# Expected value inequality

1. Oct 24, 2011

### c299792458

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$p=-i\hbar{d\over dx}$

3. The attempt at a solution
clearly, $\langle \psi|p^2|\psi\rangle = \langle p\psi|p\psi\rangle \geq 0$ since p is Hermitian. But why the strict inequality??

2. Oct 26, 2011

### Thaakisfox

When would that inner product with zero? and what would happen then? ;)

3. Oct 26, 2011

### dextercioby

Well, if there exists $\psi \in D(p)$, so that $\langle \psi,p^2 \psi\rangle$ = 0, then $||p\psi||$ has 0 norm. Which vector has 0 norm ? Is it normalizable ?