Perturbation Calculation for Relativistic Energy Shift

Rewriting \bgroup\color{black}$H_1=-{1\over 8}{p_e^4\over m^3c^2}$\egroup as \bgroup\color{black}$H_1=-{1\over 2mc^2}\left({p^2\over 2m}\right)^2$\egroup we calculate the energy shift for a state \bgroup\color{black}$\psi_{njm_j\ell s}$\egroup. While there is no spin involved here, we will need to use these states for the spin-orbit interaction

\begin{eqnarray*}
\left< \psi_{njm_j\ell s}\left\vert H_1\right\vert \psi_{njm_j...
...er r}\right>_{n}
+ e^4\left<{1\over {r^2}}\right>_{nl} \right]
\end{eqnarray*}


where we can use some of our previous results.

\begin{eqnarray*}
E_n&=&-{1\over 2}\alpha^2mc^2/n^2={-e^2\over {2a_0n^2}} \\
\l...
...}}\right>&=&\left({1\over {a^2_0n^3(\ell + {1\over 2})}}\right)
\end{eqnarray*}


\begin{eqnarray*}
\left< \psi_{njm_j\ell s}\left\vert H_1\right\vert \psi_{njm_j...
... \over {2mc^2}} \left[ 3 - {4n \over{\ell + {1\over 2}}} \right]
\end{eqnarray*}


Since this does not depend on either \bgroup\color{black}$m_\ell$\egroup or \bgroup\color{black}$j$\egroup, total \bgroup\color{black}$j$\egroup states and the product states give the same answer. We will choose to use the total \bgroup\color{black}$j$\egroup states, \bgroup\color{black}$\psi_{njm_j\ell s}$\egroup, so that we can combine this correction with the spin-orbit correction.



Jim Branson 2013-04-22