Eq. (27) in http://arxiv.org/abs/1110.2662 says I can construct the Weyl spinor according to
$$\Psi_{ABCD} = \frac 14 C{}_{\mu\nu\lambda\rho} \left( \sigma^\mu \right){}_A{}^{\dot A} \left( \sigma^\nu \right){}_{B\dot A}\left( \sigma^\lambda \right){}_C{}^{\dot C}\left( \sigma^\rho \right){}_{D\dot C}\tag{27}$$
I understand that the partial contraction in the two index pairs $\mu\leftrightarrow\nu$ and $\lambda\leftrightarrow\rho$ induces a unique way to extract either two left-handed or two right-handed spinor indices. Moreover, the above spinor is completely symmetrical, $\Psi{}_{ABCD} = \Psi{}_{(ABCD)}$ as it must to encompass all ten components of the Weyl tensor.
But why does the above work?