# Spin-orbit coupling and jj scheme

The hamiltonian representing spin-orbit interaction is

{\displaystyle {\begin{aligned}&{\hat {H}}_{SO}={\frac {1}{2mc^{2}}}\sum _{i}{\hat {{\vec {S}}_{i}}}{\hat {{\vec {L}}_{i}}}{\frac {1}{r_{i}}}{\frac {\partial {\hat {V}}\left(r_{i}\right)}{\partial r_{i}}}\\&{\hat {V}}\left(r\right)={\frac {-Ze^{2}}{4\pi \epsilon _{0}r}}+{\hat {V}}_{H}\left(r\right)\end{aligned}}}

We can treat ${\displaystyle {\hat {H}}_{SO}}$ as a perturbation to ${\displaystyle {\hat {H}}_{H.F}}$. If we're not dealing with other perturbation of the same order ( which means spin-orbit corrections are the leading ones), we simply have to find a basis which is able to diagonalize ${\displaystyle {\hat {H}}_{SO}}$ and ${\displaystyle {\hat {H}}_{H.F}}$ simultaneously. That basis is ${\displaystyle \left|n_{i},l_{i},j_{i},m_{j_{i}}\right\rangle }$, where ${\displaystyle {\hat {{\vec {J}}_{i}}}={\hat {{\vec {L}}_{i}}}+{\hat {{\vec {S}}_{i}}}}$ represent the total angular momentum of each electron. We can pass from the previous basis to the new one through Clebsch-Gordan coefficients:

${\displaystyle \left|n_{i},l_{i},j_{i},m_{j_{i}}\right\rangle =\sum _{m_{l_{i}},m_{s_{i}}}\underbrace {\left(\dots \right)} _{\mbox{ Clebsch-Gordan}}\left|n_{i},l_{i},m_{l_{i}},m_{s_{i}}\right\rangle }$

The mathematical need for this basis represents the physical fact that when spin-orbit interaction becomes strong (in heavy atoms, ${\displaystyle Z\geq 40}$), each orbital angular momentum tends to couple with its own spin:${\displaystyle {\hat {{\vec {J}}_{i}}}={\hat {{\vec {L}}_{i}}}+{\hat {{\vec {S}}_{i}}}}$. The ${\displaystyle {\hat {{\vec {J}}_{i}}}}$ then combines to give the total ${\displaystyle {\hat {\vec {J}}}=\sum _{i}{\hat {{\vec {J}}_{i}}}}$ (jj coupling)

Individual spin orbitals are now labeled with the new good quantum numbers:

${\displaystyle [n][l]_{[j]}}$

The effect of spin-orbit coupling consists in a partial removal of degeneracy according to ${\displaystyle j-}$value:

${\displaystyle \left\langle {\hat {H}}_{SO}\right\rangle =E^{\left(1\right)}\left(\{n_{i},l_{i}\},j_{i}\right)}$
For example ${\displaystyle 2p}$ level, characterized by ${\displaystyle l=1,s={\frac {1}{2}}}$, is split in ${\displaystyle 2p_{\frac {1}{2}}2p_{\frac {3}{2}}}$.

If we want to fill ${\displaystyle 2p}$ level with electrons we simply have to add an apix ${\displaystyle \alpha }$ representing the occupation number. After placing all the electrons in the jj levels compute the possible values of the total J: ${\displaystyle \left(nl_{j}^{\alpha }n'l_{j'}^{'\alpha '}\dots \right)_{J_{1},J_{2},\dots }}$

When we take into account non-spherical corrections on these states ${\displaystyle m_{j_{i}}}$ is no longer a good quantum number and we need to label states with ${\displaystyle \left|\{n_{i},l_{i},j_{i}\},J,M_{J}\right\rangle }$. Degeneracy of configurations with same ${\displaystyle J}$ is removed:

${\displaystyle \left\langle \{n_{i},l_{i},j_{i}\},J,M_{J}\left|{\hat {H}}_{1}\right|\{n_{i},l_{i},j_{i}\},J,M_{J}\right\rangle =E\left(\{n_{i},l_{i},j_{i}\},J\right)}$
Each configuration with different value of ${\displaystyle J}$ has now different energy. The lowest energy configuration is given by Hund's third rule [1]

Remark
• ${\displaystyle [n][l]_{[j]}}$ can have a maximum of ${\displaystyle 2j+1}$ electrons
• If a level ${\displaystyle [n][l]}$, which can contain ${\displaystyle 2\left(2l+1\right)}$ electrons is split in ${\displaystyle [n][l]_{[j_{1}]},[n][l]_{[j_{2}]}}$ the equality ${\displaystyle 2\left(2l+1\right)=\left(2j_{1}+1\right)+\left(2j_{2}+1\right)}$ must hold. For example ${\displaystyle 2p}$ level is split in ${\displaystyle 2p_{\frac {1}{2}}2p_{\frac {3}{2}}}$ and we have ${\displaystyle 2\left(2\cdot 1+1\right)=6=\underbrace {\left(2\cdot {\frac {1}{2}}+1\right)} _{=2}+\underbrace {\left(2\cdot {\frac {3}{2}}+1\right)} _{=4}}$

1. For a detailed discussion about how to find atomic terms read JJ scheme