Square of a Weyl spinor bilinear
= Square of a Weyl spinor bilinear
{title2=$(\chi\psi)^2=-\tfrac12(\psi\psi)(\chi\chi)$}
For four independent odd <Grassmann variables> $a,b,c,d$, $(bc-ad)^2=-2abcd$, whereas $(\psi\psi)(\chi\chi)=(-2cd)(-2ab)=4abcd$. This proves the displayed two-component identity and fixes its sign. It is not valid with the same interpretation for commuting spinors, whose self-contractions vanish.