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?

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    $\begingroup$ possible duplicates: physics.stackexchange.com/q/264991/84967, physics.stackexchange.com/q/306426/84967 $\endgroup$ Oct 16, 2020 at 15:32
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    $\begingroup$ Yes, of course. (Incompletely localized) eigenstates of momentum in QFT are particles, speeding chunks, or excitations... Tony doesn't make that clear? Particles move in momentum space, relativistically. $\endgroup$ Oct 16, 2020 at 15:33
  • $\begingroup$ @CosmasZachos Tony? which page of his book? $\endgroup$
    – Sven2009
    Oct 17, 2020 at 6:56
  • $\begingroup$ Tony stands for Anthony. Unfamiliar with that book. Try a mainstream QFT text, like Tong. $\endgroup$ Oct 18, 2020 at 20:23


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