quantum mechanics - Clarification in deriving the radial momentum operator $p_r$

In deriving an expression for $p_r$, a particle's radial momentum, I am unsure what is happening at a certain step. The derivation given in The Physics of Quantum Mechanics by Binney and Skinner is as follows: $$$$ $$p_r=\frac{1}{2}(\hat{\mathbf{r}}\cdot\mathbf{p}+\mathbf{p}\cdot\hat{\mathbf{r}})$$ $$=\frac{1}{2}(\frac{\mathbf{r}}{r}\cdot\mathbf{p}+\mathbf{p}\cdot\frac{\mathbf{r}}{r})$$ because $\hat{\mathbf{r}}=\frac{\mathbf{r}}{r}$. Putting in $\mathbf{p}=-i\hbar\vec{\nabla}$ we get $$p_r=\frac{1}{2}(\frac{\mathbf{r}}{r}\cdot-i\hbar\vec{\nabla}+-i\hbar\vec{\nabla}\cdot\frac{\mathbf{r}}{r})$$ or $$p_r=-\frac{i\hbar}{2}(\frac{1}{r}\mathbf{r}\cdot\vec{\nabla}+\vec{\nabla}\cdot\mathbf{r}\frac{1}{r})$$ Now here is the part that confuses me: Because $r\frac{\partial}{\partial r}=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}=\mathbf{r}\cdot\vec{\nabla}$ and $\vec{\nabla}\cdot\mathbf{r}=3$ we can say $$p_r=-\frac{i\hbar}{2}(\frac{\partial}{\partial r}+\frac{3}{r}-\frac{r}{r^2}+\frac{\partial}{\partial r})$$ I can clearly see where the first two terms of that last equation come from, but I don't see where the $-\frac{r}{r^2}+\frac{\partial}{\partial r}$ comes into play.

The only step after that last equation is to simplify and you get $$p_r=-i\hbar(\frac{\partial}{\partial r}+\frac{1}{r})$$ which I know is correct. Could someone please clarify that middle step?

Answer

To show that the operator $-\frac{i\hbar}{2}(\frac{1}{r}\mathbf{r}\cdot\vec{\nabla}+\vec{\nabla}\cdot\mathbf{r}\frac{1}{r})$ equals the operator $-\frac{i\hbar}{2}(\frac{\partial}{\partial r}+\frac{3}{r}-\frac{r}{r^2}+\frac{\partial}{\partial r})$ you first note they are functions so you have to show that the send the same vectors in the Hilbert Space to the same vectors in the Hilbert Space.

So let $|A\rangle$ be an arbitrary wavefunction (in the domain of both operators) and show the two operators send $|A\rangle$ to the same wavefubction. Don't forget the product rule, and this is really just what it means to show two operators are the same.

It's like verifying that two matrices are the same by comparing each column to each column.