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

[Kamikaze_951]

Homework Statement

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

$K = \beta(\Sigma . L + \overline{h})$
where $\Sigma_i = \left( <br /> \begin{array}{cc}<br /> \sigma_i && 0 \\<br /> 0 && \sigma_i \\<br /> \end{array}\right)$

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

$K^2 = L^2 + i\Sigma . (L × L) + 2\overline{h}\Sigma . L + \overline{h} = L^2 + \overline{h}\Sigma . L + \overline{h}^2$

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

Homework Equations

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 $\vec{L} = \vec{x}× (-i\overline{h}\nabla)$. If we had vectors instead of operators, we would have $\vec{L} × \vec{L} = 0$, but what does the vector product of 2 identical operators mean?

Thanks for the help.

Kami

 

[dextercioby]

The misleading/confusing notation L x L c/should be a shorthand from $[L_i, L_j]$, i.e. from the commutator.

 

[Kamikaze_951]

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.