Exponential Potential Born Approximation

The potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \!} is spherically symmetric, so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{\text{Born}}(\theta)= -\frac{2m}{\hbar^2} \int_0^\infin dr'\,V(r') \frac{\sin(qr')}{qr'} {r'}^2.}

Substituting in the given potential, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{\text{Born}}(\theta) = - \frac{2mV_0}{\hbar^2 q} \int_0^\infin dr'\, r' \sin(qr') e^{-r'/a}. }

Integrating by parts, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f_{\text{Born}}(\theta) &= -\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 e\left (\int_0^\infin 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}^{\infin} \\ &= -\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}}

The differential cross section is therefore

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.}

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