The interacting Green function as a path integral with respect to the free ground state

Question

In all the books on qft that include a presentation of path integrals, the interacting Green function is expressed as a path integral with respect to the non-interacting (i.e., free) ground state of the system. Many books just simply write this formula without offering a proof, there are others that try hand-wavingly to justify this formula, and then there is the wonderful book "Quantum Many-particle Systems" by J.W. Negele and H. Orland, that on page 141 attempts to prove in minute details this formula, as their Eq. (3.8).

Free Hamiltonian:

$$H_{0}\left\vert \Phi _{n}\right\rangle =W_{n}\left\vert \Phi _{n}\right\rangle$$

Full Hamiltonian:

$$H\left\vert \Psi _{n}\right\rangle =\left( H_{0}+V\right) \left\vert \Psi _{n}\right\rangle =E_{n}\left\vert \Psi _{n}\right\rangle$$

In Eq. $\left( 3.8\right)$ from page 141 of the book "Quantum Many-particle Systems" by J.W. Negele and H. Orland, there is an unfortunate error, namely

$$\begin{eqnarray*} &&\lim_{T_{0}\rightarrow \infty }\frac{\left( -i\right) ^{n}\zeta ^{P}}{% \left\vert \left\langle \Phi _{0}|\Psi _{0}\right\rangle \right\vert ^{2}e^{-iE_{0}T_{0}}}\sum_{l,m}\left\langle \Phi _{0}|\Psi _{l}\right\rangle \left\langle \Psi _{l}\right\vert e^{-i\left( \frac{1}{2}T_{0}-t_{P\left( 1\right) }\right) H}\widetilde{a}_{\alpha _{P\left( 1\right) }}e^{-i\left( t_{P\left( 1\right) }-t_{P\left( 2\right) }\right) H}\widetilde{a}_{\alpha _{P\left( 2\right) }} \\ &&\times \ldots \widetilde{a}_{\alpha _{P\left( 2n\right) }}e^{-i\left( t_{P\left( 2n\right) }-\frac{1}{2}T_{0}\right) H}\left\vert \Psi _{m}\right\rangle \left\langle \Psi _{m}|\Phi _{0}\right\rangle \end{eqnarray*}$$
is equal to
$$\begin{eqnarray*} &&\lim_{T_{0}\rightarrow \infty }\frac{\left( -i\right) ^{n}\zeta ^{P}}{% \left\vert \left\langle \Phi _{0}|\Psi _{0}\right\rangle \right\vert ^{2}e^{-iE_{0}T_{0}}}\sum_{l,m}\left\langle \Phi _{0}|\Psi _{l}\right\rangle \left\langle \Psi _{m}|\Phi _{0}\right\rangle  \\ &&\times e^{-iT_{0}\frac{\left( E_{l}-E_{m}\right) }{2}}\left\langle \Psi _{l}\right\vert \prod\limits_{i=1}^{2n}\widetilde{a}_{\alpha _{P\left( i\right) }}^{\left( H\right) }\left( t_{P\left( i\right) }\right) \left\vert \Psi _{m}\right\rangle \end{eqnarray*}$$
and not to
$$\begin{eqnarray*} &&\lim_{T_{0}\rightarrow \infty }\frac{\left( -i\right) ^{n}\zeta ^{P}}{% \left\vert \left\langle \Phi _{0}|\Psi _{0}\right\rangle \right\vert ^{2}e^{-iE_{0}T_{0}}}\sum_{l,m}\left\langle \Phi _{0}|\Psi _{l}\right\rangle \left\langle \Psi _{m}|\Phi _{0}\right\rangle  \\ &&\times e^{-iT_{0}\frac{\left( E_{l}+E_{m}\right) }{2}}\left\langle \Psi _{l}\right\vert \prod\limits_{i=1}^{2n}\widetilde{a}_{\alpha _{P\left( i\right) }}^{\left( H\right) }\left( t_{P\left( i\right) }\right) \left\vert \Psi _{m}\right\rangle \end{eqnarray*}$$

One can see that $E_{l}$ and $E_{m}$ enter the formula correctly as $\left( E_{l}-E_{m}\right)$ and not as $\left( E_{l}+E_{m}\right)$.

However, if one uses the correct formula, then one cannot obtain as a final result the interacting Green function

$$\begin{eqnarray*} &&\left( -i\right) ^{n}\left\langle \Psi _{0}\right\vert \mathbf{T}\left[ a_{\alpha _{1}}^{\left( H\right) }\left( t_{1}\right) \cdots a_{\alpha _{n}}^{\left( H\right) }\left( t_{n}\right) a_{\alpha _{n+1}}^{\left( H\right) \dagger }\left( t_{n+1}\right) \cdots a_{\alpha _{2n}}^{\left( H\right) \dagger }\left( t_{2n}\right) \right] \left\vert \Psi _{0}\right\rangle  \\ &=&G_{n}\left( \alpha _{1}t_{1},\ldots ,\alpha _{n}t_{n}|\alpha _{2n}t_{2n},\ldots ,\alpha _{n+1}t_{n+1}\right) \end{eqnarray*}$$

How can one then obtain the interacting Green function for zero temperature as a path integral with respect to the free ground state?