Hard-Sphere Scattering

Let us try out the scheme outlined in the previous section 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 $\psi({\bf x})$ is zero in the region $r<a$ , which implies that $u_l =0$ for all $l$ . Thus,

$$\beta_{l-} = \beta_{l+} = \infty$$ (10.103)

for all $l$ . Equation (10.97) yields

$$\tan \delta_l = \frac{j_l(k\,a)}{\eta_l(k\,a)}.$$ (10.104)

In fact, this result is most easily obtained from the obvious requirement that $A_l(a)=0$ . [See Equation (10.95).]

Consider the $l=0$ partial wave, which is usually referred to as the S-wave. Equation (10.105) gives

$$\tan\delta_0 = \frac{\sin (k\,a)/k\,a}{-\cos (k\,a)/ka} = -\tan (k\,a),$$ (10.105)

where use has been made of Equations (10.60) and (10.61). It follows that

$$\delta_0 = -k\,a.$$ (10.106)

The S-wave radial wave function is

$$A_0(r) = \exp(-{\rm i}\, k\,a) \left[\frac{\cos (k\,a) \,\sin (k\,r) -\sin (k\,a) \,\cos( k\,r)}{k\,r}\right]$$

$$=\exp(-{\rm i}\, k\,a)\, \frac{ \sin[k\,(r-a)]}{k\,r}.$$ (10.107)

[See Equation (10.95).] The corresponding radial wavefunction for the incident wave takes the form

$$\tilde{A}_0(r) = \frac{ \sin (k\,r)}{k\,r}.$$ (10.108)

[See Equations (10.79), (10.80), (10.94), and (10.107).] It is clear that 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 consider the low- and high-energy asymptotic limits of $\tan\delta_l$ . Low energy corresponds to $k\,a\ll 1$ . In this limit, the spherical Bessel functions and Neumann functions reduce to

$$j_l(k\,r) \simeq \frac{(k\,r)^{\,l}}{(2\,l+1)!!},$$ (10.109)

$$\eta_l(k\,r) \simeq -\frac{(2\,l-1)!!}{(k\,r)^{\,l+1}},$$ (10.110)

where $n!! = n\,(n-2)\,(n-4)\cdots 1$ [1]. It follows that

$$\tan\delta_l = \frac{-(k\,a)^{\,2\,l+1}}{(2\,l+1) \,[(2\,l-1)!!]^{\,2}}.$$ (10.111)

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 (10.28), (10.81), and (10.107) that

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{\sin^{\,2} (k\,a)}{k^{\,2}} \simeq a^{\,2}$$ (10.112)

for $k\,a\ll 1$ . Note that the total cross-section,

$$\sigma_{\rm total} = \oint d{\mit\Omega}\,\frac{d\sigma}{d{\mit\Omega}} = 4\pi \,a^{\,2},$$ (10.113)

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 de Broglie 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, by analogy with classical scattering, the scattered particles with the largest angular momenta about the origin have angular momenta $\hbar\,k\,a$ (i.e., the product of their incident momenta, $\hbar\, k$ , and their maximum possible impact parameters, $a$ ). Given that particles in the $l$ th partial wave have angular momenta $\sqrt{l\,(l+1)}\,\hbar$ , we deduce that all partial waves up to $l_{\rm max} \simeq k\,a$ contribute significantly to the scattering cross-section. It follows from Equation (10.90) that

$$\sigma_{\rm total} = \frac{4\pi}{k^{\,2}} \sum_{l=0,l_{\rm max}} (2\,l+1)\,\sin^2\delta_l.$$ (10.114)

Making use of Equation (10.105), as well as the asymptotic expansions (10.62) and (10.63), we find that

$$\sin^2\delta_l =\frac{\tan^2\delta_l}{1+\tan^2\delta_l} = \frac{j_l^{\,2}(k\,a)}{j_l^{\,2}(k\,a)+\eta_l^{\,2}(k\,a)} =\sin^{\,2}(k\,a-l\,\pi/2).$$ (10.115)

In particular,

$$\sin^2\delta_l + \sin^2\delta_{l+1} = \sin^2(k\,a-l\,\pi/2)+\cos^2(k\,a-l\,\pi/2)= 1.$$ (10.116)

Hence, it is a good approximation to write

$$\sigma_{\rm total} \simeq \frac{2\pi}{k^{\,2}} \sum_{l=0,l_{\rm max}} (2\,l+1)= \frac{2\pi}{k^{\,2}} \,(l_{\rm max}+1)^{\,2} \simeq 2\pi\,a^{\,2}$$ (10.117)

[59]. This is twice the classical result, which is somewhat surprising, because we might expect to obtain the classical result in the short-wavelength limit. In fact, for hard-sphere scattering, all incident particles with impact parameters less than $a$ are deflected. However, in order to produce a shadow behind the sphere, there must be scattering in the forward direction (recall the optical theorem) to produce destructive interference with the incident plane wave. The effective cross-section associated with this forward scattering is $\pi \,a^{\,2}$ , which, when combined with the cross-section for classical reflection, $\pi \,a^{\,2}$ , gives the actual cross-section of $2\pi \,a^{\,2}$ [95].