Exponential Potential Born Approximation

(Submitted by team 1)

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Using the Born approximation, find the differential cross section for the next exponential potential:

$V(r)=-V_{0}e^{-{\frac {r}{a}}}$

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If the potential $V\!$ is spherical symmetric we can use the equation:

$f_{born}(\theta )=-{\frac {2m}{\hbar ^{2}}}\int _{0}^{\infty }dr'V(r'){\frac {\sin(qr')}{qr'}}{r'}^{2}$

So,

$f_{born}(\theta )=-{\frac {2mV_{0}}{\hbar ^{2}q}}\int _{0}^{\infty }r'\sin(qr')e^{-{\frac {r'}{a}}}dr'$

Solving this integral by parts,

${\begin{aligned}f_{born}(\theta )&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\int _{0}^{\infty }\cos(qr')e^{-{\frac {r'}{a}}}dr'\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}Re\left[\int _{0}^{\infty }e^{iqr'}e^{-{\frac {r'}{a}}}dr'\right]\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}Re\left[{\frac {e^{(iq-{\frac {1}{a}})r'}}{iq-{\frac {1}{a}}}}\right]_{_{0}}^{^{\infty }}\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}Re\left[{\frac {1}{{\frac {1}{a}}+iq}}\right]\end{aligned}}$

$f_{born}(\theta )={\frac {4mV_{0}}{\hbar ^{2}a}}({\frac {1}{{\frac {1}{a^{2}}}+q^{2}}})^{2}$

So, the differential cross section,

${\frac {d\sigma }{d\theta }}=\left|f_{born}(\theta )\right|^{2}={\frac {16m^{2}V_{0}^{2}}{\hbar ^{4}a^{2}}}\left({\frac {1}{{\frac {1}{a^{2}}}+q^{2}}}\right)^{4}$