Let's have $$ \varepsilon^{\alpha \beta} = \varepsilon^{\dot {\alpha }\dot {\beta }} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\alpha \beta}, \quad \varepsilon^{\alpha \beta} = -\varepsilon_{\alpha \beta}, $$ which represents the metric of 2-spinor $\psi_{\alpha}$: $$ \psi^{\alpha \dot {\beta }} = \varepsilon^{\alpha \gamma}\varepsilon^{\dot {\beta }\dot {\delta}}\psi_{\gamma \dot {\delta}}. $$
Then let's have Pauli matrices: $$ \hat {\sigma}_{0} = \hat {\mathbf E}, \quad \hat {\sigma}_{1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \hat {\sigma}_{2} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \hat {\sigma}_{3} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$ Then - let's have the definition $$ (\tilde {\sigma}^{\mu})^{\dot {\beta } \beta } = \varepsilon^{\beta \gamma}\varepsilon^{\dot {\beta} \dot {\gamma}}(\sigma^{\mu})_{\gamma \dot {\gamma }} = \left( \hat {\mathbf E }, -\hat {\mathbf \sigma } \right). $$ So, how to show that $$ \varepsilon^{\dot {\alpha} \dot {\beta }}(\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta }} = (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha \beta}? $$
Addition.
I made it (look at the end of the question): https://physics.stackexchange.com/questions/75360/spinor-indices-and-antisymmetric-tensor .