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Yes, it is possible to represent the fermionic operators as matrices with the caveat that the fermionic Fock space of states is a super vector space, and the matrices are super matrices. If we have two creation operators $\hat{c}^{\dagger}_{\sigma}$, $\sigma\in \{\uparrow,\downarrow\}$, then there are: 2 bosonic states (1 vacuum state ...


3

There is no way to represent Grassmann variables using matrices ! Actually, this is the big obstacle that hinders the use of the so-called quantum state diffusion approach for systems placed in Fermionic baths. You may find many papers on this by googling this topic. Also, individual Grassmann variable has no physical meaning. It is something invented mostly ...


9

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 ...


8

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} ...


1

First, the second equation starting with $S^a\propto\dots $ should probably say $S^b\propto\dots$. Now, the first two equations for the operators $S^a$ and $S^b$ which are the relevant parts of $S=S^a+S^b$ have the positive plus sign – the additional factors that are omitted don't differ by any extra sign because there is a well-defined factor (and sign) ...



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