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Let's have two symmetric (by the indices) spinor tensors $F_{ab}, F_{\dot {a}\dot {b}}$ and conditions $$ F_{ab}, \partial^{\dot {a} a}F_{ab} = 0, \quad F_{\dot {a}\dot {b}}, \partial^{\dot {a}a}F_{\dot {a}\dot {b}} = 0, \qquad (1) $$ $$ \partial^{a \dot {a}} = (\tilde {\sigma}_{\mu})^{\dot {a} a}\partial^{\mu}, \quad (\tilde {\sigma}_{\mu})^{\dot {a} a} = \varepsilon^{\dot {a}\dot {b}}\varepsilon^{a b}(\sigma_{\mu})_{b \dot {b}} = (\sigma_{0}, -\sigma_{i}). $$ by which these tensors may represent massless field with helicity $1, -1$ respectively. If our field is real, we must take the direct sum of these representations. It may be described through antisymmetrical 4-tensor rank 2.

I know that there is the correspondence between an arbitrary antisymmetric tensor $F_{\mu \nu}$ and two symmetric spinor tensors $F_{ab} , F_{\dot {a} \dot {b}}$: $$ F_{\mu \nu} = \frac{1}{2}(\sigma_{\mu \nu})^{a b}F_{ab} - \frac{1}{2}(\tilde \sigma_{\mu \nu})^{\dot a \dot b}F_{\dot {a} \dot {b}}, $$ $$ F_{ab} = (\sigma^{\mu \nu})_{ab}F_{\mu \nu}, \quad F_{\dot {a} \dot {b}} = (\tilde {\sigma}^{\mu \nu})_{\dot {a}\dot {b}}F_{\mu \nu}. \qquad (2) $$ Then it obvious that I must substitute $(2)$ into $(1)$ and get Maxwell's equations. But I can't do it, because I have the problem with convolution of sigma-matrices.

Can you help me?

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  • $\begingroup$ For instance, see formulae $(2.47) \to (2.53)$, page $16$ in this paper $\endgroup$ – Trimok Dec 25 '13 at 19:40
  • $\begingroup$ @Trimok : thank you! Do you know how to prove eq. $(2.52)$? $\endgroup$ – Andrew McAddams Dec 25 '13 at 20:30
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    $\begingroup$ By brute force, I think you may separate, in cases where you have all indices equal to zero, $2$ indices equal to zero, $1$ indice equal to zero, all indices different of zero. $\endgroup$ – Trimok Dec 26 '13 at 11:58

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