Elastic and Inelastic Scattering

According to the analysis of Section 10.5, for the case of a spherically symmetric scattering potential, the scattered wave is characterized by

$$f(\theta)= \sum_{l=0,\infty}(2\,l+1)\,f_l\,P_l(\cos\theta),$$ (10.138)

where

$$f_l=\frac{\exp(\,{\rm i}\,\delta_l)}{k}\,\sin \delta_l= \frac{S_l-1}{2\,{\rm i}\,k}$$ (10.139)

is the amplitude of the $l$ th partial wave, whereas $\delta_l$ is the associated phase-shift. Here,

$$S_l= {\rm e}^{\,{\rm i}\,2\,\delta_l}.$$ (10.140)

Moreover, the fact that $\vert S_l\vert=1$ ensures that the scattering is elastic (i.e., that the number of particles is conserved). Finally, according to the analysis of Section 10.6, the net elastic scattering cross-section can be written

$$\sigma_{\rm elastic} = \frac{4\pi}{k^{\,2}}\sum_{l=0,\infty} (2\,l+1)\,\sin^2\delta_l=4\pi\sum_{l=0,\infty} (2\,l+1)\,\vert f_l\vert^{\,2}.$$ (10.141)

It turns out that many scattering experiments are characterized by the absorption of some of the incident particles. Such absorption may induce a change in the quantum state of the target, or, perhaps, the emergence of another particle. Note that scattering that does not conserve particle number is known as inelastic scattering. We can take inelastic scattering into account in our analysis by writing

$$S_l = \eta_l\,{\rm e}^{\,{\rm i}\,2\,\delta_l},$$ (10.142)

where the real parameter $\eta_l$ is such that

$$0\leq \eta_l\leq 1.$$ (10.143)

It follows from Equation (10.140) that

$$f_l = \frac{\eta_l\,\sin(2\,\delta_l)}{2\,k} + {\rm i}\left[\frac{1-\eta_l\,\cos(2\,\delta_l)}{2\,k}\right].$$ (10.144)

Hence, according to Equation (10.142), the net elastic scattering cross-section becomes

$$\sigma_{\rm elastic} = 4\pi\sum_{l=0,\infty} (2\,l+1)\,\vert f_l\vert^{\,2}$$

$$=\frac{\pi}{k^{\,2}}\sum_{l=0,\infty}(2\,l+1)\left[1+\eta_l^{\,2}-2\,\eta_l\,\cos (2\,\delta_l)\right].$$ (10.145)

The net inelastic scattering (i.e., absorption) cross-section follows from Equations (10.24) and (10.88):

$$\sigma_{\rm inelastic} = \frac{\oint d{\mit\Omega}\,r^{\,2}\,(-j_r)}{\vert{\bf j}_{\rm i... ...\frac{\pi}{k^{\,2}}\sum_{l=0,\infty}(2\,l+1)\left(1-\vert S_l\vert^{\,2}\right)$$

$$=\frac{\pi}{k^{\,2}}\sum_{l=0,\infty}(2\,l+1)\left(1-\eta_l^{\,2}\right).$$ (10.146)

Thus, the total cross-section is

$$\sigma_{\rm total} = \sigma_{\rm elastic}+ \sigma_{\rm inelastic}$$

$$= \frac{2\pi}{k^{\,2}}\sum_{l=0,\infty} (2\,l+1)\left[1-\eta_l\,\cos (2\,\delta_l)\right].$$ (10.147)

Note, from Equations (10.139), (10.140), and (10.143) that

$${\rm Im}[f(0)] =\frac{1}{2\,k} \sum_{l=0,\infty} (2\,l+1)\left[1-\eta_l\,\cos (2\,\delta_l)\right].$$ (10.148)

In other words,

$$\sigma_{\rm total} = \frac{4\pi}{k}\,{\rm Im}[f(0)].$$ (10.149)

Hence, we deduce that the optical theorem, described in Section 10.6, still applies in the presence of inelastic scattering.

If $\eta_l=1$ then there is no absorption, and the $l$ th partial wave is scattered in a completely elastic manner. On the other hand, if $\eta_l=0$ then there is total absorption of the $l$ th partial wave. However, such absorption is necessarily accompanied by some degree of elastic scattering. In order to illustrate this important point, let us investigate the special case of scattering by a black sphere. Such a sphere has a well-defined edge of radius $a$ , and is completely absorbing. Consider short-wavelength scattering characterized by $k\,a\gg 1$ . In this case, we expect all partial waves with $l\leq l_{\rm max}$ , where $l_{\rm max} \simeq k\,a$ , to be completely absorbed (because, by analogy with classical physics, the impact parameters of the associated particles are less than $a$ --see Section 10.8), and all other partial waves to suffer neither absorption nor scattering. In other words, $\eta_l=0$ for $0\leq l_{\rm max}$ , and $\eta_l=1$ , $\delta_l=0$ for $l > l_{\rm max}$ . It follows from Equations (10.146) and (10.147) that

$$\sigma_{\rm elastic} = \frac{\pi}{k^{\,2}}\sum_{l=0,l_{\rm max}}(2\,l+1)= \frac{\pi}{k^{\,2}}\,(1+l_{\rm max})^{\,2}\simeq \pi\,a^{\,2},$$ (10.150)

and

$$\sigma_{\rm inelastic} = \frac{\pi}{k^{\,2}}\sum_{l=0,l_{\rm max}}(2\,l+1)= \frac{\pi}{k^{\,2}}\,(1+l_{\rm max})^{\,2}\simeq \pi\,a^{\,2}$$ (10.151)

[59]. Thus, the total scattering cross-section is

$$\sigma_{\rm total}=\sigma_{\rm elastic}+\sigma_{\rm inelastic}= 2\pi\,a^{\,2}.$$ (10.152)

This result seems a little strange, at first, because, by analogy with classical physics, we would not expect the total cross-section to exceed the cross-section presented by the sphere. Nor would we expect a totally absorbing sphere to give rise to any elastic scattering. In fact, this reasoning is incorrect. The absorbing sphere removes flux proportional to $\pi \,a^{\,2}$ from the incident wave, which leads to the formation of a shadow behind the sphere. However, a long way from the sphere, the shadow gets filled in. In other words, the shadow is not visible infinitely far downstream of the sphere. The only way in which this can occur is via the diffraction of some of the incident wave around the edges of the sphere. Actually, the amount of the incident wave that must be diffracted is the same amount as was removed from the wave by absorption. Thus, the scattered flux is also proportional to $\pi \,a^{\,2}$ [53].

Consider low-energy scattering by a hard-sphere potential. As we saw in Section 10.8, this process is dominated by S-wave (i.e., $l=0$ ) scattering. Moreover, the phase-shift of the S-wave takes the form

$$\delta_0 = -k\,a,$$ (10.153)

where $k$ is the wavenumber of the incident particles, and $a$ is the radius of the sphere. Note that the low-energy limit corresponds to $k\,a\ll 1$ . It follows that

$$S_0 = {\rm e}^{\,{\rm i}\,2\,\,\delta_0} \simeq 1 - 2\,{\rm i}\,k\,a.$$ (10.154)

We can generalize the previous analysis to take absorption into account by writing

$$S_0 \simeq 1 - 2\,{\rm i}\,k\,\alpha,$$ (10.155)

where $\alpha$ is complex, $k\,\vert\alpha\vert\ll 1$ , and ${\rm Im}(\alpha)<0$ . According to Equations (10.146) and (10.147),

$$\sigma_{\rm elastic} \simeq \frac{\pi}{k^{\,2}}\,\vert S_0-1\vert^{\,2} \simeq 4\pi\,\vert\alpha\vert^{\,2},$$ (10.156)

$$\sigma_{\rm inelastic} \simeq \frac{\pi}{k^{\,2}}\left(1-\vert S_0\vert^{\,2}\right) \simeq \frac{4\pi\,{\rm Im}(-\alpha)}{k}.$$ (10.157)

We conclude that the low-energy elastic scattering cross-section is again independent of the incident particle velocity (which is proportional to $k$ ), whereas the inelastic cross-section is inversely proportional to the particle velocity. Consequently, as the incident particle velocity decreases, inelastic scattering becomes more and more important in comparison with elastic scattering [8].