# Proof of two Lorentz-algebra identities

I am currently working through the QFT introduction text by Peskin and Schroeder and try to fill in two identities that I wasn't able to prove (it should be fairly simple, but my experience with this kind of calculations is limited).

1. $$[\gamma^{\mu}, S^{\rho\sigma}] = (\mathcal{J}^{\rho\sigma}~)^{\mu}{}_{\nu}\gamma^{\nu}$$ where $\gamma^{\mu}$ are the gamma-matrices in Weyl-representation (don't think the specific representation matters though), $S^{\rho\sigma} = \frac{i}{4}[\gamma^{\rho}, \gamma^{\sigma}~]$ and $(\mathcal{J}^{\rho\sigma}~)_{\mu\nu} = i(\delta^{\rho}{}_{\mu}~\delta^{\sigma}{}_{\nu} - \delta^{\rho}{}_{\nu}~\delta^{\sigma}{}_{\mu})$ a specific representation of the Lorentz algebra.
2. $$\left(1+\frac{i}{2}\omega_{\rho\sigma}~S^{\rho\sigma}\right)\gamma^\mu\left(1-\frac{i}{2}\omega_{\rho\sigma}~S^{\rho\sigma}\right) = \left(1-\frac{i}{2}\omega_{\rho\sigma}~\mathcal{J}^{\rho\sigma}\right)^{\mu}{}_{\nu}~\gamma^{\nu}$$ using the result of 1. I especially do not understand where the term $$\dfrac{1}{4}\omega_{\rho\sigma}~S^{\rho\sigma}\gamma^{\mu}\omega_{\rho\sigma}~S^{\rho\sigma}$$ is going to?
• For 1, have you tried the following: Write our $S$ in terms of gamma martices, fully write out the commutators, then shuffle around the gamma matrices using their anticommutator to make terms cancel with eachother until only terms with one gamma matrix are left? – Gerben Jun 25 '16 at 22:58
• @SimonFromme Regarding your last point, $\omega$ is small, so you just set the $\omega^2$ term to zero. – Brian Moths Jun 26 '16 at 3:50
• I could simplify the first identity to $[\gamma^{\mu}, S^{\rho\sigma}] = i(g^{\mu\rho}\gamma^{\sigma}-g^{\mu\sigma}\gamma^{\rho})$ but didn't know how to proceed further. – Simon Fromme Jun 26 '16 at 17:43
• Ok, it's actually quite trivial to proceed further. I got it now :) – Simon Fromme Jun 26 '16 at 17:57

## 1 Answer

Since I finally got the first one right, I might as well answer my own question.

\begin{align*} [\gamma^{\mu}, S^{\rho\sigma}] &= \dfrac{i}{4}[\gamma^{\mu}, [\gamma^{\rho}, \gamma^{\sigma}]] \\ &= \dfrac{i}{4}\left( \gamma^{\mu}\gamma^{\rho}\gamma^{\sigma} - \gamma^{\mu}\gamma^{\sigma}\gamma^{\rho} - \gamma^{\rho}\gamma^{\sigma}\gamma^{\mu} + \gamma^{\sigma}\gamma^{\rho}\gamma^{\mu}\right) \\ &= \dfrac{i}{4}\left( [2g^{\mu\rho} - \gamma^{\rho}\gamma^{\mu}]\gamma^{\sigma} - [2g^{\mu\sigma} - \gamma^{\sigma}\gamma^{\mu}]\gamma^{\rho} - \gamma^{\rho}\gamma^{\sigma}\gamma^{\mu} + \gamma^{\sigma}\gamma^{\rho}\gamma^{\mu}\right) \\ &= \dfrac{i}{4}\left(-\gamma^{\rho}[2g^{\mu\sigma}-\gamma^{\sigma}\gamma^{\mu}] + \gamma^{\sigma}[2g^{\mu\rho}-\gamma^{\rho}\gamma^{\mu}] - \gamma^{\rho}\gamma^{\sigma}\gamma^{\mu} + \gamma^{\sigma}\gamma^{\rho}\gamma^{\mu}\right) + \dfrac{i}{2}\left( g^{\mu\rho}\gamma^{\sigma} - g^{\mu\sigma}\gamma^{\rho}\right)\\ &= i\left( g^{\mu\rho}\gamma^{\sigma} - g^{\mu\sigma}\gamma^{\rho}\right) \\ &= i \left( \delta^{\rho}_{~\mu}\delta^{\sigma}_{~\nu} -\delta^{\sigma}_{~\mu}\delta^{\rho}_{~\nu} \right)g^{\mu\nu}\gamma^{\nu} \\ &= \left( \mathcal{J}^{\sigma\rho} \right)^{\mu}_{~\nu}\gamma^{\nu} \end{align*}

When taking into consideration that $\omega$ is infinitesimal the second identity is trivial. Neglecting higher order terms in $\omega$ and using $(1)$ yields \begin{align*} \left(1+\frac{i}{2}\omega_{\rho\sigma}~S^{\rho\sigma}\right)\gamma^\mu\left(1-\frac{i}{2}\omega_{\rho\sigma}~S^{\rho\sigma}\right) &= \gamma^\mu + \frac{i}{2}\omega_{\rho\sigma}[S^{\rho\sigma}, \gamma^\mu]\\ &= \left(1-\frac{i}{2}\omega_{\rho\sigma}~\mathcal{J}^{\rho\sigma}\right)^{\mu}{}_{\nu}~\gamma^{\nu} \end{align*}

• There is something wrong with your indices. – AMS Jun 27 '16 at 7:52
• Well, without being more specific your comment is not very helpful – Simon Fromme Jun 27 '16 at 14:20