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I've been trying (for a while) to prove that $S^{\mu\nu}:=\frac{i}{4}\left[\gamma^\mu,\,\gamma^\nu\right]$ is a representation of the Lorentz Lie algebra, that is, to prove that it satisfies the commutation relations of equation (3.17): $$ \left[J^{\mu\nu},\,J^{\rho\sigma}\right]=i\left(\eta^{\nu\rho}J^{\mu\sigma}-\eta^{\mu\rho}J^{\nu\sigma}-\eta^{\nu\sigma}J^{\mu\rho}+\eta^{\mu\sigma}J^{\nu\rho}\right)$$ where $\gamma$ are the gamma matrices defined by the relation $\left\{\gamma^\mu,\,\gamma^\nu\right\}=2\eta^{\mu\nu}\mathbb{1}_{n\times n}$. Peskin says "by repeated use of the defining relation of the gamma matrices it is easy to verify that these matrices satisfy the commutation relations." But I failed at that:my attempt

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  • $\begingroup$ You're on the right way! Just try again. Ideally, try to see how to use the anticommutation relation to create "chains" of four gamma matrices with equal ordering of the indices and opposite signs. $\endgroup$
    – Neuneck
    Nov 23, 2013 at 12:04
  • $\begingroup$ What does your symbol $\gamma^{\mu\nu\rho\sigma}$ stand for? $\endgroup$
    – user26143
    Nov 23, 2013 at 12:32
  • $\begingroup$ @user26143, that symbol was a shorthand for $\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma$. $\endgroup$
    – PPR
    Nov 23, 2013 at 14:57

2 Answers 2

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I struggled with this one as well and once I found I have written it in LaTeX which I will copy here below. Do note that I am using slightly different conventions than P&S, however it should still work out the same. \begin{equation} \begin{aligned} S^{\mu \nu} & = - \frac{i}{4}[\gamma^\mu,\gamma^\nu] \\& = - \frac{i}{4}(\gamma^\mu \gamma^\nu - \gamma^\nu \gamma^\mu) \\& = - \frac{i}{4}(\gamma^\mu \gamma^\nu - \left\{ \gamma^\mu, \gamma^\nu \right\} + \gamma^\mu \gamma^\nu) \\& = - \frac{i}{4}(2 \gamma^\mu \gamma^\nu - 2 g^{\mu \nu}) \\& = - \frac{i}{2}(\gamma^\mu \gamma^\nu - g^{\mu \nu}) \end{aligned} \end{equation} Subsequently, it will be convenient to first look at: \begin{equation} \begin{aligned} & [ S^{\mu \nu}, \gamma^\rho] = -\frac{i}{2}[(\gamma^\mu \gamma^\nu - g^{\mu \nu}),\gamma^\rho ] \\& =- \frac{i}{2} [\gamma^\mu \gamma^\nu,\gamma^\rho ] \\& = - \frac{i}{2}(\gamma^\mu \gamma^\nu \gamma^\rho - \gamma^\rho \gamma^\mu \gamma^\nu) \\& = - \frac{i}{2}(\gamma^\mu \gamma^\nu \gamma^\rho - \gamma^\rho \gamma^\mu \gamma^\nu - (\gamma^\mu \gamma^\rho \gamma^\nu - \gamma^\mu \gamma^\rho \gamma^\nu) - (\gamma^\mu \gamma^\rho \gamma^\nu - \gamma^\mu \gamma^\rho \gamma^\nu)) \\& = -\frac{i}{2}( \gamma^\mu \left\{ \gamma^\nu, \gamma^\rho \right\} - \left\{ \gamma^\rho, \gamma^\mu \right\} \gamma^\nu) \\& = - i( \gamma^\mu g^{\nu \rho} - g^{\rho \mu} \gamma^\nu) \end{aligned} \end{equation} and now: \begin{equation} \begin{aligned} & [S^{\mu \nu},S^{\rho \lambda}] = \left[S^{\mu \nu},-\frac{i}{2}(\gamma^\rho \gamma^\lambda - g^{\rho \lambda})\right] \\& = -\frac{i}{2} [S^{\mu \nu},\gamma^\rho \gamma^\lambda] \\& = - \frac{i}{2}(S^{\mu \nu} \gamma^\rho \gamma^\lambda - \gamma^\rho \gamma^\lambda S^{\mu \nu}) \\& = - \frac{i}{2}(S^{\mu \nu} \gamma^\rho \gamma^\lambda - \gamma^\rho S^{\mu \nu} \gamma^\lambda - \gamma^\rho \gamma^\lambda S^{\mu \nu}+\gamma^\rho S^{\mu \nu} \gamma^\lambda) \\& = - \frac{i}{2}\left( [S^{\mu \nu}, \gamma^\rho] \gamma^\lambda + \gamma^\rho [S^{\mu \nu},\gamma^\lambda] \right) \\& = -\frac{1}{2}\left( ( \gamma^\mu g^{\nu \rho} - g^{\rho \mu} \gamma^\nu) \gamma^\lambda + \gamma^\rho( \gamma^\mu g^{\nu \lambda} - g^{\lambda \mu} \gamma^\nu) \right) \\& = -\frac{1}{2}\left( g^{\nu \rho} \gamma^\mu \gamma^\lambda - g^{\rho \mu} \gamma^\nu \gamma^\lambda + g^{\nu \lambda} \gamma^\rho \gamma^\mu - g^{\lambda \mu} \gamma^\rho \gamma^\nu \right) \end{aligned} \end{equation} Now, we can write the first equation as: \begin{equation} \gamma^\mu \gamma^\nu = 2i S^{\mu \nu} + g^{\mu \nu} \end{equation} in order to write: \begin{equation} \begin{aligned} & [S^{\mu \nu},S^{\rho \lambda}] = -\frac{1}{2}\bigl( g^{\nu \rho} (2i S^{\mu \lambda} + g^{\mu \lambda}) - g^{\rho \mu} (2i S^{\nu \lambda} + g^{\nu \lambda}) \\& + g^{\nu \lambda} (2i S^{\rho \mu} + g^{\rho \mu}) - g^{\lambda \mu} (2i S^{\rho \nu} + g^{\rho \nu}) \bigr) \\& =- i g^{\nu \rho} S^{\mu \lambda} + i g^{\rho \mu} S^{\nu \lambda} - i g^{\nu \lambda} S^{\rho \mu} + i g^{\lambda \mu} S^{\rho \nu} \end{aligned} \end{equation} I hope/think this is the correct solution but I do not know if it is the quickest method.

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  • $\begingroup$ Thanks a lot! I used your "tricks" and was able to prove it for Peskin's sign conventions. Due to the differences I wasn't able to confirm if your final result is right; it seems like the last term should have a minus sign according to your convention. $\endgroup$
    – PPR
    Nov 23, 2013 at 17:28
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It is proven in Tong's QFT script http://www.damtp.cam.ac.uk/user/dt281/qft.html section 4.1. in a quite nice fashion.

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