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?