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Derivation of Spin-orbit Coupling

  1. Apr 21, 2013 #1
    I was reading Griffiths's book on quantum mechanics.

    In chapter 6, he tried to derive the spin-orbit coupling using a classical approach.
    [tex]H=-\vec{\mu}\cdot \vec{B}[/tex]

    1. Finding the relation between [itex]\vec{μ}[/itex] and [itex]\vec{S}[/itex]
    He consider a spinning charged ring with mass [itex]m[/itex], radius [itex]r[/itex], total charge [itex]e[/itex] and period [itex]T[/itex].
    Magnetic Dipole Moment: [itex]\vec{\mu}=i\vec{A}=\frac{e\pi r^2}{T}\hat{n}[/itex]
    (Spin) Angular Momentum: [itex]\vec{S}=\tilde{I}\vec{\omega}=\frac{2m\pi r^2}{T}\hat{n}[/itex]
    Dirac: [itex]\frac{e}{2m}\rightarrow\frac{e}{m}[/itex]

    2. Find the magnetic field [itex]\vec{b}[/itex]
    He consider the rest frame of electron.
    [tex]\vec{B}=\frac{\mu_0 i}{2r}\hat{n}=\frac{1}{c^2\epsilon_0}\frac{e/T}{2r}\hat{n}=\frac{1}{4\pi\epsilon_0}\frac{e}{mc^2r^3}\vec{L}[/tex]

    3. Combining both terms

    Then he point out that when we choose the electron's rest frame, there is one problem as it is a non-inertial frame. Therefore, we need to consider the Thomas's precession, which gives a factor of 1/2. Finally, it cancels with the 2 we multiplied to the classically derived gyromagnetic ratio.

    My problem is when we choose an accelerating frame, obviously we need some modifications. But, how should we modify? and how can we get the factor of 1/2 from Thomas's precession?

    Thanks in advance.
  2. jcsd
  3. Apr 22, 2013 #2


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  4. Apr 22, 2013 #3


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    kiwakwok, Unfortunately, almost all quantum books slide over the spin-orbit issue with this approach, in which quantum mechanics is temporarily abandoned, the electron is imagined to travel in a definite orbit about the nucleus, and the interaction is treated in a noninertial frame.

    Doing it correctly is not at all difficult, although it involves a short calculation with the Dirac Equation. The QM book by Shiff has a good one-page derivation.
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