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advanced27 min studyLesson 670 in Physics

Extract scattering from physical external poles

Specify state normalization, residues, amputation, and the scope of asymptotic particles.

Before you begin
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The question for today

What you’ll understand

  • Identify a stable one-particle pole in a time-ordered correlator.
  • Remove external pole factors with the correct residue normalization.
  • Explain why unstable or confined fields are not ordinary LSZ external particles.

Generating functionals produce time-ordered field correlations. Scattering experiments instead compare widely separated incoming and outgoing particles. LSZ reduction connects these descriptions when suitable stable asymptotic particle states exist. The bridge is an isolated one-particle pole and the overlap of an interpolating field with its normalized particle. Amputation removes external propagation, while residue factors remove the arbitrary normalization of that field. Neither operation turns every off-shell correlation function into an observable, and neither bypasses the need for a well-defined renormalized continuum theory.

Declare the particle and Fourier conventions

Assume a relativistic scalar theory with a stable isolated mass shell, a suitable vacuum, and asymptotic wavepacket states. Use ℏ=c=1, metric signature plus minus minus minus, and p²=(p⁰)²−|p|². Normalize one-particle momentum states with the Lorentz-invariant factor 2Ep times a three-dimensional delta function. Choose the field’s one-particle matrix element to be the positive square root of a residue Zφ. These declarations fix the external normalization rather than leaving it to a diagram’s appearance.

The Fourier-transformed exact time-ordered two-point function then has a pole iZφ divided by p²−m²+i0, plus less singular terms. The pole mass is physical; Zφ depends on the chosen field. A regulated finite system helps define correlators, but scattering also needs appropriate infinite-volume and asymptotic-time limits. One cannot identify a discrete finite-box level with a continuum scattering amplitude without supplying that additional limit and normalization machinery.

⟨p′∣p⟩=2Ep(2π)3δ(3)(p′−p),⟨Ω∣ϕ(0)∣p⟩=Zϕ,G~2(p)∼iZϕp2−m2+i0\langle\mathbf p'|\mathbf p\rangle=2E_{\mathbf p}(2\pi)^3\delta^{(3)}(\mathbf p'-\mathbf p),\qquad \langle\Omega|\phi(0)|\mathbf p\rangle=\sqrt{Z_\phi},\qquad \widetilde G_2(p)\sim\frac{iZ_\phi}{p^2-m^2+i0}
The Fourier convention is Gtilde₂(p)=∫d⁴x exp(ip·x)⟨Tφ(x)φ(0)⟩ with +--- metric. m is the stable pole mass, Zφ>0 the overlap residue, and i0 specifies the Feynman boundary value.↔ Scroll wide equations horizontally.

Amputate before the on-shell limit

Near all external one-particle poles, a connected multipoint correlator factorizes into one external factor i√Zφ divided by each pole denominator, multiplied by the normalized connected scattering amplitude. Multiply by the inverse of each external factor, then take the physical on-shell limits. Setting every denominator to zero before this cancellation would produce singular expressions rather than an amplitude. Less singular external contributions disappear in the pole extraction under the stated hypotheses.

Use Sfi equal to the identity contribution plus i times the momentum-conserving delta function times M for the connected process. With all external momenta incoming in the correlator, crossing identifies the physical incoming and outgoing assignments. Connectedness removes spectator products, while amputation removes external propagation; these are separate operations. Proper vertices from Γ can help construct the result, but internal exchange contributions may also be required for a full connected scattering amplitude.

G~N,c∼(2π)4δ(4) ⁣(∑jpj)iM∏j=1NiZjDj,Dj=pj2−mj2+i0\widetilde G_{N,c}\sim(2\pi)^4\delta^{(4)}\!\left(\sum_jp_j\right)i\mathcal M\prod_{j=1}^{N}\frac{i\sqrt{Z_j}}{D_j},\qquad D_j=p_j^2-m_j^2+i0
This is the leading simultaneous external-pole structure, with all momenta incoming and residues matched to the declared external states. Removing the displayed factors yields iM, not M without its convention-dependent i.↔ Scroll wide equations horizontally.

Check the asymptotic-state hypothesis

An unstable resonance does not provide an isolated real stable-particle pole and cannot simply be inserted as an ordinary external LSZ state. Confined quark or gluon fields do not label freely separated detected hadrons. Long-range massless interactions introduce infrared difficulties that can require inclusive observables or dressed states rather than naive charged-particle Fock states. These limitations concern the existence and interpretation of asymptotic states, not merely whether enough Feynman diagrams have been drawn.

The scattering lab can check a coupling power only. Reset and set coupling to 0.5 and angle to 90°; predict schematic weight 1/16. Reset and set coupling to 1 and angle to 90°; predict weight 1, sixteen times larger. The model is g⁴(1+cos²θ), with no poles, external-state residues, phase space, or units of cross section. Its term diagram therefore cannot validate an LSZ reduction or turn this weight into a measured process rate.

Lint=−λ4!ϕ4⟹iMtree=−iλ,Mtree=−λ\mathcal L_{\rm int}=-\frac{\lambda}{4!}\phi^4\quad\Longrightarrow\quad i\mathcal M_{\rm tree}=-i\lambda,\qquad \mathcal M_{\rm tree}=-\lambda
This simple four-scalar contact result uses residue one at tree level, the declared S-matrix convention, and canonically normalized real scalar fields. It is an amplitude, with phase space and flux still needed for a cross section.↔ Scroll wide equations horizontally.

Work it through

Cancel an arbitrary field normalization

For four identical stable scalar external particles, define C=lim[∏j(Dj/i)]Gtilde₄,c after removing the overall momentum delta function. Suppose C=−0.05i and Zφ=0.25. Find M, then check the field rescaling φ′=2φ.

  1. The pole factorization gives C=iM(√Zφ)⁴=iM Zφ². Here Zφ²=1/16, so dividing C by this factor gives iM=−0.8i.

  2. Thus M=−0.8 in the declared convention. Dividing by Zφ only once would have left a spurious dependence on the normalization of the interpolating field.

  3. Under φ′=2φ, the two-point residue becomes Z′φ=4Zφ=1 and the four-point coefficient becomes C′=16C=−0.8i. Their ratio C′/(Z′φ)² again gives iM=−0.8i, as a physical amplitude must.

The result

The amplitude is M=−0.8 and remains unchanged by the stated field rescaling.

The living notebook

Explore the model

Interactive lab

Interactive lab · scattering

Check a schematic coupling power

Reset and set g=0.5,θ=90°; predict weight 1/16. Reset and set g=1,θ=90°; predict weight 1, a factor 16 larger.

Check a schematic coupling powerA schematic term diagram with dimensionless weight g⁴(1 + cos²θ). Lines denote terms, not particle paths; this is not a process cross-section or loop calculation. Current readings: Coupling factor g⁴: 0.24; Angular factor 1 + cos²θ: 1.329; Illustrative weight: 0.319. All parameters are adjustable with the labeled controls below.Incoming termsOutgoing termsθSchematic interaction term · no particle trajectories
Incoming termsOutgoing termsAngle θ
Coupling factor g⁴
0.24
Angular factor 1 + cos²θ
1.329
Illustrative weight
0.319
t = 0 s

A schematic term diagram with dimensionless weight g⁴(1 + cos²θ). Lines denote terms, not particle paths; this is not a process cross-section or loop calculation. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe weight is dimensionless g⁴(1+cos²θ). It has no external poles, residues, phase-space integration, or physical cross-section normalization.

Pause & reason

Check your understanding

What makes a scalar field suitable for ordinary massive-particle LSZ reduction?

Your turn

Try it yourself

Use interaction −0.30φ⁴/4! with canonically normalized real scalar fields and Zφ=1 at tree level. Write the connected four-point external-pole contribution and identify M. Does its numerical value alone give a cross section?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Attach four factors i/(pj²−m²+i0) to the contact vertex and use S−1=iδM for the connected convention.

Compare with the solution

After removing the overall momentum delta function, the tree correlator is [∏j i/(pj²−m²+i0)](−0.30i). Amputating leaves iM=−0.30i, so M=−0.30. A cross section additionally needs physical kinematics, initial flux, final-state phase space, and any identical-particle factors; the bare amplitude value alone is insufficient.

Keep the idea

Take it with you

  • LSZ isolates stable external poles with explicit state normalization.
  • Residue factors cancel the arbitrary normalization of interpolating fields.
  • Scattering amplitudes require asymptotic states and remain distinct from cross sections.

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