University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.3

Physical mass and field-strength renormalisation Z

Every amplitude from here on is read off a pole: its position is the mass we quote, its residue the √Z on each external leg. This lesson fixes both from first principles, shows why the canonical commutator caps Z at 1, and computes Z from the self-energy in a model where the answer comes out finite.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 4.02 wrote the exact two-point function as a positive sum of free propagators; this lesson reads off what the one-particle states contribute. Near p² = m² the exact propagator is iZ/(p² − m² + iε) plus terms smooth there, because every multiparticle state lies above a threshold. Two numbers describe the pole. Its position m is the mass of the stable particle, what a spectrometer returns, and it is the zero of p² − m₀² − Σ(p²), not the Lagrangian's m₀.

Its residue Z = |⟨Ω|φ(0)|p⟩|² says how strongly φ makes that particle from the vacuum, and Lorentz invariance makes it the same for every p. The canonical commutator, untouched by interactions without derivatives, fixes the total spectral weight at 1; positivity splits it into Z for the pole and 1 − Z for the continuum, so 0 ≤ Z ≤ 1, with Z = 1 only for a field making nothing but single particles. Expanding the inverse propagator about its zero gives 1/Z = 1 − dΣ/dp². The model has edges.

A derivative coupling changes the conjugate momentum and the sum rule with it; an unstable particle, or a massless one at a threshold, has no isolated pole; and with dimensionless couplings in four dimensions perturbation theory finds the continuum weight logarithmically divergent, so Z exists only with a regulator. That is harmless: Z belongs to the field chosen, not to the particle, and nothing measures it, whereas m is measured.

Simple definition: The physical mass m is the position of the exact propagator's one-particle pole, and the field-strength renormalisation Z is its residue, Z = |⟨Ω|φ(0)|p⟩|², the share of the field's total spectral weight that goes into making a single particle.
Example: In the scalar Yukawa model with M = m = 1.00 GeV and g = 5.0 GeV, ψψ̄ pairs above 4M² = 4.00 GeV² take 1 − Z = 0.033 of the weight at order g², so Z = 0.967 and ⟨Ω|φ(0)|p⟩ = √Z = 0.983 at every momentum.

The one-particle pole of the exact propagator∫ d⁴x e^(ip⋅x) ⟨Ω|Tφ(x)φ(0)|Ω⟩ = iZ/(p² − m² + iε) + terms regular at p² = m²

Derived from topic 4.02's spectral form: one-particle states give the pole, and every multiparticle state lies above a threshold, so nothing else is singular at m².

p and the physical mass m in GeV, the left side in GeV⁻², Z dimensionless; ħ = c = 1, metric (+, −, −, −)

The residue as a vacuum-to-particle amplitudeZ = |⟨Ω|φ(0)|p⟩|², the same for every p

Derived: U(Λ) leaves Ω and the scalar φ(0) alone and sends |p⟩ to |Λp⟩, so no frame is special. The free field's mode expansion gives exactly 1.

[φ] = 1 and [|p⟩] = −1 in mass units, with ⟨p|q⟩ = (2π)³ 2Eₚ δ³(p − q), so Z is dimensionless

Spectral sum ruleZ + ∫ (ds/2π) ρ_c(s) = 1, integrated from sₜₕ to ∞

Derived from [φ(t, x), ∂ₜφ(t, y)] = iδ³(x − y): each mass μ in the spectral sum contributes iδ³ times its weight. With ρ_c ≥ 0 it gives 0 ≤ Z ≤ 1.

ρ_c ≥ 0 the multiparticle density in GeV⁻², sₜₕ its threshold in GeV²; canonical φ, non-derivative couplings only

The physical massm² − m₀² − Σ(m²) = 0

Derived as the zero of the summed propagator's denominator. m is what is measured; m₀² = m² − Σ(m²) shifts whenever the regulator does.

m₀ the Lagrangian mass; Σ(p²) topic 3.08's self-energy, propagator i/(p² − m₀² − Σ(p²) + iε); every term in GeV²

The residue from the self-energy1/Z = 1 − dΣ/dp², evaluated at p² = m²

Derived by expanding p² − m₀² − Σ(p²) to first order about its zero. A p-independent Σ, such as φ⁴'s one-loop tadpole, leaves Z = 1.

dΣ/dp² is dimensionless; the exact spectral form forces dΣ/dp² = 1 − 1/Z ≤ 0 at the pole

Scalar Yukawa model at order g²1 − Z = (g²/16π²) ∫₀¹ dx x(1 − x)/[M² − x(1 − x)m²]

Derived from the ψ loop: Σ's divergence is a p²-independent constant and drops out of dΣ/dp². Finite because [g] = 1, and unbounded as m approaches the threshold 2M.

g in GeV from −gψ†ψφ; M the ψ mass and m the φ mass, below 2M; tends to g²/(96π²M²) as m → 0

The pole, its position and its residue

Work in natural units, ħ = c = 1, with the metric (+, −, −, −), and let φ be a real scalar with the canonical ½∂_μφ∂^μφ in its Lagrangian. Topic 4.02 showed, without perturbation theory, that the Fourier transform of ⟨Ω|Tφ(x)φ(0)|Ω⟩ is ∫(dμ²/2π) ρ(μ²) i/(p² − μ² + iε) with ρ ≥ 0. A stable particle of mass m that φ can create puts 2πZδ(μ² − m²) into ρ, hence iZ/(p² − m² + iε) into the propagator. Everything else sits elsewhere: any other particle φ can make has a pole at its own mass, and states of two or more particles begin at sₜₕ, above m², so near p² = m² the rest is smooth and the pole is isolated. m² is the value of P^μP_μ on the one-particle state, so m is the mass a spectrometer measures. The residue is Z = |⟨Ω|φ(0)|p⟩|², and since U(Λ) fixes Ω and the scalar φ(0) while sending |p⟩ to |Λp⟩, one Z serves every momentum. Dimensions: [φ] = 1 and [|p⟩] = −1, so Z is a pure number and the transform carries GeV⁻², as i/(p² − m²) does. Free limit: the mode expansion with [a(p), a†(q)] = (2π)³2Eₚ δ³(p − q) gives ⟨0|φ(0)|p⟩ = 1, so Z = 1 and ρ is one δ function.

The canonical commutator fixes the total weight at 1

With no derivatives in Lᵢₙₜ the momentum conjugate to φ is still π = ∂ₜφ, so the Heisenberg field obeys [φ(t, x), ∂ₜφ(t, y)] = iδ³(x − y) exactly. Take its vacuum expectation value and use the spectral form for both orderings: ⟨Ω|[φ(x), φ(y)]|Ω⟩ becomes ∫(dμ²/2π) ρ(μ²) times the free commutator function of mass μ, the Pauli-Jordan function of UAQM-09.04, and by UAQM-09.02 its time derivative at equal times is iδ³(x − y) whatever μ is. Matching the two sides gives ∫(dμ²/2π) ρ(μ²) = 1. Split ρ into its pole and its continuum and Z = 1 − ∫(ds/2π) ρ_c(s). Each term of ρ_c is a squared modulus |⟨Ω|φ(0)|n⟩|², so ρ_c ≥ 0 and Z ≤ 1, while Z ≥ 0 is itself a squared modulus. Z = 1 needs ρ_c = 0, that is ⟨n|φ(0)|Ω⟩ = 0 for every multiparticle state n: φ makes nothing but single particles, which is the free field. Any interaction that lets φ reach two or three particles pushes Z below 1. The proof used π = ∂ₜφ; a derivative coupling changes π, and the right-hand side of the sum rule with it.

The physical mass, and Z from the self-energy

Topic 3.08 summed the one-particle-irreducible insertions into i/(p² − m₀² − Σ(p²) + iε). Its pole is not at m₀². The denominator vanishes where m² − m₀² − Σ(m²) = 0, and that root is the m of the spectral form, the measured mass. Expand about it: p² − m₀² − Σ(p²) = (p² − m²)[1 − Σ′(m²)] + O((p² − m²)²), with Σ′ = dΣ/dp². Near the pole the propagator is therefore i/([1 − Σ′(m²)](p² − m²)), and matching iZ/(p² − m²) gives 1/Z = 1 − Σ′(m²). Read the other way, the exact spectral form forces Σ′(m²) = 1 − 1/Z ≤ 0: positivity forbids the self-energy from rising through the pole, a sign check on any calculation. Two limits. A constant Σ, which is φ⁴'s one-loop tadpole, only shifts the mass and leaves Z = 1, as topic 4.05 uses; φ⁴'s first Z ≠ 1 comes at order λ² from the sunset diagram. And the derivative is taken at m², not m₀²; the difference changes Z only at the next order. The mass and the residue are properties of the exact propagator, while m₀ and the split of the denominator into m₀² and Σ depend on the regulator.

A finite case: the scalar Yukawa bubble

In the scalar Yukawa model, Lᵢₙₜ = −gψ†ψφ with complex ψ of mass M and real φ of mass m below 2M, φ's order-g² self-energy is topic 3.06's ψ loop, −iΣ(p²) = (−ig)² ∫d⁴k/(2π)⁴ [i/(k² − M²)][i/((k + p)² − M²)], with no ½ because ψ is complex. It diverges logarithmically, but only through a p²-independent constant, which moves m₀ and nothing else. So differentiate first. Feynman parameters, the Wick rotation and the quoted ∫d⁴l/(2π)⁴ (l² − Δ)⁻³ = −i/(32π²Δ) give −Σ′(p²) = (g²/16π²) ∫₀¹dx x(1 − x)/Δ, with Δ = M² − x(1 − x)p², finite and positive below threshold. Dimensions: g² in GeV² over Δ in GeV², a pure number. As g → 0, Z → 1; as m → 0 the integral tends to 1/(6M²). For M = m = 1.00 GeV and g = 5.0 GeV, g²/16π² = 0.158 GeV² and the integral is 0.209 GeV⁻², so 1 − Z = 0.033. As m → 2M the integrand at x = ½ grows like 1/(4M² − m²) and the integral like 1/√(4M² − m²): at fixed g, 1/Z = 1 − Σ′ grows without bound and Z falls towards zero, the field's weight draining into ψψ̄ pairs just above the pole.

Divergent in four dimensions, and never measured

Why was that finite? Count dimensions. The sum rule makes ρ a mass⁻², and far above every threshold the only scale left in ρ_c(s) is s. A coupling of mass dimension 1, like g, lets ρ_c fall as g²/s²: here ρ_c(s) = (g²/8π)β/(s − m²)² with β = √(1 − 4M²/s), and the weight above s is about g²/(16π²s). A dimensionless coupling allows only 1/s, and ∫ds/s diverges. In φ⁴, three-body phase space (quoted) gives ρ_c ≈ λ²/(1536π³s) at large s, so with a cutoff Λ, 1 − Z ≈ [λ²/(12(16π²)²)] ln(Λ²/m²), the standard two-loop result: 1.5 × 10⁻⁴ for λ = 1 and Λ/m = 10¹⁰, small but unbounded. QED's electron and photon fields do the same; Unit 6 absorbs such factors into the counterterm δ_Z, and Unit 12 treats QED's Z₂ and Z₃ that way. No observable is touched, because Z is not one: rescale φ → cφ and the residue becomes c²Z while every S-matrix element stays put, since LSZ divides each leg by the √Z the correlation function carries. The mass is measured. The charged pion's 139.570 MeV and the electron's 0.51099895 MeV are pole positions; no measurement returns the Z of a field.

Change one variable at a time

Make the relationship visible.

Interactive model

Coupling g/M6.0

Mass ratio m/M1.50

Set g/M to 0 for a free field: one jump of exactly 1. Raise it and weight leaves the pole for the continuum. Push m/M towards 2: the pole crowds the threshold, Z falls, the dashed share piles up near 4M², and the two readouts for 1 − Z, from ρ and from dΣ/dp², stay equal.

POLE RESIDUE Z0.928

CONTINUUM WEIGHT ∫ρ_c ds/2π0.072

−dΣ/dp² AT p² = m²0.072

SHARE OF 1 − Z ABOVE 100 M²3.2 %

Live interpretationPOLE RESIDUE Z: 0.928. CONTINUUM WEIGHT ∫ρ_c ds/2π: 0.072. −dΣ/dp² AT p² = m²: 0.072. SHARE OF 1 − Z ABOVE 100 M²: 3.2 %

Catch the common trap

Explain before calculating.

Misconception check

The canonical commutator survives the interaction unchanged, so it is tempting to conclude that the interacting field still creates its particle with unit strength: Z = 1, and the free-field factors carry straight into LSZ. The commutator does survive, but what it fixes is the total, ∫(dμ²/2π) ρ(μ²) = 1, summed over everything φ(0) can make from the vacuum. Once φ couples to anything, φ(0)|Ω⟩ also has components along multiparticle states, each with positive weight, and the one-particle pole keeps only what they leave, Z = 1 − ∫(ds/2π) ρ_c. In the scalar Yukawa model with M = m = 1.00 GeV and g = 5.0 GeV that is 0.967, not 1. The opposite slip is as common: reading 1 − Z as a probability that the particle fails to be produced, so that rates should carry a factor Z. They do not. LSZ divides out each external leg's √Z, and Z itself changes if φ is rescaled, which no rate does.

In a theory of one real scalar with non-derivative interactions, the full propagator is written i/(p² − m₀² − Σ(p²) + iε). A student's one-loop calculation gives dΣ/dp² = +0.05 at the physical pole p² = m². What should the student conclude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA canonically normalised real scalar with only non-derivative self-couplings and a φ → −φ symmetry has a stable particle of mass m = 0.80 GeV, so its continuum begins at s₀ = (3m)². Model that continuum by the toy density ρ_c(s) = c/s² for s ≥ s₀, with c = 0.90 GeV², which falls faster than the 1/s a dimensionless φ⁴ coupling gives and so has finite weight. Check the dimensions of ρ_c, then find Z, dΣ/dp² at the pole, and the largest c the sum rule allows.

Dimensions: c/s² carries GeV² ÷ GeV⁴ = GeV⁻², which is what ∫(ds/2π) ρ = 1 demands of a spectral density. The threshold is s₀ = (3 × 0.80 GeV)² = 5.76 GeV², well above m² = 0.64 GeV², so the pole is isolated.
Continuum weight: the integral of (ds/2π) c/s² from s₀ to ∞ is c/(2πs₀) = 0.90/(2π × 5.76) = 0.90/36.19 = 0.0249.
Sum rule: Z = 1 − 0.0249 = 0.975. Lorentz invariance makes this the residue for the particle at rest and at any momentum alike.
From 1/Z = 1 − Σ′(m²): Σ′(m²) = 1 − 1/Z = 1 − 1.0255 = −0.0255. It is negative, as positivity requires, and the first-order estimate Σ′ ≈ −(1 − Z) = −0.0249 differs from it only at second order.
Bound: Z ≥ 0 requires c/(2πs₀) ≤ 1, so c ≤ 2π × 5.76 GeV² = 36.2 GeV². At c = 0, Z = 1, the free field.

AnswerZ = 0.975 and dΣ/dp² = −0.0255 at p² = m²; the sum rule allows c ≤ 36.2 GeV², and ρ_c carries GeV⁻² as a spectral density must.

MediumIn the scalar Yukawa model, Lᵢₙₜ = −gψ†ψφ, take M = 1.00 GeV for ψ and g = 6.0 GeV. The φ particle is measured at m = 1.50 GeV. Find dΣ/dp² at the pole and Z to order g², then repeat for m = 1.90 GeV, and check the m → 0 limit. Why does the divergence of Σ not enter?

Thresholds: φ reaches ψψ̄ from s = 4M² = 4.00 GeV², above m² = 2.25 GeV², so the pole is isolated. The φφ continuum, from 4m² = 9.00 GeV², enters ρ only at order g⁶, through a ψ triangle.
Σ's divergence is a constant independent of p², so it fixes m₀² = m² − Σ(m²) and vanishes from the derivative: −Σ′(m²) = (g²/16π²) ∫₀¹dx x(1 − x)/[M² − x(1 − x)m²], finite.
With r = m/M, completing the square in x gives the integral as [4 arcsin(r/2)/(r√(4 − r²)) − 1]/(r²M²). For r = 1.50: arcsin 0.75 = 0.84806 and √1.75 = 1.32288, so the integral is (1.70953 − 1)/2.25 = 0.31535 GeV⁻².
g²/16π² = 36.0/157.91 = 0.22797 GeV², so Σ′(m²) = −0.22797 × 0.31535 = −0.0719. Then 1/Z = 1.0719 and Z = 0.933, which agrees with 1 − 0.0719 = 0.928 to order g²: Z ≈ 0.93.
For r = 1.90: arcsin 0.95 = 1.25324 and √0.39 = 0.62450, so the integral is (4.22481 − 1)/3.61 = 0.89330 GeV⁻² and Σ′(m²) = −0.204. Now 1/1.2036 = 0.8308 and 1 − 0.2036 = 0.7964 differ by 0.034, the size of the uncomputed order-g⁴ terms: Z ≈ 0.8, and perturbation theory is straining near threshold.
Limit m → 0: the bracket tends to 1/6, giving Σ′ = −g²/(96π²M²) = −0.0380, smaller in size than both results, as the growth towards threshold requires.

Answerm = 1.50 GeV: dΣ/dp² = −0.072 and Z ≈ 0.93; m = 1.90 GeV: dΣ/dp² = −0.20 and Z ≈ 0.8, with order-g⁴ terms near 0.03. Σ's divergence is independent of p² and moves only m₀.

HardSame model, M = 1.00 GeV, m = 1.50 GeV, g = 6.0 GeV. (a) Build the order-g² continuum ρ_c(s) from ⟨Ω|φ(0)|ψψ̄⟩ and two-body phase space, and check its dimensions. (b) Integrate it to get 1 − Z from the sum rule and compare with −dΣ/dp² = 0.0719 from the Feynman-parameter route. (c) How much of that weight lies above s = 100 GeV², and why would the matching weight in φ⁴ have no limit?

At order g one vertex and a φ propagator give ⟨Ω|φ(0)|ψ(k₁)ψ̄(k₂)⟩ = (−ig) × i/(s − m²) = g/(s − m²), with s = (k₁ + k₂)². Summing the pair over phase space, ∫d³k₁ d³k₂ (2π)⁴δ⁴(p − k₁ − k₂)/[(2π)⁶ 4E₁E₂] = β/8π with β = √(1 − 4M²/s), gives ρ_c(s) = (g²/8π)β/(s − m²)² for s ≥ 4M²: GeV² × GeV⁻⁴ = GeV⁻², as required.
Substitute v = β: s = 4M²/(1 − v²), ds = 8M²v dv/(1 − v²)² and s − m² = (4M² − m² + m²v²)/(1 − v²). The weight becomes 1 − Z = (g²/16π²) ∫₀¹ 8M²v² dv/(m²v² + 4M² − m²)², a rational integral.
With a = m² = 2.25 and b = 4M² − m² = 1.75 (GeV², M = 1 GeV), ∫v²dv/(av² + b)² = [arctan(v√(a/b))/√(ab) − v/(av² + b)]/(2a), so from 0 to 1 the integral is (4/2.25)[0.84806/1.98431 − 0.25000] = 1.77778 × 0.17738 = 0.31535 GeV⁻². Here arctan(1.5/1.32288) = 0.84806 is the bubble's arcsin 0.75.
Times g²/16π² = 0.22797 GeV²: 1 − Z = 0.0719, equal to −dΣ/dp² to every digit. That is 1/Z = 1 − dΣ/dp² at order g² seen from the spectral side: integrating the continuum and differentiating the self-energy are two routes to one weight.
Up to s = 100 GeV², v = √0.96 = 0.97980 and the integral is (4/2.25)[0.83792/1.98431 − 0.97980/3.91000] = 1.77778 × 0.17169 = 0.30522 GeV⁻². The tail is 0.31535 − 0.30522 = 0.01013 GeV⁻², a weight 0.22797 × 0.01013 = 0.0023, or 3.2% of 1 − Z. For s ≫ 4M² the density is g²/(8πs²), whose tail g²/(16π²s) = 0.0023 agrees.
In φ⁴ the coupling λ is dimensionless, so the large-s density can only fall as 1/s; massless three-body phase space, quoted, gives ρ_c ≈ λ²/(1536π³s). The weight between S and Λ² is then [λ²/(3072π⁴)] ln(Λ²/S), which grows without limit as Λ → ∞: 7.7 × 10⁻⁵ for λ = 1 and Λ² = 10¹⁰ S, and twice that at 10²⁰ S.

Answerρ_c = (g²/8π)β/(s − m²)² in GeV⁻²; 1 − Z = 0.0719 by both routes; 0.0023 of it, 3.2%, lies above 100 GeV² and falls as 1/s, while φ⁴'s 1/s density gives a weight growing as ln Λ².