Far from a potential that falls off fast enough, a stationary scattering state looks like the incident plane wave plus an outgoing spherical wave whose strength depends on direction: $\psi\approx e^{ikz}+f(\theta,\phi)e^{ikr}/r$. The function $f$, with the dimensions of a length, is the scattering amplitude, and it holds everything a detector far away can learn. The $1/r$ makes the scattered intensity fall as $1/r^{2}$, so the number of particles crossing a sphere round the target is the same whatever its radius, and computing the current through a small patch of that sphere gives the central formula of the subject: $d\sigma/d\Omega=|f(\theta,\phi)|^{2}$. For a central potential $f$ depends on $\theta$ alone.

Drop a stone into a pond crossed by steady waves: far away the old waves still pass, and a new circular ripple spreads out, stronger in some directions than in others. The scattering amplitude is that ripple’s strength, direction by direction.

Crests of the incident plane wave, dashed, moving to the right, and of the outgoing spherical wave $f(\theta)\,e^{ikr}/r$, solid, spreading from the target at the centre. Its intensity falls as $1/r^{2}$, so the same number of particles crosses every sphere round the target.

For a potential that falls faster than $1/r$, the stationary state of energy $E=\hbar^{2}k^{2}/2m$ far from the target is

• $f$ has the dimensions of a length; for a central potential it depends on $\theta$ alone.
• The incident current is $\hbar k/m$ along $z$; the scattered current is radial, $\frac{\hbar k}{m}\frac{|f|^{2}}{r^{2}}$, up to terms that fall faster.

Compute the probability current of each part. The plane wave carries $\hbar k/m$ per unit area. The spherical wave’s current is radial and falls as $1/r^{2}$, so through the patch $r^{2}d\Omega$ it carries a fixed rate. Dividing the rate by the incident flux gives the cross section; the cross terms between the two waves oscillate with angle and matter only straight ahead.

The probability current of the dynamics chapter.

For $e^{ikz}$, $\operatorname{Im}(e^{-ikz}\,ik\,e^{ikz})=k$.

For $\psi_{\text{sc}}=f\,e^{ikr}/r$; the angular derivatives bring only terms falling as $1/r^{3}$.

The rate through the patch $r^{2}d\Omega$ of a sphere, the same at every radius.

Divide by the incident flux.

Far away the scattering state is the plane wave plus $f(\theta,\phi)e^{ikr}/r$, and the cross section in each direction is $|f|^{2}$.

The scattered intensity falls as $1/r^{2}$ while a sphere’s area grows as $r^{2}$, so the same number of particles crosses every sphere.

The plane wave is a pure number and $e^{ikr}/r$ an inverse length, so the amplitude $f$ is a length.

A central potential looks the same from every side of the beam axis, so $f$ cannot depend on $\phi$.

Great work! You now know how to:

• write the asymptotic form of a scattering state
• derive $d\sigma/d\Omega=|f|^{2}$ from the probability current
• say what a detector can and cannot measure