Elastic Scattering

In elastic scattering, the initial and final atomic states are the same, as are the initial and final photon energies.

$$\bgroup\color{black} {d\sigma_{elastic}\over d\Omega} = \left... ...rangle \over \omega_{ji}+\omega} \right] \right\vert^2 \egroup$$

With the help of some commutators, the $\delta_{ii}$ term can be combined with the others.

The commutator $[\vec{x},\vec{p}]$ (with no dot products) can be very useful in calculations. When the two vectors are multiplied directly, we get something with two Cartesian indices.

$$\bgroup\color{black} x_ip_j-p_jx_i=i\hbar\delta_{ij} \egroup$$

The commutator of the vectors is $i\hbar$ times the identity. This can be used to cast the first term above into something like the other two.

$$\begin{eqnarray*} x_ip_j-p_jx_i&=&i\hbar\delta_{ij} \\ \hat{\epsilon}\cdot\hat{... ... -(\hat{\epsilon}'\cdot\vec{p})(\hat{\epsilon}\cdot\vec{x}) \\ \end{eqnarray*}$$

Now we need to put the states in using an identity, then use the commutator with $H$ to change $\vec{x}$ to $\vec{p}$.

$$\begin{eqnarray*} 1&=&\sum\limits_j\langle i\vert j\rangle \langle j\vert i\ran... ...}\cdot\vec{p})_{ij}(\hat{\epsilon}'\cdot\vec{p})_{ji}\right] \\ \end{eqnarray*}$$

(Reminder: $\omega_{ij}={E_i-E_j\over\hbar}$ is just a number. $(\hat{\epsilon}\cdot\vec{p})_{ij}=\langle i\vert\hat{\epsilon}\cdot\vec{p}\vert j\rangle$ is a matrix element between states.)

We may now combine the terms for elastic scattering.

$$\begin{eqnarray*} {d\sigma_{elas}\over d\Omega} &=&\left({e^2\over 4\pi mc^2}\ri... ...\over \omega_{ji}(\omega_{ji}+\omega)} \right] \right\vert^2 \\ \end{eqnarray*}$$

This is a nice symmetric form for elastic scattering. If computation of the matrix elements is planned, it useful to again use the commutator to change $\vec{p}$ into $\vec{x}$.

$${d\sigma_{elas}\over d\Omega} =\left({e^2\ov... ...\vec{x} \vert i\rangle \over \omega_{ji}+\omega} \right] \right\vert^2$$