# Grassmann numbers in the dual space

I'm reading the section on Grassmann numbers in QFT for the Gifted Amateur and I'm confused by something said therein: First, they define a coherent state for fermions $\rvert \eta \rangle$ as \begin{align} \rvert \eta \rangle &= e^{-\eta \hat{c}^\dagger} \rvert 0 \rangle \\ &= \rvert 0 \rangle - \eta \rvert 1 \rangle \tag{28.12} \end{align} where $c^{\dagger}$ is the fermion creation operator and $\eta$ is a Grassmann number. Also, $$\hat{c}\rvert \eta \rangle = \eta \rvert \eta \rangle.$$ Now here's the part that confuses me:

We can also define a state $\langle\bar{\eta} \lvert \hat{c}^{\dagger} = \langle \bar{\eta} \lvert \bar{\eta}$ where $$\langle \bar{\eta} \lvert = \langle 0 \lvert - \langle 1 \lvert \bar{\eta} = \langle 0 \lvert + \bar{\eta} \langle 1 \lvert. \tag{28.15}$$ Note that $\bar{\eta}$ is not the complex conjugate of $\eta$ and $\langle \bar{\eta} \lvert$ is not the adjoint of $\rvert \eta \rangle$. With these definitions it follows that the value of an inner product is $$\langle \bar{\zeta} \lvert \eta \rangle = e^{\bar{\zeta}\eta}. \tag{28.16}$$

Here the $\eta$ and $\zeta$ are both Grassmann variables. My question is, what exactly is $\bar{\eta}$? This probably isn't some quirky notation of the book, since in chapter 21 of his QM book, Shankar also uses this same notation and also emphasizes that it's not complex conjugation that he's talking about.

I feel as if this fact that this is different from complex conjugation is emphasized just to make it crystal clear that Grassmann numbers aren't real numbers nor complex numbers, so there's no well-defined way to complex conjugate them or take adjoints of their kets. But then in other textbooks like Srednicki's QFT book, I've seen mention of complex conjugates of Grassmann numbers, something that doesn't seem permitted according to QFT for the Gifted Amateur and Shankar. Could somebody clarify the distinction between barred and un-barred Grassmann numbers and states? Is writing the Grassmann inner product in this bra-ket notation just a suggestive notation for defining an inner product?

In Ref. 1 the variables $\eta$ and $\bar{\eta}$ are independent variables. Be aware that other authors may use different notations, such as e.g., $\eta$ and $\eta^{\ast}$.
• $\uparrow$ Yes. – Qmechanic Dec 7 '15 at 19:36