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Grassmann-odd variables provide a classical description of Grassmann-odd quantum operators in the same way that Grassmann-even variables provide a classical description of Grassmann-even quantum operators. The classical super-Poisson bracket $$\{\psi^i,\psi^j\}_{PB} ~=~ -i (T^{-1})^{ij} \tag{A}$$ is related to the super-commutator$^1$ $$\hat{\psi}^i\hat{\... 3 Yes. OP is right. There is a minus. Since by convention the complex conjugation obeys$$ (z w)^{\ast} ~=~ w^{\ast}z^{\ast}~=~(-1)^{|z|~|w|} z^{\ast}w^{\ast} \tag{1}$$for any two supernumbers z, w (of definite Grassmann parities |z|,|w|), we should also have$$ (A f)^{\ast} ~=~(-1)^{|A| ~|f|} A^{\ast}f^{\ast} \tag{2} for an operator $A$ and ...

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$$b_{j} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi pj/n}\beta_{p}\qquad b_{j+1} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi q(j+1)/n}\beta_{q}$$ Then b_{j}^{\dagger}b_{j+1}^{\dagger} = \frac{1}{n}\sum_{p,q}e^{\pi i(p+q)j/n}e^{\pi iq/n}\beta_{q}^{\dagger}\beta_{p}^{\dagger} = \sum_{p} e^{-\pi ip/n}\beta_{-p}^{\dagger}\beta_{p}^{...

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It is equivalent to ask the operators on different sites to commute or anticommute. Namely, there is always a so-called Klein transformation changing the commutation between different sites. If they anticommute one says they have natural commutation relations.

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On the mere level of "second quantization" there is nothing wrong with fermionic operators commuting with other fermionic operators. They don't "know" that they are operators for "the same fermion" on different sites, so they could as well commute. But the deeper reason that fermionic operators on different sites anticommute is that they are just modes of ...

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According to the CPT thoerem, the choice of positive parity for particles and negative parity for antiparticles is just as arbitrary as the choice of positive charge for protons and negative charge for electrons.

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1) The reduction they are referring to is explain in the middle of page 13: "We now write (4.5) in the invariant sector as a sum of $n/2$ commuting $2×2$ Hamiltonians that we can diagonalize." 2) They are "reducing" the difficult problem of diagonalizing a very large matrix to the much easier problem of separately diagonalizing $n/2$ different $2 \times 2$ ...

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Use the definition $<v_x> = \int d^3 v v_x (exp(\frac{E}{k_B T})+1)^{-1}/ Z$ with sum of all states $Z$ and $E = \frac{m}{2}(v_x^2+v_y^2+v_z^2)$, introduce spherical coordinates and make the substitution $z = |\vec{v}|/(\sqrt{2 k_B T/m})$. What is if $T=0$?

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