i want to show that the following equation holds: $$ \frac{1}{8}\left[\gamma^{\mu},\omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu} ] \right] = \omega^{\mu}_{~~~\nu}\,~ \gamma^{\nu} $$

$\gamma^{\mu}$ are the Dirac matrices and $\omega $ is the part of the infinitesimal Lorentz transformation that is responsible for boosts and rotations. $$L^{\mu}_{~~~\nu} = \delta^{\mu}_{\nu}+ \omega^{\mu}_{~~~\nu} $$

I didn't get very far. I just replaced the commutator of the Dirac matrices with anticommutator $[\gamma^{\mu},\gamma^{\nu}]=2\gamma^{\mu}\gamma^{\nu}-2g^{\mu\nu}$ and ended up with: $$ \frac{1}{4}\left[ \gamma^{\mu}\omega_{\mu\nu}\gamma^{\mu}\gamma^{\nu}-\gamma^{\mu} \omega_{\mu\nu} \eta^{\mu\nu}-\omega_{\mu\nu}\gamma^{\mu}\gamma^{\nu}\gamma^{\mu} +\omega_{\mu\nu}\eta^{\mu\nu}\gamma^{\mu}\right] $$ for the left hand side.

Another idea was to use the leibniz rule for commutators $[A,BC]=[A,B]C+B[A,C]$: $$ \frac{1}{8}\left[\gamma^{\mu},\omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu} ] \right] =\frac{-i}{2}\left( \left\{ \gamma^{\mu},\omega_{\mu\nu}\right\} S^{\mu\nu}-\omega_{\mu\nu}\left\{\gamma^{\mu},S^{\mu\nu} \right\}\right)$$

Both ideas seem to fail since i don't know how to proceed. Can somebody please help me? Links and References to literature are welcome, too :-)

Thanks in Advance, mechanix


You should first prove that $\omega_{\mu\nu}$ is antisymmetric (from the properties of Lorentz transforms). Then your equality is obviously correct if $\mu=\nu$ (both parts vanish). Then you can assume that $\mu\ne\nu$. Then you can use, e.g., your third formula (by the way, please use either $g$ or $\eta$, not both), and use $(\gamma^\mu)^2=g^{\mu\mu}$, $\gamma^\mu\gamma^\nu=-\gamma^\nu\gamma^\mu$. I am not sure that the correct factor is $\frac{1}{8}$ though.

  • $\begingroup$ Ok, Great! Thanks for your answer! The antisymmetry is rather easy to show. I just plugged the infinitesimal lorentz transform into: $\eta_{\mu \nu}L^{\mu}_{~~~\rho}L^{\nu}_{~~~\sigma}=g_{\rho\sigma}$ and then I could see it. $\endgroup$
    – Mechanix
    Apr 1 '15 at 0:58
  • $\begingroup$ Two terms of my equation nr. 3 get cancelled, because I $\eta^{\mu\nu}=0$ for $\mu\neq\nu$. So I'm left with $\gamma^{\mu}\omega_{\mu \nu}\gamma^{\mu}\gamma^{\nu}+\omega^{\mu}_{~~~\nu}\gamma^{\nu}$ How can I get rid of the other term? $\endgroup$
    – Mechanix
    Apr 1 '15 at 1:10
  • $\begingroup$ Well, also remember that if $S_{\mu\nu}$ is symmetric and $A^{\mu\nu}$ is antisymmetric, then $A^{\mu\nu}S_{\mu\nu}=0$ identically...which should simplify considering $\omega_{\mu\nu}[\gamma^{\mu},\gamma^{\nu}]$ if you plug in the metric tensor plus term quadratic in $\gamma$...recalling that the metric tensor is symmetric...and $\omega$ antisymmetric... $\endgroup$ Apr 1 '15 at 2:29
  • $\begingroup$ @Mechanix: As I said, use the formula for $(\gamma^\mu)^2$ in my answer. Components of omega are just c-numbers. $\endgroup$
    – akhmeteli
    Apr 1 '15 at 2:59
  • $\begingroup$ @akhmeteli you say that $\omega$-components are complex numbers... do you mean that $\gamma$ and $\omega$ commute? Can I do the following? $\gamma^{\mu}\omega_{\mu\nu}\gamma^{\mu}\gamma^{\nu}= \omega_{\mu\nu}(\gamma^{\mu})^2\gamma^{\nu} = \omega_{\mu\nu}\eta^{\mu\mu}\gamma^{\nu} = 0 $ for $\mu\neq\nu$ ? $\endgroup$
    – Mechanix
    Apr 1 '15 at 9:30

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