How do we prove that the expectation value of momentum operator in any bound state is zero ?
$$ \langle \hat{P} \rangle_{\text{bound state}}=0 $$
and what about the position expectation value ?
How do we prove it for a general case. How do we get a physical intuition from the corresponding wavefunctions ?
I think for 1-dimensional bound states, this is the proof :
The expectation value of the momentum operator,
$\langle \hat{P} \rangle$=$\langle \psi|\hat{P} \psi\rangle$=$\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)dx$=$\frac{\hbar}{i}\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\partial}{\partial x}\psi(x)dx$=$\frac{\hbar}{2i}\int_{-\infty}^{+\infty}\frac{\partial}{\partial x}({\psi}^2(x))dx$ =$\frac{\hbar}{2i}{\psi}^2(x) \Big|_{-\infty}^{+\infty}$= 0
Since $\psi(x)$=0 at x=$\pm \infty$ for bound states.
and similarly for position expectation value in the bound state,
$\langle \hat{x} \rangle$ = 0