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# 14.6: Hard-Sphere Scattering

Let us test out this scheme using a particularly simple example. Consider scattering by a hard sphere, for which the potential is infinite for $$r<a$$, and zero for $$r>a$$. It follows that $$\psi({\bf r})$$ is zero in the region $$r<a$$, which implies that $$u_l =0$$ for all $$l$$. Thus, $\beta_{l-} = \beta_{l+} = \infty,$ for all $$l$$. Equation ([e17.82]) thus gives $\label{e17.90} \tan \delta_l = \frac{j_l(k\,a)}{y_l(k\,a)}.$

Consider the $$l=0$$ partial wave, which is usually referred to as the $$S$$-wave. Equation ([e17.90]) yields $\tan\delta_0 = \frac{\sin (k\,a)/k\,a}{-\cos (k\,a)/ka} = -\tan (k\,a),$ where use has been made of Equations ([e17.58a]) and ([e17.58b]). It follows that $\label{e17.92} \delta_0 = -k\,a.$ The $$S$$-wave radial wave function is [see Equation ([e17.80])] \begin{aligned} {\cal R}_0(r) &= \exp(-{\rm i}\, k\,a)\, \frac{[\cos (k\,a) \,\sin (k\,r) -\sin (k\,a) \,\cos (k\,r)]}{k\,r}\nonumber\\[0.5ex] &= \exp(-{\rm i}\, k\,a)\, \frac{ \sin[k\,(r-a)]}{k\,r}.\end{aligned} The corresponding radial wavefunction for the incident wave takes the form [see Equation ([e15.49])] $\tilde{\cal R}_0(r) = \frac{ \sin (k\,r)}{k\,r}.$ Thus, the actual $$l=0$$ radial wavefunction is similar to the incident $$l=0$$ wavefunction, except that it is phase-shifted by $$k\,a$$.

Let us examine the low- and high-energy asymptotic limits of $$\tan\delta_l$$. Low energy implies that $$k\,a\ll 1$$. In this regime, the spherical Bessel functions reduce to: \begin{aligned} j_l(k\,r) &\simeq \frac{(k\,r)^l}{(2\,l+1)!!},\\[0.5ex] y_l(k\,r) &\simeq -\frac{(2\,l-1)!!}{(k\,r)^{l+1}},\end{aligned} where $$n!! = n\,(n-2)\,(n-4)\cdots 1$$ . It follows that $\tan\delta_l = \frac{-(k\,a)^{2\,l+1}}{(2\,l+1) \,[(2\,l-1)!!]^{\,2}}.$ It is clear that we can neglect $$\delta_l$$, with $$l>0$$, with respect to $$\delta_0$$. In other words, at low energy, only $$S$$-wave scattering (i.e., spherically symmetric scattering) is important. It follows from Equations ([e15.17]), ([e17.73]), and ([e17.92]) that $\frac{d\sigma}{d{\mit\Omega}} = \frac{\sin^2 k\,a}{k^{\,2}} \simeq a^{\,2}$ for $$k\,a\ll 1$$. Note that the total cross-section $\sigma_{\rm total} = \int\frac{d\sigma}{d{\mit\Omega}}\,d{\mit\Omega} = 4\pi \,a^{\,2}$ is four times the geometric cross-section $$\pi \,a^{\,2}$$ (i.e., the cross-section for classical particles bouncing off a hard sphere of radius $$a$$). However, low energy scattering implies relatively long wavelengths, so we would not expect to obtain the classical result in this limit.

Consider the high-energy limit $$k\,a\gg 1$$. At high energies, all partial waves up to $$l_{\rm max} = k\,a$$ contribute significantly to the scattering cross-section. It follows from Equation ([e17.75]) that $\label{e17.99} \sigma_{\rm total} \simeq \frac{4\pi}{k^{\,2}} \sum_{l=0,l_{\rm max}} (2\,l+1)\,\sin^2\delta_l.$ With so many $$l$$ values contributing, it is legitimate to replace $$\sin^2\delta_l$$ by its average value $$1/2$$. Thus, $\sigma_{\rm total} \simeq \sum_{l=0,k\,a} \frac{2\pi}{k^{\,2}} \,(2\,l+1) \simeq 2\pi \,a^{\,2}.$ This is twice the classical result, which is somewhat surprising, because we might expect to obtain the classical result in the short-wavelength limit. For hard-sphere scattering, incident waves with impact parameters less than $$a$$ must be deflected. However, in order to produce a “shadow” behind the sphere, there must also be some scattering in the forward direction in order to produce destructive interference with the incident plane-wave. (Recall the optical theorem.) In fact, the interference is not completely destructive, and the shadow has a bright spot (the so-called “Poisson spot” ) in the forward direction. The effective cross-section associated with this bright spot is $$\pi \,a^{\,2}$$ which, when combined with the cross-section for classical reflection, $$\pi \,a^{\,2}$$, gives the actual cross-section of $$2\pi \,a^{\,2}$$ .

# Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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