University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.4
The LSZ reduction formula derived
Unit 3 taught you to compute ⟨Ω|Tφ⋯φ|Ω⟩ to any order, and an experiment reports S-matrix elements. This lesson builds the bridge: in and out operators made from the interacting field itself, limits that hold only weakly and only at the physical mass, and the reduction formula that every rate in this course quietly stands on.
Build the model
Connect the measurement to the mechanism.
The LSZ formula rests on one physical claim and one identity. The claim, the asymptotic condition, is that long before or long after a collision the interacting field, smeared over a packet f that solves the Klein-Gordon equation with the physical mass m, acts like √Z times a free annihilation operator. It comes from the spectral analysis of topics 4.02 and 4.03: φ reaches a single particle with amplitude √Z and a time dependence the packet cancels exactly, and reaches multiparticle states whose masses form a continuum, so their phases run at rates that vary continuously and average away.
The limit is weak: matrix elements converge, the operator does not, because the continuum part keeps its norm while turning orthogonal to every fixed state. The identity is that the time derivative of the Klein-Gordon product is i ∫ d³x f*(□ + m²)φ, so out minus in is a spacetime integral of the field's failure to be free. Peeling every particle out of ⟨β out|α in⟩ this way, with time ordering placing each limit for free, leaves S as Klein-Gordon operators applied to ⟨Ω|Tφ⋯φ|Ω⟩, derived here from that condition.
It stops where its inputs do: it needs an isolated one-particle pole, particles that never decay, and outgoing momenta distinct from incoming ones, and it is silent about the forward pieces in which a particle passes untouched.
- Simple definition
- The LSZ reduction formula writes every S-matrix element as the vacuum correlation function ⟨Ω|Tφ⋯φ|Ω⟩ with each external leg acted on by (□ + m²) at the physical mass, multiplied by i/√Z and Fourier-transformed with that particle's momentum.
- Example
- For 2 → 2 scattering it takes ⟨Ω|Tφ(x₁)φ(x₂)φ(y₁)φ(y₂)|Ω⟩, applies four Klein-Gordon operators, weights the outgoing legs by e(+ip⋅x) and the incoming ones by e(−ik⋅y), and multiplies by (i/√Z)⁴ = 1/Z², which is 1.23 when Z = 0.90.
Defined, then evaluated: for a free field of mass m it equals a(f) = ∫ d³k/((2π)³2Eₖ) f̃*(k) a(k) at every t, the natural candidate for aᵢₙ and aₒᵤₜ.
f = ∫ d³k/((2π)³2Eₖ) f̃(k) e(−ik⋅x), Eₖ = √(|k|² + m²) at the physical mass m in MeV; f and φ carry mass dimension 1
Derived. Zero for a free field of mass m and non-zero in any interacting theory: the drift is where the scattering is stored.
□ = ∂₀² − ∇², of mass dimension 2; uses (□ + m²)f = 0 and one integration by parts in space
Derived for matrix elements from the vacuum: the one-particle term is constant and the continuum dephases. Its extension to every pair of states is argued from separated packets, not proved.
Z = |⟨Ω|φ(0)|p⟩|² from topic 4.03, 0 ≤ Z ≤ 1, dimensionless; |α⟩, |β⟩ are fixed normalisable states
Derived by integrating the drift over all time. Its conjugate gives a†ᵢₙ − a†ₒᵤₜ with f in place of f*. A free field of mass m gives zero, so S = 1.
d⁴x runs over all spacetime; for a packet with ∫ d³k/((2π)³2Eₖ)|f̃|² = 1 both sides are dimensionless
Derived by peeling each particle out of ⟨β out|α in⟩, the t → ∓∞ limits putting every field in one T product with each □ outside it. Valid when no p equals any k.
N = n + l particles; one e(+ip⋅x) per outgoing momentum p, one e(−ik⋅y) per incoming k, all on shell in MeV
Derived by parts. A term survives the on-shell limit only through a pole 1/(p² − m²) in that leg: the residue topic 4.05 turns into amputated diagrams.
F̃(p) = ∫ d⁴x e(ip⋅x) F(x); p² = E² − |p|² in MeV²; the factor vanishes at p² = m²
What is given, what is wanted, and the packet
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Wanted: Sβα = ⟨β out|α in⟩ of topic 4.01. Given: the correlation functions ⟨Ω|Tφ⋯φ|Ω⟩ of Unit 3, and from topic 4.03 that ⟨Ω|φ(x)|p⟩ = √Z e(−ip⋅x), with m the pole mass. Plane waves are not normalisable, so smear with f(x) = ∫ d³k/((2π)³2Eₖ) f̃(k) e(−ik⋅x), a positive-frequency solution of (□ + m²)f = 0 at the physical m. Put Unit 1's free field of mass m into φf(t) = i ∫ d³x [f*∂₀φ − (∂₀f*)φ]: each a(k) term comes with (Eₖ + Eₖ′)(2π)³δ³(k − k′), which cancels the measure's 2Eₖ, while each a†(k) term comes with Eₖ − Eₖ′ at k′ = −k and vanishes. So φf(t) = ∫ d³k/((2π)³2Eₖ) f̃*(k) a(k) = a(f) at every t, derived. That constancy is what an interacting field loses, and it is why a(f) is worth chasing.
The drift is the Klein-Gordon operator
Differentiate under the integral. The ∂₀f* ∂₀φ terms cancel, leaving dφf/dt = i ∫ d³x [f* ∂₀²φ − (∂₀²f*) φ]. The packet obeys ∂₀²f* = (∇² − m²)f*, and moving ∇² onto φ by parts costs nothing for a packet that vanishes at large |x|, so dφf/dt = i ∫ d³x f*(□ + m²)φ, derived. For a free field of mass m that is zero, as it must be. In φ⁴ theory the field equation is (□ + m₀²)φ = −(λ/3!)φ³, so (□ + m²)φ = (m² − m₀²)φ − (λ/6)φ³: the drift has two sources, the self-interaction and the gap between the Lagrangian mass m₀ and the pole mass m. Integrate over all time and, if the limits exist, φf(+∞) − φf(−∞) = i ∫ d⁴x f*(□ + m²)φ. Everything now hangs on what φf(t) does as t → ±∞.
The limits hold weakly, and only at the pole
With ⟨Ω|φ(x)|p⟩ = √Z e(−ip⋅x) and a packet of the same mass the phases cancel: ⟨Ω|φf(t)|p⟩ = √Z f̃*(p) = √Z⟨Ω|a(f)|p⟩ for every t, derived. A multiparticle state |n⟩ of momentum p and invariant mass μ ≥ μₜₕ gives instead f̃*(p)[(E + Eₙ)/2E] e(i(E − Eₙ)t)⟨Ω|φ(0)|n⟩, with Eₙ = √(|p|² + μ²) above E. Its phase runs at a rate that varies continuously with μ, so against any normalisable state the continuum's contribution tends to zero as t → ±∞, by the Riemann-Lebesgue lemma, derived. That is a weak limit and no more: only the phases of the continuum part depend on t, so it keeps its norm while turning orthogonal to every fixed vector. For a packet at rest in φ⁴, where μₜₕ = 3m, each continuum phase turns at least once per 2π/(2m); at m = 150 MeV that is 2πħ/(300 MeV) = 1.38 × 10⁻²³ s. What survives is the δ-function in ρ, so the pole must be isolated: where the continuum reaches down to m, as massless photons make it for the electron, the pole becomes a branch point and nothing is left to survive (topic 4.09).
Out minus in, and why it is plausible for every state
Dividing by √Z, the weak limits define aᵢₙ(f) and aₒᵤₜ(f), and the integrated drift becomes aₒᵤₜ(f) − aᵢₙ(f) = (i/√Z) ∫ d⁴x f*(□ + m²)φ, derived. Its conjugate is a†ᵢₙ(f) − a†ₒᵤₜ(f) = (i/√Z) ∫ d⁴x f(□ + m²)φ. The vacuum matrix elements above prove the limits only between Ω and one or many quanta of φ. For general |α⟩ and |β⟩ one argues from geometry: in the far past the packets are far apart, so near packet f the field sees a lone particle, and the same split into a constant one-particle part and a dephasing remainder goes through. This is the asymptotic condition, assumed here; Haag-Ruelle scattering theory establishes it, given a mass gap and an isolated one-particle shell, and is named only. Two checks. A free field of mass m has (□ + m²)φ = 0, so aₒᵤₜ = aᵢₙ and S = 1: no interaction, no scattering. And dimensions: for a normalised packet f carries mass dimension 1, so ∫d⁴x (−4), f* (+1), □ + m² (+2) and φ (+1) sum to 0, matching the dimensionless a(f).
Peeling the particles out of ⟨β out|α in⟩
Write |α in⟩ = a†ᵢₙ(f₁)|α′ in⟩ and use the conjugate formula: ⟨β out|α in⟩ = ⟨β out|a†ₒᵤₜ(f₁)|α′ in⟩ + (i/√Z) ∫ d⁴y f₁(□ + m²)⟨β out|φ(y)|α′ in⟩. The first term takes a particle in packet f₁ out of the out state, so it survives only if particle 1 crosses untouched: a forward piece, excluded for momenta in general position and assigned by topic 4.05 to the 1 in S or to a lower-point process with spectators. Outgoing particles come off the left the same way, from aₒᵤₜ(f′) = aᵢₙ(f′) + (i/√Z) ∫ d⁴x f′*(□ + m²)φ. The next particle meets a field already in the matrix element, and time ordering settles where it goes: a†ᵢₙ(f₂) is the t → −∞ limit of a Klein-Gordon product, which a T product puts to the right of φ(y), and a†ₒᵤₜ(f₂) is the t → +∞ limit, which it puts to the left. Their difference is again ∫dt d/dt, now outside the T product. Repeat, then let each packet become a plane wave, f → e(−ik⋅y) incoming and f* → e(+ip⋅x) outgoing: the LSZ formula, derived.
Checks: dimensions, free fields and the poles
With ⟨p|k⟩ = 2E(2π)³δ³(p − k), each one-particle state has mass dimension −1, so an N-particle S-matrix element has −N. On the right each leg brings ∫d⁴x (−4), □ + m² (+2) and a field (+1), also −1 per leg; removing (2π)⁴δ⁴ (−4) leaves [M] = 4 − N, as topic 4.01 found. For free fields the four-point function is a sum of pairs such as DF(x₁ − y₁)DF(x₂ − y₂). Since (□ + m²)DF(x − y) = −iδ⁴(x − y), quoted from Unit 1, one operator in each pair yields a δ-function and its partner then leaves a factor such as m² − k², zero on shell: no connected part, and S = 1 comes entirely from the forward pieces. In momentum space each (□ + m²) becomes −(p² − m²), which vanishes on shell, so a term survives only through a pole 1/(p² − m²) in every external leg. S-matrix elements are residues of correlation functions at their external poles, the form in which topic 4.05 uses them. No experiment tests LSZ alone; every cross-section of Units 5 to 14 tests it.
Change one variable at a time
Make the relationship visible.
Set Z to 1 and the curve is flat: a free field's product is a(f) at every t. Then lower Z and slide mt up: the overlap shrinks like t(−3/2) while the norm² readout never moves. Raise the gap and the continuum turns faster without settling sooner.
PARTICLE PLATEAU Z0.70
CONTINUUM OVERLAP AT mt, REAL PART0.0186
ITS ENVELOPE0.0261
CONTINUUM NORM², EVERY t0.30
Live interpretationPARTICLE PLATEAU Z: 0.70. CONTINUUM OVERLAP AT mt, REAL PART: 0.0186. ITS ENVELOPE: 0.0261. CONTINUUM NORM², EVERY t: 0.30
Catch the common trap
Explain before calculating.
A scalar field has Z = 0.81. A packet f solving the Klein-Gordon equation with the physical mass is used to form φf(t) = i ∫ d³x [f*∂₀φ − (∂₀f*)φ]. What happens as t → −∞?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn the scalar Yukawa model take m = 600 MeV for φ and M = 400 MeV for ψ, and suppose φ's spectral weight has Z = 0.81. A packet at rest is used to form φf(t). Find the value its one-particle matrix element takes relative to a free field, where φ's continuum begins, and the longest period with which any continuum term turns.
- From topic 4.03, ⟨Ω|φ(x)|p⟩ = √Z e(−ip⋅x). A packet built with the same m cancels this phase, so ⟨Ω|φf(t)|p⟩ = √Z f̃*(p) at every t: the ratio to a free field is √0.81 = 0.90, constant rather than merely a limit.
- Topic 4.02 puts φ's continuum at the smaller of 2m and 2M. Here 2M = 800 MeV lies below 2m = 1200 MeV, so μₜₕ = 800 MeV, the lightest ψψ̄ pair; m below 2M also keeps φ stable, as LSZ needs.
- For a packet at rest a continuum state of mass μ turns at the rate μ − m, slowest at threshold: μₜₕ − m = 800 − 600 = 200 MeV. This gap is what isolates the pole.
- Restore ħ = 6.582 × 10⁻²² MeV s: the longest period is 2πħ/(200 MeV) = 2.07 × 10⁻²³ s, so in 10⁻²¹ s even the slowest continuum term turns 48 times.
AnswerThe one-particle term is 0.90 of the free-field value at every t; the continuum begins at 2M = 800 MeV, and its slowest phase turns once every 2.07 × 10⁻²³ s.
MediumApply the LSZ formula to the order-λ four-point function of φ⁴ theory to find M for φφ → φφ, with incoming k₁, k₂ and outgoing p₁, p₂ in general position. Evaluate it for λ = 0.50 and say what becomes of the order-λ⁰ terms.
- Unit 3 gives the connected order-λ term, −iλ ∫ d⁴z DF(x₁ − z)DF(x₂ − z)DF(y₁ − z)DF(y₂ − z), quoted. At this order Z = 1, since φ⁴'s one-loop self-energy does not depend on p² (topic 4.05), so the prefactor (i/√Z)⁴ is i⁴ = 1.
- Each leg's operator acts on its own propagator: (□ + m²)DF(x − z) = −iδ⁴(x − z), the defining property of the Feynman propagator (Unit 1, quoted). Four legs give (−i)⁴ = 1 and pin x₁, x₂, y₁ and y₂ to z.
- What remains is −iλ ∫ d⁴z e(i(p₁ + p₂ − k₁ − k₂)⋅z) = −iλ(2π)⁴δ⁴(p₁ + p₂ − k₁ − k₂). Matching S = 1 + iT with ⟨f|iT|i⟩ = (2π)⁴δ⁴(p₁ + p₂ − k₁ − k₂) iM gives iM = −iλ, so M = −λ = −0.50.
- An order-λ⁰ term such as DF(x₁ − y₁)DF(x₂ − y₂): one operator turns DF(x₁ − y₁) into −iδ⁴(x₁ − y₁), and its partner then leaves a multiple of (m² − k₁²)δ⁴(p₁ − k₁), zero on shell and zero anyway for p₁ ≠ k₁. Forward scattering lives in the terms the derivation dropped.
- Checks: [M] = 4 − N = 0 and λ is dimensionless; M is independent of momenta, as a contact interaction's must be; and M below zero is the sign of a repulsive contact force in the non-relativistic limit.
AnsweriM = −iλ, so M = −λ = −0.50, dimensionless and momentum-independent; the order-λ⁰ pairs contribute nothing when the p's differ from the k's.
HardA packet at rest makes the state ψ(t) = φf(t)†|Ω⟩, with weight Z = 0.90 on the particle and 0.10 on a continuum that begins at μₜₕ = 2m with density proportional to √(μ − μₜₕ) e(−(μ − μₜₕ)/m), the two-body threshold law cut off over m. Find the continuum part of ⟨ψ(0)|ψ(t)⟩, its size at t = 5/m and 50/m, its large-t law, and t = 50/m in seconds for m = 140 MeV, and compare it with the continuum's norm².
- Each continuum mass μ enters with its weight times e(i(μ − m)t). Put ν = μ − μₜₕ, so the phase is e(i(μₜₕ − m)t) e(iνt) and the normalised weight is √ν e(−ν/m)/(Γ(3/2) m√m), with Γ(3/2) = √π/2 = 0.8862.
- ∫₀^∞ dν √ν e(−ν(1/m − it)) = Γ(3/2)(1/m − it)(−3/2), Euler's integral for Γ(3/2) continued to a complex rate with positive real part. With 1 − Z = 0.10 of the weight in the continuum and μₜₕ − m = m, the overlap is C(t) = 0.10 e(imt)(1 − imt)(−3/2).
- Its size is |C| = 0.10 (1 + m²t²)(−3/4). At t = 5/m: 0.10 × 26(−3/4) = 0.10 × 0.0869 = 8.7 × 10⁻³. At t = 50/m: 0.10 × 2501(−3/4) = 2.8 × 10⁻⁴. At t = 0 it is 0.10: every mass in phase.
- For mt ≫ 1, |C| → 0.10 (mt)(−3/2): ten times longer buys 10(−3/2) = 0.032, and indeed 2.83 × 10⁻⁴/8.69 × 10⁻³ = 0.033. The square-root edge sets this power; the gap μₜₕ − m only sets how fast the phase e(imt) turns.
- In seconds, ħ/(140 MeV) = 6.582 × 10⁻²² MeV s/140 MeV = 4.70 × 10⁻²⁴ s, so t = 50/m is 2.35 × 10⁻²² s.
- The continuum part of the state never shrinks: only its phases depend on t, so its norm² stays 0.10 while its overlap with the t = 0 state falls by a factor of about 350. Overlaps with fixed states die and the length does not: a weak limit.
Answer|C(t)| = 0.10 (1 + m²t²)(−3/4): 8.7 × 10⁻³ at t = 5/m and 2.8 × 10⁻⁴ at t = 50/m, which is 2.35 × 10⁻²² s for m = 140 MeV, falling as t(−3/2) while the continuum's norm² stays 0.10.