quantum mechanics - How to construct the radial component of the momentum operator?

I'm having trouble doing it. I know so far that if we have two Hermitian operators $A$ and $B$ that do not commute, and suppose we wish to find the quantum mechanical Hermitian operator for the product $AB$, then

$$\frac{AB+BA}{2}.$$

However, if I have to find an operator equivalent for the radial component of momentum, I am puzzled. It does not come out to be simply

$$\frac{\vec{p}\cdot\frac{\vec{r}}{r}+\frac{\vec{r}}{r}\cdot\vec{p}}{2},$$

where $\vec{r}$ and $\vec{p}$ are the position and the momentum operator, respectively. Where am I wrong in understanding this?

Answer

You would have to use the fact that the momentum operator in position space is $\vec{p} = -i\hbar\vec{\nabla}$ and use the definition of the gradient operator in spherical coordinates:

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

So the radial component of momentum is

$$p_r = -i\hbar\hat{r}\frac{\partial}{\partial r}$$

However: after a bit of investigation prompted by the comments, I found that in practice this is not used very much. It's more useful to have an operator $p_r'$ that satisfies

$$-\frac{\hbar^2}{2m}\nabla^2 R(r) = \frac{p_r'^2}{2m} R(r)$$

This lets you write the radial component of the time-independent Schrödinger equation as

$$\biggl(\frac{p_r'^2}{2m} + V(r)\biggr)R(r) = E R(r)$$

The action of the radial component of the Laplacian in 3D is

$$\nabla^2 R(r) = \frac{1}{r^2}\frac{\partial}{\partial r}\biggl(r^2\frac{\partial R(r)}{\partial r}\biggr)$$

and if you solve for the operator $p'_r$ that satisfies the definition above, you wind up with

$$p'_r = -i\hbar\biggl(\frac{\partial}{\partial r} + \frac{1}{r}\biggr)$$

This is called the "radial momentum operator." Strictly speaking, it is different from the "radial component of the momentum operator," which is, by definition, $p_r$ as I wrote it above, although I wouldn't be surprised to find people mixing up the terminology relatively often.