Average value of creation and annihilation operators - are these two expressions the same?

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?

You can reconcile the formulas by specifying the boundary conditions of your path integral. For (2), the time boundaries (at $$0,\beta$$) are assumed to be in state $$|\Omega\rangle$$ but to retrieve (1), you only assume periodic boundary solutions. Things usually get clearer once you've actually computed a concrete case (which is doable when your hamiltonian is quadratic since you only have gaussian integrals to calculate).