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I want to prove that if $\Lambda=\exp\Omega$, with $\Omega=\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}$, is a proper orthochronous Lorentz transformation and $D$ is the $(1/2,0)$ representation of $\mathrm{SL}(2,\mathbb{C})$ then \begin{equation} D(\Lambda)^{-1}\bar{\sigma}^\mu D(\Lambda)={\Lambda^\mu}_{\nu}\bar{\sigma}^\nu.\tag{1} \end{equation} In this representation the generators of the Lorentz group are given in terms of the Pauli matrices by \begin{equation} J^{\mu\nu}= \frac{i}{4}(\bar{\sigma}^\mu\sigma^\nu-\bar{\sigma}^\nu\sigma^\mu). \end{equation} I tried to use the equation \begin{equation} \exp(-A)B\exp A= B+[A,B]+\frac12[A,[A,B]]+\frac1{3!}[A,[A,[A,B]]]+\dotsb \end{equation} but I'm stuck calculating the commutator. For (1) to hold, I think that it should be $[\Omega,\bar{\sigma}^\rho]={\omega^\mu}_\nu\bar{\sigma}^\nu$. I have that \begin{equation} [\Omega,\bar{\sigma}^\mu]= -\frac14\omega_{\nu\rho}[\bar{\sigma}^\nu\sigma^\rho,\bar{\sigma}^\mu]= -\frac14\omega_{\nu\rho}(\bar{\sigma}^\nu\sigma^\rho\bar{\sigma}^\mu-\bar{\sigma}^\mu\bar{\sigma}^\nu\sigma^\rho): \end{equation} I know that \begin{equation} \bar{\sigma}^\nu\sigma^\rho\bar{\sigma}^\mu= -\eta^{\nu\rho}\bar{\sigma}^\mu+\eta^{\nu\mu}\bar{\sigma}^\rho-\eta^{\rho\mu}\bar{\sigma}^\nu+i\epsilon^{\nu\rho\mu\tau}\bar{\sigma}_\tau \end{equation} but I don't know any similar identities that could help me with the other term. Does someone know a way to "solve" that term?

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    $\begingroup$ Hint: $\{\sigma^\mu, \sigma^\nu \} = 2 \eta^{\mu\nu}$. $\endgroup$ – Oktay Doğangün Jul 4 '18 at 21:35
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I found the solution, starting from the anticommutation relations of the matrices as suggested by Oktay (thanks!). Since

  • $\bar{\sigma}^0\sigma^0+\sigma^0\bar{\sigma}^0=2 I_2$,
  • $\bar{\sigma}^0\sigma^i+\sigma^i\bar{\sigma}^0=\sigma^i-\sigma^i$,
  • $\bar{\sigma}^i\sigma^j+\sigma^j\bar{\sigma}^=-\{\sigma^i,\sigma^j\}=-2\delta^{ij} I_2$

we have $\{\bar{\sigma}^\mu,\sigma^\nu\}=-2\eta^{\mu\nu}I_2$ (I use the metric convention $\eta=\operatorname{diag}(-1,1,1,1)$) therefore \begin{equation} \bar{\sigma}^\mu\bar{\sigma}^\nu\sigma^\rho= \bar{\sigma}^\mu(-\sigma^\rho\bar{\sigma}^\nu-2\eta^{\nu\rho})= -\bar{\sigma}^\mu\sigma^\rho\bar{\sigma}^\nu-2\eta^{\nu\rho}\bar{\sigma}^\mu, \end{equation} and I can apply again the identity with the three Pauli matrices like I did in the other term, obtaining in the end the correct result.

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