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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Nov 8, 2017 at 23:42
  • $\begingroup$ They are already supernumber-linear. $\endgroup$
    – Qmechanic
    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. $\endgroup$ Nov 9, 2017 at 11:50
  • $\begingroup$ I assume that this is a property that fermionic creation / annihilation operators do have as well? $\endgroup$ Nov 9, 2017 at 22:23
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Nov 9, 2017 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.