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K^2 = J^2 + 1/4 for the central force problem of the Dirac equation

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    To whom it may concern,

    I am trying to understand the central force problem of the Dirac equation. In particular, I am following Sakurai's Advanced Quantum Mechanics book. There (section 3.8, p.122), it is shown that there is an operator

    [itex]K = \beta(\Sigma . L + \overline{h})[/itex]
    where [itex]\Sigma_i = \left(
    \begin{array}{cc}
    \sigma_i && 0 \\
    0 && \sigma_i \\
    \end{array}\right)[/itex]

    This operator commutes with the hamiltonian H and the total angular momentum J. At the top of page 123, it is shown that [itex]K^2 = J^2 + \overline{h}/4[/itex]. At some point, we get

    [itex]K^2 = L^2 + i\Sigma . (L × L) + 2\overline{h}\Sigma . L + \overline{h} = L^2 + \overline{h}\Sigma . L + \overline{h}^2[/itex]

    I don't understand this last equality. Why is [itex]i \Sigma . (L × L) = -\overline{h}\Sigma.L [/itex]?

    2. Relevant equations



    3. The attempt at a solution
    I looked a lot in the literature, but I didn't find any more precise calculation. I don't seem to grasp what L × L means. I know that [itex]\vec{L} = \vec{x}× (-i\overline{h}\nabla)[/itex]. If we had vectors instead of operators, we would have [itex]\vec{L} × \vec{L} = 0[/itex], but what does the vector product of 2 identical operators mean?

    Thanks for the help. Best regards,

    Kami
     
  2. jcsd
  3. Jan 15, 2013 #2

    dextercioby

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    Science Advisor
    Homework Helper

    The misleading/confusing notation L x L c/should be a shorthand from [itex] [L_i, L_j] [/itex], i.e. from the commutator.
     
  4. Jan 16, 2013 #3
    Hi dextercioby,

    Thank you a lot for your reply, that was exactly what I needed. I did the calculation with that in mind and it worked. In fact, I am ashamed not having seen this by myself.
     
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