# 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.

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