Where does a fermionic coherent state live (which Hilbert space)?

There have been a couple of questions on fermionic coherent states, but I didn't find any that covered the following question:

If I define a coherent fermionic state in the 2-level-system spanned by $|0\rangle$ and $|1\rangle$, I will write it as \begin{align} |\gamma\rangle=e^{a^\dagger\gamma-\overline{\gamma}a}|0\rangle\,, \end{align} where $\gamma$ is a Grassmann variable and $a^\dagger$ and $a$ are fermionic creation and annihilation operators. Such a state has the property that the expectation value of $a^\dagger$ is given by $\gamma$ and the one of $a$ by $\overline{\gamma}$. How can a regular operator have an expectation value that is not a complex number?

To me it seems that we formally extend our Hilbert space to something where vectors cannot just have complex numbers as coefficients, but also polynomials of Grassmann variables. What's the best way to think of this? Can I use such a state to describe a concrete physical state? What happens if the number operator has an expectation value containing Grassmann variables?

• – Qmechanic Jul 24 '16 at 18:06
• I want to point out one thing. Physical quantities will be bilinear with two $a$ or $a\dagger$s. So they will be Grassman even objects. So I don't think there should be a problem in it being a physical state. Number operator too will have two of them. But I'm with you on a regular operator having a non complex number expectation value, I think that part needs a better explanation. – BoundaryGraviton Jul 24 '16 at 18:08
• I agree that this is how one makes physically sense out of it. I don't recall the details for Grassmann even operators - is there a natural identification of Grassmann even objects with real or complex numbers? – LFH Jul 24 '16 at 18:10
• BTW: The reference that Qmechanic added cleared up a lot of things. I assume that one needs to extend the Hilbert space to a super Hilbert space or something like that - where coefficients can be Grassmann-valued. – LFH Jul 24 '16 at 18:18
• For this, you can see the question link shared by QMechanic. – BoundaryGraviton Jul 24 '16 at 18:18