= Solution
For nonzero $\chi$, the outer product $X=\chi\chi^\dagger$ is rank-one Hermitian positive semidefinite, so $\det X=0$ and $ct=\chi^\dagger\chi/2>0$. 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 $X=-\chi\chi^\dagger$, rather than the positive outer product alone.
For the printed normalized <spinor>, direct multiplication yields
$$
X=\frac12\begin{pmatrix}
1+\cos\theta&\sin\theta\,e^{-i\phi}\\
\sin\theta\,e^{i\phi}&1-\cos\theta
\end{pmatrix}.
$$
Comparing with the printed $x+iy$ convention gives
$$
\boxed{(ct,x,y,z)=\frac12
(1,\ \sin\theta\cos\phi,\ -\sin\theta\sin\phi,\ \cos\theta).}
$$
\b[Its usual spatial <azimuth> is $-\phi$, not $\phi$.] 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> $\phi$. 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.
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