Phy5645/Cross Section Relation

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The differential cross section is related to the scattering amplitude through

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} \sigma (\theta)}{\mathrm{d} \Omega} = |f_{k}(\theta)|^2}

Since

we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\sigma (\theta)}{d\Omega} \geq (\Im m[f_{k}(\theta)])^{2}.}

On the other hand, from the optical theorem we have

For a central potential, the scattering amplitude is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_k(\theta) = \frac{1}{k}\sum_{l = 0}^{\infty}(2l + 1) e^{i\delta _{l}} \sin\delta _{l} P_{l} (\cos \theta),}

and thus the differential cross section is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\sigma (\theta)}{d\Omega} = \frac{1}{k^2}\sum_{l = 0}^{\infty}\sum_{l^{\prime} = 0}^{\infty}(2l + 1)(2l^{\prime} + 1) e^{i(\delta _{l}- \delta _{l^{\prime}})} \sin\delta _{l}\sin\delta _{l'} P_{l} (\cos \theta)P_{l'} (\cos \theta)}

The total cross section is then

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{l}(1)= 1\!} we can write

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\sigma (0)}{d\Omega} = \frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) e^{i\delta _{l}} sin\delta _{l} \right ]^2=\frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) \sin\delta _{l}\cos\delta _{l} + i\sin^2\delta _{l} \right ]^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) \sin\delta _{l}\cos\delta _{l}\right ]^2 +\frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) \sin^2\delta _{l} \right ]^2\geq \frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) \sin^2\delta _{l} \right ]^2 = \frac{k^2\sigma ^{2}}{16\pi ^{2}}.}

From this, it follows that

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