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Orbital magnetic moments

  1. Sep 15, 2012 #1
    Hi

    I am reading about quantization of orbital angular momentum and, as I read, I am trying to compare it to the theory I already know on how atoms behave in magnetic fields. Here is the situation so far (I only look at 1-electron atoms to keep it simple for now): I assume the quantization axis is along z, i.e. a magnetic field points this way. Say the atom is in a state such that m=1 in the magnetic field (an excited state), i.e. the projection of its angular momentum onto the Lz-axis is (h/2π) -- so this means that there is a nonzero angle between the dipole moment of the atom and the B-field.

    If I now turn off the magnetic field, the orientation of μ (the magnetic moment) is lost. If I then apply the magnetic field again (non-adiabatically), then what determines which state the atom ends up in?

    Best,
    Niles.
     
  2. jcsd
  3. Sep 15, 2012 #2
    Why do you say that the orientation of mu is lost?

    All values of m are Eigenstates of the Hamiltonian, with or without magnetic field. This means that even without field the atom can remain in that state without changes.

    In practice there are always small stray fields that will lead to precession unless there is a large dominant "guide field".

    The situation is the same for orbital, spin or total (l+2s) magnetic moments. A practical application is found in polarized neutron scattering (neutron has spin=1/2 and no orbital moment). Techniques like Spin Echo make a fine art of spin alignment and precession.
     
  4. Sep 15, 2012 #3
    Thanks for replying. OK, so the direction of mu is not lost (my bad..), but the state is not well-defined any longer, as there is no magnetic field to project mu onto (so the magnetic quantum number cannot be determined). This must be correct?
     
  5. Sep 15, 2012 #4
    Define "well-defined".

    There is no more energy difference, so changes can be induced by any epsilon perturbation.

    It is still well defined on paper becasue - field or no field - it is an Eigenstate and has no a-priori reason to change into something else.
     
  6. Sep 15, 2012 #5
    Can I ask what you mean by this statement? By "changes" do you mean changes in the state? And what perturbations can cause this?

    By "well-defined" I just meant that since there is no B-field, it doesn't make sense to talk about projection the mu onto it. But then again, as you say, the atom is still in that same state.
     
  7. Sep 15, 2012 #6
    Yes, changes from m=1 to m=0 or m=-1 for example. Perturbations can be a small magnetic field (have you properly screen out the earth's magnetic field?) or whatever.

    In the presence of a finite magnetic field there is a finite energy difference between m=1, m=0 and m=-1. For the atom to go into another state "the energy difference has to come from somewhere" (Nothing is free, the most fundamental law of them all). This is of course most relevant for the ground state, not the exited state you are refering to.
     
  8. Sep 15, 2012 #7
    Thanks. I have spent the past 2 hours trying to find information about quantization axes and σ+/- light, but I can't find a completely paper/book/chapter describing it all in detail. I think I've also confused myself quite a bit by now. So here is the point:

    If an atom is in a state state with magnetic quantum number mF (no magnetic field applied so far), then regardless of what quantization axis I project the magnetic moment onto, the result will be hbar*mF. Now I apply a magnetic field, and projecting onto this will still give me hbar*mF since - as you mentioned - nothing has pushed the atom out of the state (assuming no "background"/residual perturbations).

    If I want to induce a change Δm=+1, then I have to apply light polarized with +hbar. This is where my confusion arises: Do I need a magnetic field in order to define light to have +hbar-polarization?
     
  9. Sep 16, 2012 #8
    Forget about my post #7, it is not relevant now.

    I'm not sure I agree fully with your statement:
    If that is the case, then why do Majorana spin-flips occur in quadropole magnetic traps? They occur when the magnetic field changes faster than the precression frequency, i.e. the relative orientation of the B-field and the magnetic moment are lost. This directly relates the maximum-allowed change of the B-field. See page 12 of this (as an example): http://arxiv.org/pdf/cond-mat/9904034v2.pdf

    EDIT: Just to clarify, what I mean is that for a given quantization axis defined by a magnetic field B, then the shift in energy is given by the projection of F onto B, which equals mB. It B changes direction (but has the same magnitude), then m must change, i.e. the atom has changed state.
     
    Last edited: Sep 16, 2012
  10. Sep 17, 2012 #9
    Note that there may be a zero average magnetic field, but an oscillating magnetic field still is a magnetic field. A simple way to think of it is that the spin oscillates between high- and low-energy when the spin stays constant, say up. Whenever the field is antiparallel to the spin, there is a chance for a spin-flip. (Watch out, I am sure that this icture is too simple to be a starting point for any useful calculations!!!)

    In the adiabatic case the field changes slowly, so the spin has time to react, e.g. by precession.

    As an exercise, you can compare the energy scales: hbar*omega of the electromagnetic field , and the Zeeman energy mu_B*B

    BTW, a photon travelling || z carries an angular momentum of +/-1 with quantization axis ||z. +/-1 for circular polarization (sign depends on convention), linear is a superposition and can carry both. If the direction of the photon is not the same as the quantization axis of the spin/atom, then it is easiest (for me) to change the basis of the spin/atom.

    The thesis (?) you quote mentions that such flips exist, but not much more. This paper here seems to give more detail, even though I have not read all of it:

    http://arxiv.org/pdf/quant-ph/0610135v1.pdf
     
  11. Sep 17, 2012 #10
    Thanks. OK, I have a number of things I need clarification on.

    This statement I need to discuss with you. Say the atom is in a state with mF=+1 with respect to a magnetic field B, so the projection of F onto B yields +1*B. Assuming the state is magnetically trappable, this will be the case for a magnetic trap.

    Now I look at the case where the magnetic field changes direction so fast that the atomic momentum cannot follow. Lets say that e.g. the field changes direction by 150 degrees with respect to its previous direction. As we discussed above, the atom is still in mF=+1 -- but this means that the projection of F onto B must still equal +1*B. How can this equality be satisfied now? And where does the spin flip come into play in all of this?

    I will have to think about this.

    Best,
    Niles.
     
  12. Sep 17, 2012 #11
    I know nothing about traps, but let's see where this goes...

    Hmmm, this is where I don't follow you.

    For me, if the atom is still in m=+1 and the field has changed, then mu.B has changed sign (mu being fixed).

    I think the problem lies in the concept of quantization axis. In a steady equilibrium state, you can assume your atom to be in an Eigenstage of energy. Then the quantization axis is important, but it actually arises from the solution of the Schrödinger (or Pauli or Dirac) equation.

    For non-equilibrium states you just have to settle on a coordinate system. The spherical harmonics form a complete basis. This means that any moment direction can be written in *any* coordinate system as a combination of m=-l...m=+l.

    Now if the magnetic field suddenly changes, then there are probabilities that the previous state turns into a different state. In the present case this transition could be a "spin flip".

    Check out "sudden approximation" and "Fermi's golden rule".
     
  13. Sep 18, 2012 #12
    Thanks for your post. It is a direct answer to my question/doubts, but I am a little slow, so I think I need some clarification. But this discussion is a tremendous help.

    When there is no applied B-field, I guess any spatial direction constitutes a quantization axis, however an applied B-field defines it if present. In steady state, the projection F.B gives what the magnetic quantum number of the current eigenstate has. So far so good.


    This I don't understand. The field doesn't change adiabatically, so the projection F.B can't be the same as it would be in steady state. Thus the quantum state must have changed.


    Thanks, this is a really good explanation. But from it I understand that the "spin-flip" can happen from any m to any m'.


    Best,
    Niles.
     
  14. Sep 19, 2012 #13
    No problem, that's what this forum is for
    Yes.

    But because the basis set is complete you can choose any basis=orientation of coordinate axes you like.

    If you chose the z-axis || B, then your solutions will be simple.
    If you chose another axis, then each Eigenstate will be contain (in general) all values of m.

    You can then define an operator L_B that will be a combination of Lx, Ly and Lz (or L+, L- and Lz) that measures the magnetic moment parallel to B. The Eigenvalues of that operator will always be (for L=1) +1 hbar, 0, and -1 hbar. That is why the direction of B defines the quantization axis.

    Note that all along we imply that the system is in an Eigenstate=nice and steady equilibrium.

    This is my view:

    You have changed the directon B rapidly, so fast that the atom cannot follow (by definition of non-adiabatic).

    Therefore you have changed the quantization axis. Because it is slow, the system is not in an Eigenstate of the new quantization axis. It is still in an Eigenstate of the old quantization axis.

    In the old coordinate system the system has not changed, it is stil lin m=+1 (but that is not an Eigenstate anymore).

    In the new coordnate system the system is also not in an Eigenstate. The wave function will contain all values of m. To find the wave function you have to do a coordinate transformation of the old Eigenstates to new ones, e.g. using Wigner D functions.

    I guess it can, but the probabilities depend on the matrix elements. I am too much an experimentalist to be able to say that is possible (non-zero) and what is not.


    Index "z" means old coordinate system, index "B" means new coordinate system.

    [itex]
    \left| \psi \right>_z = \left |-1\right>_z \\

    \left| \psi \right>_B = \sum_m \alpha_m \left|m\right>_B \\

    \alpha_m = \left< m \right|_B \cdot \left| -1 \right>_z
    [/itex]

    |α(-1)|^2 will give the probability for a spin-flip.

    If you describe the change of coordinate systems by Euler angles, then alpha is just the Wigner D coefiicient.

    http://en.wikipedia.org/wiki/Wigner_D-matrix

    (edit: itex error)
     
    Last edited: Sep 19, 2012
  15. Sep 19, 2012 #14
    Thanks for the above post, that was really enlightening. That answered practically most of the questions I had on the topic, so I'm really grateful for that. Naturally I need to read more about it, but the framework has been established.

    I thought of an example myself to test that I have understood the concept. It is the usual situation: A magnetic field is applied to an atom, and it defines the quantization axis. The atom is in a state mF. Now the magnetic field adiabatically decreases to zero, from which it increases - but in the opposite direction relative to before. So basically the z-axis has been flipped 180 degrees adiabatically.

    The atom is still in the eigenstate of the old z-axis. But this must be -mF of the new axis, so the atom is now in -mF. Can it really be true that we can "optically pump" atoms by changing the B-field direction like this?

    Best,
    Niles.
     
    Last edited: Sep 19, 2012
  16. Sep 20, 2012 #15
    I suppose yes. The atom can end up in a high-energy state in the new basis. It will then eventually drop to the ground state by emitting a photon.

    Where does the energy come from? I guess it takes work to change the field direction.
     
  17. Sep 20, 2012 #16
    Thanks, it was extremely kind of you to help me. I learned a lot from this. I have some questions regarding a comment you made in post #9 about changing the basis when [itex]E\perp B[/itex], but I'll create a new thread for that. I have to think about it first.

    Best wishes,
    Niles.
     
    Last edited: Sep 20, 2012
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