Dirac image and free fields in QFT
In QFT, when we take the scattering matrix $S$ and work it out to compute amplitudes, it is usually said that $S$ is written in the interaction (Dirac) image such that the fields $\phi_I$ that $S$ depend on evolve in time in following way:
$$ \phi_I(\vec{x}, t) = U_0^\dagger (t, t_0)\phi(\vec{x}, t_0) U_0(t, t_0) \tag1 $$
and therefore, this fields in Dirac image are free fields. This is good for us because once this is considered we can use all our knowledge about the free theory to compute amplitude in scattering (i.e., interacting) processes.
$U_0$ is the time evolution operator built with only the free Hamiltonian, that can be time dependent in <PERSON> picture.
Nevertheless my problem is the following. In principle, to start with a Lagrangian that can be time-dependent in <PERSON> picture and then, in order to use Dirac image you rotate the states in such a way
$$ |\psi (t)\rangle_I = U_0^\dagger (t, t_0) U(t, t_0) |\psi(t_0)\rangle $$
and the operators as
$$ \phi_I(\vec{x}, t) = U_0^\dagger (t, t_0)\phi(\vec{x}, t) U_0(t, t_0) \tag2 $$
Notice the difference with respect to Eq. (1): there's $t$-dependence in the field $\phi$ because Lagrangian and Hamitonians can be time dependent in <PERSON> picture.
However, in QFT books you always see Eq. (1) as the expression for the fields in Dirac image and since $\phi(\vec{x}, t_0)$ is time-independent, then that expression for the Dirac image field represents a free field (evolves only with free Hamiltonian). So, what am I missing? We shouldn't assume that Eq. 281
(1) is the correct one and therefore using free theory after writing $S$ in terms of Eq. (1) fields, because the correct Dirac image is Eq. (2) that due to the time dependence of $\phi$ does not evolve in time only via free Hamiltonian. This makes the Eq. (2)'s $\phi_I$ not equivalent to a free field.
The only solution I find is to consider that the interaction time scale is so small that the interaction Hamiltonian $H_{int}$ (from which the fields $\phi$ where coming from) is almost constant:
$$ H_{int}(t) = f[\phi(t)], {\rm\ for\ only\ } t\in (t_0 - \delta t, t_0 + \delta t) {\rm \ with\ } \delta t \rightarrow 0^{+} $$
$f$ is some functional of the fields $\phi(\vec{x}, t)$; and since there's an integral in $\vec{x}$ that erases the dependence on this variable, we'll omit it when talking about the Hamiltonian. Thereby, we can simply take
$$ H_{int}(t) = H_{int}(t_0) = f[\phi(t_0)], $$
Then this Hamiltonian in Dirac image is
$$ H_{int, I}(t) = U_0^\dagger (t, t_0)H_{int}(t_0) U_0(t, t_0) = f[U_0^\dagger (t, t_0)\phi(t_0) U_0(t, t_0)] = f[\phi_I(t)] $$
And now the fields $\phi_I(t)$, out of which $H_{int, I}$ is made of, are free fields indeed just by comparing with Eq. (1).
What do you think? 281