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Current density operator

  1. Sep 6, 2014 #1

    Dale

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    Is there a current density operator or something equivalent? If so, how does it relate to other operators like momentum and angular momentum?

    Basically, the classical picture of a magnetic moment is a little loop of current, I would like to understand the quantum analog.
     
  2. jcsd
  3. Sep 6, 2014 #2

    ShayanJ

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    I think there is no such a thing but I need more thought for explaining it.
    But about the quantum picture of a magnetic moment. Consider energy eigenstates of hydrogen atom. These are stationary states which means the average values are time-independent. But the wavefunctions themselves do depend on time. That variation means that we may have a probability current.
    Now if we assume that multiplying the charge of the particle by its probability density gives us a charge density, then its easy to accept that multiplying the charge by the probability current density can give us an electric current density and this current density can be used in explaining the magnetic moment of the atom. Its obvious that we don't need a current density operator here.

    EDIT: Looks like I was wrong. Take a look at here.
    Anyway, my explanation about the magnetic moment is still untouched. That's a valid way anyway!
     
    Last edited: Sep 6, 2014
  4. Sep 6, 2014 #3

    WannabeNewton

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    Yes it is given by ##j^{\mu} = \bar{\psi}\gamma^{\mu}\psi## where ##\psi## is a Dirac field; this is the Noether current of the Dirac lagrangian under a global phase shift. Furthermore ##j^{\mu} = i(\varphi \partial^{\mu} \varphi^{\dagger} - \varphi^{\dagger}\partial^{\mu}\varphi)## is the current density for a complex scalar field ##\varphi##; in the non-relativistic limit this is the usual probability current density from QM.

    I don't quite understand your second question. What is the classical analogue of the relationship between momentum/angular momentum and current density that you have in mind? Are you asking about the quantum analogue of something like ##\vec{j} = \rho \vec{v}## where ##\vec{v}## is the velocity field of a charged fluid?
     
  5. Sep 6, 2014 #4
    Even more generally than WannabeNewton's answer, you can define a charged current density from the action as [itex]j^{\mu} = \delta S/\delta A_{\mu}[/itex] (divide by charge q to get the regular probability current density). From gauge invariance, this current must be conserved. If you aren't considering coupling to a gauge field, you can just consider this to be a conserved current associated with a global U(1) symmetry.

    For specific situations, you can derive the probability current by just computing this - some examples are in this Wiki article. For a particle with spin, you'll either want either the "Spin-s particle in an electromagnetic field" example from the link for the non-relativistic case, or WannabeNewton's example for the relativistic case. They are derived from the Schrödinger-Pauli and QED lagrangians respectively.
     
  6. Sep 6, 2014 #5

    strangerep

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    Are you ok with the subject of quantized angular momentum and intrinsic spin? (If not, it's probably best to study their origins first, i.e., unitary irreducible representations of the rotation group -- cf. Ballentine ch7.)

    The intrinsic magnetic moment then makes an appearance as an extra term in the Hamiltonian proportional to ##B \cdot S##, where ##S## is the particle's intrinsic spin vector and ##B## is an external magnetic field vector.

    [BTW,... since you're obviously trying to learn QM,... which resources are you using?]
     
  7. Sep 7, 2014 #6

    Dale

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    Thanks, that reference was quite helpful.
     
  8. Sep 7, 2014 #7

    Dale

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    Yes, that is what I wanted to know. Classically ##\vec{j} = \rho \vec{v}##, and according to the reference Shyan posted essentially the same relationship holds in QM also where v is the momentum operator divided by m.
     
  9. Sep 7, 2014 #8

    Dale

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    Yes, I am OK with that. In fact, my understanding of that is what leads me to question the usual "little current loops" model from classical EM.

    OK, this is very interesting. So, how much of this B.S term can be attributed to the current density? I mean, classically the same term would be ##\mu \cdot B## where ##\mu = \frac{1}{2}\int r \times j \; dV##. Since there is a quantum current density we can calculate something corresponding to ##\mu##. What is the relationship between that and ##S##.

    Google and YouTube :smile:
     
  10. Sep 7, 2014 #9

    atyy

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    Magnetism in real materials is mainly due to the intrinsic spin of the electron. The intrinsic spin and orbital angular momentum should both contribute to the magnetism of materials, but in practice the contribution of the orbital angular momentum is negligible.


    http://web.mit.edu/course/6/6.732/www/6.732-pt3.pdf

    You can find some famous effects due to the movement of an electron in an external magnetic field by googling "Landau Levels" or "Aharonov-Bohm effect" (which shows we must use the vector potential in in the Hamiltonian if we want a "local" Hamiltonian).

    Googling also suggests phenomena like the "Zeeman effect" http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html and "spin-orbit coupling" http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html. An interesting heuristic for spin-orbit coupling is that spin is intrinsic, and exists only in the "frame of the electron". For the hydrogen atom, we write a static electric field produced by the proton in the frame of the proton. But because the electron is moving, and spin is intrinsic, the spin will see the electric field as a magnetic field in its own frame, leading to spin-orbit coupling. Actually, I don't know whether this heuristic is really correct, but the effect can be derived from the Dirac equation applied to the hydrogen atom.

    One word about the Dirac equation - for the hydrogen atom, it is usually treated as an equation for particles, like the Schroedinger equation. However, it can only be consistently quantized as a field describing more than 1 identical particle. The Schroedinger equation can be treated consistently both ways, equivalently.

     
    Last edited by a moderator: Sep 25, 2014
  11. Sep 7, 2014 #10

    DrDu

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    Non-relativistically, ##j=\frac{e}{2m}\{p-eA,\delta(r)\}##
     
  12. Sep 7, 2014 #11

    strangerep

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    None. If you're thinking of little current loops (i.e., circular movement through space), then you're thinking of orbital angular momentum. However, intrinsic angular momentum has nothing to do with movement through space, but only with how the object transforms under rotations.

    Can you not afford a QM textbook? My favorite one is reasonably cheap. :biggrin:
     
  13. Sep 7, 2014 #12

    Dale

    Staff: Mentor

    So the current density described above is entirely "extrinsic" (I.e. due to orbital angular momentum or bulk motion). There is no current density associated with intrinsic spin?
     
  14. Sep 7, 2014 #13

    strangerep

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    Correct.

    Caveats: Actually, I'd better be careful not to oversimplify. In the nonrelativistic case the orbital and intrinsic parts of total angular momentum make sense separately -- i.e., they transform separately under the Galilei group. The ##B\cdot S## stuff I mentioned before is for the nonrelativistic case.
    In the relativistic case, things are murkier since the orbital and intrinsic parts mix together under Lorentz boosts, in general. But (e.g.,) the Dirac equation with minimal coupling still yields such a term in the nonrelativistic limit, though in general there are higher order terms in the Hamiltonian (e.g., Darwin term, and others).

    If you can access a copy of Misner, Thorne & Wheeler, there's a Box 5.6(iirc) which explains some of these subtleties of orbital and intrinsic angular momenta in a general relativistic context.]
     
  15. Sep 8, 2014 #14

    DrDu

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    Of course there is. Any spin leads to a magnetic moment and due to ##\mathrm{rot }\, M=j##, it is linked to current density.
     
  16. Sep 8, 2014 #15

    Dale

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    Non-relativistic is fine for now. DrDu and strangerep seem to contradict each other, is there some subtelty I am missing or a genuine disagreement?
     
  17. Sep 8, 2014 #16

    DrDu

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    I can also offer an alternative derivation: ##j=\frac{\delta H}{\delta A}## (please don't try to pin me down on the sign) with ##B=\text{rot }A## it can be seen that the BS term will make a contribution ##\sim \text{rot } S##.
     
  18. Sep 8, 2014 #17
  19. Sep 8, 2014 #18

    strangerep

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    Probably, I misunderstood your question. I thought you were asking whether the intrinsic spin & magnetic moment arise from a classical-type motion. But you used the word "associated", so maybe that's not what you were asking. Of course the intrinsic spin gives rise to a part of a current in the canonical context, as expressed in that Wiki article mentioned by King Vitamin.
     
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