We know the following:

$$\boldsymbol\nabla = \hat r \frac{\partial }{\partial r} + \hat \theta \frac1r \frac{\partial }{\partial \theta} + \hat\phi\frac{1}{r\sin\theta} \frac{\partial }{\partial \phi}$$

It's clear that: $ \hat{r}·\vec{p} \; \phi =-i\hbar \frac{\partial \phi}{\partial r}$


$$\begin{align}\vec p\cdot \hat r \Psi & = -~i\hbar \boldsymbol \nabla\cdot (\hat r \Psi)\\&= -~i\hbar ((\boldsymbol \nabla \Psi)\cdot \hat r +(\boldsymbol\nabla \cdot \hat{r})\Psi)\\ &=-~i\hbar \left(\frac{\partial}{\partial r}\Psi +\frac2r \Psi\right)\\ &\ne \hat r\cdot \vec p \cdot \Psi \end{align}$$

My question is, why $\nabla · \hat{r} = \frac{2}{r}$ ?


2 Answers 2


You could explicitly check what $\nabla\cdot \hat{r}$ is in Cartesian coordinates:

$$\partial_x \frac{x}{\sqrt{x^2+y^2+z^2}}+\partial_y \frac{y}{\sqrt{x^2+y^2+z^2}}+\partial_z \frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{2}{r}$$

I suspect that your confusion is arising from naive application of $\nabla$ written as above on the function $\hat{r}$. In curvilinear coordinates, the divergence of a function is given by

$$\nabla\cdot A = \frac{1}{r^2} \partial_{r} (r^2 A_r) + \frac{1}{r\sin\theta }\partial_\theta (\sin\theta A_\theta) +\frac{1}{r\sin\theta}\partial_\phi A_\phi$$

Applying this to $\hat{r}$ gives $2/r$ directly.

Note that this expression is different from what you get by taking a 'dot product of $A$ with $\nabla$'.

You can arrive at the above expression for divergence in many ways, simplest is perhaps to start from the definition of divergence:

$$\nabla\cdot A (\vec{r}) =\lim_{V\to 0} \frac{1}{V} \int_{S(V)} A\cdot\mathrm ds$$

where $V$ is a volume around the point $\vec{r}$, $S(V)$ is the surface of that volume.


This is shown by using $\vec{r} = (x,y,z)$ and $\hat{r} = \frac{\vec{r}}{||r||}$ and then taking the derivatives $\partial /\partial x, \partial /\partial y, \partial /\partial z$. For example, you can find the calculation here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.