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I am reading Zee's QFT in a nutshell, 2nd ed. On pg. 97 below eq. 14 he writes:

$$ S \gamma^{\lambda } S^{-1} = \omega_{\,\, \mu }^{\lambda } \gamma ^{\mu }+\gamma ^{\lambda }. $$

Building up a finite Lorentz transformation by compounding infinitesimal transformations (just as in the standard discussion of the rotation group in quantum mechanics), we have $$S \gamma ^{\lambda } S^{-1}= \Lambda ^{\lambda }_{\,\,\mu } \gamma ^{\mu}.$$

I am having trouble seeing how this second equation follows from the first. I would very much appreciate some help on what are the in between steps.

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    $\begingroup$ Write ${\Lambda^\lambda}_\mu = {\delta^\lambda}_\mu+{\omega^\lambda}_\mu +O(\omega^2)$. In other words, $\omega$ is an infinitesimal Lorentz transformation. $\endgroup$ – suresh Sep 25 '14 at 6:14
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As suggested by @suresh in the comments, he is considering the infinitesimal form of $\Lambda^\mu_\nu$: $$ \Lambda^\mu_\nu = \delta^\mu_\nu + \omega^\mu_\nu + \mathcal{O}(\omega^2). $$ Consequently you have $$ S(\Lambda) \gamma^\lambda S^{-1}(\Lambda) = \Lambda^\lambda_{\,\,\mu} \gamma^\mu = ( \delta^\lambda_{\,\,\mu} + \omega^\lambda_{\,\,\mu})\gamma^\mu = \gamma^\lambda + \omega^\lambda_{\,\,\mu} \gamma^\mu, $$ where $S(\Lambda)$ is the spinor representation of the Lorentz transformation with vector representation $\Lambda = 1 + \omega$.

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