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The optical theorem

The differential scattering cross-section $d\sigma/d\Omega$ is simply the modulus squared of the scattering amplitude $f(\theta)$. The total cross-section is given by
$\displaystyle \sigma_{\rm total}$ $\textstyle =$ $\displaystyle \int \vert f(\theta)\vert^2 d\Omega$  
  $\textstyle =$ $\displaystyle \frac{1}{k^2} \oint d\varphi \int_{-1}^{1} d\mu
\sum_l \sum_{l'} (2 l+1) (2 l'+1)
\exp[ {\rm i} (\delta_l-\delta_{l'}]$  
    $\displaystyle \mbox{\hspace{1cm}}\times \sin\delta_l  \sin\delta_{l'} 
P_l(\mu)  P_{l'}(\mu),$ (966)

where $\mu = \cos\theta$. It follows that
\begin{displaymath}
\sigma_{\rm total} = \frac{4\pi}{k^2} \sum_l (2 l+1) \sin^2\delta_l,
\end{displaymath} (967)

where use has been made of Eq. (953). A comparison of this result with Eq. (965) yields
\begin{displaymath}
\sigma_{\rm total} = \frac{4\pi}{k}  {\rm Im}\left[f(0)\right],
\end{displaymath} (968)

since $P_l(1) = 1$. This result is known as the optical theorem. It is a reflection 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 this direction.

It is usual to write

\begin{displaymath}
\sigma_{\rm total} = \sum_{l=0}^\infty \sigma_l,
\end{displaymath} (969)

where
\begin{displaymath}
\sigma_l = \frac{4\pi}{k^2}  (2 l+1)  \sin^2\delta_l
\end{displaymath} (970)

is the $l$th partial cross-section: i.e., the contribution to the total cross-section from the $l$th partial wave. Note that the maximum value for the $l$th partial cross-section occurs when the phase-shift $\delta_l$ takes the value $\pi/2$.


next up previous
Next: Determination of phase-shifts Up: Scattering theory Previous: Partial waves
Richard Fitzpatrick 2006-02-16