I'll risk a quick off-topic answer here, since I think it's straightforward QM, not vague "interpretation" stuff. :oldbiggrin:
In QM (e.g., Ballentine p81), for a free particle, ##H = \frac12 \, M\, V\cdot V + E_0##. So in the ground state ##|E_0\rangle## we have ##H|E_0\rangle = E_0...