Basic strong Zeeman effect question

[vaart]

I have a question about the Zeeman effect and pertubation theory. I read in Griffiths that with the strong Zeeman effect the total angular momentum is not conserved but Lz and Sz are. I don't really understand why this is in a physical sense, because I thought that angularmomentum always was conserved. What makes it more confusing is that I didn't expect that Lz and Sz are conserved compared with the unpertubated system, because the magnetic field lies along the z-axis so I expected an increase in Lz and Sz.

Could someone please help me with this faulty physical picture?
Thanks,
Vaart

 

[Bill_K]

I thought that angular momentum always was conserved.

Total angular momentum is conserved if and only if the Hamiltonian is rotationally invariant. That is true for an isolated atom, but here we're talking about an atom immersed in an external field. Since the Hamiltonian contains B_z, as Griffiths so inelegantly puts it, the atom "experiences a torque".

The system is still cylindrically symmetric about the z-axis, so J_z is a good quantum number even though J² is not.

What makes it more confusing is that I didn't expect that Lz and Sz are conserved compared with the unpertubated system, because the magnetic field lies along the z-axis so I expected an increase in Lz and Sz.

The additional part of the Hamiltonian is B_z(L_z + 2S_z). What commutes with this? Clearly L_z and S_z do! Also it's easy to realize that L² commutes with it, since [L², L_z] = 0.

 

[vaart]

Ah, I think I understand now. Let me rephrash in my own words.
1) Total anguar momentum is not always conserverd because in a non rotational invariant Hamiltonian rotational energy will be transvererd to potential energy and vice versa.
2) Jz is still a good quantum number because Jz=Lz+Sz and the last two are still good quantum numbers because of the cylindrical symmetry. And while Lz and Sz are still good quantum numbers this doesn't mean the total energy of the new system doesn't changes.

I think I understand it better now, atleast I hope!
Thanks

 

[The_Duck]

2) Jz is still a good quantum number because Jz=Lz+Sz and the last two are still good quantum numbers because of the cylindrical symmetry.

Rather, Jz is a good quantum number because of the cylindrical symmetry, and Lz and Sz are also good quantum numbers only if there is no spin-orbit coupling.

 

[aphys]

2) Jz is still a good quantum number because Jz=Lz+Sz and the last two are still good quantum numbers because of the cylindrical symmetry. And while Lz and Sz are still good quantum numbers this doesn't mean the total energy of the new system doesn't changes.

No, in the weak B_ext limit J is a "good" number not because S and L are "good", but because J is conserved.