# Are interacting and free field operators equal in the Schroedinger picture?

Taking for example a scalar hermitian field (which in the free case would obey to the Klein-Gordon equation), is it true that in the Schroedinger picture the following expression hold true both in a free and in an interacting theory?

$$\hat \phi_S(\vec x)= \int \frac{d\vec k}{2 \omega_k (2\pi)^3} \big( \hat a_S(\vec k) e^{i \vec k \cdot \vec x} +a_S^\dagger (\vec k) e^{-i \vec k \cdot \vec x} \big)$$

where $[\hat a_S(\vec k),a_S^\dagger (\vec{ k }')]= (2\pi)^3 2 \omega_k \delta(\vec k - \vec k')$ while all the other commutators are null.

The Schrodinger-picture field operators $\phi(x)$ shouldn't change when you add interactions. You start with some Hamiltonian like $$H_0=\int d\vec{x} \; \pi^2 + (\nabla \phi)^2 + m^2 \phi^2,$$ and then you add an interation like $$H=H_0+ \lambda \int d\vec{x} \; \phi(x)^4 H=H_0 + V(\phi)$$ using the same field operators $\phi(x)$.
However, in the new theory, $\phi(x)$ acting on the new vauum will no longer create single-particle excitations. And if you choose to express the field operator in terms of ladder operators as you did, those ladder operators will no longer create/destroy particles. For instance, the Kallen-Lehmann spectral density function describes how, in an interacting theory, the Schrodinger field operator acting on the interacting vacuum will create multi-particle states.