+\begin{proof}
+This can be shown by rewriting the $LS$ operator
+\begin{equation}
+J=L+S \Leftrightarrow J^2=L^2+S^2+2LS \Leftrightarrow
+LS=\frac{1}{2}\left(J^2-L^2-S^2\right)
+\text{ ,}
+\end{equation}
+which, if used in equation~\eqref{eq:solid:so_bs2}, gives the same (diagonal) matrix elements
+\begin{align}
+\bra{l\pm\frac{1}{2},m\pm\frac{1}{2}}V(\vec{r})
+\ket{l\pm\frac{1}{2},m\pm\frac{1}{2}}&=
+\bar{V}_l(\vec{r})+V^{\text{SO}}_l(\vec{r})
+\frac{1}{2}\left(l(l+1)-j(j+1)-\frac{3}{4}\right) \\
+&= \bar{V}_l(\vec{r})+\frac{1}{2}V^{\text{SO}}_l(\vec{r})
+\left\{\begin{array}{rl}
+-\left(l+\frac{3}{2}\right) & \text{ for } j=l+\frac{1}{2}\\
+\left(l-\frac{1}{2}\right) & \text{ for } j=l-\frac{1}{2}
+\end{array}\right.
+\end{align}
+as equation~\eqref{eq:solid:so_bs1}
+\begin{equation}
+\text{ .}
+\end{equation}
+
+\end{proof}