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9.5: Optical Theorem

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    08:40, 16 Nov 2014
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The differential scattering cross-section $ d\sigma/d{\mit\Omega}$ is simply the modulus squared of the scattering amplitude $ f(\theta)$ . The total cross-section is given by

 

$\displaystyle \sigma_{\rm total}$ $\displaystyle = \oint d{\mit\Omega}\,\vert f(\theta)\vert^{\,2}$    
  $\displaystyle = \frac{1}{k^2} \oint d\varphi \int_{-1}^{1} d\mu \sum_l \sum_{l'... ...elta_l-\delta_{l'})]\, \sin\delta_l \,\sin\delta_{l'}\, P_l(\mu)\, P_{l'}(\mu),$ (980)

 

 

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

 

$\displaystyle \sigma_{\rm total} = \frac{4\pi}{k^2} \sum_{l=0,\infty} (2\,l+1)\,\sin^2\delta_l,$ (981)

 

 

where use has been made of Equation (967). A comparison of this result with Equation (979) yields

 

$\displaystyle \sigma_{\rm total} = \frac{4\pi}{k}\, {\rm Im}\left[f(0)\right],$ (982)

 

 

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

 

$\displaystyle \sigma_{\rm total} = \sum_{l=0,\infty} \sigma_l,$ (983)

 

 

where

 

$\displaystyle \sigma_l = \frac{4\pi}{k^2}\, (2\,l+1)\, \sin^2\delta_l$ (984)

 

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$ .

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