University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.2

The Källén-Lehmann representation

Diagrams hand you correlation functions; detectors hand you particles with physical masses. Before the two can be joined, you need to know what an exact propagator must look like. Insert every state the theory has, and it shows its particles as poles, its bound states as spikes and pair production as a cut, all with positive weight.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Diagrams build correlation functions order by order; this topic asks what the exact two-point function must look like before any is drawn. Take topic 2.02's Heisenberg field and interacting vacuum, assume ⟨Ω|φ|Ω⟩ = 0, and insert a complete set of eigenstates of the full H and P. Translation invariance turns each matrix element into a phase times its value at the origin, and Lorentz invariance boosts every state to rest, so the dynamics collapses into one non-negative number per rest-frame state, |⟨Ω|φ(0)|λ₀⟩|².

The result, the Källén-Lehmann representation, writes the exact propagator as a superposition of free Feynman propagators of every mass μ, weighted by a spectral density ρ(μ²) ≥ 0. Reading ρ is reading the spectrum. A stable particle gives a delta function, so an isolated pole at its physical mass with residue Z; a bound state gives another delta; multiparticle states give a continuum from the lowest threshold φ can reach, which symmetry decides: 4m² generically, 9m² in φ⁴, the smaller of 4m² and 4M² for the Yukawa model's φ.

Above threshold Im Δ = −ρ/2, where iΔ is the exact propagator. No perturbation theory is used, but the argument assumes a stable particle below every threshold, the free theory's Lorentz covariance and a positive-norm state space; an unstable particle, a massless one and a covariantly quantised photon each break one.

Simple definition: The Källén-Lehmann representation writes the exact two-point function of a scalar field as an integral over mass μ of free Feynman propagators, weighted by a spectral density ρ(μ²) ≥ 0 that measures how strongly φ(0) creates states of invariant mass μ from the vacuum.
Example: A free scalar of mass 0.50 GeV has ρ(μ²) = 2πδ(μ² − 0.25 GeV²) and nothing else. Give it a φ⁴ coupling, with m₀ retuned so that the physical mass stays 0.50 GeV, and the spike sits there, now with weight Z, while ρ stays exactly zero from there up to 9m² = 2.25 GeV², where three-particle states begin.

One number per state, fixed at rest⟨Ω|φ(x)|λₚ⟩ = e^(−ip⋅x) ⟨Ω|φ(0)|λ₀⟩, with p⁰ = Eₚ(λ) = √(|p|² + m_λ²)

Derived: e^(iP⋅x) supplies the phase, and a boost that leaves Ω and the scalar φ(0) unchanged carries |λₚ⟩ to rest. One complex number per state is all the dynamics left.

λ labels a rest-frame eigenstate of the full H and P; m_λ is its invariant mass in GeV, p in GeV, x in GeV⁻¹

Källén-Lehmann representation⟨Ω|Tφ(x)φ(y)|Ω⟩ = ∫₀^∞ (dμ²/2π) ρ(μ²) D_F(x − y, μ²)

Derived without perturbation theory from completeness, translations, Lorentz invariance and ⟨Ω|φ|Ω⟩ = 0. The free field is the case ρ = 2πδ(μ² − m²).

D_F(x − y, μ²) is the free Feynman propagator of mass μ, in GeV²; μ² in GeV²; ρ in GeV⁻²

Spectral densityρ(μ²) = Σ_λ 2π δ(μ² − m_λ²) |⟨Ω|φ(0)|λ₀⟩|² ≥ 0

Derived: a sum of squared moduli cannot be negative. Interactions move spectral weight from one mass to another, but never subtract any.

the sum runs over every rest-frame state, with continuous labels integrated; ρ in GeV⁻², zero below the lightest state φ reaches

Pole plus cut in momentum space∫d⁴x e^(ip⋅x) ⟨Ω|Tφ(x)φ(0)|Ω⟩ = iZ/(p² − m² + iε) + i∫ (dμ²/2π) ρ_c(μ²)/(p² − μ² + iε), over μ² ≥ sₜₕ

Derived: the one-particle delta becomes an isolated pole with residue iZ, the continuum ρ_c a branch cut along real p² ≥ sₜₕ. Topic 4.03 pins Z between 0 and 1.

m is the physical mass, not m₀; Z = |⟨Ω|φ(0)|p⟩|² is dimensionless; sₜₕ in GeV² is the lowest multiparticle threshold

The continuum is the imaginary partρ(s) = −2 Im Δ(s + iε), where iΔ(p²) is the transformed two-point function above

Derived from 1/(x + iε) = P(1/x) − iπδ(x), P the principal value. The jump across the cut is ρ itself, so the imaginary part of a loop diagram is the continuum.

s = p² real, in GeV²; Δ in GeV⁻²; below sₜₕ and away from the pole, Δ is real

Scalar Yukawa φ: the ψψ̄ continuum at order g²ρ(s) = g² √(1 − 4M²/s) / [8π (s − m²)²] for s ≥ 4M²

Derived here from ⟨Ω|φ(0)|ψψ̄⟩ = g/(s − m²) and two-body phase space β/(8π). It rises as √(s − 4M²) and falls as g²/(8πs²).

g in GeV; M the ψ mass and m the φ mass in GeV; s in GeV²; ρ in GeV⁻²; valid while 2M exceeds m

Insert every state the interacting theory has

Work in natural units ħ = c = 1 with the metric (+, −, −, −). Topic 2.02 supplies the ingredients, quoted here: the Heisenberg field φ(x) = e^(iP⋅x) φ(0) e^(−iP⋅x), the interacting vacuum |Ω⟩ with P^μ|Ω⟩ = 0, and the correlation function ⟨Ω|Tφ(x)φ(y)|Ω⟩. Nothing below is expanded in a coupling. H and P commute, so their joint eigenstates span the space. Sort them by their rest frame: |λ₀⟩ has zero momentum and energy m_λ, and |λₚ⟩ is its boost, with Eₚ(λ) = √(|p|² + m_λ²). Then 1 = |Ω⟩⟨Ω| + Σ_λ ∫ d³p/((2π)³ 2Eₚ(λ)) |λₚ⟩⟨λₚ|, where λ runs over one-particle states, bound states and multiparticle states, whose relative momenta make Σ_λ hide integrals. Insert this between φ(x) and φ(y), with x⁰ the later time. The vacuum term contributes |⟨Ω|φ(0)|Ω⟩|², a constant; take it to be zero, because a symmetry forbids it or because φ has been shifted by its expectation value.

Translations give a phase, boosts give one number per state

Translation invariance moves the field to the origin: ⟨Ω|φ(x)|λₚ⟩ = ⟨Ω|e^(iP⋅x) φ(0) e^(−iP⋅x)|λₚ⟩ = e^(−ip⋅x)⟨Ω|φ(0)|λₚ⟩, since Ω carries no momentum and λₚ carries p. Lorentz invariance removes the p. Assume, as for the free field of topic 1.03, that the full theory's U(Λ) leaves Ω unchanged and acts on the scalar field as U(Λ)φ(0)U(Λ)⁻¹ = φ(0). Choosing the boost with U|λₚ⟩ = |λ₀⟩ gives ⟨Ω|φ(0)|λₚ⟩ = ⟨Ω|U⁻¹ Uφ(0)U⁻¹ U|λₚ⟩ = ⟨Ω|φ(0)|λ₀⟩. So, with x⁰ the later time, ⟨Ω|φ(x)φ(y)|Ω⟩ = Σ_λ |⟨Ω|φ(0)|λ₀⟩|² ∫ d³p/((2π)³ 2Eₚ(λ)) e^{(−ip⋅(x−y))}. The integral is the free function D(x − y) of mass m_λ from Advanced QM's quantised scalar field. Each state propagates freely with its own invariant mass, weighted by how strongly φ(0) creates it from the vacuum.

Time ordering and the spectral density

When y⁰ is the later time, the same steps give the sum with x and y exchanged, and the two orderings assemble, mass by mass, into the free Feynman propagator, exactly as they did for the free field. Collect the masses with a delta function, ρ(μ²) = Σ_λ 2πδ(μ² − m_λ²)|⟨Ω|φ(0)|λ₀⟩|², and the result is ⟨Ω|Tφ(x)φ(y)|Ω⟩ = ∫₀^∞ (dμ²/2π) ρ(μ²) D_F(x − y, μ²). Three properties follow at once. ρ ≥ 0, being a sum of squared moduli. ρ = 0 for negative μ², because physical states have p² ≥ 0 and p⁰ ≥ 0, the spectral condition this course assumes. And the dimensions close: φ has mass dimension 1, so the left side and D_F carry mass², which leaves ρ with mass⁻², as 2πZδ(μ² − m²) has. The free-field limit is ρ = 2πδ(μ² − m²), one state with Z = 1, and the representation returns D_F itself. In momentum space each D_F becomes i/(p² − μ² + iε).

What sits in ρ: a pole, bound states and a threshold

A stable particle of physical mass m contributes 2πZδ(μ² − m²), with Z = |⟨Ω|φ(0)|p = 0⟩|², and so an isolated pole iZ/(p² − m² + iε). Its position is the energy of the one-particle state at rest, whatever m₀ the Lagrangian carried. A bound state that φ can create adds another delta below its constituents' threshold. Multiparticle states have continuous invariant mass: two particles of mass m with momenta ±k in their rest frame have μ = 2√(m² + k²), any value from 2m up, so ρ turns smooth from μ² = 4m². That is the generic answer, not a law. In φ⁴ theory the unbroken symmetry φ → −φ is carried by a unitary U that leaves Ω alone and multiplies an n-particle state by (−1)ⁿ. Then ⟨Ω|φ(0)|n⟩ = ⟨Ω|U⁻¹(Uφ(0)U⁻¹)U|n⟩ = −(−1)ⁿ⟨Ω|φ(0)|n⟩, which vanishes for every even n, to all orders. φ reaches only odd numbers of quanta, and the continuum starts at 9m².

The Yukawa φ: its threshold, and its continuum computed

With interaction −gφψ†ψ, φ carries no ψ-number, so it creates only states with as many ψ as ψ̄; a single ψ never appears in its ρ. The lightest candidates are ψψ̄ from 2M and φφ from 2m, and no symmetry forbids either, since the coupling is odd in φ. The continuum therefore starts at the smaller of 4M² and 4m², though φφ needs a ψ loop and enters ρ only at order g⁶. Compute the ψψ̄ part. Sandwich the field equation (□ + m₀²)φ = −gψ†ψ between Ω and |ψψ̄⟩ of total momentum P: □ brings down −s, and ⟨0|ψ†ψ(0)|ψψ̄⟩ = 1 at lowest order, where m₀ = m, so ⟨Ω|φ(0)|ψψ̄⟩ = g/(s − m²). Summing over pairs at rest, ∫ d³k/((2π)³ 4Eₖ²) 2πδ(√s − 2Eₖ) = β/(8π) with β = √(1 − 4M²/s). So ρ(s) = g²β/[8π(s − m²)²]. Checks: mass⁻², a √(s − 4M²) rise at threshold, a g²/(8πs²) tail, and nothing at all as g → 0.

Pole and cut in the complex p² plane, and where it fails

Write the transform as iΔ(p²), with Δ(p²) = Z/(p² − m² + iε) + ∫(dμ²/2π) ρ_c(μ²)/(p² − μ² + iε) over μ² ≥ sₜₕ. As a function of complex p², Δ is analytic except for the pole at m² and a cut along the real axis from sₜₕ, across which it jumps by −iρ. On the real axis, 1/(x + iε) = P(1/x) − iπδ(x), with P the principal value, gives Im Δ(s + iε) = −ρ(s)/2. Below threshold the exact propagator is real; above it, its imaginary part is the continuum, which is how a loop diagram's imaginary part is read later in the course. The derivation assumed a stable particle below every threshold. Raise the Yukawa φ above 2M and it decays to ψψ̄: its delta dissolves into the continuum as a resonance bump, the case topic 4.09 takes up. A massless particle among the states drags the threshold down onto the pole. And positivity needs a state space of positive norm, which the covariantly quantised photon does not have.

Change one variable at a time

Make the relationship visible.

Interactive model

Mass ratio M/m0.80

Coupling g/m2.0

Slide M/m down towards 0.55: the cut closes in on the pole and the continuum swells. Push it past 1 and φφ at 4m² becomes the lowest threshold although ψψ̄ still carries the order-g² weight. Doubling g/m lifts the whole curve by 0.6 of a decade, because ρ ∝ g².

LOWEST THRESHOLD √sₜₕ / m1.60

ψψ̄ THRESHOLD 4M²/m²2.56

GAP (sₜₕ − m²)/m²1.56

ψψ̄ WEIGHT ∫ρ ds/2π0.0097

Live interpretationLOWEST THRESHOLD √sₜₕ / m: 1.60. ψψ̄ THRESHOLD 4M²/m²: 2.56. GAP (sₜₕ − m²)/m²: 1.56. ψψ̄ WEIGHT ∫ρ ds/2π: 0.0097

Catch the common trap

Explain before calculating.

Misconception check

The commonest slip is to treat 4m² as the place every continuum starts, as if a field could always make two of its own quanta. The threshold is the lowest invariant mass among the states φ(0) can actually reach from the vacuum, and selection rules decide which those are. In φ⁴ theory the unbroken φ → −φ symmetry makes ⟨Ω|φ(0)|two particles⟩ vanish exactly, at every order, so nothing at all happens at 4m²: the continuum begins at 9m², where the imaginary part of the two-loop sunset diagram switches on. In the scalar Yukawa model the threshold can sit below 4m², at 4M² when M is less than m, because φ makes ψψ̄ pairs; and it is never at M², because ψ-number forbids a single ψ. A second slip is to put the pole at the Lagrangian's m₀. The representation is exact, so the pole sits at the physical mass, the energy of the one-particle state at rest.

In the scalar Yukawa model, take m = 1.00 GeV for φ and M = 0.70 GeV for ψ. Where does the continuum in the spectral density of the φ two-point function begin, and at what order in g does it first appear there?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn QCD with electromagnetism and weak decays switched off, the pseudoscalar density O_π = ūiγ⁵u − d̄iγ⁵d creates a π⁰ from the vacuum. Taking the measured m_π = 134.977 MeV, find where the pion's delta and the continuum sit in the spectral density of O_π, say which selection rule removes two-pion states, and compare how fast the pole and the continuum fall off in space.

One particle: the π⁰ shares the quantum numbers of O_π, J^P = 0⁻ and isospin 1. So ρ contains 2πZ_π δ(μ² − m_π²), a spike at μ² = m_π² = (0.134977 GeV)² = 0.018219 GeV².
Two pions: in a J = 0 state they have no orbital angular momentum, so their parity is (−1)(−1) = +1, opposite to the −1 of O_π. Parity is exact in QCD, so ⟨Ω|O_π|ππ⟩ = 0 to all orders. Parity plays the part that φ → −φ plays in φ⁴.
Three pions in relative S-waves have parity (−1)³ = −1 and can carry isospin 1, so they are allowed. The continuum starts at μ = 3m_π = 0.40493 GeV, μ² = 9m_π² = 0.16397 GeV², not at 4m_π² = 0.072875 GeV² as a careless two-particle count would say.
The gap is 9m_π² − m_π² = 8m_π² = 0.14575 GeV², so the pion pole is well isolated. The measured π(1300), at about 1.69 GeV², lies inside the continuum and decays to three pions, so it appears as a broad bump, not a delta.
At large spacelike separation r each D_F(x − y, μ²) falls as e^(−μr) times powers of r. Restoring ħc = 197.327 MeV fm, the pion term has range ħ/(m_π c) = 1.462 fm and the continuum at most ħ/(3m_π c) = 0.4873 fm; at r = 3.00 fm the extra factor is e^{(−2m_π c r/ħ)} = e^(−4.104) = 0.0165.

AnswerPion delta at μ² = 0.01822 GeV², continuum from 9m_π² = 0.1640 GeV² (3m_π = 404.9 MeV) because parity forbids two pions; at 3.00 fm the continuum is suppressed relative to the pole by e^(−4.104) = 0.0165 before prefactors.

MediumIn the scalar Yukawa model take m = 0.50 GeV for φ, M = 0.40 GeV for ψ and g = 1.0 GeV. Find where the φ continuum starts, evaluate ρ and Im Δ at s = 1.00 GeV², and test the result against the threshold limit at s = 0.65 GeV² and the high-energy limit at s = 25 GeV².

Thresholds: φ has no ψ-number and nothing makes it odd, so the candidates are ψψ̄ from 4M² = 0.640 GeV² and φφ from 4m² = 1.00 GeV². The continuum opens with ψψ̄ at √s = 0.800 GeV, at order g². The pole at m² = 0.250 GeV² lies below it, since 2M exceeds m.
Matrix element: sandwiching (□ + m₀²)φ = −gψ†ψ between Ω and |ψψ̄⟩ gives (m² − s)⟨Ω|φ(0)|ψψ̄⟩ = −g, using ⟨0|ψ†ψ(0)|ψψ̄⟩ = 1 and m₀ = m at this order. So ⟨Ω|φ(0)|ψψ̄⟩ = g/(s − m²).
At s = 1.00 GeV²: β = √(1 − 0.640/1.00) = 0.600, the pair's rest-frame speed k*/E* = 0.300 GeV/0.500 GeV. Then ρ = g²β/[8π(s − m²)²] = (1.0 GeV²)(0.600)/(8π × 0.750² GeV⁴) = 0.0424 GeV⁻², and Im Δ = −ρ/2 = −0.0212 GeV⁻².
Threshold limit: at s = 0.65 GeV², 1.6% above 4M², β = 0.124 and ρ = 0.0308 GeV⁻². ρ climbs from zero as β ∝ √(s − 4M²), the opening of a two-body S-wave, not as a step.
High-energy limit: at s = 25 GeV², β = 0.987 and ρ = 6.41 × 10⁻⁵ GeV⁻², within 0.7% of g²/(8πs²) = 6.37 × 10⁻⁵ GeV⁻², so the tail falls as 1/s² and ∫ρ ds converges. Dimensions: GeV² × GeV⁻⁴ = GeV⁻², as ρ must carry.

AnswerContinuum from s = 0.640 GeV² (√s = 0.800 GeV), from ψψ̄ at order g²; ρ(1.00 GeV²) = 0.0424 GeV⁻² and Im Δ = −0.0212 GeV⁻²; ρ vanishes as √(s − 4M²) at threshold and approaches g²/(8πs²) at high s.

HardIn φ⁴ theory with Lᵢₙₜ = −(λ/4!)φ⁴, λ = 1.2 and physical mass m, show that no two-particle state enters ρ, derive the lowest-order three-particle matrix element, and, using the quoted massless three-body phase space Φ₃ = s/(256π³), find ρ at s = 400m² and the continuum's weight ∫ρ ds/2π between s = 100m² and 10⁴m². Contrast the growth of that weight with the Yukawa φ's tail.

Symmetry: L is even in φ, so a unitary U with UφU⁻¹ = −φ and U|Ω⟩ = |Ω⟩ multiplies an n-particle state by (−1)ⁿ. Then ⟨Ω|φ(0)|n⟩ = −(−1)ⁿ⟨Ω|φ(0)|n⟩, zero for n = 2, 4, … at every order. After the pole at m², ρ vanishes until 9m².
Matrix element: the field equation is (□ + m₀²)φ = −(λ/6)φ³. Between Ω and |k₁k₂k₃⟩ of total momentum P, □ → −s, and at lowest order ⟨0|φ³(0)|k₁k₂k₃⟩ = 3! = 6 ways of matching fields to particles. So (m² − s)⟨Ω|φ(0)|k₁k₂k₃⟩ = −λ, and ⟨Ω|φ(0)|k₁k₂k₃⟩ = λ/(s − m²).
Spectral density: identical particles carry 1/3! in the completeness sum, and the matrix element is constant over phase space, so ρ(s) = λ²Φ₃(s)/[6(s − m²)²]. For s ≫ 9m² the quoted Φ₃ = s/(256π³) gives ρ ≈ λ²s/[1536π³(s − m²)²], tending to λ²/(1536π³s).
At s = 400m²: ρ = 1.44 × 400/(1536π³ × 399² m²) = 7.60 × 10⁻⁸/m², 0.5% above the asymptotic 7.56 × 10⁻⁸/m². That 0.5% is only the (s − m²)² factor: a numerical phase-space integral that keeps the three masses gives 9% less, 6.92 × 10⁻⁸/m², so the massless Φ₃ supports two figures here and is worse nearer threshold. It falls only as 1/s, against 1/s² for the Yukawa φ, because three-body phase space grows as s while two-body phase space tends to a constant.
Weight: ∫ρ ds/2π ≈ [λ²/(3072π⁴)] ln(s₂/s₁) = 4.81 × 10⁻⁶ × ln 100 = 2.2 × 10⁻⁵ between 100m² and 10⁴m² (2.1 × 10⁻⁵ with the masses kept in Φ₃), and each further factor of 100 adds the same again: the integral to infinity diverges logarithmically. The Yukawa ψψ̄ tail above s₁ adds only g²/(16π²s₁), 6.33 × 10⁻⁵ above 100m² for g = m and 100 times less above 10⁴m², which converges.
Check: per unit ln s the weight is λ²/(3072π⁴) = λ²/[12(16π²)²], the quoted two-loop anomalous dimension of the φ⁴ field, which the renormalisation of φ⁴ meets as a logarithm of the cutoff. Topic 4.03 shows why this integral is 1 − Z.

AnswerNo two-particle states (φ → −φ); ⟨Ω|φ(0)|k₁k₂k₃⟩ = λ/(s − m²); ρ(400m²) = 7.6 × 10⁻⁸/m² with the massless Φ₃ (6.9 × 10⁻⁸/m² with the masses kept); the weight between 100m² and 10⁴m² is 2.2 × 10⁻⁵ and grows by that much per factor of 100 in s without limit, while the Yukawa tail converges.