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\heading{E786: Calculation of the cross section for the Yukawa potential} 
\auname{Brigit Ben Ami}


{\bf The problem:}

\Dn

Find the differential cross section and the total cross section by the Born approximation for Yukawa potential : ${ U(r)=U_0\frac{e^{-(\frac{r}{a})}}{r} }$ . Define the conditions for the use of the Born approximation. \\  \\

\Dn


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{\bf The solution:}

\Dn

From the Fermi golden rule we have the formula for the differential cross section called the Born approximation \\

$ \sigma (\Omega)= \frac{1}{2\pi^{2}} \left(\frac{k_{E}}{v_{E}}\right)^{2} \left|\tilde{U} \left(\vec{k_{\Omega}}- \vec{k_{0}}\right)\right|^{2} $ \\

In order to use the Born formula we define the system coordinates: \\
The incident wave propagates in the z direction. The scattering direction is $ \Omega = \left( \theta ,\varphi \right)$. We also define $ \vec{q} = \vec{k_{\Omega}} - \vec{k_{0}} $ \; as the difference between the $  k_{\Omega} $ of the scattered wave and the $ k_{0} $ of the incident wave. \\
$ k_{E} = \left| \vec {k_{\Omega}} \right| = \left| \vec{k_{0}} \right| $  \ (ellastic scattering). \\ 


The approximation is that the scattering potential can be treated as a perturbation, allowing us to use the Fermi golden rule, from which we get the formula above. \\ \\  


$ \tilde{U}(\vec {q}) $ is the Fourier transform of the scattering potential U(r): \\
 
$ \tilde{U} (\vec {q} ) = \int\int\int U(r) e^{-i \vec {q} \vec {r} }  d^{3}r = U_{0} \int^{2\pi}_{0} d\varphi  \int^{\pi}_{0} d\theta  \int^{\infty}_{0} dr \cdot r^{2} sin\theta e^{ -i \vec{q} \vec{r} cos\theta } \frac{e^{-\frac{r}{a}}}{r} $ \\


The given potential is spherically symmetric, so we can rotate coordinate system for the calculation of this integral. We choose $ \vec {q} $ in the z direction, which means $ \theta =0 $. \\ 
 
$ \tilde{U} (\vec {q})  = 2\pi U_{0} \int^{\infty}_{0} dr \cdot r \cdot e^{-\frac{r}{a}} \int^{1}_{-1} d[cos\theta] e^{ -i q r cos\theta } = $ \\ \\ \\
 $ 2\pi U_{0} \int^{\infty}_{0} dr \cdot r \cdot e^{-\frac{r}{a}} \frac{e^{ -i q r cos\theta }}{-i q r } |^{1}_{-1}=  2\pi U_{0} \int^{\infty}_{0} dr \cdot e^{-\frac{r}{a}} \frac{e^{ -i q r } - e^{ i q r }}{-i q } = $  \\ \\ \\
 $  \frac{2 \pi U_{0}}{i q}  \int^{\infty}_{0} dr \left[  e^{\left( iq -\frac{1}{a} \right) r} -  e^{\left( -iq -\frac{1}{a} \right) r }  \right] = \frac{2 \pi U_{0}}{i q}  \left[ \frac{e^{\left(iq -\frac{1}{a}\right) r }}{iq -\frac{1}{a}} - \frac {e^{\left(-iq -\frac{1}{a}\right) r} }{-iq -\frac{1}{a}} \right] |^{\infty}_{0}= $ \\ \\ \\
 $ \frac{2 \pi U_{0}}{i q}  \left[ \frac{-1}{iq -\frac{1}{a}} + \frac{-1}{iq + \frac{1}{a}} \right] =  \frac{2 \pi U_{0}}{i q} \frac{2iq}{q^{2} + \frac{1}{a^{2}} } = $ \\ \\ \\ 
 $  \frac{4\pi U_{0}}{q^{2} + \frac{1}{a^{2}} }   $   \\ \\ \\


$ \frac{k_{E}}{v_{E}} = m $ (non-relativistic dispersion relation) so we get: \\

 $ \sigma (\Omega)= \frac{m^{2}}{4\pi^{2}} \left|\tilde{U} (q) \right|^{2} =  \frac{4 m^{2} U_{0}^{2} }{\left(q^{2} + \frac{1}{a^{2}} \right)^{2} } $ \\ 
 
now we choose $ \theta $ to be the angle between $ \vec{k_{0}} $ and $ \vec{k_{\Omega}} . $ \\

since $ k_{E} = \left| \vec {k_{\Omega}} \right| = \left| \vec{k_{0}} \right| $  we have $ q = 2k_{E} sin \left( \frac{\theta}{2} \right) $ \\ 
 

now $ \sigma (\Omega)=  \frac{4 m^{2} U_{0}^{2} }{ \left( \frac{1}{a^{2}} + 4 k_{E}^{2} sin^{2}\left( \frac{\theta}{2}\right) \right) ^{2} } $ \\ 

By taking the limit $ a\rightarrow \infty $ the cross section of the Yukawa potential approaches the Rutherford scattering cross section: \\

$ \sigma (\Omega) \left[ a\rightarrow \infty  \right]=  \frac{4 m^{2} U_{0}^{2} }{ \left( 4 k_{E}^{2} sin^{2}\left( \frac{\theta}{2}\right) \right) ^{2} } $ \\  \\  \\

Now, let's calculate the total cross section for the Yukawa potential: \\

$ \sigma_{total} = \int\int \sigma (\Omega) d\Omega = $ \\ \\ \\
\
$  \sigma_{total} = \int^{2\pi}_{0} d\varphi \int^{\pi}_{0} sin \theta d\theta \cdot \sigma (\theta) =  2\pi \int^{\pi}_{0} \sigma (\theta) sin \theta d\theta $ \\ \\  

It will be easier to use the expression for q : \\

$  q = 2k_{E} sin \left( \frac{\theta}{2} \right) \ \Rightarrow dq = k_{E} cos \left( \frac{\theta}{2} \right) d\theta \cdot \frac{q}{q}  = \frac{k_{E}^{2}}{q} sin \theta d\theta  \ \ \Rightarrow  sin \theta d\theta = \frac{q\cdot dq}{k_{E}^{2}} $ \\ \\ \\

$ \sigma_{total} =  \frac{4 m^{2} U_{0}^{2} } {k_{E}^{2} } 2\pi \int^{2k_{E}}_{0} dq \cdot q \frac{1}{\left( q^{2} + \frac{1}{a^{2}}   \right)^{2} } $ \\ \\

$z=q^{2} \ \ dz=2q \cdot dq $ \\ \\

$ \Rightarrow \sigma_{total} =  \frac{4 m^{2} U_{0}^{2} \pi } {k_{E}^{2} } \int^{4k_{E}^{2} }_{0} \frac{dz}{ \left( \frac{1}{a^{2}} + z \right) ^{2} } =  \frac{4 m^{2} U_{0}^{2} \pi } {k_{E}^{2} }  \frac{-1}{ \frac{1}{a^{2}} + z } |^{4k_{E}^{2}}_{0} = $ \\ \\

$ \frac{4 m^{2} U_{0}^{2} \pi } {k_{E}^{2} } \left( a^{2} - \frac{a^{2}}{1 + 4 a^{2}k_{E}^{2}}  \right) =  4 m^{2} U_{0}^{2} \pi \frac{4 a^{4}}{1 + 4 a^{2}k_{E}^{2}} $ \\ \\ 

$ \Rightarrow  \sigma_{total} =  \frac{16 a^{4} m^{2} U_{0}^{2} \pi}{ 1 + 4 a^{2}k_{E}^{2}} $ \\ \\ \\



 
 
 
 
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