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| <math>\frac{d\sigma (\theta)}{d\Omega} = |f_{k}(\theta)|^2.</math> | | <math>\frac{d\sigma (\theta)}{d\Omega} = |f_{k}(\theta)|^2.</math> |
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| Since <math>| f |^2 = (\Re e f )^2 + (\Im m f )^2 \geq (\Im m f )^2,</math> | | Since <math>| f |^2 = (\Re e\,f )^2 + (\Im m\,f )^2 \geq (\Im m f )^2,</math> |
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| we obtain | | we obtain |
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| <math> \frac{d\sigma (\theta)}{d\Omega} \geq (\Im m[f_{k}(\theta)])^{2}.</math> | | <math> \frac{d\sigma (\theta)}{d\Omega} \geq (\Im m[f_{k}(\theta)])^{2}.</math> |
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| On the other hand, from the optical theorem we have | | On the other hand, the optical theorem states that |
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| <math> \sigma =\frac{4\pi}{k} \Im m[f_{k}(\theta)].</math> | | <math> \sigma =\frac{4\pi}{k} \Im m[f_{k}(0)],</math> |
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| For a central potential, the scattering amplitude is
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| <math>f_k(\theta) = \frac{1}{k}\sum_{l = 0}^{\infty}(2l + 1) e^{i\delta _{l}} \sin\delta _{l} P_{l} (\cos \theta),</math>
| | <math>\frac{d\sigma (0)}{d\Omega}\geq \frac{k^2\sigma ^{2}}{16\pi ^{2}}.</math> |
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| and thus the differential cross section is
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| <math>\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)</math>
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| The total cross section is then
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| <math>\sigma = \frac{4\pi ^2}{k^2}\sum_{l = 0}^{\infty}(2l + 1) \sin^2\delta _{l}.</math>
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| Since <math> P_{l}(1)= 1\!</math> we can write
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| <math>\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</math> | |
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| <math>=\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}}.</math>
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| From this, it follows that <math>\sigma\leq \frac{4\pi}{k}\sqrt{\frac{d\sigma (0)}{d\Omega}}.</math> | | From this, it follows that <math>\sigma\leq \frac{4\pi}{k}\sqrt{\frac{d\sigma (0)}{d\Omega}}.</math> |
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| Back to [[Central Potential Scattering and Phase Shifts]] | | Back to [[Central Potential Scattering and Phase Shifts#Problems|Central Potential Scattering and Phase Shifts]] |
Latest revision as of 13:50, 18 January 2014
The differential cross section is related to the scattering amplitude through
Since 
we obtain
![{\displaystyle {\frac {d\sigma (\theta )}{d\Omega }}\geq (\Im m[f_{k}(\theta )])^{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb3e932d4b3d2da6c7aec3ef3d2c6954e723b0f4)
On the other hand, the optical theorem states that
![{\displaystyle \sigma ={\frac {4\pi }{k}}\Im m[f_{k}(0)],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65449244b1134bd3afbf6d5b0e7732fd91a045cb)
so that
From this, it follows that 
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