Elastic Scattering

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

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

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

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

\begin{displaymath}\bgroup\color{black} x_ip_j-p_jx_i=i\hbar\delta_{ij} \egroup\end{displaymath}

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

x_ip_j-p_jx_i&=&i\hbar\delta_{ij} \\
-(\hat{\epsilon}'\cdot\vec{p})(\hat{\epsilon}\cdot\vec{x}) \\

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

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] \\

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

We may now combine the terms for elastic scattering.

{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 \\

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 \bgroup\color{black}$\vec{p}$\egroup into \bgroup\color{black}$\vec{x}$\egroup.
{d\sigma_{elas}\over d\Omega}
...\vec{x} \vert i\rangle
\over \omega_{ji}+\omega}
\right] \right\vert^2

Jim Branson 2013-04-22