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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} ...


As Lubos Motl mentions in a comment, for all practical purposes, the sought-for equation (v1) is proved via Wick's Theorem. It is interesting to try to generalize Wick's Theorem and to try to minimize the number of assumptions that goes into it. Here we will outline one possible approach. I) Assume that a family $(\hat{A}_i)_{i\in I}$ of operators ...


You need to be a bit careful about the counting of supercharges. In four dimensions, the smallest spinor representation is four-dimensional (over the real numbers), so $\mathcal{N}=1$ supersymmetry has four supercharges. When there are more supercharges, there are simply more states combined into a single 'multiplet'. For example, in $\mathcal{N}=1$ ...


A "multiplet" refers to an irreducible representation of some Lie group/algebra. If it is the representation of some internal symmetry group, you get a multiplet of particles eg: the meson octet or the baryon decuplet are irreducible representations of SU(3) flavour symmetry. What that means is that under SU(3) flavour transformations, these ...


If the integral $I:=\int d\theta$ on the algebra ${\cal A} $ of superfunctions $f(\theta)=\theta a + b$ should be 1) a (graded) linear operation, 2) translation invariant, i.e., $\int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta)$, 3) and if the output $\int d\theta~ f(\theta)$ should not depend on the integration variable $\theta$, then it is ...


I think that this Wikipedia article will tells this all. The only problem is that for $n$ (I mean $\theta_1,\theta_2,...\theta_n$) Grassmann numbers you will need to use $2^n\times 2^n$ matrices.


The bosonic subalgebra of $F(4)$ is $SO(5,2)\times SU(2)$. This certainly can be embedded into the bosonic subalgebra of $OSp(8|4)$ which is $SO(6,2)\times SO(5)$. The tricky bit is indeed to figure out the fermionic generators.


Grassman $d\theta$ has opposite mass dimension to $\theta$, which is why the notation is not 100% optimal, it confuses on this issue. But if you know how to evaluate the integral, that it goes like the derivative, then you know how change of scale works, and it's the opposite of normal change of scale: $$\int d(k\theta) f(k\theta) = {1\over k} \int d\theta ...


So I asked Jeff Harvey about this, and this is what he told me. "I suppose one could figure it out by first looking at whether there is an embedding of the bosonic subalgebra and if that works going on to look at the fermionic generators and checking if one can piece the two parts of the embedding together."


Apparently the answer to the question is "no", as shown in Appendix C.4.1 of this paper http://arxiv.org/pdf/0810.1484.pdf


Here I will just make a couple of general remarks. 1) Graded algebras usually refer to $\mathbb{Z}$- or $\mathbb{N}$-graded algebras, while superalgebras are $\mathbb{Z}_2$-graded algebras. 2) Grassmann numbers are oddly graded supernumbers. Please click on the links for further information, important properties and references. References: 1) Bryce ...

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