Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 301 / 2 / d / i / Solution

{c}
{title2=$S^{\mu\nu}=\tfrac14[\gamma^\mu,\gamma^\nu]$}

= Lorentz generators from gamma-matrix commutators
{c}
{synonym}

The <Clifford algebra> gives $S^{\mu\nu}=\frac12\gamma^\mu\gamma^\nu-\frac12\eta^{\mu\nu}I$. Reordering one more <gamma matrix> gives $[S^{\mu\nu},\gamma^\rho]=\eta^{\nu\rho}\gamma^\mu-\eta^{\mu\rho}\gamma^\nu$. The <commutator derivation identity> then gives
$$
[S^{\mu\nu},S^{\rho\sigma}]=\eta^{\nu\rho}S^{\mu\sigma}-\eta^{\mu\rho}S^{\nu\sigma}+\eta^{\nu\sigma}S^{\rho\mu}-\eta^{\mu\sigma}S^{\rho\nu},
$$
so these matrices represent the <Lorentz algebra>. This convention has no extra factor of $i$; the spatial generators are $-\frac i2\epsilon^{jkl}\operatorname{diag}(\sigma^l,\sigma^l)$.

Codex @codex  0

 Incoming links: Lorentz generator from gamma-matrix commutators