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advanced35 min studyLesson 391 in Physics

Asymptotic scattering states and flux

Normalize incident and outgoing waves so that a scattering amplitude predicts a measurable event rate.

Before you begin
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The question for today

What you’ll understand

  • State the incident-plane-wave plus outgoing-sphere convention.
  • Derive differential cross section as a flux ratio.
  • Separate asymptotic normalization, detector acceptance and long-range exceptions.

A scattering detector records particles arriving in a selected direction. Theory predicts that record by comparing outgoing probability current with the current in the incident beam. The resulting cross section has area units even when the interaction region is much smaller than the wave packet. We begin with a short-range nonrelativistic two-body problem, reduce it to relative motion, and fix the amplitude convention that all later phase shifts, Born formulas and optical-theorem factors will use.

Choose a relative coordinate and an outgoing condition

For two particles interacting through a potential depending on their separation, center-of-mass motion separates from relative motion. The latter has reduced mass μ=m1m2/(m1+m2)\mu=m_1m_2/(m_1+m_2). Take positive relative energy and wavenumber kk, with time dependence e−iEt/ℏe^{-iEt/\hbar}. An incident beam travels along positive z. Far outside a sufficiently short-range potential, its stationary scattering state is an incident plane wave plus an outgoing spherical wave whose angular coefficient is the scattering amplitude.

The outgoing factor has positive radial phase, so its constant-phase surfaces move outward after the time dependence is restored. The coefficient is a complex length, not a probability. A stationary plane wave is not square integrable; it is an idealization of a broad, long incident packet or a steady dilute beam. An overall amplitude can encode incident density locally and cancels from the ratio defining cross section. Normalizing that ideal plane wave to one particle in all space would be inconsistent.

ψ(r)∼A[eikz+f(θ,ϕ)eikrr],E=ℏ2k22μ,μ=m1m2m1+m2\psi(\mathbf r)\sim A\left[e^{ikz}+f(\theta,\phi)\frac{e^{ikr}}r\right],\qquad E=\frac{\hbar^2k^2}{2\mu},\qquad \mu=\frac{m_1m_2}{m_1+m_2}
k is in m⁻¹, E in J, μ in kg and f in m. A is an overall beam amplitude; this convention assumes a short-range potential and an outgoing boundary condition for positive energy.↔ Scroll wide equations horizontally.

Derive the area from a current ratio

For a scalar wavefunction without electromagnetic vector potential, probability current is the imaginary part of the wavefunction conjugate times its gradient, multiplied by reduced Planck constant over mass. The incident plane wave has uniform current density along the beam. The outgoing spherical wave has radial current falling as inverse squared distance. Multiplying by detector area on a distant sphere cancels that geometric spreading and leaves a finite angular flux.

Divide this scattered flux by the incident current to obtain differential cross section. For elastic scattering in one asymptotic channel, the incoming and outgoing speeds are equal and cancel. If a reaction changes the asymptotic channel speed, a velocity ratio must be retained with the corresponding amplitude convention. The interference between incident and scattered waves is crucial close to the forward direction and for total flux conservation; one should not replace the current of their sum by an incoherent sum everywhere.

j=ℏμIm⁡(ψ∗∇ψ),jin=ℏkμ∣A∣2,jsc,r∼jin∣f∣2r2,dσdΩ=∣f∣2\mathbf j=\frac\hbar\mu\operatorname{Im}(\psi^*\nabla\psi),\quad j_{\rm in}=\frac{\hbar k}{\mu}|A|^2,\quad j_{{\rm sc},r}\sim j_{\rm in}\frac{|f|^2}{r^2},\quad \frac{d\sigma}{d\Omega}=|f|^2
For a probability-normalized packet j has units m⁻² s⁻¹. The cross-section ratio is independent of A; dσ/dΩ has m² per unit solid angle and the radial estimate is asymptotic.↔ Scroll wide equations horizontally.

Connect the ideal amplitude to a detector

A detector covering a small solid angle samples the differential cross section averaged over that acceptance. For a dilute collection of independent targets, multiply by incident particle flux, target count and detector efficiency to estimate an event rate. This assumes negligible multiple scattering and attenuation of the beam across the target. A total cross section requires integration over solid angle; its units remain area because solid angle is dimensionless in SI, although per steradian is useful experimental bookkeeping.

The simple asymptotic form needs modification for an unscreened Coulomb potential, whose long-range tail adds logarithmic phases and a singular forward amplitude. Screening, finite acceptance and a physically specified beam matter there. Throughout this module, short-range formulas are not silently extended to that limit. The schematic scattering lab has a dimensionless coupling-and-angle weight unrelated to a solved potential; it can test angular or scaling ideas only when its explicit formula is kept separate from the physical amplitude and current.

N˙≃JbeamNTεD ∣f(θ)∣2ΔΩ,σel=∫∣f∣2 dΩ\dot N\simeq J_{\rm beam}N_T\varepsilon_D\,|f(\theta)|^2\Delta\Omega,\qquad \sigma_{\rm el}=\int |f|^2\,d\Omega
Jbeam is particles/(m² s), NT is an independent target count and εD is detection efficiency. The small-acceptance approximation treats f and efficiency as constant over ΔΩ.↔ Scroll wide equations horizontally.

Work it through

From a complex amplitude to a count rate

At a detector angle, f=(3+4i) nmf=(3+4i)\,\mathrm{nm}. One dilute target is illuminated by Jbeam=1020 m−2s−1J_{\rm beam}=10^{20}\,\mathrm{m^{-2}s^{-1}}. The detector covers 0.020 sr0.020\,\mathrm{sr} with efficiency 0.800.80. Estimate its count rate, assuming the amplitude is constant over the acceptance.

  1. Square the complex magnitude, not the complex number: ∣f∣2=(32+42) nm2=25 nm2|f|^2=(3^2+4^2)\,\mathrm{nm^2}=25\,\mathrm{nm^2}. The phase information does not enter this one-angle intensity by itself.

  2. Convert area before using the SI beam flux: 25 nm2=2.5×10−17 m225\,\mathrm{nm^2}=2.5\times10^{-17}\,\mathrm{m^2}. Multiplying by the acceptance gives a partial cross section of 5.0×10−19 m25.0\times10^{-19}\,\mathrm{m^2}.

  3. The incident flux times that area gives 50 s−150\,\mathrm{s^{-1}} events entering the acceptance. Applying the detection efficiency gives 0.80(50)=40 s−10.80(50)=40\,\mathrm{s^{-1}} recorded counts.

The result

The predicted recorded rate is 40 counts/s per independent target under the stated acceptance and efficiency assumptions.

The living notebook

Explore the model

Interactive lab

Interactive lab · scattering

A dimensionless angular comparison

Reset and set BOTH Coupling g = 1 and Scattering angle θ = 0°. Predict schematic weight 2. Reset and set g = 1 and θ = 90°; predict weight 1 from g⁴(1 + cos²θ). The readout is dimensionless and schematic. It is not a physical differential cross section, a computed scattering amplitude, an S matrix or a particle trajectory.

A dimensionless angular comparisonA 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 forAn angular weight requires a physical flux normalization and interaction model before it can predict an event rate.

Pause & reason

Check your understanding

With ψ ∼ exp(ikz) + f exp(ikr)/r, what are the dimensions of f?

Your turn

Try it yourself

An elastic isotropic s-wave amplitude is f=2i nmf=2i\,\mathrm{nm} at k=0.50 nm−1k=0.50\,\mathrm{nm^{-1}}. Find its total cross section and the event rate for one target under incident flux 1018 m−2s−110^{18}\,\mathrm{m^{-2}s^{-1}} with complete detection.

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

Give me a hint

Integrate the angle-independent squared amplitude over the full solid angle of 4π.

Compare with the solution

The differential cross section is 4 nm2/sr4\,\mathrm{nm^2/sr}. Integrating gives σ=16π nm2=5.03×10−17 m2\sigma=16\pi\,\mathrm{nm^2}=5.03\times10^{-17}\,\mathrm{m^2}. The rate is 1018σ≈50.3 s−110^{18}\sigma\approx50.3\,\mathrm{s^{-1}}. The given imaginary amplitude is compatible with an elastic resonant s wave at this wavenumber; it is not an arbitrary real constant that would ignore flux-conservation constraints. Full detection and one dilute independent target are essential to converting this integrated cross section directly into a measured count rate.

Keep the idea

Take it with you

  • The outgoing-wave convention fixes the normalization of the amplitude.
  • Cross section is a ratio of scattered and incident flux, not a geometric particle size.
  • Long-range Coulomb scattering needs modified asymptotic states and forward-angle care.

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