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 $\varepsilon_{\alpha \beta}\varepsilon^{\gamma \beta} = -\delta^{\quad \gamma}_{\alpha}$: $$ \varepsilon_{\gamma \delta}\varepsilon^{\delta \beta }h_{(\alpha \beta )} = h_{(\alpha \gamma )} = -\frac{1}{4}\left( (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha \gamma} - (\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha \gamma }\right)M_{\mu \nu}. $$


1 Answer 1


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 $_{\alpha\beta}$

  • $\begingroup$ Excuse me, I gave false defs of $h_{(\alpha \beta)}, h_{(\dot {\alpha }\dot {\beta})}$. The correct expressions are written in the end of question body. $\endgroup$
    – user8817
    Aug 27, 2013 at 9:43
  • $\begingroup$ Why did you write, that $\tilde {\sigma}^{\nu}$ has lower indices? It has an upper indices by the definition. $\endgroup$
    – user8817
    Aug 27, 2013 at 9:55
  • $\begingroup$ Also, my def $(\tilde {\sigma}^{\nu})^{\beta \dot {\beta}} = \varepsilon^{\beta \alpha}\varepsilon^{\dot {\beta }\dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\alpha}}$ is totally incorrect, considering approval $\tilde {\sigma}^{\beta \dot {\beta }} = (\hat {\mathbf E} , -\hat {\mathbf \sigma})$. $\endgroup$
    – user8817
    Aug 27, 2013 at 10:23
  • 1
    $\begingroup$ @PhysiXxx : Uppering or lowering indices do no change the problem, the indices of $(\sigma^{\mu}\tilde {\sigma }^{\nu })_{\alpha \beta }$ could not be $_{\alpha \beta }$. You have to insert somewhere $\epsilon_{\gamma \delta}$ terms to have it. $\endgroup$
    – Trimok
    Aug 27, 2013 at 10:33
  • $\begingroup$ By multiplying $h_{(\alpha \beta)}$ by $\varepsilon^{\delta \beta}$ I got $$ \varepsilon^{\delta \beta}h_{(\alpha \beta)} = -\frac{1}{8}\varepsilon^{\delta \beta}\varepsilon^{\dot {\alpha}\dot {\beta}}\left( (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} + (\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }} - (\sigma^{\mu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\beta \dot {\alpha }}\right)M_{\mu \nu} = $$ $\endgroup$
    – user8817
    Aug 27, 2013 at 10:55

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