Cheat sheets/Quantum mechanical scattering

Schrodinger equation

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

If we put

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

then

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

Recall

$\mathrm{\nabla }={\partial }_r\hat{r}+\frac{1}{r}{\partial }_{\theta }\hat{\theta }+\frac{1}{r\sin \theta }{\partial }_{\varphi }\hat{\varphi }$

One important result

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

Green's function

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

Simple energy relationships

$E=\frac{|\overrightarrow{p}{|}^2}{2m}=\frac{{\hslash }^2}{2m}|\overrightarrow{k}{|}^2=\frac{1}{2}m|\overrightarrow{v}{|}^2$

Wavefunction

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

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

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

Scattering Amplitude: f(k,φ,θ)

$f(k,\varphi ,\theta )=-\frac{1}{4\pi }\int e^{-i{k}^{\prime }r}u(r){\psi }_k(r)dr$

$f=-\frac{m}{2\pi {\hslash }^2}⟨{\varphi }_{k^{\prime }}|V|{\psi }_k⟩$

Plane Wave: ${\varphi }_{k^{\prime }}(r)=e^{i{k}^{\prime }r}$

Differential cross section

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

Total cross section

$\int d\mathrm{\Omega }\sigma (\mathrm{\Omega })=\underset{0}{\overset{2\pi }{\int }}d\varphi \underset{0}{\overset{\pi }{\int }}\sin \theta d\theta |f(\theta ,\varphi ){|}^2$