In quantum field theory, It’s often said that the interacting annihilation operator (defined by the Klein Gordon inner product between the interacting field and a plane wave) behaves like the free annihilation operator in the limit of asymptotic times. In other words,
$a(k)|k_{W}\rangle = |\Omega\rangle$ as $t \rightarrow \infty$
Where $|k_{W}\rangle$ is a wave packet formed by smearing the eigenstates of the momentum operator.
What’s the justification for this?
Under suitable assumptions, such results are established via Haag-Ruelle-type theorems, see ref. 1 for the mathematically precise statement and proof.
In more down-to-earth terms, this is best taken as a working hypothesis rather than a theorem. The usual way to justify this is by using the common trick of turning-off of coupling constants. Namely, we take the interaction to be $e^{-\epsilon|t|}H_\text{int}$ for some small $\epsilon$. Now, at large times the states become free and the claim trivially follows. See e.g. Gell-Mann and Low theorem for an example of this philosophy.
See also ref 2.
References:
Bogolubov, Anatoly A. Logunov, A.I. Oksak, I. Todorov - General principles of quantum field theory, Part IV (specifically, chapter 12).
Ticciati, R. - Quantum Field Theory for Mathematicians, section 10.4.
The interacting annihilation operator is defined by
$$a(k) = \int d^4x \, e^{ikx} \phi(x)$$
where $\phi(x)$ is the interacting field operator and $k$ is a four-momentum. The Klein-Gordon inner product between $\phi(x)$ and a plane wave $e^{-ikx}$ is given by
$$\langle \phi(x), e^{-ikx} \rangle = i \int_\Sigma d\Sigma^\mu \left( \phi^*(x) \partial_\mu e^{-ikx} - e^{-ikx} \partial_\mu \phi^*(x) \right)$$
where $\Sigma$ is a spacelike hypersurface. The interacting annihilation operator $a(k)$ can be shown to satisfy the commutation relation
$$[a(k), a^\dagger(k')] = (2\pi)^3 2\omega_k \delta^{(3)}(\vec{k} - \vec{k}')$$
where $\omega_k = \sqrt{\vec{k}^2 + m^2}$ and $m$ is the mass of the particle.
The claim that $a(k)|k_W\rangle = |\Omega\rangle$ as $t \to \infty$ means that in the asymptotic future, the interacting state $|k_W\rangle$, which is a wave packet formed by smearing the eigenstates of the momentum operator, becomes indistinguishable from the vacuum state $|\Omega\rangle$. This implies that the particle with momentum $k$ has been annihilated by the interaction.
The justification for this claim is based on the assumption that the interaction Hamiltonian $H_I$ vanishes sufficiently fast as $t \to \pm \infty$. This allows one to define the in and out states as
$$|k_W\rangle_{\pm} = U_0^\dagger(\pm\infty) U(\pm\infty) |kW\rangle$$
where $U(t)$ is the time evolution operator for the full Hamiltonian $H = H_0 + H_I$ and $U_0(t)$ is the time evolution operator for the free Hamiltonian $H_0$. The in and out states are eigenstates of the free Hamiltonian with momentum $k$, i.e.
$$H_0 |k_W\rangle_{\pm} = E_k |k_W\rangle_{\pm}$$
where $E_k = \omega_k$. The asymptotic condition then states that
$$U_0(t) |k_W\rangle_{\pm} - U(t) |k_W\rangle = 0$$
as $t \to \pm \infty$. This means that in the far past or future, the interacting state evolves like a free state with momentum $k$. In particular, as $t \to +\infty$, we have
$$U_0(t) |k_W\rangle_{+} - U(t) |k_W\rangle = 0$$
Using the definition of $|k_W\rangle_{+}$, this implies that
$$U_0(t) U_0^\dagger(+\infty) U(+\infty) |k_W\rangle - U(t) |k_W\rangle = 0$$
Taking the limit as $t \to +\infty$, we get
$$U_0^\dagger(+\infty) U(+\infty) |k_W\rangle - |\Omega\rangle = 0$$
where we have used the fact that $U_0(+\infty)|k_W\rangle = |\Omega\rangle$ since it is a free state with positive energy. Multiplying both sides by $a(k)$, we obtain
$$a(k) U_0^\dagger(+\infty) U(+\infty) |k_W\rangle - a(k)|\Omega\rangle = 0$$
Since $a(k)|\Omega\rangle = 0$, we finally get
$$a(k)|k_W\rangle_{+} = a(k) U_0^\dagger(+\infty) U(+\infty) |k_W\rangle = |\Omega\rangle$$
as desired.
