[Physics] Commutator of Dirac gamma matrices

Question

Quick question…For some reason I'm having trouble finding an identity or discussion for the commutator of the gamma matrices at the moment…i.e $$\gamma^u\gamma^v-\gamma^v \gamma^u$$ but I am not finding this anywhere. I have an idea of what it may be, but then again I'm not always right. Can anyone fill me in here fill me in? (I already know the anticommutator, i.e $$\gamma^u\gamma^v+\gamma^v\gamma^u=2g^{uv}I.)$$

Answer 1

Although the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}$ is the most famous, there is an expression for the commutator:

$$[\gamma^\mu,\gamma^\nu] = 2\gamma^\mu \gamma^\nu - 2 \eta^{\mu\nu}$$

The matrix defined by $[\gamma^\mu,\gamma^\nu]$ actually has a purpose: it forms a representation of the Lorentz algebra. If we define $S^{\mu\nu}$ as $1/4$ the commutator, then we have,

$$[S^{\mu\nu},S^{\rho\sigma}] = \eta^{\nu\rho}S^{\mu\sigma} - \eta^{\mu\rho}S^{\nu\sigma} + \eta^{\mu\sigma}S^{\nu\rho} - \eta^{\nu\sigma}S^{\mu\rho}= \eta^{\rho[\nu}S^{\mu]\sigma} + \eta^{\sigma [\mu}S^{\nu]\rho}$$

which is the Lorentz algebra. One can verify this by simply using the first commutator, and the rule for the commutator involving a product.

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There is a particularly important use for the commutator, namely defining $\sigma^{\mu\nu} = \frac{i}{2} [\gamma^\mu,\gamma^\nu]$, the action of a spin-$\frac32$ particle is given by,

$$\mathcal L = -\frac{1}{2}\bar{\psi}_\mu \left( \varepsilon^{\mu\lambda \sigma \nu} \gamma_5 \gamma_\lambda \partial_\sigma -im\sigma^{\mu\nu}\right)\psi_v,$$

which can be used to describe the superpartner to the graviton, namely the gravitino, thus making it necessary for supergravity theories.