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Dirac equation

  1. May 16, 2012 #1
    The Schrodinger equation solved for the hydrogen atom gave good agreement with spectral lines, except for line doublets.

    To account for these electron spin theory was grafted onto the theory, despite the problem of electron being a point particle.

    In 1928 Dirac gives his different answer to this doublet problem with the Dirac equation giving the correct spin for the electron. The spin being connected to relativity.

    My question is does the Dirac equation when solved for the Hydrogen atom give all the other correct predictions for spectral lines as the Schrodinger solution does ?

    Or are there many other problems/difficulties/unknowns introduced by the Dirac solution ?
     
  2. jcsd
  3. May 16, 2012 #2
    The Pauli equation (Schroedinger + spin and a few other relativistic corrections) can be derived as the non-relativistic limit of the Dirac equation. As such it correctly gives the same results as the Schroedinger/Pauli equation for the spectral lines of the Hydrogen atom.

    The Dirac equation gives 4-component spinors which are usually interpreted as spin-up particle, spin-down particle, spin-up antiparticle and spin-down antiparticle. It kind of allows the creation of particle/antiparticle pairs out of pure energy. In the hydrogen atom, these effect can be neglected., but in a more general context, it requires the introduction of the "second quantization" that systematically treats the creation and annihilation of particles. That is a big can of worms with infinities creeping up that require "renormalization" and other similar candies.

    We all know by now that antiparticles exist, so having predicted that is a huge success for the Dirac equation. But mathematically, renormalization theory still seems to be a mess even if many hugely successful predictions have come out.

    I am sure someone with better knowledge will have a more qualified opinion on this :-)
     
  4. May 16, 2012 #3

    Bill_K

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    No need to make a nonrelativisitic approximation, the full Dirac equation for a central Coulomb potential can be solved exactly. (See one of the standard QM references, such as Messiah vol II.) The resulting energy levels include most of the known corrections such as hyperfine splitting (spin-orbit coupling) and the relativistically correct kinetic energy, which becomes important for inner electrons in high-Z atoms. What it does not include are QFT corrections such as the Lamb shift.

    More generally, treating the Dirac equation as an equation for a single particle in an external potential is a valid approximation as long as V(x) << mc2. At that point pair production becomes important.
     
  5. May 16, 2012 #4
    OK thanks for that.

    pretty impressive then.
     
  6. May 16, 2012 #5

    dextercioby

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    The hyperfine splitting is not a consequence of the Dirac equation. The fine splitting is.
     
  7. May 16, 2012 #6

    Jano L.

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    Some time ago on this forum we had a discussion about the stability of the atom according to the quantum theory, and one contributor made an interesting remark that due to the fact that in Dirac's eq. the energy is linear in momenta, according to this eq. models of many-electron atoms become unstable, in contrary to models based on Schr. eq, where energy is quadratic in momenta. Maybe in quantum field theory they become stable again, but I think the consequences of the equations are so hard to infer that we do not know for sure.

    Bill, this makes an impression that the model is fully relativistic and can be solved exactly. But "exactly" above means only exactly within the approximate model. The approximation is that the electron moves in central electrostatic field of the proton. In relativistic theory, this is necessarily an approximation to field worse than 1/c^2 (accompanied by total neglect of readiation from the proton and retardation of its field). Better approximation has to be achieved in other way, maybe somehow in quantum field theory or otherwise.

    Can you please explain this? What is V(x)?
     
  8. May 16, 2012 #7
    So the dirac equation isn't a form of QFT? I've been told different things, but that being said, I haven't had a course in QFT yet (I'm now having a course in elementary particles using Griffiths' book, which does discuss the Dirac equation and such)
     
  9. May 16, 2012 #8

    dextercioby

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    Assumed of infinite mass, which is an approximation done in this case specifically, but is unjustified in the non-specially-relativistic Schroedinger equation. The electrostatic approximation for the e-m interaction electron/proton in the Schroedinger equation is justified, of course, because the particles' dynamics is non-specially-relativistic as well.
     
    Last edited: May 16, 2012
  10. May 16, 2012 #9

    dextercioby

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    If the Dirac field were a true classical field (like the e-m field or the gravitational field), then it would have a Lagrangian whose Euler-Lagrange equations would be the Dirac equations.
     
  11. May 16, 2012 #10
    So it is a form of QFT? (I'm sorry if I'm completely missing your point, I'm not familiar enough with these topics)
     
  12. May 16, 2012 #11

    dextercioby

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    By my version of QFT, it's not, it's a form of relativistic quantum mechanics, that is quantum mechanics (like the one of Schroedinger or Heisenberg) + special relativity instead of Galilean relativity.
     
    Last edited: May 16, 2012
  13. May 16, 2012 #12
    Interesting, although I'm slightly confused, since Griffiths seems to mention a Lagrangian for the Dirac equation, while you seem to say it doesn't exist?
     
  14. May 16, 2012 #13

    dextercioby

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    What Griffiths and many others are using is a <fabrication> of a CLASSICAL Lagrangian for a theory which is purely quantum mechanical. It has no physical meaning, for classical physics cannot account for the existence of the Dirac field.

    Surely, one can build classical theories of electrons, but not with spinors.
     
  15. May 16, 2012 #14
    What is a good source for enlightenment on these matters?
     
  16. May 16, 2012 #15

    dextercioby

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    I would reccomend starting with a course on formal QFT, then you can ask yourself questions like I did. A solid background in mathematics is a must, some answers to your present & future questions are impossible to grasp without it.
     
  17. May 17, 2012 #16

    Bill_K

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    dextercioby, No one will deny you the right to make the assumption that the proton has infinite mass, but it is pointless and counterproductive. The idea in replacing the Schrodinger equation with the Dirac equation is to make a better approximation, not a worse one. Failure to use the reduced mass means you won't even be able to reproduce the Bohr-Sommerfeld energy levels.
     
  18. May 17, 2012 #17

    dextercioby

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    Replacing me by μ is indeed a useful trick and is reccomended by Messiah (the only one to my knowledge) to account for the error introduced by the assumption that the nucleus mass is infinite, however, due to the difficuly of separating the 2 particle motion into COM motion and relative "dummy" particle motion in the context of SR*, it remains as a trick, not as a result of specially relativistic dynamics.

    *Here again I refer to Greiner's text on field equations as in the other recent thread on this subject (H atom and Dirac's eqn.)
     
  19. May 17, 2012 #18
    Lol, this is a standard procedure in every textbook on Classical Mechanics for a two-body problem interacting with a central force.
     
  20. May 17, 2012 #19

    dextercioby

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    Yes, because in Newtonian mechanics it is well justified. In special relativity, it's not.
     
  21. May 17, 2012 #20
    Of course. But, then again, there is no concept of potential energy in SR.
     
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