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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.

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As far as I can see you answered your own question. The inequality is not strict in the case of periodic boundary conditions. It is strict on the real line. Because in the first case the constant function is normalizable and in the second it is not. –  BebopButUnsteady Oct 24 '11 at 16:55
    
For a free particle it can be made infinitesimally small anyway. –  Vladimir Kalitvianski Oct 24 '11 at 18:23
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up vote 6 down vote accepted

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.

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