University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.5

Amputation: iℳ from amputated diagrams, √Z per leg

LSZ says an S-matrix element hides in the poles of a correlation function. This lesson turns that statement into the working recipe for the rest of the course: strip off every external leg, self-energies and all, keep one √Z per particle, discard the pieces in which someone never interacts, and read off iℳ.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 4.04 proved an exact statement that is awkward to use: an S-matrix element is the coefficient of the product of external poles in a Fourier-transformed correlation function. Diagrams make that coefficient easy to read. Follow any external point of a diagram inward and the line first runs through a chain of self-energy insertions before it reaches anything that involves another particle; summed, that chain is the full propagator, which near p² = m² is iZ/(p² − m² + iε) plus terms with no pole.

LSZ multiplies each leg by (i/√Z)(m² − p²), which cancels that pole, kills the regular terms and leaves √Z. So iℳ is (√Z)ⁿ times the sum of connected, amputated diagrams for n external particles. Pieces in which a particle passes untouched have fewer independent poles than legs and give only the 1 in S, or a smaller process with a spectator.

At tree level Z = 1, and in φ⁴ the one-loop tadpole carries no external momentum, so it moves the mass and leaves Z = 1 to order λ. The recipe leans on an isolated pole. It needs repair for a particle that decays, which has no pole on the real axis, and for an electron dressed by massless photons, whose continuum starts at the pole itself and makes Z infrared divergent.

Simple definition: An amputated diagram is a Feynman diagram with each external leg's full propagator, every self-energy on that leg included, cut away; iℳ is the sum of the connected, amputated diagrams times one √Z for each external particle.
Example: In φ⁴ at order λ the connected four-point function is −iλ times four leg propagators i/(pₖ² − m²). Cut the four legs, multiply by (√Z)⁴ = 1, and iℳ = −iλ, so λ = 0.25 gives ℳ = −0.25 for φφ → φφ at every energy and angle.

One external leg through LSZ(i/√Z)(m² − p²) × iZ/(p² − m² + iε) → √Z as p² → m²

Derived: i × i × (−1) = +1, so the pole cancels, the residue's Z is divided by √Z, and one √Z is all the leg leaves behind.

p and m in GeV; Z dimensionless, nonzero and at most 1 at a true pole; the leg's terms without a pole are multiplied by zero

Amputation formulaiℳ = (√Z)ⁿ × [sum of connected, amputated diagrams], overall (2π)⁴δ⁴ removed

Derived by matching the n-fold pole leg by leg. With two species each leg carries its own factor, √Z_φ or √Z_ψ.

n counts every external particle, in and out; ℳ has mass dimension 4 − n, a pure number for 2 → 2

Z from the slope of the self-energyZ = 1/(1 − dΣ/dp²), evaluated at p² = m²

A Σ that does not depend on p² moves the pole, through m² = m₀² + Σ(m²), but leaves its residue at exactly 1.

Σ(p²) in GeV² is the 1PI self-energy in i/(p² − m₀² − Σ(p²) + iε); quoted from topic 4.03

The φ⁴ tadpole−iΣ = (−iλ/2) ∫ d⁴ℓ/(2π)⁴ i/(ℓ² − m² + iε)

Derived: the external p never enters the loop, so dΣ/dp² = 0 and Z = 1 + O(λ²). The divergent constant only shifts m₀² to m²; UQFTI-06.05 evaluates it.

λ dimensionless; loop momentum ℓ in GeV; Σ in GeV²; ½ is the symmetry factor of topic 3.02

Tree amplitude for φφ → φφiℳ = −iλ, so ℳ = −λ

Derived: one vertex, four amputated legs, (√Z)⁴ = 1 at this order. Dimensionless as 4 − n = 0 requires, and zero as λ → 0, where S = 1.

From Lᵢₙₜ = −(λ/4!)φ⁴: four external fields meet one vertex in 4! ways, cancelling the 1/4!

Scalar Yukawa rulesvertex −igψ line i/(q² − M² + iε)φ line i/(q² − m² + iε)external line 1

Vertex and lines quoted from topic 3.06; the external 1 is derived here, times √Z_ψ or √Z_φ = 1 + O(g²). One exchange gives ℳ of order g²/(q² − m²), dimensionless.

Lᵢₙₜ = −gψ†ψφ with g in GeV; M and m are the ψ and φ masses in GeV; arrows carry ψ number

Cut each external leg where it first meets the rest

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Fourier-transform a connected n-point function ⟨Ω|T φ(x₁)…φ(xₙ)|Ω⟩ and follow any external point of any diagram inward. Before the line meets anything that involves another particle, it may loop back on itself through self-energy insertions. Make the cut precise: on each leg, cut at the propagator farthest from the tip whose removal separates that one external point, and nothing else, from the rest of the diagram. Summed over all diagrams, what lies outside the cut on a leg is every chain of 1PI insertions, which topic 3.08 resummed into the full propagator D(p²) = i/(p² − m₀² − Σ(p²) + iε). What lies inside is an amputated diagram. Every diagram splits this way and every chain appears on every leg, so the sum factorises exactly: G̃(p₁, …, pₙ) = D(p₁²) D(p₂²) ⋯ D(pₙ²) × Ã(p₁, …, pₙ), where Ã is the sum of connected, amputated diagrams with the overall (2π)⁴δ⁴ set aside. Internal lines stay inside Ã, even those a single cut would sever.

LSZ cancels each pole and leaves √Z

Near p² = m², topic 4.03 gives D(p²) = iZ/(p² − m² + iε) + R(p²), with R regular at the pole. Topic 4.04's formula, read in momentum space, multiplies each leg by (i/√Z)(m² − p²), the Fourier image of (i/√Z)(∂² + m²), and sends p² → m². On the pole term that gives (i × i)(m² − p²)Z/[√Z(p² − m²)] = √Z exactly; on R it gives a factor (p² − m²) that sends it to zero. So every leg contributes √Z and nothing else, however complicated its self-energy. Comparing coefficients says the same thing: G̃ contains ∏ iZ/(pₖ² − m²), LSZ says it contains ∏ i√Z/(pₖ² − m²) times the S-matrix element, and the ratio is (√Z)ⁿ. Writing S = 1 + iT and ⟨f|iT|i⟩ = (2π)⁴δ⁴(Σk − Σp) iℳ, with k incoming and p outgoing as in topic 4.01, gives the amputation formula iℳ = (√Z)ⁿ Ã on shell. With Z = 0.95, each leg gives 0.975 and a 2 → 2 amplitude carries Z² = 0.9025.

Self-energies on a leg build the pole, not ℳ

Why not keep a self-energy bubble on an external leg as one more loop correction? On a leg the chain is the bare line i/(p² − m₀²) multiplied by powers of r = (−iΣ)(i/(p² − m₀²)) = Σ/(p² − m₀²). Expanded about the bare mass, one insertion gives iΣ/(p² − m₀²)², a double pole at m₀², and two give a triple pole: the signature of a pole that has moved, not of a small correction. At the physical mass the pole condition m² − m₀² = Σ(m²) makes r exactly 1, so every term of the chain is as large as the first and no truncation means anything; only the sum, with its pole at m² and residue Z, does. That is why amputation takes the whole chain away. What survives is information LSZ already holds: the external momenta sit on the physical shell, pₖ² = m², not m₀², and each leg leaves √Z. The loop corrections that do belong in ℳ live inside Ã: vertex corrections, boxes, and self-energies on internal lines.

Disconnected pieces have too few poles

The free four-point function shows the mechanism, as topic 3.04 found. Its term D_F(x₁ − y₁) D_F(x₂ − y₂), transformed with outgoing p₁, p₂ and incoming k₁, k₂, becomes (2π)⁴δ⁴(p₁ − k₁) i/(k₁² − m²) × (2π)⁴δ⁴(p₂ − k₂) i/(k₂² − m²). The δ-functions tie each outgoing momentum to an incoming one, so four legs share two poles. LSZ extracts the coefficient of four independent poles, and this term contributes nothing to it; what it carries is 2E₁ 2E₂ (2π)⁶ δ³(p₁ − k₁) δ³(p₂ − k₂) plus the exchanged pairing, which is ⟨p₁p₂|k₁k₂⟩, the 1 in S = 1 + iT. With more particles the same count sorts the pieces. In a 3 → 3 function, a term in which two particles scatter while the third passes untouched has five poles for six legs: it is the 2 → 2 amplitude times a δ³ for the spectator, a different process. Vacuum bubbles were already divided out by the Gell-Mann-Low denominator in topic 3.03. Only connected diagrams build the n-particle iℳ.

Z = 1 at tree level, and still 1 with the φ⁴ tadpole

At tree level there are no loops and so no self-energy: D is the free propagator, m = m₀ and Z = 1, which is why topic 2.09's first decay amplitude needed no √Z. In φ⁴ the only one-loop self-energy is the tadpole, −iΣ = (−iλ/2)∫d⁴ℓ/(2π)⁴ i/(ℓ² − m² + iε). Momentum conservation at its single vertex reads p + ℓ = p + ℓ, so ℓ runs free and p never enters the loop: Σ is one divergent number. It moves the pole, m² = m₀² + Σ, but dΣ/dp² = 0, so Z = 1 + O(λ²). The first p-dependent self-energy is the two-loop sunset, three propagators between two vertices with symmetry factor 1/6, so (√Z)⁴ alters ℳ for φφ → φφ only at order λ³. The scalar Yukawa model differs: its one-loop self-energies are bubbles that p flows through. Quoted ahead of UQFTI-06.02's Feynman parameters, dΣ_φ/dp² at p² = m² is −(g²/16π²)∫₀¹dx x(1 − x)/(M² − x(1 − x)m²): negative, so Z_φ falls below 1 as positivity requires. For g = M and m = M/2 it gives Z_φ = 0.9989.

The rules for iℳ, collected and checked

For φ⁴, with Lᵢₙₜ = −(λ/4!)φ⁴: draw every connected, amputated diagram with the n external particles; write −iλ at each vertex and i/(q² − m² + iε) on each internal line; conserve momentum at every vertex and integrate ∫d⁴ℓ/(2π)⁴ over each loop momentum left free; divide by the symmetry factor; multiply by (√Z)ⁿ, which is 1 until order λ². For φφ → φφ at order λ the four external fields meet the vertex's four fields in 4! ways, cancelling the 1/4!, so iℳ = −iλ and ℳ = −λ. For the scalar Yukawa model, Lᵢₙₜ = −gψ†ψφ: vertex −ig, ψ lines i/(q² − M² + iε) with arrows conserving ψ number, φ lines i/(q² − m² + iε), no sign for a closed ψ loop, and φ shifted by a constant so that ⟨Ω|φ|Ω⟩ = 0, which removes the one-point tadpoles LSZ cannot accept. Checks: ℳ = −λ is dimensionless, as [ℳ] = 4 − n demands for n = 4; it vanishes as λ → 0, leaving S = 1; and no experiment measures λ, so the measured tests of these rules wait for QED in Units 9 and 10.

Change one variable at a time

Make the relationship visible.

Interactive model

Residue Z of the pole0.80

External particles n4

Drag Z down from 1: the top branches hug the pole more tightly while the bottom curve stays finite and crosses the pole at √Z, not Z. Then raise n and watch the filled bar (√Z)ⁿ pull away from the open bar Zⁿ that dropping LSZ's 1/√Z would give. At Z = 1 both bars reach the tree-level line.

RESIDUE Z0.80

PER LEG √Z0.894

ALL LEGS (√Z)ⁿ0.640

MASS DIMENSION OF ℳ0

Live interpretationRESIDUE Z: 0.80. PER LEG √Z: 0.894. ALL LEGS (√Z)ⁿ: 0.640. MASS DIMENSION OF ℳ: 0

Catch the common trap

Explain before calculating.

Misconception check

Amputated is not the same as one-particle irreducible. Having learned to cut lines off, students often discard every diagram that falls apart when a single line is cut, keeping only 1PI pieces such as the φ⁴ vertex and its loop corrections. The cut that defines amputation must separate one external point from everything else. A line whose cut splits the diagram into two parts that each still hold several external particles is internal, and it stays. The exchanged φ in ψψ → ψψ is such a line, and so is the line joining the two vertices of any φ⁴ tree with six external particles: discard those diagrams and the order-g² Yukawa amplitude and the order-λ² amplitude for φφ → φφφφ both vanish, although both processes happen. 1PI pieces are the building blocks of self-energies and vertex functions; ℳ needs the whole connected, amputated sum.

In the scalar Yukawa model, suppose the one-loop self-energy of ψ gives its pole the residue Z = 0.96. The connected, amputated diagrams for ψψ → ψψ sum to A₀ at order g². Through order g⁴, by what factor do the four external legs multiply A₀ in iℳ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn an interacting scalar theory the Källén-Lehmann continuum holds 10% of the spectral weight, so the one-particle pole has Z = 0.90. At one on-shell kinematic point the connected, amputated diagrams for a 2 → 2 process sum to −0.30i. Find ℳ, compare it with what keeping Z per leg or no factor at all would give, and find the leg factor a 2 → 3 amplitude carries.

Each leg: LSZ's (i/√Z)(m² − p²) times the leg's iZ/(p² − m² + iε) leaves √Z = √0.90 = 0.9487.
A 2 → 2 process has n = 4 external particles, so the factor is (√Z)⁴ = Z² = 0.8100 and iℳ = 0.8100 × (−0.30i) = −0.243i. So ℳ = −0.243, dimensionless as [ℳ] = 4 − 4 = 0 requires.
Keeping the residue Z on each leg would give Z⁴ = 0.6561 and ℳ = −0.197, 19% too small in magnitude; dropping the legs' factors altogether gives −0.300, 23% too large.
A 2 → 3 process has n = 5, so it carries (√Z)⁵ = 0.90⁵⁄² = 0.7684, and its ℳ has mass dimension 4 − 5 = −1.
Limit: as the continuum's weight goes to zero, Z → 1 and every one of these factors returns to the tree-level value 1.

Answerℳ = −0.243, from √Z = 0.949 per leg and Z² = 0.810 for four legs; a 2 → 3 amplitude would carry (√Z)⁵ = 0.768.

MediumIn φ⁴ theory with λ = 0.60, the one-loop tadpole cut off at Euclidean loop momentum Λ, with Λ/m = 10, is Σ = (λ/32π²)[Λ² − m² ln(1 + Λ²/m²)], quoted here. Find Σ, the bare mass m₀ and Z to order λ. Then sort the connected order-λ² diagrams of the four-point function into those that enter iℳ for φφ → φφ and those amputation removes, and show why the latter cannot be kept as corrections.

Σ = (0.60/315.83)(100 − ln 101) m² = 0.0018998 × 95.385 m² = 0.1812 m². The pole condition m² = m₀² + Σ then gives m₀² = 0.8188 m², so m₀ = 0.905 m.
The loop momentum ℓ leaves and re-enters the one vertex, so the external p never flows through the loop: Σ is the same number at every p², dΣ/dp² = 0, and Z = 1 to order λ.
The connected order-λ² diagrams number seven: the s-, t- and u-channel bubbles, each with symmetry factor ½, which join all four particles and are already amputated; and four diagrams with a tadpole on one external leg.
On a leg the tadpole multiplies the bare line by Σ/(p² − m₀²). At the physical pole that is 0.1812/(1 − 0.8188) = 1.000: every further insertion is as large as the first, so the chain is the pole at m², not a correction, and amputation removes all of it.
So iℳ = −iλ + (s, t and u bubbles) + O(λ³), with (√Z)⁴ = 1 at this order. The tadpole's only trace is that the external momenta sit on the physical shell p² = m².

AnswerΣ = 0.181 m², m₀ = 0.905 m and Z = 1 to order λ; of the seven connected order-λ² diagrams, the three bubbles enter iℳ and the four leg tadpoles are amputated.

HardIn φ⁴ theory with λ = 0.60 and m = 0.80 GeV, two particles collide head-on at √s = 4m = 3.2 GeV, exactly the threshold of φφ → φφφφ, so the four products are at rest. Find ℳ from the connected, amputated tree diagrams, check its mass dimension, and say why neither √Z nor any disconnected piece changes it at order λ².

At order λ² a tree has two vertices joined by one internal line, each vertex holding three external particles: one diagram for each split of the six particles into two triples, C(6,3)/2 = 10. The 2!(4!)² ways of attaching fields cancel the 1/(2!(4!)²) of the expansion, so each gives (−iλ)² i/(q² − m² + iε) and ℳ = −λ² Σ 1/(q² − m²).
Kinematics: k₁ = (2m, 0, 0, √3m) and k₂ = (2m, 0, 0, −√3m), with k² = 4m² − 3m² = m², and each pᵢ = (m, 0, 0, 0). Four splits put both incoming particles on one vertex: q = k₁ + k₂ − pᵢ = (3m, 0, 0, 0), q² = 9m², each term 1/(8m²). Six put one incoming particle on each vertex: q = k₁ − pᵢ − pⱼ = (0, 0, 0, √3m), q² = −3m², each term −1/(4m²).
Sum: 4/(8m²) − 6/(4m²) = −1/m², so ℳ = +λ²/m². The four timelike exchanges give −λ²/(2m²) = −0.281 GeV⁻² and the six spacelike ones +3λ²/(2m²) = +0.844 GeV⁻², a total of 0.5625 GeV⁻².
Dimension: [ℳ] = 4 − 6 = −2, matched by λ²/m² with λ dimensionless, and ℳ → 0 as λ → 0. All ten diagrams are one-particle reducible and all ten stay: their internal line is not an external leg.
Z = 1 + O(λ²), first from the two-loop sunset, so (√Z)⁶ moves ℳ only at order λ⁴. Disconnected pieces vanish on shell: each piece has an even number of legs, so the six split 4 + 2 or 2 + 2 + 2, and every split needs either one particle turning into three, which needs m ≥ 3m, or a free line joining two outgoing or two incoming particles, whose momenta would have to sum to zero, impossible for positive energies.

Answerℳ = λ²/m² = 0.56 GeV⁻² at threshold, the six spacelike exchanges outweighing and reversing the four timelike ones; mass dimension −2, as 4 − n requires for n = 6.