I'm reading the LSZ reduction formula in the wikipedia: https://en.wikipedia.org/wiki/LSZ_reduction_formula
To make the argument simple, let $$\mathcal{L}=\frac{1}{2}(\partial \varphi)^2 - \frac{1}{2}m_0^2 \varphi^2 - \frac{1}{4!}\lambda \varphi^4,$$ where $m_0$ is the bare mass.
Let $m$ be the physical mass and $x = (t, \vec{x})$ for some $t$. Let $\omega_{\vec{k}} := \sqrt{m^2 + \vec{k}^2}$ for each $\vec{k}$ and
$$f_{\vec{k}}(x) = \frac{e^{-ikx}}{(2\pi)^{3/2} (2\omega_{\vec{k}})^{1/2}} $$
Now we define the operator $a_{\vec{k}}(t)$ as the following:
$$a_{\vec{k}}(t) := i \int {d^3 \vec{x} f^*_{\vec{k}} \partial_0 \varphi(x) - \varphi(x) \partial_0 f^*_{\vec{k}} }. $$
Now it seems natural to define $a_{\vec{k}}$ as $\lim_{t\rightarrow -\infty} a_{\vec{k}}(t)$, assuming the limit exists. Also, asymptotic creators and annihilators probably satisfies canonical quantization relations
However, if we differentiate $a_{vec{k}}(t)$ by time $t$, \begin{align*} \frac{d}{dt}a_{\vec{k}}(t) & = i\int {d^3 \vec{x} \big((\partial_0f^*_{\vec{k}} \partial_0 \varphi(x) - \varphi(x) \partial_0^2 f^*_{\vec{k}}) + (f^*_{\vec{k}}\partial_0^2\varphi(x) - \partial_0\varphi(x) \partial_0 f^*_{\vec{k}} )\big) } \\ &= \omega_{\vec{k}} a_{\vec{k}}(t) + i\int{ d^3 \vec{x} f^*_{\vec{k}} \partial^2_0 \varphi(x) - \partial_0 \varphi(x) \partial_0 f^*_{\vec{k}} } \\ &= \omega_{\vec{k}}a_{\vec{k}}(t) + i \int{ d^3 \vec{x} f^*_{\vec{k}} [H, \partial_0 \varphi(x)] - [H, \varphi(x)] \partial_0 f^*_k } \\ &= \omega_{\vec{k}} a_{\vec{k}} (t) + [H, a_{\vec{k}}(t)]. \\ \end{align*} In the last line, I used the fact $f^*_{\vec{k}}$ is just complex-valued, hence $f^*_{\vec{k}}$ and $H$ commute.
Sending $t\rightarrow -\infty$, we get $0 = \omega_{\vec{k}}a_{\vec{k}} + [H, a_{\vec{k}}] $. In other words, $[H,a_{\vec{k}}^\dagger] = \omega_{\vec{k}}a_{\vec{k}}^\dagger$.
This means $a_{\vec{k}}| \Omega \rangle$ is an $H$-eigenstate with eigenvalue bumped by $w_{\vec{k}}$. I believe this is what we call "1-particle state".
However, if we consider 2-particle states, we still get \begin{align*} Ha_{\vec{p}}^\dagger a_{\vec{q}}^\dagger|\Omega\rangle &= [H, a_{\vec{p}}^\dagger] a_{\vec{q}}^\dagger |\Omega\rangle + a_{\vec{p}}^\dagger [H, a_{\vec{q}}^\dagger] | \Omega \rangle + a_{\vec{p}}^\dagger a_{\vec{p}}^+ H| \Omega \rangle \\ &= (\omega_{\vec{p}} + \omega_{\vec{q}} + base) a_{\vec{p}}^\dagger a_{\vec{q}}^\dagger |\Omega\rangle \\ \end{align*} which means $|\vec{p} \vec{q} \rangle$ is another $H$-eigenstate.
This absurd because 2-particles would never scatter; $S$ matrices will be rather trivial.
Where did I make a mistake? Please help me.
