 Solution

Choose the active boost producing $p_z=m\sinh \rho$ and $E=m\cosh \rho$. The supplied two-component boost $L=\exp (-\rho {\sigma }_3/2)$ acts on the left-chiral block $\eta$; the right-chiral block $\xi$ transforms by its inverse. Applying them to b givesUse $m{e}^{\pm \rho }=E\pm p_z$ and ${\sigma }_3{\chi }_{\pm }=\pm {\chi }_{\pm }$ to obtainBoth are particle spinors, with spin respectively parallel and antiparallel to the positive $z$ axis. A compact equivalent expression is $\xi =\sqrt{E+p_z{\sigma }_3}\chi$, $\eta =\sqrt{E-p_z{\sigma }_3}\chi$, using the positive matrix square roots.

The products of corresponding nonzero upper and lower components equal $\sqrt{E^2-p_z^2}=m$, so ${\bar{u}}_{\pm }{u}_{\pm }=2m$ and $u_{\pm }^{†}{u}_{\pm }=2E$. Substitution also verifies $(E-p_z{\sigma }_3)\xi =m\eta$ and $(E+p_z{\sigma }_3)\eta =m\xi$. These checks fix which boost belongs to which component and prevent silently exchanging the chiral ordering defined by the source's $U$.