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Consider a flat space QFT, the Lagrangian (in general interacting) has Poincare symmetry, and $\lvert\Omega\rangle$ is the ground state (or just merely no insertion at the far boundaries, from functional integral point of view). There's a "freedom" in defining the Hamiltonian (and usually, it is chosen so that the divergence term of the Hamiltonian that came from the contributions of all oscillation modes can be cancelled), hence the ground state has $P^0 \lvert\Omega\rangle = 0$). But I'm not interested in that story, what I want to see is whether the spatial momenta $\vec{P} \lvert\Omega\rangle$ are always $0$ (well, for a free theory, they certainly vanish) or not.

Is there a proof whether or not the spatial momentum of the vacuum state of an interacting theory vanishes? How can I see it?

Some useful equations:

\begin{align}\vec{P} &= \int d^D \vec{x} \Pi(\vec{x}) \nabla \Phi(\vec{x})\\\Pi(\vec{x}) &= \frac{\partial L[\Phi,\partial \Phi]}{\partial \partial_t \Phi}\\\langle\Omega\rvert \vec{P} \lvert\Omega\rangle &= \int \mathcal{D} \Phi \exp{(i\int dt L)} \vec{P}\end{align}

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This is (at least part of) a million dollar question. The existence of a unique Poincaré-invariant state $\vert0\rangle$ is one of the Wightman axioms for a QFT. Since $P^\mu$ is a vector of generators of the Poincaré group, the invariance of $\vert0\rangle$ implies that $P^\mu\vert0\rangle = 0$ (and vice versa).

The question you really want to ask then is if it is possible to realize a given interacting field theory so that it satisfies the Wightman axioms. For the physically most interesting class of field theories, Yang-Mills gague theories, this one of the steps in solving one of the Millenium Prize Problems.

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  • $\begingroup$ Thank you for your answer. Still, in principle, one can always try to check it, well, by brute-force (say, lattice QFT), the average value of $\langle \vec{P} \rangle$ in the way described above, isn't it? I mean, for the sake of seeing the rotational invariant of the ground state. About the energy, well, one can always try to shift the whole things by redefining the definition of a Halmintonian (don't think the very same trick can be applied for $\vec{P}$, though). $\endgroup$ – user109798 Mar 24 '15 at 18:34

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