The potential V {\displaystyle V\!} is spherically symmetric, so that
f Born ( θ ) = − 2 m ℏ 2 ∫ 0 ∞ d r ′ V ( r ′ ) sin ( q r ′ ) q r ′ r ′ 2 . {\displaystyle 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 Born ( θ ) = − 2 m V 0 ℏ 2 q ∫ 0 ∞ d r ′ r ′ sin ( q r ′ ) e − r ′ / a . {\displaystyle 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
f Born ( θ ) = − 2 m V 0 ℏ 2 q ∂ ∂ q ∫ 0 ∞ d r ′ cos ( q r ′ ) e − r ′ / a = − 2 m V 0 ℏ 2 q ∂ ∂ q ℜ e ( ∫ 0 ∞ d r ′ e i q r ′ e − r ′ / a ) = − 2 m V 0 ℏ 2 q ∂ ∂ q ℜ e [ e ( i q − 1 / a ) r ′ i q − 1 / a ] 0 ∞ = − 2 m V 0 ℏ 2 q ∂ ∂ q ℜ e [ 1 1 / a + i q ] = 4 m V 0 a 3 ℏ 2 ( 1 1 + q 2 a 2 ) 2 . {\displaystyle {\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
d σ d θ = | f Born ( θ ) | 2 = 16 m 2 V 0 2 a 6 ℏ 4 ( 1 1 + q 2 a 2 ) 4 . {\displaystyle {\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 Born Approximation and Examples of Cross-Section Calculations
is spherically symmetric, so that
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a283f1a2f0af178fb60f8b019eb1f5c1e31fc095)
