Exponential Potential Born Approximation

The potential $V\!$ is spherically symmetric, so that

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

Substituting in the given potential, we obtain

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

Integrating by parts, we obtain

${\begin{aligned}f_{\text{Born}}(\theta )&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\int _{0}^{\infty }dr'\,\cos(qr')e^{-r'/a}\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\Re e\left(\int _{0}^{\infty }dr'\,e^{iqr'}e^{-r'/a}\right)\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\Re e\left[{\frac {e^{(iq-1/a)r'}}{iq-1/a}}\right]_{0}^{\infty }\\&=-{\frac {2mV_{0}}{\hbar ^{2}q}}{\frac {\partial }{\partial q}}\Re e\left[{\frac {1}{1/a+iq}}\right]\\&={\frac {4mV_{0}a^{3}}{\hbar ^{2}}}\left({\frac {1}{1+q^{2}a^{2}}}\right)^{2}.\end{aligned}}$

The differential cross section is therefore

${\frac {d\sigma }{d\theta }}=\left|f_{\text{Born}}(\theta )\right|^{2}={\frac {16m^{2}V_{0}^{2}a^{6}}{\hbar ^{4}}}\left({\frac {1}{1+q^{2}a^{2}}}\right)^{4}.$

Back to Differential Cross Section and the Green's Function Formulation of Scattering