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In my previous question, I asked about how to handle Grassmann-number variations of operators. I read a book that uses those variations $\delta \Phi = c \mathbb{1}$, with $c$ being a grassmann number (which, from what I understood, ist meant to be not an operator, but just a grassmann number, and $\mathbb{1}$ being the unity operator of the the hilbert space that all the other operators do act on (which I understood to be a vector space over the complex Numbers). Since all the other operators in the field theory are simply operators over $\mathbb{C}$, I thought that $\delta \Phi$ must be an operator over $\mathbb{C}$ as well. So here is my question:

For a given Field-Operator $\Phi$ (which is an operator operating on a $\mathbb{C}$-vectorspace, can I represent grassmann-number valued variations $(c \mathbb{1})$ (with $c$ being a grassmann variable) of this operator by an other operator, operating solely on a $\mathbb{C}$-vectorspace as well?

If not, how can we vary fermionic fields in the way it is described in my last question? Is adding a $c$-valued operator to a grassmann-number-valued operator even a valid operation?

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It seems that the core of OP's question is essentially resolved by the following facts.

  1. Complex supernumbers furnish a $\mathbb{C}$-vector space.

  2. Operators in supermathematics are not only $\mathbb{C}$-linear but graded linear wrt. supernumbers.

For more information and references, see e.g. this related Phys.SE post.

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  • $\begingroup$ I guess my irritation arrises from the fact that I try to mix the operator formalism and the grassman number formalism up (as the author, Manoukian, that I read, also does). To clear it up a bit: Are fermionic field operators (the usual ones you hear of in your usual Introduction to QFT lecture) $\mathbb{C}$-linear, or are they allready supernumber-linear? $\endgroup$
    – Quantumwhisp
    Commented Nov 8, 2017 at 23:42
  • $\begingroup$ They are already supernumber-linear. $\endgroup$
    – Qmechanic ♦
    Commented Nov 9, 2017 at 0:11
  • $\begingroup$ This is resolving the issues that I had on this topic. At the same time I'm wondering why non of the books on QFT I ever had in my hands mentioned this. $\endgroup$
    – Quantumwhisp
    Commented Nov 9, 2017 at 11:50
  • $\begingroup$ I assume that this is a property that fermionic creation / annihilation operators do have as well? $\endgroup$
    – Quantumwhisp
    Commented Nov 9, 2017 at 22:23
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Commented Nov 9, 2017 at 22:38

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