14.5: Partial Waves in the Classical Limit - Hard Spheres

Consider the scattering of a small hard sphere (radius $x_m$, mass $m$) by a large hard sphere ($X_M, M$). Firstly we transform the problem to the center of mass reference frame where it becomes that of a single effective particle of mass $\mu = mM/(m + M)$ moving in a hard sphere potential $(V (r < r_H = X_M + x_m) = \infty )$. Thus the boundary condition is $R_l(r_H) = 0$.

Consider the classical limit, where the sphere radius is much larger than the de Broglie wavelength, $kr_H \gg 1$. Up to $l = Kr_H$ the phase shift is enormous and $\sin \delta_l$ could have any value. For $l > Kr_H$ the impact parameter is so large that the particles miss and $\delta_l = 0$. Thus we can write the scattering cross section:

$$\sigma = \frac{4\pi}{ K^2} \sum^{l = Kr_H}_{l = 0} (2l + 1)\frac{1}{2} \nonumber$$

where we replace $\sin^2 \delta_l$ with its average value of $\frac{1}{2}$.

Since $Kr_H$ is large, we can replace the sum by an integral and take only the leading term; $(Kr_H)^2 \gg Kr_H$:

$$\sigma \approx \frac{2\pi}{K^2} \int^{l = Kr_H}_{l = 0} (2l + 1) dl \approx 2\pi {r_H}^2 \nonumber$$

This result should send us rushing back to look for the extra factor of 2, since the cross-section of a sphere might be expected to be $\pi {r_H}^2$. In fact, though, the analysis is correct and closer analysis of the $\theta$ dependence of the wavefunction shows that half the amplitude is diffracted into the classical ‘shadow’ of the sphere to cancel the amplitude of the unscattered wave there.