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 and Attributions

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

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