- #1

Markus Kahn

- 112

- 14

## Homework Statement

Prove that the sets ##(S_{\mu\nu})_L## and ##(S_{kl})_R##, where

$$

\left( S _ { k \ell } \right) _ { L } = \frac { 1 } { 2 } \varepsilon _ { j k \ell } \sigma _ { j } = \left( S _ { k \ell } \right) _ { R } \quad\text{and}\quad \left( S _ { 0 k } \right) _ { L } = \frac { 1 } { 2 } i \sigma _ { k } = ( S ^ { 0 k }) _ { R }

$$

satisfy the commutation relation of the Lorentz group, namely

$$

\left[M_{\mu \nu}, M_{\rho \sigma}\right]=-\eta_{\mu \rho} M_{\nu \sigma}+\eta_{\mu \sigma} M_{\nu \rho}-\eta_{\nu \sigma} M_{\mu \rho}+\eta_{\nu \rho} M_{\mu \sigma}.

$$

## The Attempt at a Solution

My attempt was straight forward

$$

\begin{align*}

[(S_{kl})_L, (S_{bc})_L]

&= \frac{1}{4}\varepsilon_{jkl}\varepsilon_{abc}[\sigma_j,\sigma_a] = \frac{1}{4}\varepsilon_{jkl}\varepsilon_{abc} (2i \varepsilon_{jau}\sigma_u) = \frac{i}{2}\varepsilon_{jkl}\varepsilon_{abc} \varepsilon_{jau}\sigma_u\\

&=\frac{i}{2} (\delta_{ka}\delta_{lu}-\delta_{ku}\delta_{al})\varepsilon_{abc}\sigma_u = \frac{i}{2}\varepsilon_{kbc}\sigma_l -\frac{i}{2} \varepsilon_{lbc}\sigma_k

\end{align*}

$$

but this seems to lead to nowhere. One of my problems here is that ##(S_{kl})_L## is only defined for ##k,l\in\{1,2,3\}## but ##M_{\mu\nu}## is defined for ##\mu,\nu\in\{0,\dots,3\}##... I'm not sure how to make sense of this but I honestly also don't know where I made a mistake in the above calculation...