Given a free particle, the ground state of the system is the eigenstate with zero momentum. However, we also know that momentum operator does not have proper eigenstates, but rather it's spectrum is described in terms of projection-valued measure over the real axis. I am trying to understand how these two things can live together from a mathematical point of view. Any idea?
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Actually, the question is derived from a misunderstanding which appears right in the first sentence. The ground state of a quantum system is defined as the (possibly generalized) eigenstate of the Hamiltonian (and not of the momentum operator) for its smallest spectral value, whereas the Hamiltonian by definition is bounded from below (*). The Hamiltonian of a free particle is a positive definite linear self-adjoint operator and its spectrum (which coincides with the spectrum of the tight-rigging extension of it) has the lower bound = $0$. So the PVM from von Neumann's spectral decomposition are built over $[0,\infty)$ (and not the full real line), which is the spectrum of the free-particle's Hamiltonian.
(*) A technical result is here: https://math.stackexchange.com/questions/4124505/does-the-spectrum-of-a-bounded-from-below-self-adjoint-operator-have-a-lower-bou
DanielC
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