9.5: Optical Theorem
( \newcommand{\kernel}{\mathrm{null}\,}\)
The differential scattering cross-section f(θ) . The total cross-section is given by
=∮dΩ|f(θ)|2![$ = \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),$](http://farside.ph.utexas.edu/teaching/qm/lectures/img2256.png)
where σtotal=4πk2∑l=0,∞(2l+1)sin2δl, ???
where use has been made of Equation ???. A comparison of this result with Equation ??? yields
Pl???=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 (σtotal=∑l=0,∞σl, ???where
l th partial cross-section: i.e., the contribution to the total cross-section from the l th partial cross-section occurs when the phase-shift π/2 .Contributors
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)