Perturbation Calculation for Relativistic Energy Shift

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

$$\bgroup\color{black}\left< \psi_{njm_j\ell s}\left\vert H_1\r... ...ver {2m}}\right)^2 \right\vert\psi_{njm_j\ell s}\right> \egroup$$

$$\bgroup\color{black} = -{1\over {2mc^2}}\left< \psi_{njm_j\el... ...^2\over r}\right)^2 \right\vert\psi_{njm_j\ell s}\right>\egroup$$

$$\bgroup\color{black} = -{1\over {2mc^2}}\left< \psi_{njm_j\el... ...} +{e^4\over {r^2}} \right\vert\psi_{njm_j\ell s}\right>\egroup$$

$$\bgroup\color{black} = -{1\over {2mc^2}}\left[ E^2_n + \left<... ...ver {r^2}}\right\vert \psi_{njm_j\ell s}\right> \right] \egroup$$

$$\bgroup\color{black} = -{1\over {2mc^2}}\left[ E^2_n + 2 E_n ... ...ght>_{n} + e^4\left<{1\over {r^2}}\right>_{nl} \right] \egroup$$

Where $E_n=-{1\over 2}\alpha^2mc^2/n^2={-e^2\over {2a_0n^2}}$, $\left<{1\over r}\right>_{n}=\left({1\over {a_0n^2}}\right)$, and $\left<{1\over {r^2}}\right>_{nl}=\left({1\over {a^2_0n^3(\ell + {1\over 2})}}\right)$

$$\bgroup\color{black}\left< \psi_{njm_j\ell s}\left\vert H_1\r... ...^2}} + {e^4\over {a^2_0n^3(\ell + {1\over 2})}} \right]\egroup$$

$$\bgroup\color{black} = -{1\over {2mc^2}} {E^{(0)}_n}^2 \left[ 1 - 4 + {4n \over{\ell + {1\over 2}}}\right] \egroup$$

$$\bgroup\color{black} \left< \psi_{njm_j\ell s}\left\vert H_1\... ...mc^2}} \left[ 3 - {4n \over{\ell + {1\over 2}}} \right] \egroup$$

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