# Peskin and Schroeder Equation 3.23

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:

• 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. Nov 23, 2013 at 12:04
• What does your symbol $\gamma^{\mu\nu\rho\sigma}$ stand for? Nov 23, 2013 at 12:32
• @user26143, that symbol was a shorthand for $\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma$.
– PPR
Nov 23, 2013 at 14:57

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{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} Subsequently, it will be convenient to first look at: \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} and now: \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} Now, we can write the first equation as: $$\gamma^\mu \gamma^\nu = 2i S^{\mu \nu} + g^{\mu \nu}$$ in order to write: \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} I hope/think this is the correct solution but I do not know if it is the quickest method.