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

  • $\begingroup$ So for us non-physicists the notation is non-standard and it would help to use mathematical notation. I assume $\langle x\mid y\rangle$ is the $L^2$ inner product? $\endgroup$ Jun 8, 2018 at 15:28
  • $\begingroup$ Is $\overrightarrow{r}$ random? Because if it's uniformly distributed you can argue that there is no preferential direction for it to be and it's equally likely to be opposite of any given direction therefore it would have a net $\langle\overrightarrow{r}\rangle=0$ $\endgroup$
    – Jepsilon
    Jun 8, 2018 at 15:49
  • $\begingroup$ Notice the "weight" $e^{-2\alpha r}$ is invariant under $\vec{r} \mapsto -\vec{r}$. If you cut $V$ into two half space $V_1 + V_2$ by a plane passing through origin and change variable from $\vec{r_3} = -\vec{r_2}$ in $V_2$, you get $$\int_V \vec{r} e^{-2\alpha r} d^3r = \left(\int_{V_1} + \int_{V_2}\right) \vec{r} e^{-2\alpha r} d^3r = \int_{V_1} \vec{r_1} e^{-2\alpha r_1} d^3 r_1 - \int_{V_1} \vec{r_3} e^{-2\alpha r_3} d^3 r_3 = 0 $$ In general $\langle \psi | \vec{r} | \psi \rangle = 0$ whenever $|\psi(\vec{r})|^2$ is invariant under $\vec{r} \mapsto -\vec{r}$. $\endgroup$ Jun 8, 2018 at 15:53
  • $\begingroup$ Please incorporate some text in the titles you choose. The title doesn't have to be all text, just the inclusion of a word or two in normal text is sufficient. I like to right-click on text in a title in order to open the question in a new tab. When no text exists in the title, I have to use back arrows and such. Besides, consider search-ability factors for users who might have the same question as you do. $\endgroup$
    – amWhy
    Jun 9, 2018 at 0:43

2 Answers 2


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

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


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.

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.

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.


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