Optical Theorem

The differential scattering cross-section, $d\sigma/d{\mit\Omega}$ , is simply the modulus squared of the scattering amplitude, $f(\theta)$ . [See Equation (10.28).] The total scattering cross-section is defined as

$$\begin{multline} \sigma_{\rm total} = \oint d{\mit\Omega}\,\frac{d\sigma}{d{\mit... ... [0.5ex] \sin\delta_l \,\sin\delta_{l'}\, P_l(\mu)\, P_{l'}(\mu), \end{multline}$$

where $\mu = \cos\theta$ . It follows that

$$\sigma_{\rm total} = \frac{4\pi}{k^{\,2}} \sum_{l=0,\infty} (2\,l+1)\,\sin^2\delta_l,$$ (10.89)

where use has been made of Equation (10.65). A comparison of the preceding expression with Equation (10.81) reveals that

$$\sigma_{\rm total} = \frac{4\pi}{k}\, {\rm Im}\left[f(0)\right] =\frac{4\pi}{k}\,{\rm Im}\left[f({\bf k},{\bf k})\right],$$ (10.90)

because $P_l(1) = 1$ [1]. This result is known as the optical theorem [107,73], and is a consequence of the fact that the very existence of scattering requires scattering in the forward ($\theta=0$ ) direction, in order to interfere with the incident wave, and thereby reduce the probability current in that direction.

It is conventional to write

$$\sigma_{\rm total} = \sum_{l=0,\infty} \sigma_l,$$ (10.91)

where

$$\sigma_l = \frac{4\pi}{k^{\,2}}\, (2\,l+1)\, \sin^2\delta_l$$ (10.92)

is termed the $l$ th partial scattering cross-section: that is, the contribution to the total scattering cross-section from the $l$ th partial wave. Note that (at fixed $k$ ) the maximum value for the $l$ th partial scattering cross-section occurs when the associated phase-shift, $\delta_l$ , takes the value $\pi/2$ .