quantum field theory - A question about the energy of turning on and off interaction adiabatically in QFT

I read a saying as follows:

In a theory with no particles which decay and no bound states, the turning on and off of the interactions merely serves to limit the effective range of forces. In this case, turning the interactions on and off adiabatically will not significantly affect the evolution of any state, so the initial energy and the final energy as determined by the full Hamiltonian $H$ will equal to that determined by the free Hamiltonian $H_0$, so we find that scattering preserves the $H_0$-energy of the states:

$$\left[H_0,S \right]=0.$$

I didn't prove this saying, but I think for infinite past the state is asymptotically free, for example $|p_1 p_2\rangle$, and because the full Hamiltonian in the infinte past is equal to free Hamiltonian in adiabatically switching off, in the infinte past the energy of full Hamiltonian is equal to free Hamiltonain $E_1+E_2$. While the whole process must be conservation of energy, the energy of full Hamiltonian is equal to free one, $E_1+E_2$, in any time.

While according to Gell-Mann and Low theorem, the energy of free Hamiltonian $H_0$ is different from the full Hamiltonian $H$:

$$E = E_0 + \langle\Psi_0 | H_\epsilon-H_0 | \Psi_\epsilon\rangle$$

This also seems to be right. Because in adiabatic process, energy crossing will not occur, that is, some energy level $E_n(t_1)$ of $H(t_1)=H_0+H_I(t_1)$ will become corresponding energy level $E_n(t_2)$ of $H(t_2)=H_0+H_I(t_2)$. Therefore the energy will different in different time. But it also seems to be weird, since energy is not conserved in interacting QFT.

How to resolve this feud?