Why is m_j not a good quantum number in strong-field Zeeman effect?

[Happiness]

TL;DR Summary

When solving for the correction to the Hamiltonian due to strong-field Zeeman effect (using perturbation theory), why is m_j not a "good" quantum number, given that J_z is conserved too?

This textbook claims ##m_j## is not a "good" quantum number because the total angular momentum (of an electron of a hydrogen atom placed in a strong uniform magnetic field) is not conserved. I don't understand why ##m_j## is not a "good" quantum number.

Since ##J=L+S##, ##J_z=L_z+S_z##.
Since ##L_z## and ##S_z## are both conserved, so is ##J_z##.
##J_z## commutes with ##H'_Z## too.
So shouldn't ##m_j## be a "good" quantum number too?

The phrase "good quantum number" relates to the following theorem in perturbation theory:

The book is "Introduction to Quantum Mechanics", 2nd edition, by David Griffiths.

 

[PeterDonis]

Since ##L_z## and ##S_z## are both conserved, so is ##J_z##.

But ##J## is not, and ##m_j## is a quantum number for ##J##, not ##J_z##.

 

[Happiness]

But ##J## is not, and ##m_j## is a quantum number for ##J##, not ##J_z##.

This is the remaining part of the section in the book:

From the sentence below [6.81], we can see that eigenstates of ##S_z## and ##L_z## were used as the "good" states ##\ket{nlm_lm_s}## in the perturbation theory in [6.80].

So my question is, aren't eigenstates of ##J_z## "good" states too?

The book did not define quantum numbers explicitly. From what I understand from the book, since ##m_j## is the eigenvalue of operator ##J_z##, ie, ##J_z\psi=\hbar m_j\psi## (where ##\psi## is an eigenstate of ##J_z##), then ##m_j## is the quantum number for ##J_z##. This is how I understand it. (##m_j## is the eigenvalue apart from a factor of ##\hbar##.)