Spinor indices and antisymmetric tensor Excuse me for long prehistory. Maybe it can be useful for someone.
I was little confused with spinor indices when getting an expression relating the spinor and antisymmetric tensors. An antisymmetric tensor $M_{\mu \nu}$ have an expression (in spinor formalism)
$$
h_{ab\dot {a}\dot {b}} = \left((\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta }} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }}\right)M_{\mu \nu} = \varepsilon_{\dot {\alpha} \dot {\beta}}h_{(\alpha \beta )} + \varepsilon_{\alpha \beta}h_{(\dot {\alpha} \dot {\beta })}, \qquad (.1) 
$$
where second identity is the decomposition into irreducible spin coefficients, and
$$
\varepsilon^{\alpha \beta} = \varepsilon^{\dot {\alpha }\dot {\beta }} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\alpha \beta}, \quad \varepsilon^{\alpha \beta} = -\varepsilon_{\alpha \beta},
$$
$$
h_{(\alpha \beta)} = -\frac{1}{2}\varepsilon^{\dot {\alpha} \dot {\beta} }h_{(\alpha \beta)\dot {\alpha }\dot {\beta }}, \quad h_{(\dot {\alpha }\dot {\beta })} = -\frac{1}{2}\varepsilon^{\alpha \beta }h_{\alpha \beta (\dot {\alpha }\dot {\beta })}, \qquad (fixed)
$$
$$
h_{[\alpha \beta]\dot {\alpha }\dot {\beta }} = \frac{1}{2}\left( h_{\alpha \beta \dot {\alpha }\dot {\beta }} - h_{\beta \alpha \dot {\alpha }\dot {\beta }} \right), \quad h_{(\alpha \beta)\dot {\alpha }\dot {\beta }} = \frac{1}{2}\left( h_{\alpha \beta \dot {\alpha }\dot {\beta }} + h_{\beta \alpha \dot {\alpha }\dot {\beta }} \right).
$$
So, for $h_{(\alpha \beta)}$ I get, using $(.1)$
$$
h_{(\alpha \beta )} = -\frac{1}{8}\varepsilon^{\dot {\alpha } \dot {\beta } }\left( (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} + (\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }} - (\sigma^{\mu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\beta  \dot {\alpha }}\right)M_{\mu \nu}. \qquad (.2)
$$
Then one can show, that 
$$
\varepsilon^{\dot {\alpha } \dot {\beta } } (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} = (\sigma^{\mu}\tilde {\sigma }^{\nu })_{\alpha \beta },
$$
where
$$
(\tilde {\sigma}^{\mu})^{\dot {\beta } \beta } = \varepsilon^{\beta \gamma}\varepsilon^{\dot {\beta} \dot {\gamma}}(\sigma^{\mu})_{\gamma \dot {\gamma }} = \left( \hat {\mathbf E }, -\hat {\mathbf \sigma } \right). \qquad (fixed)
$$
But I don't understand, how to show it. Can you help me?
Addition 1. 
The correct defs of $h_{(\alpha \beta)}, h_{(\dot {\alpha} \dot {\beta})}$ are
$$
h_{(\alpha \beta)} = -\frac{1}{2}\varepsilon^{\dot {\alpha} \dot {\beta }}h_{(\alpha \beta)\dot {\alpha }\dot {\beta }}, \quad h_{(\dot {\alpha }\dot {\beta })} = -\frac{1}{2}\varepsilon^{\alpha \beta }h_{\alpha \beta(\dot {\alpha }\dot {\beta })},
$$
"thanks" to my inattention. Also I changed the def of $\tilde {\hat {\sigma}}^{\mu}$ on correct def.
Addition 2.
I got an answer, thanks to Trimok.
First, to simplify the work with each term I need to multiply $(.2)$ by $\varepsilon^{\delta \beta}$:
$$
\varepsilon^{\delta \beta }h_{(\alpha \beta )} =
$$
$$
= -\frac{1}{8}\varepsilon^{\delta \beta }\varepsilon^{\dot {\alpha } \dot {\beta } }\left( (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} + (\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }} - (\sigma^{\mu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\beta  \dot {\alpha }}\right)M_{\mu \nu}.
$$
For first and second terms, for example,
$$
\varepsilon^{\delta \beta }\varepsilon^{\dot {\alpha } \dot {\beta } }(\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} = (\sigma^{\mu} )_{\alpha \dot {\beta }}(\tilde {\sigma}^{\nu})^{\dot {\beta }\delta } = (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha}^{\quad {\delta}},
$$
$$
\varepsilon^{\delta \beta }\varepsilon^{\dot {\alpha } \dot {\beta } }(\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} = -\varepsilon^{\delta \beta }\varepsilon^{\dot {\beta } \dot {\alpha } }(\sigma^{\nu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\beta}} = -(\tilde {\sigma}^{\mu})^{\dot {\beta} \delta}(\sigma^{\nu})_{\alpha \dot {\beta}} = -(\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha}^{\quad \delta}.
$$
So
$$
\varepsilon^{\delta \beta }h_{(\alpha \beta )} = -\frac{1}{8}\left( 2 (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha}^{\quad {\delta}} - 2(\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha}^{\quad \delta} \right) M_{\mu \nu}.
$$
Second, I can multiply all expression by $\varepsilon_{\gamma \delta}$ and use identity $\varepsilon_{\alpha \beta}\varepsilon^{\gamma \beta} = -\delta^{\quad \gamma}_{\alpha}$:
$$
\varepsilon_{\gamma \delta}\varepsilon^{\delta \beta }h_{(\alpha \beta )} = h_{(\alpha \gamma )} = -\frac{1}{4}\left( (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha \gamma} - (\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha \gamma }\right)M_{\mu \nu}.
$$
 A: First, there are too much errors in the context of your question.
The last term in the expression $1$ is certainly false, because $h_{\alpha\beta\alpha'\beta'}$ is antisymmetric in the transformation $\alpha \to \beta, \alpha' \to \beta'$, while the last term is symmetric for this same transformation. Moreover, the terms $h_{(\alpha\beta)}$ are certainly zero because this a multiplication of $\epsilon^{\alpha\beta}$,  antisymmetric (in $\alpha, \beta$) , and $h_{(\alpha \beta)\dot {\alpha }\dot {\beta }}$, a symmetric quantity (in $\alpha, \beta$). 
Secondly, it is certainly false that : 
$\varepsilon^{\dot {\alpha } \dot {\beta } } (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} = \pm (\sigma^{\mu}\tilde {\sigma }^{\nu })_{\alpha \beta },$
With your notations, $\sigma^\mu$ has indices $(\sigma^{\mu})_{\alpha \dot {\alpha}}$, and $\tilde {\sigma }^{\nu }$ has indices $\tilde {\sigma }^{\nu }_{\beta \dot {\beta}}$ (the standard notation $\tilde {\sigma }^{\nu }_{\dot \beta  {\beta}}$ is preferable), anyway you cannot have a matrix  $(\sigma^{\mu}\tilde {\sigma }^{\nu })$ with indices $(\sigma^{\mu}\tilde {\sigma }^{\nu })_{\alpha \beta }$. Even if you define a matrix $\tilde {(\sigma }^{\nu })^{\beta \dot {\beta}}$ or $\tilde {(\sigma }^{\nu })^{\dot  \beta {\beta}}$, this does not work, you are unable to find 2 lower indices $_{\alpha\beta}$
