s, 0)|² = (1 + ρ²)(Im ℳ)², so at t = 0 dσₑₗ/dt = (1 + ρ²)σₜₒₜ²/(16π) = 1.01 × 284.0²/(16π) GeV⁻⁴ = 1621 GeV⁻⁴, and × 0.3894 mb GeV² this is 631 mb GeV⁻². ρ enters only as ρ² = 0.01, so the real part moves the peak by 1%.
Grey disk: aₗ = i(1 − η)/2 gives Im aₗ = (1 − η)/2 and |aₗ|² = (1 − η)²/4, so σₑₗ/σₜₒₜ = |aₗ|²/Im aₗ = (1 − η)/2. The measured 31.0/110.6 = 0.2803 gives η = 0.439, and each wave sits inside the unitarity circle by Im aₗ − |aₗ|² = (1 − η²)/4 = 0.202.
Radius: the first N waves give Σ(2l + 1) = N², so σₜₒₜ = (4π/p_cm²) N² (1 − η)/2 = 2πR²(1 − η). Then R² = 110.6 mb/(2π × 0.5606) = 31.40 mb = 3.140 fm², since 1 mb = 0.1 fm², and R = 1.77 fm.
Number of waves: N = p_cm R/(ħc) = 6500 GeV × 1.772 fm/(0.19733 GeV fm) = 5.84 × 10⁴. Checks: a black disk, η = 0, would give σₑₗ/σₜₒₜ = 1/2, so the measured 0.28 says the proton is grey; and R is close to twice the proton charge radius, 2 × 0.841 fm = 1.68 fm, quoted.

AnswerIm ℳ(s, 0) ≈ 4.8 × 10¹⁰ and dσₑₗ/dt ≈ 631 mb GeV⁻² at t = 0; the grey disk has η ≈ 0.44, R ≈ 1.77 fm and N ≈ 5.8 × 10⁴ partial waves.