Identidades de matrices de Pauli

Hay algunas definiciones y propiedades para las matrices de Pauli y sus combinaciones:

$$\varepsilon^{\alpha \beta } = \varepsilon^{\dot {\alpha} \dot {\beta} } = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_{\alpha \beta }, \quad \varepsilon_{\dot {\alpha} \dot {\beta}} = \varepsilon_{\alpha \beta} = -\varepsilon^{\alpha \beta },$$ $$(\sigma^{\mu})_{\alpha \dot {\alpha} } = (\hat {\mathbf E} , \hat {\mathbf \sigma})^{\mu}_{\alpha \dot {\alpha}}, \quad (\tilde {\sigma}^{\mu})^{\dot {\beta } \beta } = \varepsilon^{\alpha \beta}\varepsilon^{\dot {\alpha }\dot {\beta }}(\sigma^{\mu})_{\alpha \dot {\alpha}} = (\hat {\mathbf E}, -\hat {\mathbf \sigma})^{\mu , \dot {\beta }\beta },$$ $$(\sigma^{\mu \nu})_{\alpha \beta} = -\frac{1}{4}\left( (\sigma^{\mu} \tilde {\sigma}^{\nu})_{\alpha \beta } - (\sigma^{\nu} \tilde {\sigma}^{\mu})_{\alpha \beta }\right), \quad (\tilde {\sigma}^{\mu \nu})_{\dot {\alpha }\dot {\beta }} = -\frac{1}{4}\left( (\tilde{\sigma}^{\mu} \sigma^{\nu})_{\dot {\alpha} \dot {\beta} } - (\tilde {\sigma}^{\nu} \sigma^{\mu})_{\dot {\alpha} \dot {\beta} }\right),$$ $$(\tilde {\sigma}^{\mu})^{\dot {\alpha }\alpha}(\sigma_{\mu})_{\beta \dot {\beta }} = 2\delta^{\dot {\alpha}}_{\dot {\beta}}\delta^{\alpha}_{\beta}.$$ como demostrar que $$(\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = 0?$$ (Ayuda a mostrar que la representación irreducible del espinor de los generadores del grupo de Lorentz se expande en dos subgrupos del espinor).

mi intento

Traté de mostrar esto, pero solo obtuve seguimiento.

$$(\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = \frac{1}{16}\left( (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m }(\sigma^{\nu })_{m \dot {d}} - (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\nu})_{\dot {c}}^{ \quad m}(\sigma^{\mu })_{m \dot {d}}\right)$$ $$+ \frac{1}{16}\left(-(\sigma^{\beta })_{a \dot {n}}(\tilde {\sigma}^{\alpha })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m}(\sigma^{\nu })_{m \dot {d}} + (\sigma^{\beta })_{a \dot {n}}(\tilde {\sigma}^{\alpha })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\nu})_{\dot {c}}^{\quad m }(\sigma^{\mu })_{m \dot {d}} \right)g_{\alpha \mu}$$ Después de eso transformé cada sumando como $$(\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m }(\sigma^{\nu })_{m \dot {d}}g_{\alpha \mu} = (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}_{\alpha})_{\dot {c}}^{\quad m }(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} =$$ $$= \varepsilon_{\dot {c}\dot {\gamma}}(\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}_{\alpha})^{\dot {\gamma} m }(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} = 2\varepsilon_{\dot {c}\dot {\gamma }}\delta^{\dot {\gamma}}_{\dot {m}}\delta^{n}_{a}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} =$$ $$2\varepsilon_{\dot {c}\dot {m}}(\tilde {\sigma}^{\beta })^{ \dot {m}}_{\quad b}(\sigma^{\nu })_{a \dot {d}} = 2(\sigma^{\beta } )_{b \dot {c}}(\sigma^{\nu })_{a \dot {d}}.$$ Finalmente, conseguí $$(\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = \frac{1}{8}\left( (\sigma^{\beta})_{b\dot {c}}(\sigma^{\nu})_{a\dot {d}} + (\sigma^{\beta })_{b \dot {d}}(\sigma^{\nu})_{a \dot {c}} + (\sigma^{\beta })_{a \dot {c}}(\sigma^{\nu})_{b \dot {d}} + (\sigma^{\beta})_{a \dot {d}}(\sigma^{\nu})_{b \dot {c}}\right).$$ ¿Qué hacer a continuación?