Is $|irangle=sqrt{2omega_1}sqrt{2omega_2}a^dagger_{p_1}(-infty)a^dagger_{p_2}(-infty)|Omegarangle$ a momentum eigenstate?

Question

Define an asymptotic state in the far past as $$|irangle=sqrt{2omega_1}sqrt{2omega_2}a^dagger_{{vec p}_1}(-infty)a^dagger_{{vec p}_2}(-infty)|Omegarangle$$ where $|Omegarangle$ is the ground state of interacting theory and $a_{vec p}(t),a^dagger_{vec p}(t)$ are time-dependent creation and destruction operators in the expansion $$phi(vec{x},t)=intfrac{d^3{vec p}}{(2pi)^3sqrt{2omega_{vec p}}}[a_{vec p}(t)e^{-ipx}+a^dagger_{vec p}(t)e^{ipx}],~~px=vec{p}cdot{vec x}-omega_{vec p}t.$$
The state $|irangle$ 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 $[{vec P},a_{vec p}^dagger]$ is not known.

AccidentalFourierTransform · Accepted Answer

The commutator $[vec P,a^dagger_{vec p}]$ is known. Recall that (cf. this PSE post)
$$
a_{vec p}=int e^{ipx}(omega_{vec p}phi(x)+ipi(x))mathrm dvec x
$$
and therefore
$$
begin{aligned}
{}[vec P,a_{vec p}]&=int e^{ipx}(omega_{vec p}[vec P,phi(x)]+i[vec P,pi(x)])mathrm dvec x
&overset{mathrm A}=iint e^{ipx}partial_x(omega_{vec p}phi(x)+ipi(x))mathrm dvec x
&overset{mathrm B}=-i(-ivec p)int e^{ipx}(omega_{vec p}phi(x)+ipi(x)mathrm )dvec x
&=-vec pa_{vec p}
end{aligned}
$$
where I have used $[vec P,O(x)]=ipartial_x O(x)$ in $mathrm A$, and I have integrated by parts in $mathrm B $.
From this it follows that if a state $|alpharangle$ has momentum $vec k$, then the state $a^dagger_{vec p}|alpharangle$ has momentum $vec k+vec p$:
$$
vec Pa^dagger_{vec p}|alpharangle=([vec P,a^dagger_{vec p}]+a^dagger_{vec p}vec P)|alpharangleequiv (vec p+vec k)a^dagger_{vec p}|alpharangle
$$

Is $|irangle=sqrt{2omega_1}sqrt{2omega_2}a^dagger_{p_1}(-infty)a^dagger_{p_2}(-infty)|Omegarangle$ a momentum eigenstate?

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