Why are the perturbation term in Zeeman effect not diagonalized? 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?
 A: After some hours of pondering, I finally realize two things:


*

*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).

*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 
$$
K = m_J\hbar + \frac{\hbar}{2} \frac{(L\pm m_J+1/2)}{(2L+1)}-\frac{\hbar}{2} \frac{(L\mp m_J+1/2)}{(2L+1)} = m_J\hbar \Bigg( 1 \pm \frac{1}{(2L+1)}\Bigg).
$$

