Expected value inequality

Question

Why is $\langle p^2\rangle >0$ where $p=-i\hbar{d\over dx}$, (noting the strict inequality) for all normalized wavefunctions? I would have argued that because we can't have $\psi=$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?

Added: It is clear that $\langle\psi|p^2|\psi\rangle\geq0$. i am just wondering why the inequality is strict. Thanks.

Answer 1

It depends on the domain of $p$. If we take the domain of $p$ to be the Schwartz space on $\mathbb{R}$, then, by symmetry of $p$, $$ \langle p^2\rangle =\langle p\psi |p\psi \rangle =\left\| p\psi \right\| ^2 $$ This is $0$ iff $p\psi =0$ iff $\psi$ is constant. However, the only constant Schwartz function is $0$. Hence, $p^2$ is positive-definite.

This will work for any domain in which the only constant function is the $0$ function. By the way, this is just one manifestation of the fact that, for unbounded operators, the domain is a crucial part of the definition of the operator.