Define an asymptotic state in the far past as $$|i\rangle=\sqrt{2\omega_1}\sqrt{2\omega_2}a^\dagger_{p_1}(-\infty)a^\dagger_{p_2}(-\infty)|\Omega\rangle$$$$|i\rangle=\sqrt{2\omega_1}\sqrt{2\omega_2}a^\dagger_{{\vec p}_1}(-\infty)a^\dagger_{{\vec p}_2}(-\infty)|\Omega\rangle$$ where $|\Omega\rangle$ is the ground state of interacting theory and $a_p(t),a^\dagger_p(t)$$a_{\vec p}(t),a^\dagger_{\vec p}(t)$ are time-dependent creation and destruction operators in the expansion $$\phi(\vec{x},t)=\int\frac{d^3{\vec p}}{(2\pi)^3\sqrt{2\omega_{\vec p}}}[a_p(t)e^{-ipx}+a^\dagger_p(t)e^{ipx}],~~px=\vec{p}\cdot{\vec x}-\omega_{\vec p}t.$$$$\phi(\vec{x},t)=\int\frac{d^3{\vec p}}{(2\pi)^3\sqrt{2\omega_{\vec p}}}[a_{\vec p}(t)e^{-ipx}+a^\dagger_{\vec p}(t)e^{ipx}],~~px=\vec{p}\cdot{\vec x}-\omega_{\vec p}t.$$
The state $|i\rangle$ is a time-independent state. It is a momentum eigenstate or not? How do we check that? The problem is that unlike free theory, the commutator $[{\vec P},a_p^\dagger]$$[{\vec P},a_{\vec p}^\dagger]$ is not known.
