Consider the Weyl equations: \begin{align} i\sigma^{\mu} \partial_{\mu} \psi_{L} & = 0 \\ i\overline{\sigma}^{\mu} \partial_{\mu} \psi_{R} & = 0, \end{align} where $\sigma^{\mu} = \left ( \sigma^{0}, \sigma^{1}, \sigma^{2}, \sigma^{3} \right )$ and $\overline{\sigma}^{\mu} = \left ( \sigma^{0}, - \sigma^{1}, - \sigma^{2}, - \sigma^{3} \right )$, with $\psi_{L}$ and $\psi_{R}$ being left and right handed Weyl spinors such that $\psi$ is a two-component spinor: \begin{equation} \psi = \binom{\psi_{L}}{\psi_{R}}. \end{equation} These equations are Lorentz invariant. Does this mean that $\psi_{L}' = A \psi_{L}$ and $A \sigma^{\mu} A^{-1} = a_{\nu}^{\mu} \sigma^{\nu}$, similarly for $\psi_{R}$, where $A$ is a matrix and $a_{\nu}^{\mu}$ is a Lorentz transformation?

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    $\begingroup$ This question lacks context. In general, the matrices don't transform in any way, because there is nothing related to Lorentz transformation in their definition. $\endgroup$ – Prof. Legolasov Jan 11 '17 at 5:02
  • $\begingroup$ I didn't quite understand the question. What is A here? Is it the $2\times 2$ matrix representing Lorentz transformation on $\psi_L$? $\endgroup$ – SRS Jan 11 '17 at 6:59

It means that the matrices transform like $$\sigma^\mu \mapsto \Lambda^{\nu'}{}_\mu A \sigma^\mu A^{-1}.$$ The left-handed spinor belongs to the $(\frac 1 2, 0)$ representation of the Lorentz group, and the right-handed spinor to the $(0, \frac 1 2)$ representation. (Or maybe the other way around, I'm not sure right now.) This view with representations can be found in Weinberg, Vol. 1, Ch. 2. Viewing spinors in terms of their transformation properties is more the view of texts like Pirani's chapter in Lectures on General Relativity, and Penrose and Rindler's Spinor's and Spacetime. They are equivalent, of course, and I think you need both views to really get it, but you may find one easier to start with than the other.

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