University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.8
Interpolating fields and bound states
Topic 4.04 built LSZ around φ, and it is easy to leave believing the S-matrix belongs to the fields a Lagrangian lists. This topic takes that apart: we find the one property of φ the derivation spent, hand the job to φ³ and ψ†ψ, and so give scattering amplitudes to hydrogen and the deuteron, bound states with no field of their own.
Build the model
Connect the measurement to the mechanism.
Reread topic 4.04 and, beyond ⟨Ω|φ|Ω⟩ = 0, only one fact about φ is spent: it reaches the particle from the vacuum, ⟨Ω|φ(0)|p⟩ = √Z ≠ 0. Neither the canonical commutator nor the field equation enters. So run the derivation with any local operator O carrying the particle's quantum numbers. Smeared with a packet of the physical mass, O tends to ⟨Ω|O(0)|p⟩ times the same aᵢₙ(f), because in and out states belong to the Hamiltonian, not to the operator that probes them, and the multiparticle states O also makes have no pole at p² = m².
S comes out unchanged, with √ZO = |⟨Ω|O(0)|p⟩| dividing each leg in √Z's place: a residue ZO of mass dimension 2(dO − 1), not bounded by 1, regulator dependent, and cancelled. Symmetry decides which operators qualify: in φ⁴, φ³ does and φ² cannot, and the field equation gives ψ†ψ's overlap with the Yukawa φ exactly. The pay-off is bound states. Hydrogen has no field of its own in QED, nor the deuteron in a theory of nucleons, yet a product of their constituents' fields serves as O, so they have S-matrix elements, and they appear in their constituents' amplitude as a pole below threshold, just as an elementary particle would.
The S-matrix records poles and residues, not which particles were elementary. The argument needs what LSZ always needs, a stable particle on an isolated mass shell; its rigorous form is Haag-Ruelle theory.
- Simple definition
- An interpolating field for a particle is any local operator O with ⟨Ω|O(0)|Ω⟩ = 0 and ⟨Ω|O(0)|p⟩ ≠ 0; used in the LSZ formula with √ZO = |⟨Ω|O(0)|p⟩| dividing each leg, it gives the same S-matrix as the field in the Lagrangian.
- Example
- A theory of nucleons alone has no deuteron field, yet a neutron-proton product coupled to spin 1 overlaps the deuteron and so serves as O: the deuteron has S-matrix elements, and it sits in np → np as a pole at √s = 1875.61 MeV, 2.22 MeV below the threshold at 1877.84 MeV.
The only property of φ that topic 4.04 spent. O = φ returns topic 4.03's Z; any other O returns its own residue, which need not be below 1. (derived)
O local, with the particle's spin and charges; [O] = dO and [|p⟩] = −1 in mass units, so ZO has mass dimension 2(dO − 1)
Topic 4.04's formula with √Z replaced leg by leg. Only residues at pᵢ² = m² survive, so whatever else Oᵢ creates drops out and S is unchanged. (derived)
Zᵢ = |⟨Ω|Oᵢ(0)|pᵢ⟩|² for leg i; + for outgoing, − for incoming momenta in GeV; m the physical mass for every Oᵢ
A selection rule decides which operators qualify: odd ones such as φ³ may, even ones never, and φ²'s spectral density starts at 4m². (derived)
U the unitary operator of the unbroken symmetry φ → −φ; U|p⟩ = −|p⟩ because ⟨Ω|φ(0)|p⟩ ≠ 0; every order in the dimensionless λ
φ³ inherits φ's overlap through a self-contraction that grows as Λ²: the residue is regulator dependent, the S-matrix is not. (derived)
λ = 0; Λ a cutoff on |k| and m the mass, both in GeV; the overlap in GeV², so ZO = 9⟨0|φ²|0⟩² in GeV⁴
Exact, and equal to Σ(m²)√Z/g: ψ†ψ reaches the φ only because the interaction shifts its mass, so the overlap is of order g. (derived)
from (□ + m₀²)φ = −g(ψ†ψ − ⟨Ω|ψ†ψ|Ω⟩); m the physical and m₀ the Lagrangian φ mass, g in GeV; result in GeV
The same form as an elementary s-channel pole (topic 4.06), below threshold: its position is the mass, its residue a measurable coupling, unlike ZO.
m₁, m₂ the constituent masses and B the binding energy in GeV; gB the on-shell coupling in GeV for 2 → 2; stated, not derived
Reread LSZ: only one property of φ was spent
Units are natural, ħ = c = 1, with metric (+, −, −, −). Mark every line of topic 4.04 that used something about φ. The smeared operator φf(t), the Klein-Gordon product of the field with a packet f of the physical mass m, needs only a local operator to smear. Its time derivative, i∫d³x f*(□ + m²)φ, is calculus on f. The asymptotic limit is where the field earns its place: between complete sets of states the one-particle term is ⟨Ω|φ(0)|p⟩ f̃*(p), constant in time because f carries the pole mass, while multiparticle terms oscillate at continuously varying frequencies and average away. That step spent ⟨Ω|φ|Ω⟩ = 0, ⟨Ω|φ(0)|p⟩ ≠ 0, and Lorentz invariance, which makes the overlap the same for every p. It never touched the canonical commutator, which topic 4.03 spent only on the bound Z ≤ 1, nor the field equation. So any local Lorentz-scalar O with those two properties runs through the same derivation, with ⟨Ω|O(0)|p⟩ wherever √Z stood: Of(t) → ⟨Ω|O(0)|p⟩ aᵢₙ(f) weakly as t → −∞, and to ⟨Ω|O(0)|p⟩ aₒᵤₜ(f) as t → +∞. That is derived, not assumed.
The same in and out states, so the same S
Why the same answer, and not a new amplitude for each O? Because the limit lands on an operator that O does not define. aᵢₙ†(f) creates one particle in the packet f in the far past; it is fixed by the Hamiltonian's spectrum and the packet, whichever operator is used to reach it, and Sβα = ⟨β out|α in⟩ is built from such states alone. So topic 4.04's formula holds with √Zᵢ = |⟨Ω|Oᵢ(0)|pᵢ⟩| dividing leg i, the operators even differing from leg to leg, and returns the same S, up to a phase per leg that only re-phases the one-particle states. The residue is no longer pinned between 0 and 1: no commutator fixes the total weight of O's spectral density, so ZO can exceed 1, usually depends on the regulator, and has mass dimension 2(dO − 1). Check with n operators of dimension dO: the transformed correlator stripped of (2π)⁴δ⁴ has dimension n dO − 4n + 4; n factors of (p² − m²) add 2n and n of 1/√ZO remove n(dO − 1), leaving 4 − n, the dimension of ℳ. Quoted corollary: a local change of variables φ = χ + aχ³ alters every off-shell correlator and no S-matrix element.
Symmetry decides: φ³ qualifies, φ² cannot
φ⁴ theory is unchanged by φ → −φ, carried out by a unitary U with U|Ω⟩ = |Ω⟩ while the symmetry is unbroken. Since UφU† = −φ and ⟨Ω|φ(0)|p⟩ ≠ 0, the particle is odd, U|p⟩ = −|p⟩. Then ⟨Ω|φ²(0)|p⟩ = ⟨Ω|U†(Uφ²U†)U|p⟩ = −⟨Ω|φ²(0)|p⟩, which vanishes exactly, at every order in λ. Once its vacuum value is subtracted, φ² has a spectral density that begins at the two-particle threshold 4m², with no pole at m², so it cannot serve. φ³ is odd and can. At λ = 0 Wick's theorem gives φ³ = :φ³: + 3⟨0|φ²|0⟩φ, and :φ³: reaches only three-particle states, so ⟨0|φ³(0)|p⟩ = 3⟨0|φ²(0)|0⟩, which grows as 3Λ²/8π² with a cutoff Λ on |k|. At order λ the sunset diagram gives even :φ³: an overlap of its own. φ² would serve for a bound state of two particles below 2m, even under φ → −φ, if one existed; the contact force, repulsive for positive λ, binds none at weak coupling. The general rule: O must share every conserved quantum number of the particle, its charges, spin and parities.
ψ†ψ makes the Yukawa φ, and the field equation says how well
In the scalar Yukawa model ψ†ψ is a Lorentz scalar with zero ψ number, the quantum numbers of φ, so it qualifies, and the field equation fixes its overlap exactly. With −gψ†ψφ in the Lagrangian, (□ + m₀²)φ = −g(ψ†ψ − ⟨Ω|ψ†ψ|Ω⟩), the subtraction coming from the counterterm that keeps ⟨Ω|φ|Ω⟩ = 0. Translation invariance gives ⟨Ω|φ(x)|k⟩ = e(−ik⋅x)√Z, on which □ acts as −k² = −m². Hence ⟨Ω|ψ†ψ(0)|k⟩ = (m² − m₀²)√Z/g = Σ(m²)√Z/g, by topic 4.03's pole condition. Checks: [ψ†ψ] = 2, and GeV² ÷ GeV leaves GeV, matching dO − 1 = 1. Σ is of order g², so the overlap is of order g and vanishes in the free limit, where ψ†ψ only makes ψψ̄ pairs. Σ(m²) diverges logarithmically, so the overlap exists only with a regulator, as composite operators generally need; LSZ divides it out. Correlators of ψ†ψ and of −(□ + m₀²)φ/g differ by contact terms, where derivatives strike the θ functions of the time ordering; those sit at coincident points, lack the external pole and never reach S.
Bound states have an S-matrix without a field
Hydrogen has no field in QED, nor the deuteron in a theory of nucleons, yet both are stable and both are reached from the vacuum by a product of their constituents' fields: ψₑψₚ with spins coupled as in hydrogen's ground state, and a neutron-proton product coupled to spin 1 for the deuteron. Nothing above needed O to be fundamental, so each product is an interpolating field, and LSZ hands the bound state S-matrix elements of its own: np → dγ, the first step of Big Bang nucleosynthesis, or the photodisintegration γd → np. The overlap belongs to the operator, not to the atom. In non-relativistic normalisation ⟨0|ψₑψₚ(0)|H⟩ is the wavefunction at contact, up to spin factors, with |ψ(0)|² = 1/(πa₀³) = 2.15 × 10³⁰ m⁻³ for a₀ = 52.9 pm; smear O over a few a₀ and the overlap changes while S does not. One caveat: both systems couple to photons, so strictly an X + γ continuum starts right at each pole. Being neutral, they couple to soft photons only through their moments, which keeps that continuum weak at threshold; topic 4.09 and Unit 14 take up massless particles.
Poles below threshold, and no label for elementary
Stated, not derived here: a bound state also appears inside its constituents' amplitude, as a real pole in s below threshold. For the deuteron, the spin-1 S wave of np → np has it at √s = mₚ + mₙ − B = 1875.61 MeV, 2.22 MeV under threshold; hydrogen's Coulomb amplitude has poles at Eₙ = −13.6 eV/n², crowding up to threshold as a force of infinite range must. Set this beside the scalar Yukawa φ with m below 2M: it appears in ψψ̄ → ψψ̄ as g²/(m² − s), a pole below 4M² (topic 4.06). One particle had a Lagrangian field and the other did not, and the S-matrix shows both the same way, a pole whose position is a mass and whose residue is a coupling. Fields are variables: a nucleon contact interaction can be rewritten with an auxiliary deuteron field without changing S. The rigorous version is Haag-Ruelle scattering theory, named only. One refinement, quoted: for a state very near threshold, Weinberg showed in 1965 that the scattering length and effective range measure its bare elementary content, and the deuteron's positive 1.75 fm range says it has almost none.
Change one variable at a time
Make the relationship visible.
Start at the deuteron, B = 2.22 MeV and r = 1.75 fm: the readout gives a = 5.42 fm, the measured triplet scattering length. Slide B and the pole and −a move together; slide r at fixed B and the pole stays put while a and the residue change.
SCATTERING LENGTH a5.42 fm
POLE AT √s1875.62 MeV
RESIDUE 2κ/(1 − rκ)0.778 fm⁻¹
TAIL LENGTH 1/κ4.32 fm
Live interpretationSCATTERING LENGTH a: 5.42 fm. POLE AT √s: 1875.62 MeV. RESIDUE 2κ/(1 − rκ): 0.778 fm⁻¹. TAIL LENGTH 1/κ: 4.32 fm
Catch the common trap
Explain before calculating.
In φ⁴ theory a student redoes the 2 → 2 LSZ reduction with O = φ³ in place of φ on all four legs. Which statement about the result is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn φ⁴ theory with m = 0.50 GeV, work at λ = 0 with a cutoff Λ = 2.0 GeV on |k|. Show that φ² cannot be an interpolating field for the particle, then find ⟨0|φ³(0)|p⟩ and ZO for O = φ³, check their units, and say what doubling Λ does.
- φ → −φ leaves the vacuum alone and flips the particle, because ⟨Ω|φ(0)|p⟩ ≠ 0. φ² is even, so ⟨Ω|φ²(0)|p⟩ = −⟨Ω|φ²(0)|p⟩ = 0, at λ = 0 and at every order: φ² cannot serve.
- Wick's theorem: φ³ = :φ³: + 3⟨0|φ²|0⟩φ. The normal-ordered term needs three quanta, so ⟨0|:φ³:(0)|p⟩ = 0, leaving ⟨0|φ³(0)|p⟩ = 3⟨0|φ²(0)|0⟩⟨0|φ(0)|p⟩ = 3⟨0|φ²(0)|0⟩, since ⟨0|φ(0)|p⟩ = 1 for the free field.
- ⟨0|φ²(0)|0⟩ = ∫d³k/((2π)³2Eₖ) = (1/4π²)∫₀Λ k²dk/√(k² + m²) = (1/8π²)[Λ√(Λ² + m²) − m² ln((Λ + √(Λ² + m²))/m)].
- With √(Λ² + m²) = 2.0616 GeV: Λ√(Λ² + m²) = 4.1231 GeV², and m² ln(4.0616/0.50) = 0.25 × 2.0947 = 0.5237 GeV². The difference, 3.5994 GeV², over 8π² = 78.957 gives ⟨0|φ²(0)|0⟩ = 0.04559 GeV².
- So ⟨0|φ³(0)|p⟩ = 0.1368 GeV² and ZO = 0.01870 GeV⁴. Units: [φ³] = 3 and [|p⟩] = −1 give GeV², and ZO has mass dimension 2(3 − 1) = 4. At Λ = 4.0 GeV the same steps give ⟨0|φ²(0)|0⟩ = 0.1954 GeV² and ZO = 0.344 GeV⁴, 18 times larger.
Answerφ² has zero overlap by symmetry. ⟨0|φ³(0)|p⟩ = 0.137 GeV² and ZO = 0.0187 GeV⁴ at Λ = 2.0 GeV, rising 18-fold at 4.0 GeV: a regulator-dependent residue that LSZ divides out.
MediumIn the scalar Yukawa model take m = 0.600 GeV, M = 0.500 GeV and g = 1.50 GeV, with a cutoff of a few GeV that makes the Lagrangian mass m₀ = 0.640 GeV. Find Z from topic 4.03's order-g² formula, then ⟨Ω|ψ†ψ(0)|k⟩ and ZO for O = ψ†ψ, and show that reducing ψψ̄ → φφ with ψ†ψ on both φ legs returns the same ℳ.
- Topic 4.03: 1 − Z = (g²/16π²)∫₀¹ dx x(1 − x)/[M² − x(1 − x)m²]. The integral, done numerically, is 0.9462 GeV⁻², and g²/16π² = 0.014248 GeV², so 1 − Z = 0.01348, Z = 0.9865 and √Z = 0.9932.
- Field equation: (□ + m₀²)φ = −g(ψ†ψ − ⟨Ω|ψ†ψ|Ω⟩). Between ⟨Ω| and |k⟩, □ acts on e(−ik⋅x)√Z as −m², so ⟨Ω|ψ†ψ(0)|k⟩ = (m² − m₀²)√Z/g.
- m² − m₀² = 0.3600 − 0.4096 = −0.0496 GeV², so ⟨Ω|ψ†ψ(0)|k⟩ = −0.0496 × 0.9932/1.50 = −0.03284 GeV and ZO = 1.079 × 10⁻³ GeV². The minus sign is a phase, absorbed into the one-particle state.
- In ⟨Ω|Tψψ†OO|Ω⟩ the double pole at k₁² = k₂² = m² carries ⟨Ω|O(0)|k₁⟩⟨Ω|O(0)|k₂⟩ where the φ correlator carries Z: a ratio ((m² − m₀²)/g)² = 1.093 × 10⁻³ GeV². LSZ divides each of those legs by √ZO instead of √Z, the ratio cancels, and ℳ = g²[1/(M² − t) + 1/(M² − u)] as in topic 4.06.
- Checks: [ψ†ψ] = 2, so the overlap is in GeV and ZO has dimension 2(2 − 1) = 2. As g → 0, m² − m₀² = Σ(m²) falls as g², so the overlap falls as g: free ψ†ψ only makes ψψ̄ pairs.
AnswerZ = 0.987; ⟨Ω|ψ†ψ(0)|k⟩ = −0.0328 GeV and ZO = 1.08 × 10⁻³ GeV². The double-pole residue is 1.09 × 10⁻³ GeV² times φ's, LSZ removes exactly that factor, and ℳ is unchanged.
HardSpin-1 np scattering has measured aₜ = 5.419 fm and rₜ = 1.753 fm, so near threshold the S-wave amplitude is f(k) = 1/(−1/aₜ + rₜk²/2 − ik), with ℳ = 8π√s f. Continue f below threshold, locate the bound-state pole, and predict the deuteron's binding energy, the √s of the pole and its residue. Compare with the measured B = 2.2246 MeV and with a zero-range model.
- Below threshold E = −(ħcq)²/2μ, so set k = iq: f = 1/(q − 1/aₜ − rₜq²/2), real, because no channel is open to make it complex. A pole needs (rₜ/2)q² − q + 1/aₜ = 0; the root that tends to 1/aₜ as rₜ → 0 is κ = [1 − √(1 − 2rₜ/aₜ)]/rₜ.
- 2rₜ/aₜ = 0.6470 and √0.3530 = 0.5942, so κ = 0.4058/1.753 fm = 0.2315 fm⁻¹ and ħcκ = 45.68 MeV.
- μ = mₚmₙ/(mₚ + mₙ) = 469.46 MeV, so B = (ħcκ)²/2μ = 2087.1/938.92 = 2.2228 MeV, 0.08% under the measured 2.2246 MeV. The zero-range guess κ = 1/aₜ gives ħcκ = 36.41 MeV and B = 1.41 MeV, 37% low: the force's range matters.
- The pole sits at √s = mₚ + mₙ − B = 1877.837 − 2.223 = 1875.614 MeV, so s = 3.5179 GeV² against the threshold s = 3.5263 GeV²: real and below the physical region, where an elementary particle's s-channel pole would sit.
- Near k = iκ, f ≈ −A²/(k² + κ²) with A² = 2κ/(1 − rₜκ) = 0.4630/0.5942 = 0.779 fm⁻¹, so A = 0.883 fm(−1/2), against the measured asymptotic normalisation 0.8846 fm(−1/2). Since k² + κ² ≈ μ(s − md²)/md, ℳ ≈ gd²/(md² − s) per spin state, with gd² = 8πmd²(ħcA²)/μ = 2.9 × 10⁷ MeV², gd ≈ 5.4 GeV.
- Checks: κ in fm⁻¹ and B in MeV. B → 0 sends κ → 0 and aₜ → ∞, the pole onto threshold; rₜ → 0 returns the zero-range model, κ = 1/aₜ.
Answerκ = 0.2315 fm⁻¹, so B = 2.223 MeV (measured 2.2246 MeV) and the pole sits at √s = 1875.61 MeV, 2.22 MeV below threshold, with residue A² = 0.779 fm⁻¹; zero range gives only B = 1.41 MeV.