1
$\begingroup$

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

Improve this question
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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... $\endgroup$
    – Alex Nelson
    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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.