# Does Schwarz Inequality guarantee that the expectation value for an operator exists and is finite?

For two given square-integrable wave functions $$\phi(x)$$ and $$\psi(x)$$, Schwarz Inequality states that $$|\int_a^bdx\phi^*\psi|\le\sqrt{\int_a^bdx\phi^*\phi\int_a^bdx\psi^*\psi}.$$ This guarantees that the inner product between the functions exists and is finite. What about the expectation value? Does the relation guarantee the same thing when I am working instead with the integral form of expectation value for some operator $$A$$?

I think it does, since we are just taking the inner product between a wave function and that wave function worked on by $$A$$. But is my thought correct?

• What if the operator grows very fast with $x$?
– d_b
Jun 16 at 2:46

It is, however, easy to see that this is true when the observable is a bounded operator, i.e. one in which for any $$\vert\psi\rangle$$ with $$\Vert\psi\Vert<\infty$$, we have $$\Vert A\vert \psi\rangle\Vert <\infty$$. This is simply because \begin{align} \vert \langle \psi\vert A \vert \psi\rangle\vert \leq \big\Vert\psi\big\Vert~ \big\Vert A\vert \psi\rangle \big\Vert, \end{align} and $$\Vert \psi\Vert < \infty$$, $$\Vert A\vert \psi\rangle\Vert <\infty$$, with the latter being guaranteed since $$A$$ is bounded.