Dirac image and free fields in QFT

Question

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 Schrödinger picture.

Nevertheless my problem is the following. In principle, to start with a Lagrangian that can be time-dependent in Schrödinger 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 Schrödinger 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. (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.

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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?

Answer 1

The dynamics of a quantum field in the Heisenberg picture is
$\phi (\vec x, t) = S^\dagger (t, t_0) \phi (\vec x, t_0) S(t, t_0)$
where $S(t, t_0)$ is the time-evolution operator (S-matrix) satisfying
$i \partial_t S(t, t_0) = S(t, t_0) H(t)$
and
$H(t) = S^\dagger (t, t_0) H(t_0) S(t, t_0)$

In time-dependent perturbation theory you write the Hamiltonian as
$H(t) = H_0 + V(t)$
where the interaction $V(t)$ is small compared to $H_0$ which is time independent and could be the free Hamiltonian.

If you change to the interaction picture, the fields evolve only with $H_0$:
$\phi_0 (\vec x, t) = e^{i H_0 (t - t_0)} \phi (\vec x) e^{-i H_0 (t - t_0)}$
where $\phi (\vec x)$ is the Schroedinger picture field, which does not change with time.
The free fields are equal to the Schroedinger picture fields and to the Heisenberg picture fields at a reference time $t_0$.

The Heisenberg picture fields are related to the free fields by
$\phi (\vec x, t) = U^\dagger (t, t_0) \phi_0 (\vec x, t) U(t, t_0)$
The operator $U(t, t_0) = e^{i H_0 (t - t_0)} S(t, t_0)$ relates the Heisenberg picture fields to the free fields at the same time $t$. The evolution begins from the time $t_0$ where the fields in the two pictures (and the Schroedinger picture) are equal.