Suppose we want to study many-fermions quantum mechanics on a lattice, so we start with a finite-dimensional Hilbert space $\mathcal{H}$ and go to its Fock space $\mathcal{F}_{\text{fer}}(\mathcal{H})$. If $H$ is the Hamiltonian of the one-particle system in $\mathcal{H}$, let $d\Gamma(H)$ be the Hamiltonian on $\mathcal{F}_{\text{fer}}(\mathcal{H})$. If $A$ is an operator on $\mathcal{F}_{\text{fer}}(\mathcal{H})$, then the basic hypothesis of the theory is that the average value of $A$ is given by: $$\langle A \rangle_{\beta,\mu} := \frac{1}{Z}_{\beta,\mu}\text{Tr} \bigg{(}e^{-\beta (d\Gamma(H)-\mu N)}A\bigg{)} \tag{1}\label{1}$$ where $\beta > 0$ is the inverse temperature, $\mu$ is the chemical potential, $N$ the number operator and $Z$ the partition function.
The above is saying that the equilibrium state is described by the operator $e^{-\beta(d\Gamma(H)-\mu N)}$.
Now, set $K:= d\Gamma(H)-\mu N$. On page 20 of this book, the following formula is displayed: $$(-1)^{p}\langle \Omega, \mathbb{T}\prod_{l=1}^{p}\bar{\varphi}(x_{l},\sigma_{l})\Omega \rangle = \frac{\int \prod_{l=1}^{p}\bar{\psi}_{x_{l},\sigma_{l}}e^{\mathcal{A}(\bar{\psi},\psi)}\prod_{x,\sigma}d\bar{\psi}_{x,\sigma}d\psi_{x,\sigma}}{\int e^{\mathcal{A}(\bar{\psi},\psi)}d\bar{\psi}_{x,\sigma}d\psi_{x,\sigma}} \tag{2}\label{2}$$ where $\mathbb{T}$ denotes the time-ordering operator and $\bar{\varphi}$ on the left hand side can be either a creation operator $\varphi^{\dagger}(x_{l},\sigma_{l})$ or annihilation operator $\varphi(x_{l},\sigma_{l})$, while the $\bar{\psi}$ stands for the associated Grassmann variable. From the text, $\Omega$ is a eigenstate of $K$, but it is not clear if it is the ground state of $H$ as well. Thus, the right hand side of (\ref{2}) is just a path integral representation.
I want to reconcile formulas (\ref{1}) and (\ref{2}). I recognize the right hand side of (\ref{2}) as the mean value of the product $\prod_{l=1}^{p}\bar{\varphi}(x_{l},\sigma_{l})$ written in terms of a path integral. But then, the right hand side of (\ref{2}) is: $$\langle \prod_{l=1}^{p}\bar{\varphi(x_{l},\sigma_{l})}\rangle_{\beta,\mu} \tag{3}\label{3}$$ How can (\ref{3}) be also the average with respect to the eigenstate $\Omega$? The expression on (\ref{3}) carries a sum over all basis states, how can it be that it is written just with respect to $\Omega$ as well? What is wrong with my reasoning?
