No. There would be no direction for the vacuum expectation value of the momentum since all directions in a vacuum are equivalent.


The relation $[A,B]=i\hbar$ for self-adjoint operators $A,B$ with a common dense domain implies (in the physically relevant case where the operators generate a Heisenberg group) by the Stone-von Neumann theorem that both operators $A$ and $B$ have the whole real line as spectrum. Thus such a relation is impossible with $A$ or $B$ being the translation ...


The total momentum at any point will be zero but as there are minute fluctuations in electric and magnetic fields, it can be stated that vacuum has momentum. Do check out the existence of virtual particles. This might help you understand deeper about properties of vacuum


This might be a bit helpful. The uncertainty in the momentum of an electron in an atom is defined as: $(\Delta P )^{2}= \langle P^{2}\rangle-\langle P\rangle^{2}$. An electron bounded by the nucleus, has an average momentum of zero, which means ($\Delta P)^2 = {\langle P^{2}\rangle}$, but again, Feynman's explain, as explained in the comment, was a really ...

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