We have introduced a modified single-particle Hamiltonian $$ \hat{h}(x_i) = \hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i) +\xi(\boldsymbol{r})\boldsymbol{ls}=\hat{h}_0(x_i)+\xi(\boldsymbol{r})\boldsymbol{ls}. $$ We can calculate the expectation value of the latter using the fact that $$ \xi(\boldsymbol{r})\boldsymbol{ls}=\frac{1}{2}\xi(\boldsymbol{r})\left(\boldsymbol{j}^2-\boldsymbol{l}^2-\boldsymbol{s}^2\right). $$ For a single-particle state with quantum numbers \( nlj \) (we suppress \( s \) and \( m_j \)), with \( s=1/2 \), we obtain the single-particle energies $$ \varepsilon_{nlj} = \varepsilon_{nlj}^{(0)}+\Delta\varepsilon_{nlj}, $$ with \( \varepsilon_{nlj}^{(0)} \) being the single-particle energy obtained with \( \hat{h}_0(x) \) and $$ \Delta\varepsilon_{nlj}=\frac{C}{2}\left(j(j+1)-l(l+1)-\frac{3}{4}\right). $$