Phase Shifts

We now turn the partial-wave machinery into an actual number. The cleanest example is the hard sphere: an impenetrable ball of radius $a$ . It has no free parameters beyond $a$ , its phase shifts are exact, and it exhibits the famous low-energy result $\sigma = 4\pi a^2$ .

The radial equation

For a central potential, write $\psi = R_\ell(r)\,P_\ell(\cos\theta)$ and substitute $u_\ell(r) = r R_\ell(r)$ . The radial Schrödinger equation becomes a 1D problem with a centrifugal barrier:

-\frac{\hbar^2}{2m}u_\ell'' + \Big[V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2m r^2}\Big]u_\ell = E\,u_\ell , \qquad u_\ell(0) = 0 .

The phase shift $\delta_\ell$ is fixed entirely by how this $u_\ell$ joins onto the free asymptotic form $\sin(kr - \ell\pi/2 + \delta_\ell)$ at large $r$ .

Hard sphere, s-wave

Take $\ell = 0$ , where the centrifugal term drops out. Outside the sphere ( $r > a$ ) the potential vanishes, so $u_0'' = -k^2 u_0$ and the general solution is

u_0(r) = C\,\sin(kr + \delta_0).

The sphere is impenetrable, so the wavefunction must vanish on its surface: $u_0(a) = 0$ . That single condition gives

\sin(ka + \delta_0) = 0 \quad\Longrightarrow\quad \boxed{\;\delta_0 = -ka\;}

(choosing the branch continuous with $\delta_0\to 0$ as $a\to 0$ ). The phase shift is negative — the hard core pushes the wave out, a repulsive signature. This result is exact at all energies, not an approximation.

The low-energy cross-section

The s-wave contribution to the total cross-section is

\sigma_0 = \frac{4\pi}{k^2}\sin^2\delta_0 = \frac{4\pi}{k^2}\sin^2(ka).

In the low-energy limit $ka \ll 1$ , $\sin(ka)\approx ka$ , so

\sigma_0 \;\xrightarrow{\,ka\to 0\,}\; \frac{4\pi}{k^2}(ka)^2 = 4\pi a^2 .

This is four times the classical geometric cross-section $\pi a^2$ . The factor of four is purely quantum: at low energy the wave diffracts around the entire sphere, "seeing" its full surface area $4\pi a^2$ rather than just its silhouette.

The high-energy crossover

When $ka \gg 1$ , many partial waves contribute and a careful sum gives $\sigma \to 2\pi a^2$ — twice the geometric cross-section. (The "extra" $\pi a^2$ beyond classical is a forward diffraction shadow, an unavoidable wave-optics effect.) The hard sphere thus interpolates from $4\pi a^2$ at low energy to $2\pi a^2$ at high energy, never reaching the naive classical $\pi a^2$ — a clean illustration that quantum cross-sections are not simple geometric areas.

Try it

For a hard sphere of radius $a = 1\ \text{fm}$ probed at $k = 0.2\ \text{fm}^{-1}$ , compute the s-wave phase shift $\delta_0 = -ka$ and the s-wave cross-section $\sigma_0 = (4\pi/k^2)\sin^2\delta_0$ . Confirm it is close to $4\pi a^2$ .

Run your code to see the quantum state.