 Solution

For nonzero $\chi$, the outer product $X=\chi {\chi }^{†}$ is rank-one Hermitian positive semidefinite, so $\det X=0$ and $ct={\chi }^{†}\chi /2>0$. It therefore represents a future-directed null vector. Conversely any nonzero future-null Hermitian $X$ has eigenvalues $2ct$ and zero, so its positive eigenvector multiplied by $\sqrt{2ct}$ supplies $\chi$. The phase $\chi \mapsto e^{i\zeta }\chi$ leaves $X$ unchanged. A past-directed null vector needs $X=-\chi {\chi }^{†}$, rather than the positive outer product alone.

For the printed normalized spinor, direct multiplication yieldsComparing with the printed $x+iy$ convention givesIts usual spatial azimuth is $-\varphi$, not $\varphi$. The source's asserted usual-angle identification has a sign inconsistency. Either conjugating the two spinor phases or replacing the upper-right matrix entry by $x-iy$ restores the usual azimuth $\varphi$. Neither alteration is made silently here. A further positive scale factor sets the magnitude of an arbitrary future-null vector; the displayed normalized spinor fixes $ct=1/2$ in the chosen coordinate units.