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As Lubos Motl mentions in a comment, for all practical purposes, OP's sought-for eq. (1) 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 ...

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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 ...

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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 $$\tag{2} ... 8 You need to be a bit careful about the counting of supercharges. In four dimensions, the smallest spinor representation is four-dimensional (over the real numbers), so \mathcal{N}=1 supersymmetry has four supercharges. When there are more supercharges, there are simply more states combined into a single 'multiplet'. For example, in \mathcal{N}=1 ... 6 A "multiplet" refers to an irreducible representation of some Lie group/algebra. If it is the representation of some internal symmetry group, you get a multiplet of particles eg: the meson octet or the baryon decuplet are irreducible representations of SU(3) flavour symmetry. What that means is that under SU(3) flavour transformations, these ... 4 Now, I don't know what the word rigorous means, but here is a straight off the bat naive answer. Given$$ H = \int d^3 x \, \bar{\Psi}(i \gamma_i \partial_i +m)\Psi $$from$$ \mathcal{L} = i\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi - \bar{\Psi} m \Psi \quad \text{and}\quad H = \int d^3x \, (\pi \dot{\Psi}-\mathcal{L}) $$with (+,-,-,-). Let's use the ... 4 If the integral I:=\int d\theta on the algebra {\cal A}  of superfunctions f(\theta)=\theta a + b should be 1) a (graded) linear operation, 2) translation invariant, i.e., \int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta), 3) and if the output \int d\theta~ f(\theta) should not depend on the integration variable \theta, then it is ... 4 I think that this Wikipedia article will tells this all. The only problem is that for n (I mean \theta_1,\theta_2,...\theta_n) Grassmann numbers you will need to use 2^n\times 2^n matrices. 4 TL;DR: It is the wedge product \wedge and the exterior derivative/differential d (which squares to zero d^2=0) that give rise to Grassmann-odd elements and supersymmetry. More concretely, Ref. 1 first writes down a (non-relativistic) SUSY algebra {\cal A}$$\tag{10} \{Q_1,Q_1\}_+~=~2H~=~\{Q_2,Q_2\}_+, \qquad \{Q_1,Q_2\}_+~=~0, $$spanned by two ... 3 Formally, the properties "bosonic" and "fermionic" are captured by the relevant objects being part of a \mathbb{Z}_2 = \{1,-1\}-graded algebra (a superalgebra). The even degrees (with grading 1) of that algebra are called "bosonic", the odd degrees (with grading -1) are called "fermionic". It follows from the grading that if xy is even for y odd, ... 3 The formula that Ref. 1 uses is$$\tag{*} \exp\left(-\sum_j \eta_j a_j^{\dagger} \right) ~=~ \prod_j\exp\left( - \eta_j a_j^{\dagger} \right) ~=~\prod_j \left(1- \eta_j a_j^{\dagger}\right). $$Ref. 1 correctly applies [the Hermitian conjugate of] eq. (*) to the bra in answer (a) on p. 181. There is no mistake on p. 181. Ref. 1 does not write a ... 3 Given the Dirac Lagrangian density$$ \tag{1} {\cal L}~=~\overline{\psi}(i\sum_{\mu=0}^3\gamma^{\mu}\partial_{\mu}-m)\psi, \qquad \overline{\psi}~:=~\psi^{\dagger}\gamma^0, \qquad \{\gamma^{\mu},\gamma^{\nu}\}_{+} ~=~2\eta^{\mu\nu}{\bf 1}_{4\times 4}, $$with Minkowski signature (+,-,-,-), and \psi is a Grassmann-odd Dirac-spinor, the question is How ... 3 The idea behind the notation is that the operator F is supposed to count the number of fermions in an expression, i.e.$$[F,A_n]= n A_n$$if the operator A_n contains n fermions, what that means. Then$$[f(F),A_n]= f(n) A_n$$for a sufficiently well-behaved function f:\mathbb{C}\to \mathbb{C}. In particular, for f(x)=(-1)^x , one has ... 3 The bosonic subalgebra of F(4) is SO(5,2)\times SU(2). This certainly can be embedded into the bosonic subalgebra of OSp(8|4) which is SO(6,2)\times SO(5). The tricky bit is indeed to figure out the fermionic generators. 3 In my opinion, one cannot rigorously [*] define the bracket. Suppose that you use the Dirac field equation for arriving to the ordinary Lagrangian density$$ \mathcal{L} = c \bar{\psi} \left( i\hbar \gamma^\mu \frac{\partial}{\partial x^\mu} \right) \psi $$This is a function of the spinor components \psi_i and their adjoints \bar{\psi_i}. The problem ... 3 So I asked Jeff Harvey about this, and this is what he told me. "I suppose one could figure it out by first looking at whether there is an embedding of the bosonic subalgebra and if that works going on to look at the fermionic generators and checking if one can piece the two parts of the embedding together." 2 Apparently the answer to the question is "no", as shown in Appendix C.4.1 of this paper http://arxiv.org/pdf/0810.1484.pdf 2 It is possible to construct a Hamiltonian. In fact this is the way how Dirac initially wrote his equation. For that the time and space coordinates have to be treated differently. In the Schrödinger picture, the Hamiltonian generates the time dynamics via (\hbar =0)$$i \partial_t \psi = H \psi.$$We see that we can obtain this structure from the Dirac ... 2 Let's start from$$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$with \{f, f^\dagger\}=1, \{f, f\} = 0 and define fermionic position and momentum coordinates by$$ \psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\ \psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right) $$with the following anticommutation relations:$$ ...

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You're just not parsing the sentence correctly, or rather, you're just being too nitpicky. When one says that supersymmetry "transforms bosons into fermions and vice versa", one means that, under an infinitesimal transformation, the variation is purely fermionic for bosons, i.e. $$B\mapsto B+\delta B = B+\epsilon F$$ for some bosonic $B$ and some fermionic ...

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No measuring device in an experiment is going to measure a Grassmann-odd number, if that's what OP means by a physical meaning. A measuring device can only produce real outputs $\subseteq\mathbb{R}$. See also e.g. this Phys.SE post.

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Comments to the question (v3): A Grassmann-odd number is not a complex number. It is a complex supernumber $z=x+iy$, which can be decomposed in real and imaginary supernumbers, cf. e.g. this and this Phys.SE posts. The Berezin integral $\int\! d\theta~f(\theta)$ over supernumbers is an ordinary complex number $c=a+ib\in\mathbb{C}$, which can be ...

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The $\epsilon$ parametrizes the amount of symmetry transformation. Suppose you don't want any symmetry transformation in the field configuration. So what we want to do is to take the field congiguration and take it to itself. This is the case when $\epsilon = 0.$ So going by your definition, we will get $F = 0$ and conversely $B = 0$, which is not correct. ...

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Comments to the question (v3): On one hand, traditionally, the Batalin-Vilkovisky (BV) operator $\Delta$ in Lagrangian BRST formulation encodes geometric data of the antisymplectic phase space for the model, specifically the antisymplectic structure [i.e. the so-called antibracket $(\cdot,\cdot)$, or odd Poisson bracket] and a path integral volume density ...

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The bracket you have written is of the form $$[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu).$$ As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a ...

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Assuming that $X=X^\dagger$, $P=P^\dagger$ and $[X,P] = i\hbar$, let me try $$f = \sqrt{\frac{m\omega}{2\hbar}}\left( \alpha X + \frac{\beta}{m\ \omega } P \right)$$ where $\alpha$ and $\beta$ are complex numbers of modulus one. From this follows that  \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + ...

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They shouldn't be thought of as operators i.e. $q$-numbers; instead, they should be thought of as $c$-numbers. They're mutually anticommuting but otherwise they play exactly the same role as $\Delta x^\mu$ for translations or angles $\varphi$ for rotations. They're spinor variables which means that under a Lorentz transformation $\Lambda\in SO(3,1)$, they ...

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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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The (super-)algebra you are referring to is called $\mathfrak{osp}(1,2)$, where osp stands for orthosymplectic. I am not sure about the matrix representation, but a google search about "orthosymplectic superalgebra" will give you plenty of references.

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