# Confusion on the proof of Goldstone’s theorem

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 $$$$Q=\int d^D\vec{x}J^0(t,\vec{x}).$$$$ In the next, he gives a state ket $$|s\rangle$$ as $$$$|s\rangle=\int d^D\vec{x}e^{-i\vec{k}\cdot \vec{x}}J^0(t,\vec{x})|\Omega\rangle.$$$$ It can be shown $$|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?

• possible duplicates: physics.stackexchange.com/q/264991/84967, physics.stackexchange.com/q/306426/84967 Oct 16, 2020 at 15:32
• 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. Oct 16, 2020 at 15:33
• @CosmasZachos Tony? which page of his book? Oct 17, 2020 at 6:56
• Tony stands for Anthony. Unfamiliar with that book. Try a mainstream QFT text, like Tong. Oct 18, 2020 at 20:23