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## Homework Statement

Why is [itex]\langle p^2\rangle >0[/itex] where [itex]p=-i\hbar{d\over dx}[/itex], (noting the ***strict*** inequality) for all normalized wavefunctions? I would have argued that because we can't have [itex]\psi=[/itex]constant, but then I thought that we can normalize such a wavefunction by using periodic boundary conditions... So I don't how to argue that the inequality should be strict... Is it that otherwise it would be trivial?

## Homework Equations

[itex]p=-i\hbar{d\over dx}[/itex]

## The Attempt at a Solution

clearly, [itex]\langle \psi|p^2|\psi\rangle = \langle p\psi|p\psi\rangle \geq 0[/itex] since p is Hermitian. But why the strict inequality??