Experiment 07 · Elastic scattering

How can two partial waves reshape a scattering pattern?

Change s- and p-wave scattering phase shifts, inspect their coherent angular interference, integrate over solid angle, and verify the optical theorem and unitarity bounds.

20–30 minAngular distributionsPhase shiftsOptical theorem

Physical problem

Are angular interference and total scattering controlled by the same information?

Scattering from a central potential separates into independent angular momentum channels, each characterized at fixed energy by a phase shift. The observed scattering amplitude adds these channels coherently, so their relative phases affect the angular cross section. Integrating over all directions instead uses orthogonality of angular momentum states.

Predict: choose δ₀=90°, δ₁=0. Is scattering isotropic? Then set δ₁=60° and compare forward with backward scattering. Will their interference contribute to the integrated total?

Calculated observables

Cross sections and checks

Observable angular distribution

Partial-wave interference in dσ/dΩ

Blue: coherent total |f₀+f₁|². Orange and teal dashed: separately squared partial amplitudes. Their sum generally differs from the total at a given θ because amplitudes interfere. The point marks your selected angle.

Solid-angle measure

Why forward peaks need angular weighting

For σ>0, dP/dθ = 2πsinθ(dσ/dΩ)/σ. Its area with respect to θ in radians is one. This differs from dσ/dΩ because cones at different polar angles subtend different solid angles.

Elastic unitarity

Single-channel partial-wave bound

For a purely elastic s wave, σ₀=(4π/k²)sin²δ₀, bounded above by 4π/k². The maximum at δ₀=π/2 is the single-channel unitarity limit.

Derive all three cross sections

From partial waves to the optical theorem

For a central short-range potential, elastic channels have Sℓ=e2iδℓ, with real phase shift δℓ. The scattered amplitude is f(θ)=(1/k)Σℓ(2ℓ+1)eiδℓsinδℓ Pℓ(cosθ). We retain ℓ=0 and ℓ=1, where P₀=1 and P₁=cosθ.

f(θ)=k⁻¹[eiδ₀sinδ₀ + 3eiδ₁sinδ₁ cosθ].

The differential cross section is the absolute squared amplitude. Orthogonality ∫PℓPℓ′dΩ=4πδℓℓ′/(2ℓ+1) removes interference terms from the integrated total cross section.

σ=(4π/k²)[sin²δ₀+3sin²δ₁]=(4π/k) Im f(0).

The final equality is the optical theorem for elastic scattering in this model. It is checked numerically in the live readout. Another useful observable discounts small-angle scattering:

σtr = ∫(1−cosθ)|f(θ)|²dΩ = σ − (8π/k²)sinδ₀ sinδ₁ cos(δ₀−δ₁).
Why is elastic unitarity automatic?

Each diagonal S-matrix eigenvalue lies on the complex unit circle: Sℓ=e2iδℓ, so |Sℓ|=1. The probability flux in a partial wave is conserved. For an inelastic channel |Sℓ| can be below one, requiring additional absorption information; this laboratory intentionally models the purely elastic case.

Where does the polar-angle Jacobian come from?

The usual scattering differential is per unit solid angle, dΩ=sinθ dθ dφ. Integrating over the azimuth gives dσ/dθ=2πsinθ|f(θ)|². Divide by σ to obtain a properly normalized distribution over the polar angle. A visually large dσ/dΩ near the poles need not mean most scattering events lie in an infinitesimal polar cone.

Compute the transport cross section

The weighted integral ∫cosθ|f|²dΩ retains only the s–p interference term, since the angular integrals of cosθ and cos³θ vanish. Evaluating ∫cos²θdΩ=4π/3 produces σtr=σ−8π Re(a₀a₁*)/k² with aℓ=eiδℓsinδℓ. This is sensitive to relative phase shifts even when the total cross section is simply a sum of partial strengths.

Explore and verify

What do phase shifts actually change?

  1. Isotropy. Set δ₁=0 while δ₀ is finite. Verify an angularly constant dσ/dΩ and compute the total s-wave unitarity bound.
  2. p-wave angular nodes. Set δ₀=0 and δ₁=90°. Observe a cos²θ profile, its zero at θ=90°, and the sinθ Jacobian in the polar-angle distribution.
  3. Coherent interference. Use both waves and compare forward and backward cross sections. Verify that the total σ equals the optical-theorem value even though cross terms reshape the angular pattern.

Self-check: Why is the integrated elastic cross section a sum of separate ℓ contributions?

Validity and connection to the textbook

The truncation to s and p waves is illustrative; it is especially useful near thresholds where higher angular momenta are suppressed by centrifugal barriers. Phase shifts depend on the physical potential and energy, and arbitrary sliders need not correspond to a single fixed potential over a range of k. Angular and integrated cross sections have length-squared dimensions; their magnitude scales with 1/k² at fixed phase shifts.

Read Chapter 5 — Partial-Wave Method. Follow-up links: spherical harmonics and orbital angular momentum (Chapter 3), Green functions and Born approximations (Chapter 5).