Why are the perturbation term in Zeeman effect not diagonalized?

Question

In the case of weak field Zeeman effect (anomalous Zeeman effect) in hydrogen atom, the unperturbed Hamiltonian reads as $$ H_0 = \frac{\hat{p}^2}{2m} + \frac{C_1}{r} + f(r)\mathbf{L}\cdot\mathbf{S} $$ and the perturbation term, which comes from the external magnetic field, is $$ H' = C_2(L_z+2S_z) $$ In my problem, the exact form of $C_1$ and $C_2$ are not important. What bothers me is that every literature always uses the eigenkets of $H_0$, which can be denoted as $|n,L,J,m_j\rangle$ as the unperturbed kets. But these kets do not diagonalize the perturbation term $H'$. Aren't the zeroth-order eigenkets supposed to be chosen such that they diagonalize the perturbation in the case of degenerate levels, which is true in this problem?

Answer 1

After some hours of pondering, I finally realize two things:

1. If I were to diagonalize $H'$, I must do it in the subspace of fixed $n$ and $J$, instead of fixed $n$ and $L$. This is because the energy eigenvalues of the unperturbed Hamiltonian $H_0$ are specified by $(n,J)$ (states with same $n$ and $J$ but different $L$'s can have the same energy).
2. The matrix elements of $H'=J_z+S_z$ between states of the same $n,J$ but different $L$'s and/or $m_J$'s vanish, $$ \begin{aligned} \langle n,L',J,m_J'|H'| n,L,J,m_J\rangle &= m_J\hbar\delta_{m_Jm_J'}\delta_{LL'} + \langle n,L',J,m_J'|S_z| n,L,J,m_J\rangle\\ &= \delta_{LL'} \Bigg( m_J\hbar\delta_{m_Jm_J'} + \frac{\hbar}{2} \sqrt{\frac{(L'\pm m_J'+1/2)(L\pm m_J+1/2)}{(2L'+1)(2L+1)}} \delta_{m_J-1/2,m_J'-1/2} \ldots \\ & \ldots -\frac{\hbar}{2} \sqrt{\frac{(L'\mp m_J'+1/2)(L\mp m_J+1/2)}{(2L'+1)(2L+1)}} \delta_{m_J+1/2,m_J'+1/2} \Bigg)\\ &= \delta_{LL'} \Bigg( m_J\hbar\delta_{m_Jm_J'} + \frac{\hbar}{2} \frac{(L\pm m_J+1/2)}{(2L+1)} \delta_{m_J-1/2,m_J'-1/2} \ldots \\ & \ldots -\frac{\hbar}{2} \frac{(L\mp m_J+1/2)}{(2L+1)} \delta_{m_J+1/2,m_J'+1/2} \Bigg)\\ &= K\delta_{m_Jm_J'}\delta_{LL'} \end{aligned} $$ with $$ K = m_J\hbar + \frac{\hbar}{2} \frac{(L\pm m_J+1/2)}{(2L+1)}-\frac{\hbar}{2} \frac{(L\mp m_J+1/2)}{(2L+1)} = m_J\hbar \Bigg( 1 \pm \frac{1}{(2L+1)}\Bigg). $$