Scattering Theory Basics

Almost everything we know about the subatomic world comes from scattering: we fire a beam at a target, watch how particles deflect, and reconstruct the interaction from the angular pattern of what comes out. This module builds the quantum theory of that process and then crosses into relativistic wave equations. We begin with the central observable — the cross-section.

The experimental setup

Picture a steady, collimated beam of particles, all with the same momentum $\hbar\mathbf{k}$ , incident on a localized target potential $V(\mathbf{r})$ centered at the origin. Far away a detector subtending a small solid angle $d\Omega$ in the direction $(\theta,\phi)$ counts how many particles per second emerge along that direction.

Two rates characterize the experiment:

the incident flux $j_{\text{inc}}$ — the number of beam particles crossing unit area per unit time,
the scattered count rate $dN/dt$ into the detector's solid angle $d\Omega$ .

The differential cross-section

The detected rate is proportional both to the incident flux and to the solid angle. The constant of proportionality is the differential cross-section:

\frac{dN}{dt} = j_{\text{inc}}\,\frac{d\sigma}{d\Omega}\,d\Omega \quad\Longrightarrow\quad \frac{d\sigma}{d\Omega} = \frac{1}{j_{\text{inc}}}\,\frac{dN/dt}{d\Omega}.

Dimensionally, dividing a rate (per time) by a flux (per area per time) leaves an area. That is the key physical picture: $d\sigma/d\Omega$ is the effective target area, as seen from the beam, that funnels particles into the solid angle around $(\theta,\phi)$ . Its SI unit is m $^2$ ; nuclear and particle physicists use the barn, $1\,\text{b} = 10^{-28}\,\text{m}^2$ .

The total cross-section

Integrating over all directions gives the total cross-section

\sigma_{\text{tot}} = \int \frac{d\sigma}{d\Omega}\,d\Omega = \int_0^{2\pi}\!\!\int_0^{\pi} \frac{d\sigma}{d\Omega}\,\sin\theta\,d\theta\,d\phi.

This is the total effective area presented by the target — the rate at which particles are removed from the forward beam, divided by the incident flux. A "hard sphere" of geometric radius $a$ has, at low energy, $\sigma_{\text{tot}} = 4\pi a^2$ (four times its geometric cross-section $\pi a^2$ , a purely quantum result we will recover via partial waves).

Flux from the wavefunction

To connect this to quantum mechanics we need the flux of a wavefunction. The probability current for a state $\psi$ of a particle of mass $m$ is

\mathbf{j} = \frac{\hbar}{m}\,\operatorname{Im}\!\big(\psi^{*}\nabla\psi\big).

For an incident plane wave $\psi_{\text{inc}} = e^{i\mathbf{k}\cdot\mathbf{r}}$ this gives a uniform flux $j_{\text{inc}} = \hbar k/m = v$ , the particle speed. The scattered wave, as we will see next lesson, is an outgoing spherical wave $f(\theta,\phi)\,e^{ikr}/r$ whose radial flux through the detector, divided by $j_{\text{inc}}$ , is the differential cross-section. Quantitatively that identification yields the central formula of the whole subject,

\frac{d\sigma}{d\Omega} = |f(\theta,\phi)|^2,

relating the measurable area to a single complex function $f$ — the scattering amplitude — which the rest of the module is devoted to computing.

Assumptions worth naming

This elastic, single-particle, time-independent picture rests on a few idealizations: the potential is short-ranged (so "far away" is well defined; the Coulomb tail needs separate care), the target is fixed and heavy (we work in its rest frame, or equivalently use the reduced mass), and the beam is dilute enough that particles scatter at most once. Within these limits the cross-section is the complete, frame-independent summary of the collision.