Does the commutator of anything with itself not vanish? In a quantum mechanics exam one question was to write the commutator of a couple of operators. Everybody got points taken away since they did not write $[Q_i, Q_i] = 0$ for all the operators $Q_i$ in question. They said that they had to require this since there is something in QFT which will not make those commutators vanish.
What are they talking about? Can anything not commute with itself?
 A: 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
$$\tag{2}  [t,t]_{SC}~=~0, \qquad [t,\theta]_{SC}~=~0, \qquad [\theta,\theta]_{SC}~=~2\theta^2~=~0.$$
In eq. (2) the bracket $[\cdot,\cdot]_{SC}$ denotes the super-commutator
$$\tag{3} [A,B]_{SC}~:=~ AB-(-1)^{|A|~|B|}BA. $$
The supercommutator is the appropriate$^1$ generalization of the notion of a commutator to superalgebras. 
The super-commutator of $D$ operator (1) with itself is not zero:
$$\tag{4}  [D,D]_{SC}~=~ 2D^2~=~2\frac{d}{dt}\neq 0 .$$
II) More generally, the fact that a Grassmann-odd operator (super)commute with itself is a non-trivial condition, which encodes non-trivial information about the theory. This is e.g. used in supersymmetry and in BRST formulations.
On the other hand, the super-commutator of an arbitrary Grassmann-even operator with itself is automatically zero.
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$^1$ One may wonder why one uses the supercommutator $[\cdot,\cdot]_{SC}$ rather than the ordinary commutator 
$$\tag{5} [A,B]_{C}~:=~ AB-BA $$ 
in superalgebras? The commutator (5) satisfies a Jacobi identity, and the supercommutator (3) satisfies a super Jacobi identity, so that's a tie. :)
One physical motivation comes from canonical quantization: As is well-known, quantum mechanically, two Grassmann-graded operators may fail to commute or fail to supercommute. However classically ($\equiv$ when Planck's constant $\hbar$ is zero), for two Grassmann-graded functions $f$ and $g$, one would like that the appropriate bracket generalization $[f,g]$ vanishes.  To ensure this one has to use the supercommutator $[\cdot,\cdot]_{SC}$ rather than the commutator $[\cdot,\cdot]_{C}$. From this perspective, the canonical anticommutator relation (CAR) for fermions is merely a quite natural quantum deformation of a classical supercommuting description. Moreover, the supercommutator (3) [as opposed to the commutator (5)] provides a unified description of CCR for bosons and CAR for fermions. See also e.g. this Phys.SE post and links therein.  
A: 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 defined to be the operator $AB-BA$, and clearly this vanishes for $A,B = Q$.
The only way out of this is something of a cheat -- you have to define new "numbers" with noncommuting properties. But just because mathematicians have invented such concepts, and just because they turn out to apply to physics, doesn't mean they are the same objects as what you are talking about in class.
