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\heading{Ex 785: Calculation of the cross section for the Gauss potential} 
\auname{Nissim Ron, Ifergan Omri }


{\bf The problem : }

Find the differential cross section and the total cross section by the Born approximation for gauss potential : ${ U(r)=U_0\eexp{-\frac{(r/a)^2}{2} }}$ . Define the conditions for the using of the Born approximation . 

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{\bf The solution:} 

From the Fermi's Golden Rule we get a formula for the differential cross section called the Born approximation : \[{\sigma(\Omega)={(\frac{M}{2\pi})}^2|\tilde{U}(q)|^2 }\] where ${\tilde{U}(q)}$ is the Fourier Transform of the scattering potential ${U(r)}$ . 

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Lets calculate  the ${\tilde{U}(q)}$ for the gauss potential :

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\[\tilde{U}(q)=FT(U(r))={\int\int\int}U(r)\eexp{-iqr}d^3r={\int\int\int}U_0\eexp{-r^2/2a^2}\eexp{-iqr}d^3r= \]

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\[ ={\int\int\int}U_0\eexp{-r^2/2a^2} \eexp{-iqRcos \theta }{d\phi} d(cos\theta)R^2dR=4 \pi{\int^{\infty}_{0}}U_0\eexp{-R/a}sinc(qR)R^2dR= \]

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\[ =4{\pi}\frac{U_0}{q}\int^{\infty}_{0}R \eexp{-R^2/2a^2} sin(qR)dR=\frac{2{\pi}U_0}{qi}{\int^{\infty}_0}R \eexp{-R^2/2a^2}( \eexp{iqR}-\eexp{-iqR} )dR  \]

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Lets use a new markings:
\[-\frac{R^2}{2a^2}\pm iqR = -\frac{1}{2a^2}(R\pm iqa^2)^2-\frac{q^2a^2}{2}\]
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\[\ l=R\pm iqa^2\] 
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\[\ dl=dR\]

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\[\ FT[U(r)] = \frac{2\pi U_0}{iq}\eexp{-\frac{q^2a^2}{2}}\int^{\infty}_{0}\eexp{-\frac{l^2}{2a^2}}(l-iqa^2)-\eexp{-\frac{l^2}{2a^2}}(l+iqa^2)dl=\]
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\[\ = -4\pi U_0 a^2 \eexp{-\frac{q^2a^2}{2}}\int^{\infty}_{0}\eexp{-\frac{l^2}{2a^2}}dl = -U_0(a\sqrt{2\pi})^3\eexp{-\frac{q^2a^2}{2}}\]

So , the differential cross section is : 
\[ {\sigma}(\Omega)=2\pi a^6m^2U_0^2\eexp{-q^2a^2} \]

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Now , lets calculate the total cross section for the Gauss potential :

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\[{{\sigma}_{total}={\int}{\int}{\sigma({\Omega})d{\Omega}}=({\frac{1}{2\pi}})^2({\frac{k_E}{v_E}})^22{\pi}{\int^1_{-1}}|{FT[{U}(q)]}|^2d(cos{\theta}_{\Omega})}\]

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but because of ${q=2{k_E}sin({\theta}_{\Omega}/2)}$ , we get : \[{{\sigma}_{total}={\frac{1}{2{\pi}v_E^2}}{\int^{2k_E}_{0}}|{FT[{U}(r)]}|^2qdq}\]

Lets use the ${FT[{U}(r)]}$ that we found for given potential : 

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\[\ {\sigma}_{total}={\frac{1}{2{\pi}v_E^2}}{\int^{2k_E}_{0}}{|-U_0(a\sqrt{2\pi})^3\eexp{-\frac{q^2a^2}{2}}|^2}qdq=\]
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\[\frac{U_0^2}{4\pi v_E^2} (2\pi a^2)^3\int^{2k_E}_{0}\eexp{-q^2a^2}dq^2 =-\frac{2\pi^2a^4U_0^2}{v_E^2}\left[\eexp{-q^2a^2}\right]^{2k_E}_{0} \]
 we get the finel result :
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\[{\sigma}_{total}=\frac{2\pi^2a^4U_0^2}{v_E^2}\left[1-\eexp{-2k_E^2a^2}\right]\]

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the condition for the using of the Born approximation is derived from F.G.R (weak potential),\\
and so we demand that ${U<<\Delta}$ where ${\Delta}$ is the spacing of the "on site energies" and U  is the pertubation. 


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