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Dirac conserved current vs Klein-Gordon conserved current

  1. Aug 11, 2009 #1
    The conserved current for a field [tex]\phi[/tex] obeying the Klein-Gordon equation is (neglecting operator ordering) proportional to [tex]i\phi^{\dag}\partial_\mu \phi-i\phi\partial_\mu \phi^{\dag}[/tex].

    The conserved current for a four component field [tex]\psi[/tex] obeying the Dirac equation is [tex]\psi^{\dag}\gamma^0\gamma^\mu \psi[/tex], with no derivatives.

    The Klein-Gordon conserved current cannot be interpreted as conservation of probability because the 0-component is not non-negative and so cannot be a pdf. On the other hand, it does make sense as a conservation of charge (or particles vs anti-particles).

    The Dirac current does lend itself to an interpretation as conservation of probability.

    So far, so good? I am simplifying for the sake of brevity, but let me know if I said anything wrong.

    Now, what do we make of the fact that each of the individual components [tex]\psi_j[/tex], j=1,2,3,4 of the Dirac field separately obey the Klein-Gordon equation, and therefore also satisfy

    [tex]\partial^\mu\{i\psi^{\dag}_j\partial_\mu \psi_j-i\psi_j\partial_\mu \psi^{\dag}_j\}=0[/tex] ?

    That is, both currents are conserved--the Dirac current in terms of all four components, coupled, and the Klein-Gordon current, each component independently. I presume that the two conservations are not equivalent, since one involves derivatives and the other doesn't.

    What is the physical significance of each of these conserved currents?
     
    Last edited: Aug 12, 2009
  2. jcsd
  3. Aug 12, 2009 #2

    Demystifier

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  4. Aug 12, 2009 #3
    Thanks, D. Actually, I was reminded of this question while reading one of those papers. :smile:

    For now, I'm just curious what the standard answer to this question is.
     
  5. Aug 12, 2009 #4

    Demystifier

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    The standard answer is that psi should not be interpreted probabilistically at all. Instead, it should be interpreted as an operator in a new larger Hilbert space of quantum field theory, with its own probabilistic interpretation. I am not saying that such an interpretation is satisfying (just the opposite), I am just saying that this is what the standard interpretation is.
     
  6. Aug 13, 2009 #5
    Ok. I am familiar with the operator interpretation--though I will have to post a question about it soon. But before we get to that: I am still unclear. What is the significance and/or interpretation of the fact the the Dirac field satisfies both conserved currents?
     
  7. Aug 14, 2009 #6

    Demystifier

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    The Dirac current couples to the electromagnetic field, which implies that the Dirac current has a direct interpretation as a charge current. The physical interpretation of the Klein-Gordon current for spin-1/2 particles is less clear, but the papers I mentioned above suggest that this current describes the motion of particles (if you are willing to accept the Bohmian interpretation).
     
  8. Aug 14, 2009 #7
    Wait a minute. Are we saying here that, thought the Dirac equation has been around since the 1930s, the interpretation of a conserved current isn't clear? Of course, it depends on what you mean by "clear." One might say it is not clear what spin is, but there is still a standard or common interpretation.

    this is why I love studying quantum theory. Not only do we not really understand it, but we don't know as much as we think we know.

    I'll be getting back to you on those papers. Thanks a lot, D. Really.
     
  9. Aug 24, 2009 #8

    Demystifier

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    Yes, that's exactly what I am saying.
     
  10. Aug 24, 2009 #9

    Avodyne

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    I doubt that the KG current is conserved once interactions are included (with, say, an external EM field), though I haven't checked. The conservation of the Dirac current follows from Noether's theorem (as long as the lagrangian is invariant under a change of phase of the field), but I don't know any principle that would yield the KG current for a Dirac field.

    A free field has a huge amount of symmetry, since each momentum mode is decoupled, and so there are an infinite number of conserved currents in the free-field case.
     
  11. Aug 25, 2009 #10

    Hans de Vries

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    The standard Dirac 4-current is derived from the general Lorentz transform of the spinor.
    It's done via an operator on the spinor, no differential operator is needed to do this.

    Alternatively you could derive a 4-current from a Dirac wave-function in the same way
    as in in the KG case, by looking at the phase-change-rates using differential operators.


    The result should be the exactly the same for plane wave eigen functions. The local
    velocities derived from the Lorentz transform of the spinor should correspond 1:1 with
    velocity derived from the de Broglie frequency/wavelength.

    The result is not the same for arbitrary functions (without interaction). The Gordon
    decomposition of the Dirac vector current shows an extra term caused by the spin in the
    cases where there is a density gradient.


    Regards, Hans
     
  12. Aug 28, 2009 #11
    There is a modified form which includes the EM potential.

    The Dirac field conserves the KG current because the Dirac field itself obeys the KG equation.
     
  13. Aug 28, 2009 #12
    Hans, I admit I don't understand your post. Instead of questioning the various pieces which I don't understand, let me first just ask, are you suggesting that the two conserved currents are equivalent?
     
  14. Aug 31, 2009 #13

    Demystifier

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    The Noether's theorem derives the conserved current from the Lagrangian. However, the same equations of motion can be obtained from different Lagrangians. On the other hand, the conservation of a current is a consequence of the equation of motion. This means that the Klein-Gordon current can also be obtained as a Noether current if you choose a different Lagrangian.
     
  15. Aug 31, 2009 #14

    Hans de Vries

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    They are derived from two different properties. (The spinor and the phase change
    rates) Both of these are subject to the Lorentz transform and the current itself is
    also subject to Lorentz transform.

    Therefor both of them can be used to derive the current as seen from a particular
    reference frame.


    Regards, Hans
     
  16. Aug 31, 2009 #15
    I'm still unsure what you mean by the current--Dirac, KG or both.

    But the key for me is not spinor vs phase change, but that one current--the KG current--applies to each component of 4-spinor separately because each component separately obeys the KG equation. Just like in the Lorentz gauge in a charge-free region, each component of the vector potential independently obeys the wave equation.

    In the other current--the Dirac current--the four components of the field are coupled. This is why expect them to necessarily have different physical interpretations--one has all the components interdependent, one features them independently of each other.
     
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