I amd reading a proof Goldstone’s theorem in Zee's QFT book. On page $228$, Zee presents the proof as follows. The conserved charge $Q$ is given by \begin{equation} Q=\int d^D\vec{x}J^0(t,\vec{x}). \end{equation} In the next, he gives a state ket $|s\rangle$ as \begin{equation} |s\rangle=\int d^D\vec{x}e^{-i\vec{k}\cdot \vec{x}}J^0(t,\vec{x})|\Omega\rangle. \end{equation} It can be shown $|s\rangle$ is the eigenstate of momentum operator $P^i$(see the footnote On page 228). Finally Zee concludes that when we set $k$ to be $0$, $|s\rangle$ has zero energy and can be interpreted as particle(massless).
Here is my confusion: when can we interpret a ket $|\cdot \rangle$ as particle in QFT? Can we pick arbitary eigenstate of momentum operator $P^i$ and regard it as particle? Is there any standard definition for this point?
