I always thought that a particle is an eigenvector of $P^2=H^2-\boldsymbol P^2$ with an isolated eigenvalue. In other words, a necessary condition for $\varphi$ to be a particle is that
$$
P^2|\varphi\rangle=m^2|\varphi\rangle\tag{1}
$$
and
$$
\mathrm dE(\mu^2)=\delta (\mu^2-m^2)\mathrm d\mu^2+\mathrm d\sigma(\mu^2)\tag{2}
$$

My problem is that the answer [Particle/Pole correspondence in QFT Green's functions](https://physics.stackexchange.com/questions/122725/particle-pole-correspondence-in-qft-greens-functions/302712#302712) seems to suggest that a particle is an isolated eigenvalue of $H$, not $P^2$. An isolated eigenvalue of $H$ cannot be an isolated eigenvalue of $P^2$, because
$$
E(\boldsymbol p)^2=\boldsymbol p^2+m^2\tag{3}
$$
is continuously connected to $E(0)^2=m^2$.

Therefore, I can frame my question as follows: in order for $\varphi$ to be a particle, should it be an isolated eigenvalue of $P^2$, or $H$? or neither?