Show that $ \hat{r}·\vec{p} $ is not hermitian

Question

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}$

But:

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

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

Answer 1

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.

Answer 2

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.