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| Consider the scattering of a particle from a real spherically symmetric potential. If <math>\frac{\mathrm{d} \sigma (\theta) }{\mathrm{d} \Omega }</math> is the differential cross section and <math>\sigma</math> is the total cross section, show that
| | The differential cross section is related to the scattering amplitude through |
| <math>\sigma \leq \frac{4\pi}{k}\sqrt{\frac{\mathrm{d} \sigma (0) }{\mathrm{d} \Omega }}</math>
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| for a general central potential using the partial-wave expansion of the scattering amplitude and the cross section.
| | <math>\frac{d\sigma (\theta)}{d\Omega} = |f_{k}(\theta)|^2.</math> |
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| Solution:
| | Since <math>| f |^2 = (\Re e\,f )^2 + (\Im m\,f )^2 \geq (\Im m f )^2,</math> |
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| The differential cross section is related to the scattering amplitude through
| | we obtain |
| <math>\frac{\mathrm{d} \sigma (\theta)}{\mathrm{d} \Omega} = |f_{k}(\theta)|^2</math> | | |
| Since <math>| f |^2 = (Re f )^2 + (Im f )^2 \geq (Im f )^2</math>
| | <math> \frac{d\sigma (\theta)}{d\Omega} \geq (\Im m[f_{k}(\theta)])^{2}.</math> |
| therefore, \frac{\mathrm{d} \sigma (\theta)}{\mathrm{d} \Omega} \geq (Im f_{k}(\theta))^{2}
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| On the other hand, from the optical theorem we have
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| <math> \sigma =\frac{4\pi}{k} Im f_{k}(\theta)) \leq \frac{4\pi}{k}\sqrt{\frac{\mathrm{d} \sigma (0) }{\mathrm{d} \Omega }}</math>
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| For a central potential the scattering amplitude is
| | On the other hand, the optical theorem states that |
| <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>
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| and, in terms of this, the differential cross section is
| | <math> \sigma =\frac{4\pi}{k} \Im m[f_{k}(0)],</math> |
| <math>\frac{\mathrm{d} \sigma (\theta)}{\mathrm{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^{\prime}} P_{l} (cos \theta)P_{l^{\prime}} (cos \theta)</math>
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| The total cross section is
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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
| | so that |
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| <math>\frac{\mathrm{d} \sigma (0)}{\mathrm{d} \Omega} = \frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) e^{i\delta _{l}} sin\delta _{l} \right ]^2</math> | | <math>\frac{d\sigma (0)}{d\Omega}\geq \frac{k^2\sigma ^{2}}{16\pi ^{2}}.</math> |
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| <math>\frac{\mathrm{d} \sigma (0)}{\mathrm{d} \Omega} = \frac{1}{k^2}\left [\sum_{l = 0}^{\infty}(2l + 1) sin\delta _{l}cos\delta _{l} + isin^2\delta _{l} \right ]^2</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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| <math>\frac{\mathrm{d} \sigma (0)}{\mathrm{d} \Omega} = \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</math>
| | Back to [[Central Potential Scattering and Phase Shifts#Problems|Central Potential Scattering and Phase Shifts]] |
| <math>\Rightarrow \frac{\mathrm{d} \sigma (0)}{\mathrm{d} \Omega} \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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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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