How to show $\frac{\delta}{\delta \psi(x)}$ being a representation of the operator $\Psi(x)^{\dagger}$ for fermionic Schroedinger Functionals? I'm following the book of Brian Hatfield, Quantum Field Theory of particles and strings, page 217, eq. 10.89 and the following.
The author is looking for a representation of the operators $\Psi(x)$ and $\Psi(x)^{\dagger}$ that satisfies
$$\begin{align}
\{\Psi(x), \Psi(y)^{\dagger} \} = \delta(x-y).
\end{align}\tag{1}$$
The author then represents $\Psi(x)$ as a multiplication, acting on a functional $\Phi[\psi]$ by
\begin{align}
(\Psi(x) \Phi)[\psi] = \psi(x)\Phi[\psi].
\end{align}
From a previous chapter the author knows that for functionals on $a$-number (grassman) valued functions, $$\{\psi(x), \frac{\delta}{\delta \psi(y)} \} = \delta(x-y),$$ and thus he simply represents $\Psi(x)^{\dagger}$ as $\frac{\delta}{\delta \psi(x)}$.
Now what I want to show is that $\frac{\delta}{\delta \psi(x)}$ actually is the adjoint operator to $\psi(x)$. The book doesn't do this.
Namely, I want to show that if $\Phi_1$ and $\Phi_2$ are two abitrary functionals, then
\begin{align}
\langle \Psi(x) \Phi_1, \Phi_2 \rangle = \langle \Phi_1, \Psi(x)^{\dagger} \Phi_2(x)\rangle
\end{align}
still holds, if I use the mentioned representations. The first thing that I don't know is how to write down a scalar product for functionals - my most naive guess would be
\begin{align}
\langle \Psi(x) \Phi_1, \Phi_2 \rangle = \int \mathcal{D}\psi\ ( \psi(x) \Phi_1[\psi])^{*} \Phi_2[\psi].
\end{align}
And
\begin{align}
\langle \Phi_1, \Psi(x)^{\dagger} \Phi_2 \rangle = \int \mathcal{D}\psi\ \Phi_1[\psi]^*\ \frac{\delta}{\delta \psi(x)} \Phi_2[\psi].
\end{align}
But even if this is true, I don't know how I can show equality between those two. The functional derivative will generate a $\delta(x-y)$ term, that won't appear in the first scalar product.
So my questions would be: For the Schrödinger Wavefunctional theory, is there even the notion of scalar products and adjoint operators - namely, are Schrödinger functionals and the multiplicative $\psi(x)$ and $\frac{\delta}{\delta \psi(x)}$ a representation of the hilbert space and the field operators in the mathematical sense?
If they are, how do you show that $\frac{\delta}{\delta \psi(x)}$, acting on a functional, is indeed the adjoint operator to $\psi(x)$?
 A: Well, the standard 3-step argument goes as follows:

*

*The fundamental equal-time super-Poisson/Dirac bracket is the inverse supermatrix of the supermatrix for the symplectic two-form, cf. e.g. my Phys.SE answer here.


*The canonical anticommutation relations (CARs) are the fundamental super-Poisson brackets multiplied with $i\hbar$, cf. the QM correspondence principle.


*The Schrödinger representation of the CARs (1) is
$$ \hat{\psi}~=~\psi, \qquad \hat{\psi}^{\dagger}~=~\frac{\delta}{\delta \psi} .$$
A: The Approach to Defining the Inner Product
It is a good idea to consider examples in "introductory" quantum mechanics to help you construct math in advanced one. More clearly, given two states $|\psi_1\rangle$ and $|\psi_2\rangle$ in the Hilbert space $\mathcal{H}$, we have the inner product $\langle \cdot, \cdot \rangle: \mathcal{H} \times \mathcal{H} \rightarrow \mathbb{C}$ given by
$$\langle |\psi_1\rangle, |\psi_2 \rangle \rangle=\langle \psi_1|\psi_2 \rangle$$
Therefore, we expect to have something similar to this, like mapping taking place from two vectors in a vector space to a scalar in a field, when we construct inner product for functionals representing fermions.
Another helpful observation comes from second quantization. Suppose you have a two-site state $|10\rangle$ from the basis $\{|00\rangle,|10\rangle,|01\rangle,|11\rangle\}$, the only state which has nonzero inner product with it in the basis is $|10\rangle$. Therefore, we are motivated to try to find an expression which can locate the locations occupied by the fermions and those occupied by holes. If two states have nonzero inner product, some of their components must have the same locations occupied by fermions and same locations occupied by holes. We therefore know what we are expecting to construct the inner product in this case. Still, things can get quite complicated since Grassmann functions will be introduced.
The Definition of the Inner Product and Showing $\bf{\Psi^{\dagger}(x)={\delta \over \delta \psi(x)}}$
As a result I can think of (perhaps you can improve it by introducing simpler definition further), I put it down in the following. Let $V=\{\Phi[\psi]\}$ and $\overline{V}=\{\Phi[\overline{\psi}]\}$ be two sets of functionals of $\{\psi(x)\}$ and $\{\overline{\psi}(x)\}$, where $\{\overline{\psi}(x)\}$ is a copy of Grassmann functions $\{\psi(x)\}$. And define the map $\overline{(\cdot)}:V \rightarrow \overline {V}$
\begin{align}
\overline{1} & = 1 \\
\overline{\psi(x_1)\psi(x_2)\cdots\psi(x_n)} & = \overline{\psi}(x_n)\overline{\psi}(x_{n-1}) \cdots \overline{\psi}(x_1), \ n \in \mathbb{N} \\ 
\overline{(a\Phi_1[\psi]+b\Phi_2[\psi])} & = a^*\overline{\Phi_1[\psi]}+b^*\overline{\Phi_2[\psi]}, \ a,b \in \mathbb{C}
\end{align}
We are inspired by experience in "introductory" quantum mechanics to define the inner product $\langle \cdot, \cdot \rangle$
$$\langle \Phi_1,\Phi_2 \rangle=\bigg(\Pi_{x}{\bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x)}{\delta \over \delta \overline{\psi}(x)}\bigg)}\overline{\Phi_1[\psi]}\Phi_2[\psi]\bigg)_{\psi,\overline{\psi}=0} \tag{1}$$
where the parenthesis $(\cdot )_{\psi,\overline{\psi}=0}$ means we are only taking constant term, which is the term without any Grassmann function. Therefore, the nonzero contribution of the inner product only comes from terms in $\overline{\Phi_1[\psi]}\Phi_2[\psi]$ where $\overline{\psi}(x)$ and $\psi(x)$ must show up in pair or simultaneously be absent. This matches our experience abut how inner product of two states behave. A point, however, has to be stressed about Eq. (1): it is not mathematically rigorously defined because of the uncountability of $x$ and the appearance of delta function (this is a natural consequence arisen from the definition of derivatives on Grassmann functions), and with different number of fermions the order of the delta function will be different (see Eq. (10.100) in the book). Nevertheless, it works perfectly fine if we are considering lattice field theory where the spacetime is treated discretely and the normalization condition, therefore along with derivatives, do not involve delta function.
The problem to argue $\Psi^{\dagger}(x)={\delta \over \delta \psi(x)}$ now becomes easy. We have
\begin{align}
\langle \Psi(x)\Phi_1,\Phi_2 \rangle & = \bigg(\Pi_{x'}{\bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x')}{\delta \over \delta \overline{\psi}(x')}\bigg)}\overline{\psi(x)\Phi_1[\psi]}\Phi_2[\psi]\bigg)_{\psi,\overline{\psi}=0} \\
& = \bigg(\Pi_{x'}{\bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x')}{\delta \over \delta \overline{\psi}(x')}\bigg)}\overline{\Phi_1[\psi]}\overline{\psi}(x)\Phi_2[\psi]\bigg)_{\psi,\overline{\psi}=0} \\
& =\bigg(\Pi_{x' \neq x}{\bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x')}{\delta \over \delta \overline{\psi}(x')}\bigg)}\overline{\Phi_1[\psi]} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x)}{\delta \over \delta \overline{\psi}(x)}\bigg)\overline{\psi}(x)\Phi_2[\psi]\bigg)_{\psi,\overline{\psi}=0} \\
& = \bigg(\Pi_{x' \neq x}{\bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x')}{\delta \over \delta \overline{\psi}(x')}\bigg)}\overline{\Phi_1[\psi]}{\delta \over \delta \psi(x)}\Phi_2[\psi]\bigg)_{\psi,\overline{\psi}=0} \\
& = \bigg(\Pi_{x'}{\bigg(1+{1 \over \delta(0)}{\delta \over \delta \psi(x')}{\delta \over \delta \overline{\psi}(x')}\bigg)}\overline{\Phi_1[\psi]}{\delta \over \delta \psi(x)}\Phi_2[\psi]\bigg)_{\psi,\overline{\psi}=0} \\
& = \langle 
\Phi_1,{\delta \over \delta \psi(x)}\Phi_2 \rangle \tag{2} 
\end{align}
The second line is from the definition of $\overline{(\cdot)}$. The third and the fourth equalities are true since any term in $\overline{\Phi_1[\psi]}$ with $\overline{\psi}(x)$ and any term in $\Phi_2[\psi]$ without $\psi(x)$ cannot produce nonzero contribution to the inner product. Therefore, we have proved $\Psi^{\dagger}(x)={\delta \over \delta \psi(x)}$.
