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Basic strong Zeeman effect question

  1. Jun 15, 2014 #1
    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?
  2. jcsd
  3. Jun 15, 2014 #2


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    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 Bz, as Griffiths so inelegantly puts it, the atom "experiences a torque". :yuck:

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

    The additional part of the Hamiltonian is Bz(Lz + 2Sz). What commutes with this? Clearly Lz and Sz do! Also it's easy to realize that L2 commutes with it, since [L2, Lz] = 0.
  4. Jun 16, 2014 #3
    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!
  5. Jun 16, 2014 #4
    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.
  6. May 5, 2016 #5
    No, in the weak Bext limit J is a "good" number not because S and L are "good", but because J is conserved.
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