There are many posts related to this issue on this site, but I have found none that answer my specific questions about this matter.
I review my understanding of Weinbergs approach. There are probably some misconceptions throughout and I am very thankful if someone can clear up any of them. Also I will ask some more concrete questions.
So we have the full Hamiltonian operators $H$ and the free Hamiltonian $H_0$ and the difference is some potential $V$
$$
H = H_0 + V.\tag{3.1.8}
$$
Also we have corresponding Hilbert spaces $\mathcal H$ and $\mathcal H_0$, where $\mathcal H_0$ is a familiar Fock-space which is somehow embedded in $\mathcal H$, in other words $\mathcal H_0$ is sub-Hilbert-space of $\mathcal H$. A fundamental assumption is that the eigenvalue-spectra $E_{\alpha}$ of $H$ and $H_0$ coincide. Although this is mysterious to me, I can accept this.
Now $\mathcal H_0$ is well-defined. It is spanned by the orthonormal energy eigenstates $\Phi_{\alpha}$, $$H_0 \Phi_{\alpha} = E_{\alpha} \Phi_{\alpha},\tag{3.1.9}$$ that is any $\Phi \in \mathcal H_0$ can be written as
$$
\Phi = \int d\alpha~ g(\alpha) \Phi_{\alpha},
$$
where $g$ is a (momentum-space) wave-function (for the explanantion of the $\alpha$ notation I refer to the book). I presume that we can not write any $\Psi \in \mathcal H$ in this way, otherwise we would simply have $\mathcal H \simeq \mathcal H_0$, which is of course trivial, right?
Now consider the energy eigenstates of $H$. Since its spectrum is the same as $H_0$ we can also label them by $\alpha$. However there seems to be another fundamental assumption. Namely that there exist two sets of eigenstates $\Psi_{\alpha}^{\pm}$, with $$H\Psi_{\alpha}^{\pm} = E_{\alpha} \Psi_{\alpha}^{\pm}\tag{3.1.11},$$ and such that they obey the following identity $$ \lim_{t \rightarrow \mp \infty} \int d\alpha~ e^{-iE_{\alpha} t} g(\alpha) \Psi_{\alpha}^{\pm} = \lim_{t \rightarrow \mp \infty} \int d\alpha~ e^{-iE_{\alpha} t} g(\alpha) \Phi_{\alpha} \tag{3.1.12} $$ for all (smooth?) wave-functions $g(\alpha)$. Weinberg shows that the $\Psi_{\alpha}^{\pm}$ are also orthonormal. This seems to imply that the $\Psi_{\alpha}^{\pm}$ each also span $\mathcal H_0$ or at least two copies $\mathcal H_0^+ \simeq \mathcal H_0^- \simeq \mathcal H_0$ which are in some way embedded in $\mathcal H$.
Q1: Do $\Psi_{\alpha}^+, \Psi_{\alpha}^-, \Phi_{\alpha}$ actually span the same subspace $\mathcal H_0 \subset \mathcal H$? This seems to be necessary in order to expand them in terms of one another which is done repeatedly throughout the chapter. Can this be shown? Is it obvious? Or does it need to be assumed?
I am also confused about the physical interpretation of these definitions. I take the Heisenberg picture viewpoint. Say the quantum system is described be the state $\Psi \in \mathcal H$ as seen from some reference frame. Say the hermitian operator $\mathcal O(t)$ measures some properties related the particle content of a given state. An observer in the far past will measure the following expectation value $$ \langle \mathcal O(-\infty) \rangle = \lim_{t \rightarrow - \infty} (e^{-iHt}\Psi, \mathcal O(0) e^{-iHt}\Psi) $$ whereas an observer in the far future will measure $$ \langle \mathcal O(+\infty) \rangle = \lim_{t \rightarrow + \infty} (e^{-iHt}\Psi, \mathcal O(0) e^{-iHt}\Psi). $$ Then it makes sense to me to define the corresponding in-state $\Psi^-$ and out-state $\Psi^+$ as $$ \Psi^{\pm} \equiv \lim_{t \rightarrow \mp \infty} e^{-i H t} \Psi. $$ Q2: It also makes sense to claim that $\Psi^{\pm} \in \mathcal H_0$. Can this be shown from the previous assumptions? Is it obvious? Or does it need to be assumed?
Then, if my assertion in Q1 is valid, we may expand $\Psi^{\pm}$ in either $\Psi_{\alpha}^+, \Psi_{\alpha}^-$ or $\Phi_{\alpha}$.
Q3: The "correct" basis to choose for the expansion seems to be $\Psi_{\alpha}^+$ for $\Psi^+$ and $\Psi_{\alpha}^-$ for $\Psi^-$. $$ \Psi^{\pm} = \int d\alpha~(\Psi_{\alpha}^{\pm}, \Psi^{\pm}) \Psi_{\alpha}^{\pm}. $$ What I mean by "correct" is that $|(\Psi_{\alpha}^{\pm}, \Psi^{\pm})|^2$ can then be viewed as the probability distribution of measuring for $\mathcal O$ the particle content $\alpha$ at $t \rightarrow \mp \infty$. The reason should have something to do with the interpretation that $\Psi_{\alpha}^{\pm} $ describes a state with particle content $\alpha$ only when $\mathcal O$ is measured at $t \rightarrow \mp \infty$, which should somehow be equivalent to (3.1.12). Is this correct? If yes, how can this be seen from (3.1.12)? If no, how are $\Psi^{\pm}$ to be related to the $\Psi_{\alpha}^{\pm}$? The problem I have in trying to put this on a more rigorous footing is the lack of characterization of the state $\Psi$.
Q4: Weinberg explicitly states (p.109) that the in- and out-states can not be written as the limits of some state $\Psi(t)$ for $t \rightarrow \mp \infty$. However defining $\Psi(t) \equiv e^{-iHt} \Psi$ does exactly that. This is of course just the familiar Schrödinger picture formulation. Is this viewpoint not valid? Or does Weinberg mean something different here?
