Exchanging two fermion fields will create a minus sign, why no delta function $\delta^{3}(\vec{x}-\vec{y})$? The fermion fields are anti-commuting with each other,$$\{\Psi_{\alpha},\Psi_{\beta}\}=(\gamma^{0})_{\alpha\beta}~\delta^{3}(\vec{x}-\vec{y})\tag1$$ I was reading Srednicki's book quantum field theory, in section 40, eq.(40.43)&eq.(40.44), the charge conjugate of $~\bar{\Psi}A\Psi~$ transforms as
$$C^{-1}\bar{\Psi}A\Psi C=\Psi^{T}CAC\bar\Psi^{T} \tag2$$
Since $\Psi^{T}CAC\bar\Psi^{T}$ is a $1\times1$ matrix, we can transpose it without changing its value, so
$$C^{-1}\bar{\Psi}A\Psi C=(\Psi^{T}CAC\bar\Psi^{T})^{T}=-\bar\Psi C^{T}A^{T}C^{T}\Psi,\tag3$$
there is an extra minus sign because we have exchanged the two fermion fields $\Psi$,$\bar\Psi$ here. 
My question is according to the anti-commuting relation $(1)$, we should have
$$\Psi\bar\Psi \sim \delta^{3}(\vec{x}-\vec{y})-\bar\Psi\Psi,$$ rather than $\Psi\bar\Psi \sim -\bar\Psi\Psi,$ so why there is no term$\sim \delta^{3}(\vec{x}-\vec{y})$ in equation $(3)$? I think the fields here are operators rather than the classical fields.
 A: My knowledge of fermionic field theory might need more than just a refresher, but by evaluating the transposition, you are not using the anticommutator, are you? The eigenvalues of the fermionic fields are not complex, but Grassmann numbers. If fermionic fields themselves can be considered as operators/"matrices" over the Grassmann numbers, then an additional minus-sign is needed to exchange the Grassmann-valued-coefficients. However, I really do not remember fermionic field theory very well anymore.
A: You are right in one sense. However if you have a Fermi field $\Psi_a$ it is a pure operator. This acts on a Fock space of ladder states $\{|n\rangle\}$ to give the wave.  For this fermion operator it to have a representation in space you must act on this with $|x\rangle\langle x|$ so the wave vector is
$$
|\psi_a(x)\rangle~=~|x\rangle\langle x|\Psi_a|\sum_n|n\rangle
$$
Peeling off the sum over Fock states we then have a representation of the fermion operator with position space. The commutator $\{\Psi_a(x),~\Psi_b(y)\}$ is then easily seen to involves $\langle x|y\rangle~=~\delta^3(x-y)$. The delta function emerges from representing the operator in space.
