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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,
\end{displaymath} (981)

for all $l$. It follows from Eq. (974) that
\tan \delta_l = \frac{j_l(k a)}{\eta_l(k a)}.
\end{displaymath} (982)

Consider the $l=0$ partial wave, which is usually referred to as the $s$-wave. Equation (982) yields

\tan\delta_0 = \frac{\sin (k a)/k a}{-\cos (k a)/ka} = -\tan k a,
\end{displaymath} (983)

where use has been made of Eqs. (948)-(949). It follows that
\delta_0 = -k a.
\end{displaymath} (984)

The $s$-wave radial wave function is
$\displaystyle A_0(r)$ $\textstyle =$ $\displaystyle \exp(-{\rm i}  k a) \frac{[\cos k a  \sin k r
-\sin k a  \cos k r]}{k r}$  
  $\textstyle =$ $\displaystyle \exp(-{\rm i}  k a)  \frac{ \sin[k (r-a)]}{k r}.$ (985)

The corresponding radial wave-function for the incident wave takes the form
\tilde{A}_0(r) = \frac{ \sin k r}{k r}.
\end{displaymath} (986)

It is clear that 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 consider the low and high energy asymptotic limits of $\tan\delta_l$. Low energy means $k a\ll 1$. In this regime, the spherical Bessel functions and Neumann functions reduce to:

$\displaystyle j_l(k r)$ $\textstyle \simeq$ $\displaystyle \frac{(k r)^l}{(2 l+1)!!},$ (987)
$\displaystyle \eta_l(k r)$ $\textstyle \simeq$ $\displaystyle -\frac{(2 l-1)!!}{(k r)^{l+1}},$ (988)

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}.
\end{displaymath} (989)

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. (923), (965), and (984) that
\frac{d\sigma}{d\Omega} = \frac{\sin^2 k a}{k^2} \simeq a^2
\end{displaymath} (990)

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
\end{displaymath} (991)

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 do 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. (967) that

\sigma_{\rm total} = \frac{4\pi}{k^2} \sum_{l=0}^{l_{\rm max}}
(2 l+1) \sin^2\delta_l.
\end{displaymath} (992)

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} = \sum_{l=0}^{k a} \frac{2\pi}{k^2}  (2 l+1) \simeq
2\pi  a^2.
\end{displaymath} (993)

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 be scattering in the forward direction (recall the optical theorem) to produce destructive interference with the incident plane-wave. In fact, the interference is not completely destructive, and the shadow has a bright 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 $2\pi  a^2$.

next up previous
Next: Low energy scattering Up: Scattering theory Previous: Determination of phase-shifts
Richard Fitzpatrick 2006-02-16