The Scattering Amplitude

Last lesson ended on the claim $d\sigma/d\Omega = |f(\theta,\phi)|^2$ . Here we define the scattering amplitude $f$ precisely, show where that identity comes from, and meet the constraint — the optical theorem — that flux conservation imposes on it.

The scattering boundary condition

We solve the time-independent Schrödinger equation at energy $E = \hbar^2 k^2/2m$ ,

\Big(\!-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r})\Big)\psi = E\,\psi ,

subject to the physically motivated boundary condition that, far from the target, the wavefunction is an incident plane wave plus an outgoing scattered spherical wave:

\psi(\mathbf{r}) \;\xrightarrow{\,r\to\infty\,}\; e^{i k z} + f(\theta,\phi)\,\frac{e^{ikr}}{r}.

The coefficient $f(\theta,\phi)$ of the outgoing wave is the scattering amplitude. It has dimensions of length, carries all the angular information, and is exactly what experiment measures.

From amplitude to cross-section

Compute the radial probability current of the scattered piece $\psi_{\text{sc}} = f\,e^{ikr}/r$ . Using $\mathbf{j} = (\hbar/m)\,\operatorname{Im}(\psi^*\nabla\psi)$ and keeping the leading $1/r$ term,

j_r^{\text{sc}} = \frac{\hbar k}{m}\,\frac{|f(\theta,\phi)|^2}{r^2}.

The number of particles per second through a detector of area $r^2\,d\Omega$ is $j_r^{\text{sc}}\,r^2\,d\Omega = (\hbar k/m)\,|f|^2\,d\Omega$ . Dividing by the incident flux $j_{\text{inc}} = \hbar k/m$ gives, cleanly,

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

The $r$ -dependence cancels — as it must, since the cross-section is a property of the target, not of how far away we put the detector. (One subtlety: the incident and scattered waves interfere, but only in the exact forward direction $\theta=0$ ; off-axis the cross term is negligible, which is why the clean result holds for the detector placed at finite angle.)

The optical theorem

Probability is conserved, and that forces a relation between forward scattering and the total cross-section. Demanding that no probability accumulates at the target — the net outward flux of $|\psi|^2$ vanishes — yields the optical theorem:

\sigma_{\text{tot}} = \frac{4\pi}{k}\,\operatorname{Im} f(0),

where $f(0)$ is the amplitude in the forward direction $\theta=0$ . Physically, the beam is depleted because particles scatter out of it; that depletion is an interference effect between the incident wave and the forward-scattered wave, so it must be tied to $\operatorname{Im}f(0)$ .

A useful special case: isotropic s-wave

When only the lowest angular-momentum component contributes (the next lessons make this precise), $f$ is independent of angle: $f = \tfrac{1}{k}e^{i\delta_0}\sin\delta_0$ , controlled by a single phase shift $\delta_0$ . Then $d\sigma/d\Omega = |f|^2 = \sin^2\delta_0 / k^2$ is constant, and the total cross-section is just $4\pi$ times it,

\sigma_{\text{tot}} = \frac{4\pi}{k^2}\sin^2\delta_0 .

You can verify this respects the optical theorem: $\operatorname{Im}f(0) = \sin^2\delta_0/k$ , so $(4\pi/k)\operatorname{Im}f(0) = (4\pi/k^2)\sin^2\delta_0 = \sigma_{\text{tot}}$ exactly — unlike the Born approximation, the partial-wave amplitude is unitary by construction.

Try it

Take pure s-wave scattering at $k = 2\ \text{fm}^{-1}$ with phase shift $\delta_0 = 0.5\ \text{rad}$ . Compute the total cross-section $\sigma = 4\pi |f|^2$ with $|f|^2 = \sin^2\delta_0 / k^2$ .

Run your code to see the quantum state.