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

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

Recall

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

One important result

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

Green's function

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

Simple energy relationships

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

Wavefunction

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

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

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

Scattering Amplitude: f(k,φ,θ)

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

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

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

Differential cross section

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

Total cross section

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