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| | The potential <math> V \!</math> is spherically symmetric, so that |
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| Using the Born approximation, find the differential cross section for the next exponential potential:
| | <math> f_{\text{Born}}(\theta)= -\frac{2m}{\hbar^2} \int_0^\infin dr'\,V(r') \frac{\sin(qr')}{qr'} {r'}^2.</math> |
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| ::<math> V(r)= -V_0 e^{-\frac{r}{a}} </math>
| | Substituting in the given potential, we obtain |
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| | <math> f_{\text{Born}}(\theta) = - \frac{2mV_0}{\hbar^2 q} \int_0^\infin dr'\, r' \sin(qr') e^{-r'/a}. </math> |
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| | Integrating by parts, we obtain |
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| If the potential V is spherical symmetric we can use the equation:
| | <math> |
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| : <math> f_{born}(\theta) = \frac{-2m}{\hbar^2} \int_0^\infin dr' V(r') \frac{\sin(qr')}{qr'} {r'}^2 </math>
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| So,
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| : <math> f_{born}(\theta) = \frac{-2mV_0}{\hbar^2 q} \int_0^\infin r' sin(qr') e^{-\frac{r'}{a}} dr' </math>
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| Solving this integral by parts,
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| : <math>
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| \begin{align} | | \begin{align} |
| f_{born}(\theta) | | f_{\text{Born}}(\theta) |
| &= \frac{-2mV_0}{\hbar^2 q} \frac{\partial}{\partial q} \int_0^\infin cos(qr') e^{-\frac{r'}{a}} dr' \\ | | &= -\frac{2mV_0}{\hbar^2 q}\frac{\partial}{\partial q}\int_0^\infin dr'\,\cos(qr')e^{-r'/a} \\ |
| &= \frac{-2mV_0}{\hbar^2 q} \frac{\partial}{\partial q} Re[ \int_0^\infin e^{iqr'} e^{-\frac{r'}{a}} dr' ] \\ | | &= -\frac{2mV_0}{\hbar^2 q}\frac{\partial}{\partial q}\Re e\left (\int_0^\infin dr'\,e^{iqr'}e^{-r'/a}\right ) \\ |
| &= \frac{-2mV_0}{\hbar^2 q} \frac{\partial}{\partial q} Re[ \frac { e^{(iq - \frac{1}{a})r'} }{iq - \frac{1}{a}} ]_{_0}^{^\infin} \\ | | &= -\frac{2mV_0}{\hbar^2 q}\frac{\partial}{\partial q}\Re e\left [\frac{e^{(iq - 1/a)r'}}{iq - 1/a}\right ]_{0}^{\infin} \\ |
| &= \frac{-2mV_0}{\hbar^2 q} \frac{\partial}{\partial q} Re[ \frac { 1 }{\frac{1}{a} + iq }] \\ | | &= -\frac{2mV_0}{\hbar^2 q}\frac{\partial}{\partial q}\Re e\left [\frac {1}{1/a + iq}\right ] \\ |
| | | &= \frac{4mV_0a^3}{\hbar^2}\left (\frac{1}{1+q^2a^2}\right )^2. |
| \end{align}
| | \end{align}</math> |
| </math>
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| : <math> f_{born}(\theta) = \frac{4mV_0}{\hbar^2 a} (\frac{1}{ \frac{1}{a^2} +q^2 })^2 </math>
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| So, the differential cross section,
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| | The differential cross section is therefore |
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| :<math> \frac{d\sigma}{d \theta} = |f_{born}(\theta) |^2 = \frac{16m^2V_0^2}{\hbar^4 a^2} (\frac{1}{ \frac{1}{a^2} +q^2 })^4 </math>
| | <math> \frac{d\sigma}{d\theta}=\left|f_{\text{Born}}(\theta) \right|^2=\frac{16m^2V_0^2a^6}{\hbar^4} \left(\frac{1}{1+q^2a^2}\right)^4.</math> |
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| | Back to [[Differential Cross Section and the Green's Function Formulation of Scattering#Problems|Differential Cross Section and the Green's Function Formulation of Scattering]] |
Latest revision as of 13:48, 18 January 2014
The potential 
is spherically symmetric, so that
Substituting in the given potential, we obtain
Integrating by parts, we obtain
![{\displaystyle {\begin{aligned}f_{\text{Born}}(\theta )&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\int _{0}^{\infty }dr'\,\cos(qr')e^{-r'/a}\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\Re e\left(\int _{0}^{\infty }dr'\,e^{iqr'}e^{-r'/a}\right)\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\Re e\left[{\frac {e^{(iq-1/a)r'}}{iq-1/a}}\right]_{0}^{\infty }\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\Re e\left[{\frac {1}{1/a+iq}}\right]\\&={\frac {4mV_{0}a^{3}}{\hbar ^{2}}}\left({\frac {1}{1+q^{2}a^{2}}}\right)^{2}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a283f1a2f0af178fb60f8b019eb1f5c1e31fc095)
The differential cross section is therefore
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