As a simple counterpart of switching on interactions in QFT one can consider a harmonic oscillator to which one adds extra terms to the Hamiltonian, e.g. the $x^4$ term. In the latter case, while doing so, not only the new vacuum state is different from the old one -- the same refers to all eigenstates of the Hamiltonian. The counterpart of one-particle states in QFT in the case of the oscillator is the second lowest-energy eigenstate (after the vacuum). The explicit connection between old and new eigenstates in most cases cannot be written out exactly -- only as a series in perturbation theory. Let us now discuss a bit more what the single-particle asymptotic states in the interacting QFT are. As I mentioned before, these are the exact eigenstates of the complete Hamiltonian. Physically, these correspond to a single particle traveling trough an empty universe. It has nothing to interact with, so it has the dynamics of a free particle. It is important to keep in mind that, despite this particle is free in the sense I explained above, it is not the same as a free particle in the free theory. Besides the analogy with the oscillator, for which the eigenstates deform, one can keep in mind the analogy in the classical field theory. For example, let us consider a small sphere of bare mass $m_0$ as a model of the electron. Let us now allow it to have non-vanishing electric charge, which amounts to switching on the interaction with the electro-magnetic field. Then, the sphere creates electric field. The complete energy of a system considered as a whole includes not only the bare mass $m_0$, but also the potential energy of a sphere interacting with its own E-M field $E_U$ and the energy of the electric field $E_{E^2}$ in the space around the sphere. We can look at this system from other reference frames and it gets clear, that the system as a whole behaves as it is a free particle with an effective mass $m=m_0+E_U+E_{E^2}$. Thus, by switching on the interaction with the E-M field, the effective mass of the electron has changed. Experimentally, one always sees the electron, which is dressed with this E-M cloud and has mass $m$, not the bare mass $m_0$ that appears in the Lagrangian. So, any computation, which is set to compute the scattering of electrons -- which is exactly what the S-matrix does -- as the input and the output should feature these dressed electrons, not the bare ones. The story with multi-particle asymptotic states is a bit more tricky. It is clear intuitively that if a particle in the interacting theory is far away from anything else than it evolves as if it was a single particle in the universe. In the same way, we can consider several particles, which are spatially separated from each other and, thus, their interactions can be ignored. Their evolution is then well approximated by a free evolution in interacting theory. One way to phrase it is that the associated multi-particle asymptotic state is a tensor product (in the sense of representations of the Poincare algebra) of single-particle states. This, in particular, means that the total energy for such a state is the sum of energies of single-particle state. Rephrasing this in terms of the Hamiltonian, we have $H_{total}=H_1+\dots + H_n$, that is the total Hamiltonian is the sum of individual Hamiltonians. This highlights the fact that these particles are non-interacting: each evolves with its own Hamiltonian and there are no cross terms. However, this is only true once the particles are spatially separated and one can ignore interactions between them. When particles get closer to each other, interactions are no longer negligible and they undergo the scattering. This is how the scattering is set experimentally and this is what the $S$-matrix computes. No, we go back to your questions. **Doubt 1.** In the scattering setup one does not assume that interaction terms in the action are adiabatically switched off. Instead, one says that when particles are far away from each other, interactions between them can be neglected, so these are approximately free particles. But these are free particles in the interacting theory, which are very different from free particles in the free theory, as I explained above. This is why, the story is not at all about switching off the interactions adiabatically. In the free theory one has, indeed, \begin{equation} |p_1,p_2\rangle_0 = a^\dagger_{p_1} a^\dagger_{p_2}|0\rangle. \end{equation} In interacting theory \begin{equation} |p_1,p_2\rangle \ne a^\dagger_{p_1} a^\dagger_{p_2}|0\rangle, \qquad |p_1,p_2\rangle \ne a^\dagger_{p_1} a^\dagger_{p_2}|\Omega\rangle \end{equation} if $a^\dagger$ is understood as in the free theory. Instead $|p_1,p_2\rangle$ is some complex state made of multi-particle states in the free theory. Given the complexity of this relation, one needs to design procedures like the Kallen-Lehmann representation or the LSZ prescription to see how states like $|p_1,p_2\rangle$ can be accessed. **Doubt 2.** As I explained above, the multi-particle asymptotic states are eigenstates of the complete Hamiltonian only as long as particles are spatially separated and interactions between them can be ignored. Thus, strictly speaking, they are exact eigenstates of the complete Hamiltonian only for one-particle asymptotic state. For more than one particle, asymptotic states are only approximate eigenstates of the complete Hamiltonian and when particles get closer to each other, they do scatter.