Since we have a spin orbit force which is strong, it is easy to show that the total angular momentum operator $$ \hat{J}=\hat{L}+\hat{S} $$ does not commute with \( \hat{L}_z \) and \( \hat{S}_z \). To see this, we calculate for example $$ \begin{eqnarray} [\hat{L}_z,\hat{J}^2]&=&[\hat{L}_z,(\hat{L}+\hat{S})^2] \\ \nonumber &=&[\hat{L}_z,\hat{L}^2+\hat{S}^2+2\hat{L}\hat{S}]\\ \nonumber &=& [\hat{L}_z,\hat{L}\hat{S}]=[\hat{L}_z,\hat{L}_x\hat{S}_x+\hat{L}_y\hat{S}_y+\hat{L}_z\hat{S}_z]\ne 0, \end{eqnarray} $$ since we have that \( [\hat{L}_z,\hat{L}_x]=i\hbar\hat{L}_y \) and \( [\hat{L}_z,\hat{L}_y]=i\hbar\hat{L}_x \).