Question

Prove the optical theorem, which relates the total cross-section to the imaginary part of the forward scattering amplitude:

$\mathbf{\sigma }\mathbf{=}\frac{\mathbf{4}\mathbf{\pi }}{\mathbf{k}}\mathbf{Im}\mathbf{\left(}\mathbf{f}\mathbf{\right(}\mathbf{0}\mathbf{\left)}\mathbf{\right)}$

Short answer

$\sigma =\frac{4\pi }{k}\mathfrak{J}\left[f\left(0\right)\right]$

Hence, it’s proved.

Step-by-step solution

To prove the optical theorem.

The relationship between the total cross section and the scattering amplitude in three dimensional scattering is the optical theorem. Equations 11.47 and 11.48 are given by:

$f\left(\theta \right)=\frac{1}{k}\sum _{l=0}^{\infty }\left(2l+1\right)e^{i\delta }\sin \left({\delta }_l\right)P_l\left(\cos \theta \right)$……………….(1)

$\sigma =\frac{4\pi }{k^2}\sum _{l=0}^{\infty }\left(2l+1\right){\sin }^2\left({\delta }_l\right)$……………………..(2)

If$\theta =0$, we can use the fact that $P_l\left(1\right)=1$for all$l$, so we get:

$f\left(\theta \right)=\frac{1}{k}\sum _{l=0}^{\infty }\left(2l+1\right)e^{i\delta }\sin \left({\delta }_l\right)$

But,

$e^{i{\delta }_l}=\cos \left({\delta }_l\right)+i\sin \left({\delta }_l\right)$

So,

$f\left(0\right)=\frac{1}{k}\underset{l-0}{\overset{\infty }{\sum \left(2l+1\right)\left[\sin \left({\delta }_l\right)\cos \left({\delta }_l\right)+i{\sin }^2\left({\delta }_l\right)\right]}}$

Taking the imaginary part of$f\left(0\right)$we get:

$\mathfrak{J}\left[f\left(0\right)\right]=\frac{1}{k}\sum _{l-0}^{\infty }\left(2l+1\right){\sin }^2\left({\delta }_l\right)$

So equation (2) can be written as:

$\sigma =\frac{4\pi }{k}\mathfrak{J}\left[f\left(0\right)\right]$

Hence, it’s proved.