I'm considering the following wave function: $$\psi(r)=\sqrt{\frac{\alpha^3}{\pi}}e^{-\alpha r} \,\,\,\,\,\,\, \alpha \in \mathbb{R}$$ Where r is the distance from the origin not the radial vector: $${\langle\psi|\psi\rangle}=1$$ I calculated the mean value of the norm r $${\langle r \rangle}= \frac{3}{2\alpha}$$ I need to demostrate: $${\langle \overrightarrow{r} \rangle}= 0$$ Where $\overrightarrow{r}:\sin(\theta)\cos(\phi)\hat{x}+\sin(\theta)\sin(\phi)\hat{y}+\cos(\theta)\hat{z}$
I don't know to calculate the integral $${\langle\psi|\overrightarrow{r}|\psi\rangle}=\frac{\alpha^3}{\pi}\int_{V}\overrightarrow{r} e^{-2\alpha r}dV$$ I do not know if it is better to go to the momentum space with fourier transform and consider r as an operator