$\mathbf{Background:}$ Consider a free scalar field $\phi$ ($\mathcal{L}_0 = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi + \frac{1}{2}m^2 \phi^2$).
In the Hamiltonian viewpoint, this system has a Hilbert space $\mathcal{H}_0$ (the Fock space). We can write down a resolution of the identity in $\mathcal{H}_0$ (in the Schrodinger picture):
$\begin{align} I = |0 \rangle \langle 0 | + \int dp\ |p \rangle \langle p | + \frac{1}{2!} \int dp_1 dp_2\ |p_1 p_2 \rangle \langle p_1 p_2 | +\ \cdots \end{align}$
$\mathbf{Question:}$ If we add an interaction term to the Lagrangian (such as $\lambda \phi^4$), is $I$ still the identity operator in the new Hilbert space $\mathcal{H}$?
(To be clear, I mean exactly the same $I$, with the free vacuum and free particle states.)
$\mathbf{Motivation:}$ The analogous thing holds in QM. A non-relativistic particle moving in a potential $V$ has a Hilbert space $L^2({\mathbb{R}})$, so by Fourier transform, $\int dp\ |p \rangle \langle p |$ is a resolution of the identity. (Where $\langle x | p \rangle = \frac{1}{\sqrt{2\pi}} e^{i p x}$ are free particle states.) The Hilbert space is the same regardless of $V$.
