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Qmechanic explained a way in which something with the word "commutator" in it doesn't vanish when applied to two of the same operator. However, I feel it is necessary to point out that plain commutators, as seen in a quantum mechanics course, really, honestly, always, and without fail satisfy $[Q,Q] = 0$ for any operator $Q$. This is because $[A,B]$ is ...


I) Yes, they are probably referring to that a Grassmann-odd operator needs not (super)commute with itself. Take e.g. the 1st order Grassmann-odd differential operator $$\tag{1} D~:=~\frac{d}{d\theta}+ \theta\frac{d}{dt}. $$ In eq. (1) $t$ is a Grassmann-even variable and $\theta$ is a Grassmann-odd variable, which (super)commute $$\tag{2} ...

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