Keldysh Formalism and time symmetry

Question

Does the fact that the initial and final states : $|\phi_0\rangle$ and $S(\infty,-\infty) |\phi_0\rangle$ are not the same in a non-equilibrium setting - meaning essentially a time dependent Hamiltonian,(where $|\phi_0\rangle$ is a state of the time-independent hamiltonian part and $S(\infty,-\infty)$ being the S-matrix in the interaction picture) have something to do with the absence of a Gellman-Low like theorem for time dependent perturbations?

Gellman-Low states that both $\frac{U_{\epsilon I}(0,+\infty)|\Phi_0\rangle }{\langle \Phi_o|U_{\epsilon I}(0,+\infty)| \Phi_0 \rangle}$ and $\frac{U_{\epsilon I}(0,-\infty)|\Phi_0\rangle }{\langle \Phi_o|U_{\epsilon I}(0,-\infty)| \Phi_0 \rangle}$, in the limit $\epsilon \rightarrow 0$ are eigenstates of $H=lim_{\epsilon \rightarrow 0} (H_0+e^{-\epsilon |t|}H_1)$, $H_0$ being the non-interacting part, effectively implying their equality in the non-degenerate case.

For finite $\epsilon$, $U_{\epsilon I}(0,\pm\infty)|\Phi_0\rangle$ is an eigenstate of the full Hamiltonian with energy $E$.

Answer 1

I might be reversing your question, but the development of the Feynman-Dyson expansion in Abrikosov, Gorkov and Dzyaloshinski's book explicitly relies the Gel-Mann-Low theorem: by adiabatically turning on and the turning off the perturbation one can assure that the initial and the final states are the same (more precisely they differ by a phase factor), which enables the zero-temperature expansion formalism