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What is the speed of light doing in the spin half hamiltonian?

  1. Sep 6, 2012 #1
    I'm currently reading Sakurai's 'Modern Quantum Mechanics' (Revised Edition) and at page 76 he introduces a spin half hamiltonian

    [tex] H = - (\frac{e}{mc}) \vec S \cdot \vec B.[/tex]

    But what is c doing in this hamiltonian? Clasically the energy of a magnetic moment in a magnetic field is

    [tex] E = - \vec \mu \cdot \vec B = - \frac{e}{2m} \vec S \cdot \vec B[/tex]

    and as far as i know what one does to go to QM is to introduce the g-factor. What am I missing here?
  2. jcsd
  3. Sep 6, 2012 #2


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    Standard question, standard answer: the spin-orbit coupling is a specially relativistic effect, even though the spin notion itself can be derived without special relativity. The <c> comes from the Dirac equation.
  4. Sep 6, 2012 #3


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    Say what? If this term is spin-orbit coupling, where is L??

    The c comes about because according to Dirac the magnetic moment μ of the electron is -eħ/2mc, or -e/mc times the spin ħ/2.
  5. Sep 6, 2012 #4
    ah, thanks. I guess I have to be reading up on that then.
  6. Sep 6, 2012 #5


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    Sorry, not paying attention. Disregard my post above.

    If the <c> is there or not must be in the end a matter of convention, just like the as the case of CGS vs MKS in electromagnetism.
    Last edited: Sep 6, 2012
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