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,

$$\beta_{l-} = \beta_{l+} = \infty,$$ (1298)

for all $l$. Equation (1291) thus gives

$$\tan \delta_l = \frac{j_l(k\,a)}{y_l(k\,a)}.$$ (1299)

Consider the $l=0$ partial wave, which is usually referred to as the $S$-wave. Equation (1299) yields

$$\tan\delta_0 = \frac{\sin (k\,a)/k\,a}{-\cos (k\,a)/ka} = -\tan (k\,a),$$ (1300)

where use has been made of Eqs. (1268) and (1269). It follows that

$$\delta_0 = -k\,a.$$ (1301)

The $S$-wave radial wave function is [see Eq. (1289)]

$${\cal R}_0(r) = \exp(-{\rm i}\, k\,a) \frac{[\cos (k\,a) \,\sin (k\,r) -\sin (k\,a) \,\cos (k\,r)]}{k\,r}$$

$$= \exp(-{\rm i}\, k\,a)\, \frac{ \sin[k\,(r-a)]}{k\,r}.$$ (1302)

The corresponding radial wave-function for the incident wave takes the form [see Eq. (1277)]

$$\tilde{\cal R}_0(r) = \frac{ \sin (k\,r)}{k\,r}.$$ (1303)

Thus, the actual $l=0$ radial wave-function is similar to the incident $l=0$ wave-function, 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:

$$j_l(k\,r) \simeq \frac{(k\,r)^l}{(2\,l+1)!!},$$ (1304)

$$y_l(k\,r) \simeq -\frac{(2\,l-1)!!}{(k\,r)^{l+1}},$$ (1305)

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

$$\tan\delta_l = \frac{-(k\,a)^{2\,l+1}}{(2\,l+1) \,[(2\,l-1)!!]^{\,2}}.$$ (1306)

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 Eqs. (1245), (1285), and (1301) that

$$\frac{d\sigma}{d\Omega} = \frac{\sin^2 k\,a}{k^2} \simeq a^2$$ (1307)

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

$$\sigma_{\rm total} = \int\frac{d\sigma}{d\Omega}\,d\Omega = 4\pi \,a^2$$ (1308)

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 wave-lengths, 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 Eq. (1287) that

$$\sigma_{\rm total} \simeq \frac{4\pi}{k^2} \sum_{l=0}^{l_{\rm max}} (2\,l+1)\,\sin^2\delta_l.$$ (1309)

With so many $l$ values contributing, it is legitimate to replace $\sin^2\delta_l$ by its average value $1/2$. Thus,

$$\sigma_{\rm total} \simeq \sum_{l=0}^{k\,a} \frac{2\pi}{k^2} \,(2\,l+1) \simeq 2\pi \,a^2.$$ (1310)

This is twice the classical result, which is somewhat surprizing, since we might expect to obtain the classical result in the short wave-length 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. 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 $2\pi \,a^2$.