advanced29 min studyLesson 47 in Physics

From scattering amplitudes to cross sections

Connect asymptotic particle states to measurable rates through flux, phase space, and identical-particle counting.

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What you’ll understand

  • Explain the assumptions behind asymptotic particle scattering and LSZ reduction.
  • Convert a two-body invariant amplitude to a differential cross section.
  • Account once for identical final particles when integrating phase space.

An amplitude is not yet an event count. A scattering experiment also has an incoming flux, a set of possible final momenta, detector acceptance, and a preparation procedure. The cross section packages the dynamical probability into a quantity that can be compared across beam intensities. We will derive a simple leading-order prediction while making its particle-state and counting assumptions explicit.

What an S-matrix assumes

Use natural units and metric (+−−−)(+---). An S-matrix compares incoming and outgoing states that behave as separated stable particles far before and after a localized collision. We use states with ⟨p∣q⟩=2Ep(2π)3δ(3)(p−q)\langle p|q\rangle=2E_{\mathbf p}(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q). The connected matrix element defines M\mathcal M after removing the overall energy-momentum delta function and one factor of ii. Delta-normalized states are calculation tools; experimental beams are wave packets.

LSZ reduction relates scattering amplitudes to time-ordered correlation functions by extracting external one-particle poles and their residues. Its usual massive-particle form assumes suitable asymptotic states with isolated stable-particle poles. It is not a blanket rule that any field correlator defines any desired particle. Unstable resonances are not exact asymptotic states, and massless gauge theories introduce infrared issues that require inclusive observables or appropriately dressed states. Our massive scalar example avoids those complications.

dσ=∣M∣24(p1⋅p2)2−m12m22 dΦn,dΦn=(2π)4δ(4) ⁣(p1+p2−∑fpf)∏fd3pf(2π)32Efd\sigma=\frac{|\mathcal M|^2}{4\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2}}\,d\Phi_n,\qquad d\Phi_n=(2\pi)^4\delta^{(4)}\!\left(p_1+p_2-\sum_f p_f\right)\prod_f\frac{d^3p_f}{(2\pi)^3 2E_f}
This formula uses distinguishable final-state labels; divide by a factorial for each set of identical final particles when integrating the full labeled phase space. Spin averages and sums must be added where applicable.↔ Scroll wide equations horizontally.

Flux and phase space turn dynamics into rates

The invariant flux factor normalizes by how often the incoming beams encounter each other. The final-state phase-space measure counts allowed momenta with relativistic normalization and imposes total four-momentum conservation. These ingredients can suppress a rate near threshold even when the amplitude is nonzero. Conversely, a large number of accessible final configurations can enhance a rate without changing one individual transition’s amplitude.

For two-body scattering in the center-of-momentum frame, the remaining differential measure is a solid angle and a ratio of outgoing to incoming momentum magnitudes. For distinguishable spinless final particles the formula below applies directly. A cross section has dimensions of area, or energy to the power minus two in natural units. Checking that dimension is a fast way to catch a missing flux or energy factor when the amplitude itself is dimensionless.

dσdΩ=164π2s∣pf∣∣pi∣∣M∣2,s=(p1+p2)2\frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2s}\frac{|\mathbf p_f|}{|\mathbf p_i|}|\mathcal M|^2,\qquad s=(p_1+p_2)^2
This is the center-of-momentum formula before identical-final-state counting, with no unshown spin factors. For elastic equal-mass scattering the momentum ratio is one.↔ Scroll wide equations horizontally.

Count physical outcomes once

Two identical final scalars at opposite momenta do not form two different outcomes when their labels are exchanged. Integrating the labeled angular formula over the full sphere counts that pair twice. One remedy is a factor 1/2!1/2! over the full sphere; another is an appropriate half-domain without that factor. Applying both would undercount. The usual two-beam cross section does not acquire an extra initial-state factorial merely because the incoming particles are identical.

To compare with data, combine a cross section with integrated luminosity and efficiencies. An ideal expected count is the luminosity times the accepted cross section, not the bare amplitude squared. Real analyses also address backgrounds, resolution, selection, radiative corrections, and uncertainties. A clean tree-level scalar prediction is a baseline calculation with stated accuracy, rather than a detector simulation or an all-orders claim about nature.

The total tree-level real-scalar cross section

Take Lint=−λϕ4/4!\mathcal L_{\mathrm{int}}=-\lambda\phi^4/4!, elastic scattering above threshold, and the tree-level amplitude M=−λ\mathcal M=-\lambda. Integrate over the full sphere and account for identical final scalars.

  1. For equal masses in elastic center-of-momentum kinematics, ∣pf∣/∣pi∣=1|\mathbf p_f|/|\mathbf p_i|=1. The amplitude has no angular dependence at this order, so the labeled differential result is λ2/(64π2s)\lambda^2/(64\pi^2s).

  2. Include the final-state factor 1/21/2 when using the entire solid angle. The resulting differential counting convention is λ2/(128π2s)\lambda^2/(128\pi^2s) over the full sphere.

  3. Integrate dΩd\Omega to obtain 4π4\pi, giving σ=λ2/(32πs)\sigma=\lambda^2/(32\pi s). With λ=0.1\lambda=0.1 and s=10\sqrt{s}=10 in an energy unit, this is approximately 9.95×10−79.95\times10^{-7} inverse-square energy units.

σϕϕ→ϕϕtree=λ232πs,s>4m2\sigma_{\phi\phi\to\phi\phi}^{\mathrm{tree}}=\frac{\lambda^2}{32\pi s},\qquad s>4m^2
This result uses the full-sphere phase-space integral with the identical-final-state factor. Loop corrections and detector cuts are omitted.↔ Scroll wide equations horizontally.
The result

The leading result is isotropic and scales as λ²/s, with one final-state counting correction. Its stated coupling, normalization, and approximation are essential parts of the answer.

Explore the model

Interactive lab

Interactive lab · scattering

Separate an angular shape from a physical cross section

Vary the angle in the dimensionless toy weight g⁴(1 + cos²θ), then compare its forward-to-right-angle ratio with the isotropic φ⁴ result. Change g to observe the toy’s fourth-power scaling.

Separate an angular shape from a physical cross sectionA 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 lab intentionally illustrates a different angular function. It supplies no flux, phase-space units, identical-particle counting, or real process prediction; those must come from the calculation.

Check your understanding

For two identical final scalars, which full-angular integration convention counts each final state once?

Try it yourself

An accepted cross section is 2 fb2\,\mathrm{fb} and the integrated luminosity is 50 fb−150\,\mathrm{fb}^{-1}. A later selection keeps 40% of these signal events. Find the expected selected signal count and state what it omits.

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

Give me a hint

Area and inverse area cancel before applying the efficiency.

Compare with the solution

The expected accepted signal is 2×50=1002\times50=100 events, and the additional selection gives 4040. This is an expectation, not a guaranteed count. It omits background events, luminosity and cross-section uncertainty, uncertainty in the selection efficiency, and statistical fluctuations. None of those are determined by a Feynman diagram alone.

Take it with you

  • Scattering requires well-defined asymptotic particles, not just a field symbol.
  • Flux and phase space convert amplitudes into dimensionful cross sections.
  • Identical final outcomes must be counted once, and detector counts need further experimental input.

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