How to calculate the integral ${\langle\psi|\overrightarrow{r}|\psi\rangle}=\frac{\alpha^3}{\pi}\int_{V}\overrightarrow{r} e^{-2\alpha r}dV$

Question

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

Answer 1

Stay in position space and use

$$\begin{align} \vec r&=\hat xx+\hat yy+\hat zz\\\\ &=\hat xr\sin(\theta)\cos(\phi)+\hat yr\sin(\theta)\sin(\phi)+\hat zr\cos(\theta) \end{align}$$

Then, note that we have

$$\begin{align} \int_V \vec r e^{-2\alpha r}\,dV&=\int_0^{2\pi}\int_0^{\pi}\int_0^\infty \vec r e^{-2\alpha r}\,r^2\sin(\theta)\,dr\,d\theta\,d\phi\\\\ &=\hat x \int_0^{2\pi}\int_0^{\pi}\int_0^\infty r^2\sin(\theta)\cos(\phi) e^{-2\alpha r}\,r^2\sin(\theta)\,dr\,d\theta\,d\phi\\\\ &+\hat y \int_0^{2\pi}\int_0^{\pi}\int_0^\infty r^2\sin(\theta)\sin(\phi) e^{-2\alpha r}\,r^2\sin(\theta)\,dr\,d\theta\,d\phi\\\\ &+\hat z \int_0^{2\pi}\int_0^{\pi}\int_0^\infty r^2\cos(\theta) e^{-2\alpha r}\,r^2\sin(\theta)\,dr\,d\theta\,d\phi \end{align}$$

Finish by using the fact that

$$\begin{align} \int_0^{2\pi }\cos(\phi)\,d\phi&=0\\\\ \int_0^{2\pi }\sin(\phi)\,d\phi&=0\\\\ \int_0^\pi \cos(\theta)\sin(\theta)\,d\theta&=0 \end{align}$$

Answer 2

Since $\psi$ only depends on the distance to origin, why not use some symmetry?

For the $x$-coordinate of $\langle \vec r \rangle$, we have

$$\langle \psi|x|\psi\rangle = \int\limits_Vx\psi^2dV=\int\limits_{\mathbb R}\int\limits_{\mathbb R}\int\limits_{\mathbb R}x\psi^2dxdydz$$

Now, consider only $\int\limits_{\mathbb R}x\psi^2dx$. Here, $\psi^2$ is an even function, meaning that $\psi^2(x,y,z)=\psi^2(-x,y,z)$, while $x$ is odd, thus the product $x\psi^2$ is also odd, and the integral of an odd function over all space is zero. Thus, $\langle \psi|x|\psi\rangle=0$. The same applies to other 2 coordinates.

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Alternatively, we can use rotational invariance. If $g$ is any rotation, then, since rotations preserve length, we get $\psi(g\cdot \vec r)=\psi(\vec r)$. Now:

$$g \cdot \langle \vec r \rangle = \int\limits_V (g \cdot \vec r) \psi^2(\vec r)dV = \int\limits_V\vec r \psi^2(g^{-1}\cdot \vec r)dV = \int\limits_V \vec r\psi^2(\vec r)dV=\langle\vec r\rangle$$ (we used that for a rotation $g$, its inverse $g^{-1}$ is also a rotation)

Thus, $\langle\vec r\rangle$ is a vector invariant under any rotations, implying it is zero.

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The last argument works also for the transformation $g(\vec r) = -\vec r$ (note that in this case $g^{-1}=g$). We get that $\langle\vec r\rangle$ is invariant under $g$, but this means $\langle\vec r\rangle = -\langle\vec r\rangle$, so $\langle\vec r\rangle=0$.

Thanks to user achille hui for the insight.