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?

• It seams that you should ask this question to "them". – Steven Mathey Oct 8 '14 at 10:28
• @StevenMathey I will do that, I just wanted to have a little background for that. – Martin Ueding Oct 8 '14 at 11:47

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.

--

$^1$ 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.

• "Curiouser and curiouser", said Alice! Fascinating. – WetSavannaAnimal Oct 8 '14 at 11:52
• That's indeed fascinating. May I confirm with you, this operator space does not form a Lie algbra, does it? I learned that for Lie algebra, commutator being 0 with itself seems to be a necessary corollary of the Lie algebra axioms? Thanks – Zheng Liu Oct 8 '14 at 11:55
• @Zheng Liu ... for fields of characteristic $\neq 2$, depending on the precise definition of a Lie algebra. Right, the above answer refers instead to Lie superalgebras (as opposed to Lie algebras). – Qmechanic Oct 8 '14 at 11:59
• That sounds a bit like so many other names used in physics: Cartesian tensor, metric tensor, Dirac delta function, dwarf planet. – David Hammen Oct 8 '14 at 12:14
• Interesting, but here $[\cdot,\cdot]$ is an anticommutator. I wonder if it is possible to meaningfully define something using the regular commutator such that $XX-XX\neq 0$ (or $XY-YX \neq 0$ for $X\overset{\scriptscriptstyle\wedge}{=}Y$). – jdm Oct 8 '14 at 15:42

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.

• Linear vector spaces as a rule are non-commuting. The easiest axiom to get rid of for a non-vanishing self-commutator would be the existence of the additive inverse. That said, the heart of your argument is gold - Qmechanic's profs are being pedantic - the Grassmann-odd operator's commutator (non-super!) with itself vanishes trivially. – user121330 Oct 8 '14 at 17:37
• Anyhow, I'd love to up-vote your answer, but like I said, vector spaces are non-commutative as a rule. – user121330 Oct 8 '14 at 20:04
• @user121330: what do you mean? Sure, composition of linear operators doesn't commute in general... otherwise there would be no reason to ever consider the commutator. But how is this in dissonance with this answer? – leftaroundabout Oct 9 '14 at 8:32
• Forgive stupid old me... Perhaps you can explain how non-commuting numbers would allow for a non-self-commuting operator? – user121330 Oct 9 '14 at 14:43