Cheat sheets/Quantum mechanical scattering

(Assume potential is radially symmetric) ${\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(r)\right]\psi (r)=E\psi (r)}$

If we put

${\displaystyle U={\frac {2m}{\hbar }}V(r)}$

then

${\displaystyle \left[\nabla ^{2}+k^{2}-U(r)\right]\psi (r)=0}$

${\displaystyle \nabla =\partial _{r}{\hat {r}}+{\frac {1}{r}}\partial _{\theta }{\hat {\theta }}+{\frac {1}{r\sin \theta }}\partial _{\phi }{\hat {\phi }}}$

${\displaystyle \left(\nabla ^{2}+k^{2}\right){\frac {e^{ikr}}{r}}=-4\pi \delta ^{3}({\vec {r}})}$

${\displaystyle G({\vec {r}})=\int {\frac {d^{3}p}{(2\pi )^{3}}}e^{-i{\vec {p}}\cdot {\vec {r}}}{\frac {1}{k^{2}-p^{2}}}=-{\frac {1}{4\pi r}}e^{ikr}}$

${\displaystyle E={\frac {|{\vec {p}}|^{2}}{2m}}={\frac {\hbar ^{2}}{2m}}|{\vec {k}}|^{2}={\frac {1}{2}}m|{\vec {v}}|^{2}}$

${\displaystyle \psi _{\vec {k}}({\vec {r}})=e^{i{\vec {k}}\cdot {\vec {r}}}+f(k,\theta ,\phi ){\frac {e^{ikr}}{r}}}$

Incoming: ${\displaystyle e^{i{\vec {k}}\cdot {\vec {r}}}}$

Scattered: ${\displaystyle {\frac {e^{ikr}}{r}}}$

${\displaystyle f(k,\phi ,\theta )=-{1 \over 4\pi }\int e^{-ik'r}u(r)\psi _{k}(r)\,dr}$

${\displaystyle f=-{\frac {m}{2\pi \hbar ^{2}}}\left\langle \phi _{k'}|V|\psi _{k}\right\rangle }$

Plane Wave: ${\displaystyle \phi _{k'}(r)=e^{ik'r}}$

${\displaystyle {\frac {d\sigma }{d\Omega }}=|f(k,\theta ,\phi )|^{2}}$

${\displaystyle \int d\Omega \,\sigma (\Omega )=\int _{0}^{2\pi }d\phi \int _{0}^{\pi }\sin \theta d\theta |f(\theta ,\phi )|^{2}}$