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I have the transformation law of the Lorentz group for Pauli matrices: $$ \tag 0 (\sigma^{\mu})_{a \dot {a}}{'} = \Lambda^{\mu}_{\quad \nu} N_{a}^{\quad c}(\sigma^{\nu})_{c \dot {c}}(N^{-1})^{\dot {c}}_{\quad \dot {a}}, $$ where $$ N_{a}^{\quad c} = 1 + \frac{1}{2}\omega_{\mu \nu}\sigma^{\mu \nu}, \quad N^{\quad \dot {c}}_{\dot {a}} = 1 + \frac{1}{2}\omega_{\mu \nu}\tilde {\sigma}^{\mu \nu}. $$ Also I have the relation $$ \tag 1 \gamma_{\mu} = \begin{pmatrix} 0 & \sigma_{\mu} \\ \tilde {\sigma}_{\mu} & 0 \end{pmatrix}, \quad (\tilde {\sigma}_{\mu})^{\dot {a} a} = \varepsilon^{ab}\varepsilon^{\dot {a} \dot {b}}(\sigma_{\mu})_{d \dot {b}}. $$ How exactly can I get the transformation law $$ (\gamma_{\mu}){'} = \Lambda_{\mu}^{\quad \nu}\hat {S}\gamma_{\nu}\hat {S}^{-1}, \quad \hat {S} = 1 + \frac{1}{2}\omega_{\mu \nu}H^{\mu \nu}, $$ $$ H_{\mu \nu} = \frac{1}{4}\gamma_{[\mu} \gamma_{\nu ]} $$ from $(0)$?

I failed when I tried to write explicitly transformations of Pauli matrices in $(1)$: I can't pick out $\hat {S}^{-1}$, because I entangled in the indices.

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  • $\begingroup$ Looks like you have been asked to prove only the infinitesimal version. $\hat{S}$ should be $\exp(\frac12 \omega_{\mu\nu}H^{\mu\nu})$. Ask yourself what the infinitesimal version of $\Lambda$ should look like. $\endgroup$ – suresh Mar 7 '14 at 0:03
  • $\begingroup$ @suresh : $$ \Lambda_{\mu}^{\quad \nu} = \delta_{\mu}^{\quad \nu} + \omega_{\mu}^{\quad \nu} - \omega^{\nu}_{\quad \mu}, $$ $\endgroup$ – Andrew McAddams Mar 7 '14 at 4:39

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