University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.9
What LSZ assumes: mass gap, stability, completeness
This unit built the LSZ formula; this topic audits it. We find the line where each assumption is spent, put numbers on how long t → ∞ has to be, watch the free-state shortcut fail on an external leg, and name the physics, decays and massless photons, where LSZ has to give way.
Build the model
Connect the measurement to the mechanism.
LSZ turns a correlation function into an S-matrix element by reading a residue, one factor i√Z/(p² − m²) per external particle, and a residue can be read only at a pole that exists, sits on the real axis and stands alone. Two assumptions about the full theory guarantee that, each spent in a definite line of topic 4.04, and a third, spent in topic 4.01, makes the resulting S unitary. A mass gap: every multiparticle state the field can make has invariant mass at least μₜₕ, strictly above m, so at momentum p the continuum lies at least √(p² + μₜₕ²) − √(p² + m²) above the particle, dephases from it within a time of order the inverse of that gap, and is regular at p² = m².
Stability: the particle's own packet must still be there at t = +∞, which fails the moment m rises above a threshold its quantum numbers allow, because its δ function dissolves into the continuum. Asymptotic completeness: S†S = 1 needs the out states to span the space and SS† = 1 needs the in states to, an assumption that perturbation theory checks order by order and that has been proved only in special models. The same audit shows why topic 2.09's shortcut, the Dyson series between free states, cannot be trusted past tree level: a self-energy on an external leg multiplies a propagator on its own mass shell.
The assumptions fail in physics we need, unstable particles in topic 5.09 and massless photons, which close the gap, in Unit 14.
- Simple definition
- LSZ holds when the particle's mass is isolated from every other state its field can make (a mass gap), when the particle is stable so its wave packets survive to t = ±∞, and when the in states and the out states each span the whole space.
- Example
- In φ⁴ theory with m = 140 MeV the field's next states are three particles, from μₜₕ = 420 MeV: a gap of 280 MeV at rest, so the continuum dephases on the scale ħ/(280 MeV) = 2.4 × 10⁻²⁴ s, 7 × 10⁻¹⁶ of the 3.3 ns light takes to cross a metre.
Derived from E = √(p² + μ²). It is μₜₕ − m at rest and (μₜₕ² − m²)/2|p| once |p| ≫ μₜₕ, so a fast packet sheds its continuum more slowly.
m, μₜₕ, |p| and Δω in MeV; μₜₕ is the lowest invariant mass of states sharing the particle's conserved quantum numbers.
Estimated: every continuum term beats against the particle at least at Δω, so averaging over a few τ removes it. φ⁴ with m = 140 MeV, at rest: τ ≈ 2.4 × 10⁻²⁴ s.
Δω in MeV; τ in s with ħ = 6.582 × 10⁻²² MeV s; cτ in fm. A scale, not a sharp time.
Derived from μ² ≥ μₜₕ² across the continuum. With a gap, p² − m² times this vanishes at the pole, so LSZ reads only the particle; as μₜₕ → m the bound is lost.
ρc is the Kallen-Lehmann continuum, ∫ρc dμ² = 1 − Z with a regulator in place; Z is a pure number, both sides in MeV⁻².
Exponential decay quoted from topic 5.09. The one-particle term of φf(t) carries this factor, so its limit is zero: aₒᵤₜ(f) does not exist and the particle cannot be an external leg.
A is the amplitude that the packet still holds the particle; Γ the width in MeV, t in MeV⁻¹; γ = E/m, its Lorentz factor.
Derived. Out completeness gives S†S = 1, so probabilities from any initial state sum to one; in completeness gives SS† = 1. Proved only in special models.
Uses Σβ |β out⟩⟨β out| = 1 over every out state, bound states included; S is dimensionless, states normalised as in topic 4.01.
Shown: a propagator on its own pole, not an ultraviolet divergence, and the first term of the phase e(−iδE T) of a wrong-mass state. LSZ replaces it by √Z per leg.
Σ and k² in MeV², m₀ the Lagrangian mass, ε → 0⁺ the Feynman prescription; the product is dimensionless.
Audit topic 4.04 line by line
Natural units, ħ = c = 1, and the metric (+, −, −, −) throughout. Topic 4.04 wrote Sβα as (i/√Z)ⁿ times Fourier transforms of ⟨Ω|Tφ⋯φ|Ω⟩, each with its (□ + m²), and every step spent something. Writing ⟨Ω|φ(0)|p⟩ = √Z and dropping the vacuum term spent a unique, Poincaré-invariant vacuum with ⟨Ω|φ|Ω⟩ = 0, arranged by shifting φ if need be (topic 4.02). Replacing the Klein-Gordon product φf(t) by √Z aᵢₙ(f) as t → −∞, and by √Z aₒᵤₜ(f) as t → +∞, spent two things: every other state φ makes must drift out of phase with the packet, which needs m to be isolated from them, a mass gap; and the one-particle term must still be there, which needs a particle that does not decay. Reading ⟨β out|α in⟩ as a unitary matrix, as topic 4.01 did, spent a third, that the in states and the out states each span the space. The rest, time ordering and Fourier transforms, is algebra. So LSZ is exact wherever the three hold, and the working question for any theory is which of them it meets, and what takes over where it does not.
A gap lets the continuum let go
At momentum p the particle has energy √(p² + m²); a state of two or more particles that φ can make, with the same total momentum and invariant mass μ, has √(p² + μ²), and μ is at least μₜₕ. So in topic 4.04's limit each continuum term outruns the particle's phase by at least Δω(p) = √(p² + μₜₕ²) − √(p² + m²), derived, and averaging over a few times 1/Δω removes it, which is why the limit is weak yet exists. In φ⁴ the field reaches only odd numbers of quanta, so μₜₕ/m = 3, as for QCD's pion, whose G-parity bars two-pion states; for m = 140 MeV the gap is 280 MeV at rest, ħc/Δω = 0.70 fm and ħ/Δω = 2.4 × 10⁻²⁴ s. A fast packet sheds the continuum more slowly: at |p| = 1.40 GeV, Δω = 54.7 MeV (ħc/Δω = 3.6 fm), near the high-energy form (μₜₕ² − m²)/2|p| = 56 MeV. In momentum space the gap keeps the continuum analytic at p² = m²: with ∫ρc dμ² = 1 − Z its value there is at most (1 − Z)/(μₜₕ² − m²), derived, so LSZ's factor p² − m² removes it. Checks: Δω is an energy and the bound a mass⁻²; Z = 1, the free limit, zeroes the bound; as μₜₕ → m it and the dephasing time grow without limit.
Stability: the packet must still be there
The same limit spends stability. If m lies above a threshold open to the particle's quantum numbers, its δ function in ρ is gone and the pole has left the real p² axis for m² − imΓ, as topic 5.09 shows (quoted). The one-particle term of φf(t) then falls as e(−Γt/2γ) for a packet of Lorentz factor γ, so its limit at t → +∞ is zero: no out state, and no S-matrix element with that particle on an external leg. It survives only as a resonance in the scattering of its decay products, topic 4.06's s-channel pole once m passes 2M. Stability is the gap seen from the other side, since both demand that m lie below the lightest continuum with the same conserved quantum numbers; ψ number keeps ψ stable, its continuum starting at the lower of M + m (ψφ) and 3M (ψψψ̄). Measured cases set the scale. The ρ⁰, Γ ≈ 150 MeV, lives ħ/Γ = 4.4 × 10⁻²⁴ s, cτ = 1.3 fm, and is only ever a peak in ππ scattering. The charged pion, cτ = 7.80 m, flies 560 m on average at 10 GeV, and treating it as asymptotic is good to Γ/m = 1.8 × 10⁻¹⁶.
Asymptotic completeness: what unitarity spends
Unitarity needs in and out states that are not only there but enough. Out completeness, Σβ|β out⟩⟨β out| = 1 over every out state, bound states included (topic 4.08), gives (S†S)λα = Σβ⟨λ in|β out⟩⟨β out|α in⟩ = ⟨λ in|α in⟩, so probabilities from any initial state sum to one; in completeness gives SS† = 1 the same way, derived. Without it S is at best an isometry, and probability can leak into states that never separate into particles; topic 5.07's optical theorem rests on exactly this. Its status is sobering. For non-relativistic many-body systems with short-range forces it is a theorem (Sigal and Soffer, 1987). In relativistic field theory, Haag-Ruelle theory builds in and out states from a gap and an isolated mass shell, but completeness stays an extra assumption, established only in special low-dimensional models and never for an interacting theory in four dimensions. Perturbation theory assumes it and checks it order by order: topic 5.08's cut of the φ⁴ bubble returns exactly the two-particle states, which is out completeness at one loop.
The Dyson route breaks on an external leg
Topic 2.09 put T exp(−i∫HI d⁴x) between free states of the Lagrangian mass m₀ and contracted fields with them, and at tree level that gives the Feynman rules. At one loop, put a self-energy on an incoming line of momentum k. Momentum conservation makes the propagator between the loop and the rest of the diagram carry k itself, and a free state has k² = m₀², so the diagram contains (−iΣ(m₀²)) × i/(k² − m₀² + iε) = Σ(m₀²)/(iε): infinite. It is not an ultraviolet divergence, since a regulated, finite Σ does no better. Read it as a phase: with ε standing for 2E/T, it is −i(Σ/2E)T, the first term of e(−iδE T) with δE = Σ/2E, a state of the wrong mass slipping against the true particle for an infinite time. In φ⁴ the momentum-independent tadpole makes the trouble a pure mass shift. Put the free states at the physical mass with a counterterm, so that Σ(m²) = 0, and the factor becomes 0/0; cancelling it naively gives 1 + Σ′(m²) per leg, which is wrong.
Amputation and √Z put it right
In a correlation function the same insertions come in every number and sum, as topic 3.08 showed, to i/(p² − m₀² − Σ(p²)). Each term is singular at the old mass, as each term of 1/(x − δ) = 1/x + δ/x² + δ²/x³ + … is at x = 0, while the sum is finite there and has one simple pole at m², of residue iZ with 1/Z = 1 − Σ′(m²) (topic 4.03). LSZ multiplies each leg by (p² − m²)/(i√Z), so the whole tower of external insertions collapses to the physical mass on the line and a factor √Z = 1 + ½Σ′(m²) to first order: half what the naive 0/0 gave, and nothing infinite is left. In the free limit Σ → 0 the two routes agree, which is why the shortcut survives at tree level. Where an assumption fails, so does this repair. An unstable particle has no real pole to cancel, so its legs belong inside a larger process (topic 5.09). A massless photon puts electron-plus-soft-photon states at every invariant mass above m, closing the gap: the pole becomes a branch point, Z₂ is infrared divergent (topic 12.05), and Unit 14 trades exclusive amplitudes for inclusive rates.
Change one variable at a time
Make the relationship visible.
At μₜₕ/m = 3, φ⁴'s threshold and near the pion's in QCD alone, drag |p| from 0 to 5: the gap falls from 279 to 102 MeV. At μₜₕ/m = 1.05 only 1.4 MeV is left at |p|/m = 5. Below 1 the dashed edge sinks under the solid curve: the particle can decay, and the curve marks only its resonance peak.
μₜₕ − m AT REST279.1 MeV
PACKET MOMENTUM |p|140 MeV
GAP AT |p|, Δω(p)244.0 MeV
SHRINK Δω(p)/Δω(0)0.874
Live interpretationμₜₕ − m AT REST: 279.1 MeV. PACKET MOMENTUM |p|: 140 MeV. GAP AT |p|, Δω(p): 244.0 MeV. SHRINK Δω(p)/Δω(0): 0.874
Catch the common trap
Explain before calculating.
Computing ψψ → ψψ to order g⁴ in the scalar Yukawa model, a student draws a φψ self-energy loop on one incoming ψ leg and, with a mass counterterm placing the free states at the physical mass M, finds the factor [Σ(p²) − Σ(M²)]/(p² − M²) at p² = M². How should this diagram be treated?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn φ⁴ theory the particle has m = 0.50 GeV and, by the φ → −φ symmetry, the field's continuum begins at μₜₕ = 1.50 GeV. Find the gap and the dephasing scale at rest and for a packet with |p| = 2.0 GeV, test the fast-packet form (μₜₕ² − m²)/2|p|, and compare the dephasing time with the flight time to a detector 2.0 m away.
- At rest the gap is μₜₕ − m = 1.00 GeV, so ħ/Δω = 6.582 × 10⁻²⁵ GeV s ÷ 1.00 GeV = 6.6 × 10⁻²⁵ s and ħc/Δω = 0.197 GeV fm ÷ 1.00 GeV = 0.197 fm.
- At |p| = 2.0 GeV the particle sits at √(4.00 + 0.25) = 2.0616 GeV and the continuum edge at √(4.00 + 2.25) = 2.5000 GeV, so Δω = 0.4384 GeV: ħ/Δω = 1.50 × 10⁻²⁴ s and ħc/Δω = 0.450 fm.
- The fast-packet form gives (2.25 − 0.25)/(2 × 2.0) = 0.500 GeV, 14% high, because |p| = 2.0 GeV is not much larger than μₜₕ = 1.50 GeV; it becomes exact as |p| → ∞. Both forms have dimensions of energy.
- The packet moves at v = |p|/E = 0.9701, so a 2.0 m flight takes 2.0 m ÷ (0.9701 × 2.998 × 10⁸ m s⁻¹) = 6.88 ns, which is 4.6 × 10¹⁵ dephasing times.
AnswerΔω = 1.00 GeV at rest (6.6 × 10⁻²⁵ s) and 0.438 GeV at |p| = 2.0 GeV (1.50 × 10⁻²⁴ s, 0.450 fm); the fast-packet form gives 0.500 GeV. The 6.88 ns flight is 4.6 × 10¹⁵ dephasing times, so t → ∞ is safe.
MediumIn the scalar Yukawa model with M = 0.40 GeV and g = 1.0 GeV, take the φ mass first as m = 0.70 GeV, then as m = 0.90 GeV. For each, find the lowest threshold φ can reach and decide whether φ can be an LSZ external particle. Give the dephasing scale if it is stable; if not, use Γ = (g²/16πm)√(1 − 4M²/m²), quoted from topic 5.03, for its lifetime and its mean decay length at |p| = 2.0 GeV.
- φ couples to ψψ̄, threshold 2M = 0.80 GeV, and, with no φ → −φ symmetry in this model, to φφ at twice m: 1.40 GeV or 1.80 GeV. The lowest threshold is 0.80 GeV in both cases.
- At m = 0.70 GeV the particle sits 0.10 GeV below it: φ is stable, its pole isolated, and the continuum dephases at rest on ħ/Δω = 6.6 × 10⁻²⁴ s, ħc/Δω = 1.97 fm. LSZ applies.
- At m = 0.90 GeV it sits 0.10 GeV above, so φ → ψψ̄ is open: √(1 − 0.64/0.81) = 0.4581 and Γ = [1.0/(16π × 0.90)] × 0.4581 GeV = 0.01013 GeV = 10.1 MeV.
- The lifetime is ħ/Γ = 6.50 × 10⁻²³ s, cτ = 19.5 fm. At |p| = 2.0 GeV, γβ = |p|/m = 2.22 and the mean decay length is 43.3 fm: the one-particle term is gone long before any detector, so aₒᵤₜ(f) = 0 and φ has no out state.
- It appears instead as a Breit-Wigner peak of full width 10.1 MeV at √s = 0.90 GeV in ψψ̄ → ψψ̄. Checks: g²/m has dimensions of energy; Γ → 0 as m → 2M from above, joining the stable side; Γ/m = 0.011, so the peak is narrow.
AnswerThe threshold is 0.80 GeV both times. At m = 0.70 GeV φ is stable and LSZ applies (ħ/Δω = 6.6 × 10⁻²⁴ s). At m = 0.90 GeV, Γ = 10.1 MeV, ħ/Γ = 6.50 × 10⁻²³ s and the decay length at 2.0 GeV is 43 fm: no out state, only a resonance.
HardIn the scalar Yukawa model take M = 1.00 GeV and g = 2.0 GeV. Differentiating Unit 6's bubble with ψ in the loop gives, quoted, Σ′(p²) = −(g²/16π²)∫₀¹ x(1 − x) dx/(M² − x(1 − x)p²). Show why topic 2.09's route fails for this loop on an external φ leg. Find Z and √Z at m = 1.20 GeV against the naive 0/0 factor, then Z at m = 1.98 GeV and 1.999 GeV, and say what happens as m approaches 2M.
- With free states of mass m₀, the line from the ψ loop to the vertex carries k with k² = m₀², so the diagram holds Σ(m₀²)/(iε), infinite for any regulator; with a counterterm it is 0/0. Resummed, the leg is i/(p² − m₀² − Σ) ≈ iZ/(p² − m²) with 1/Z = 1 − Σ′(m²), and LSZ's factor (p² − m²)/(i√Z) leaves √Z.
- With a = m²/M² the integral is J/M², where J = ∫₀¹ x(1 − x) dx/(1 − a x(1 − x)) = (I − 1)/a and I = 4 arctan√(a/(4 − a))/√(a(4 − a)). At a = 1.44: √(a(4 − a)) = 1.92, arctan 0.75 = 0.6435, I = 1.3406 and J = 0.23655.
- κ = g²/(16π²M²) = 4.0/157.91 = 0.02533, so Σ′(m²) = −κJ = −5.99 × 10⁻³, Z = 1/1.00599 = 0.99404 and √Z = 0.99702, within 2 × 10⁻⁵ of 1 + ½Σ′ = 0.99700. The naive 0/0 cancellation gives 1 + Σ′ = 0.99401 per leg, twice the correction.
- At m = 1.98 GeV, a = 3.9204: √(a(4 − a)) = 0.5586, arctan 7.018 = 1.4293, I = 10.234 and J = 2.3554, so Σ′ = −0.0597 and Z = 0.944. At m = 1.999 GeV, J = 11.94 and Z = 0.768.
- As a → 4, I → π/√(4 − a), so J grows like 1/√(2M − m) and Z = 1/(1 + κJ) falls towards zero: in this one-loop estimate the residue drains into the ψψ̄ continuum as the gap closes. Near threshold κJ is no longer small, and the estimate shows only the trend.
- Checks: Σ′ is a pure number, since Σ and p² both carry mass²; as m → 0, J → 1/6 and Σ′ → −κ/6 = −4.2 × 10⁻³; g → 0 gives Z = 1 and the tree-level legs; and Z stays between 0 and 1, as topic 4.03's positivity requires.
AnswerFree-state route: Σ(m₀²)/(iε), infinite. At m = 1.20 GeV, Σ′ = −5.99 × 10⁻³, Z = 0.9940 and √Z = 0.9970 per leg; the naive 0/0 gives 0.9940, in effect Z for √Z. Z = 0.944 at 1.98 GeV and 0.768 at 1.999 GeV, tending to zero as the gap closes.