The left-handed Weyl operator is defined by the ##2\times 2## matrix(adsbygoogle = window.adsbygoogle || []).push({});

$$p_{\mu}\bar{\sigma}_{\dot{\beta}\alpha}^{\mu} = \begin{pmatrix} p^0 +p^3 & p^1 - i p^2\\ p^1 + ip^2 & p^0 - p^3 \end{pmatrix},$$

where ##\bar{\sigma}^{\mu}=(1,-\vec{\sigma})## are sigma matrices.

One can use the sigma matrices to go back and forth between four-vectors and ##2\times 2## matrices:

$$p_{\mu} \iff p_{\dot{\beta}\alpha}\equiv p_{\mu}\bar{\sigma}^{\mu}_{\dot{\beta}\alpha}.$$

Given two four-vectors ##p## and ##q## written as ##2\times 2## matrices,

$$\epsilon^{\dot{\alpha}\dot{\beta}}\epsilon^{\alpha\beta}p_{\dot{\alpha}\alpha}q_{\dot{\beta}\beta} = 2p^{\mu}q_{\mu}.$$

Given a complex ##2\times 2## matrix ##\Lambda_{L}## with unit determinant, it can be shown that the transformation ##p_{\dot{\beta}\alpha} \rightarrow (\Lambda_{L}^{-1\dagger}p\Lambda_{L}^{-1})_{\dot{\beta}\alpha}## preserves the product ##\epsilon^{\dot{\alpha}\dot{\beta}}\epsilon^{\alpha\beta}p_{\dot{\alpha}\alpha}q_{\dot{\beta}\beta}##.

How does it then follow that ##\Lambda_{L}## is a Lorentz transformation? Do we have to use the fact that ##\epsilon^{\dot{\alpha}\dot{\beta}}\epsilon^{\alpha\beta}p_{\dot{\alpha}\alpha}q_{\dot{\beta}\beta} \sim p^{\mu}q_{\mu}##? What is the Lorentz transformation for ##p^{\mu}## due to the transformation ##\Lambda_{L}## for ##p_{\dot{\alpha}\alpha}##?

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# A Transformation of the spinor indices of the Weyl operator under the Lorentz group

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