Cheat sheets/Quantum mechanical scattering

Schrodinger equation

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

If we put

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

then

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

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

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

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

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

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

Incoming:

e^{i{\vec {k}}\cdot {\vec {r}}}

Scattered:

{\frac {e^{ikr}}{r}}

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

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

Plane Wave:

\phi _{k'}(r)=e^{ik'r}

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

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

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