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Use the same four Grassmann generators and a fixed order $a,b,c,d$. Their squares vanish, soMeanwhile the two self-contractions areand moving the pair $cd$ past $ab$ gives no sign, yielding $(\psi \psi )(\chi \chi )=4abcd$. Thus the square of a Weyl spinor bilinear isThe order of individual odd components, rather than an informal commuting-spinor substitution, fixes this sign. For commuting spinors both self-contractions vanish, so no corresponding universal factorization of the generally nonzero mixed square is possible.