I have a normalized energy eigenfunction for the ground state of Hydrogen which is
$$
\Psi(r) = \frac{1}{\sqrt{\pi a_0^3}}\exp\left(-\frac{r}{a_o}\right),
$$
where $a_o$ is the Bohr radius,
I have been told that the expectation value of the electrons momentum, $ \left\langle\hat{\vec{p}}\right\rangle =\vec{0}$. I want to know why and I think I'm missing something obvious because my current understanding is not very great.
I'm thinking it has something to do with the electron being in a bound state and the orbit being spherically symmetric but I still think there should be a positive expectation value for momentum.
Could somebody throw me a hint here? I tried to search for answers but couldn't find.
Best Answer
This is actually not surprising. Even an electron orbit in the Bohr atom has a zero expectation value for the momentum if you average over whole orbits.
Think of what it would mean for there to be a finite expectation value for momentum: you'd see an electron that's escaping from the nucleus!
Perhaps what you want to calculate is the root mean square speed of the electron. In that case, you want to find the expectation value of $\vec{p}^2$, not $\vec{p}$.