How to construct the radial component of the momentum operator?

Question

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 1

I was able to figure it out, so here goes the clarification for the record.

Classically $$p_r=\hat{D}_r = \frac{\hat{r}}{r} \cdot\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial r}$$

However $\hat{D}_r$ is not hermitian. Consider the adjoint $$\hat{D}_r^\dagger= \hat{p}\cdot\frac{\hat{r}}{r} =\frac{\hbar}{i} \left ( \frac{\partial}{\partial r}+\frac{2}{r} \right ) $$

Now we know from linear algebra how to construct a hermitian operator from an operator and its adjoint: $$\hat{p}_r = \frac{\hat{D}_r^\dagger+\hat{D}_r}{2}=\frac{\hbar}{i} \left ( \frac{\partial}{\partial r}+\frac{1}{r} \right ) $$

And btw, for those who followed my initial question, don't do the following calculational mistake that I committed:

$$\hat{p}_r = \frac{1}{2}\left(\frac{1}{r}(\vec{r}\cdot\vec{p})+(\vec{p}\cdot\vec{r})\frac{1}{r}\right)\neq \frac{1}{2}\frac{1}{r}(\vec{r}\cdot\vec{p}+\vec{p}\cdot\vec{r})$$

Answer 2

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.

Answer 3

Even in one dimension the operator $p_r=-i\partial_r$ on the half line $r>0$ has deficiency indices $(0,1)$. There is thus no way to define it it as a self-adjoint operator. In practical terms this abstract mathematical statement means that there is no set of boundary conditions thta we can impose on the wavefunction $\psi(r)$ that lead to a complete set of eigenfunctions for $p_r$. For example, integration by parts to prove formal hermiticity requires that $\psi(0)=0$ but all potential eigenfunctions are of the form $\psi_k(r)=\exp\{ikr\}$ for some real or complex $k$, and no value of $k$ can make $\psi_k(0)$ be zero.

Since one needs eigenfunctions and eignvalues to assign a probability to the outcome of a measurement of $p_r$ this means that $p_r$ is not an ``observable.''

Answer 4

1D is indeed not good one for radial operator (and if one uses it then it corresponds to half line with reflection boundary condition at x=0, and then the boundary term vanishes there so again h/i*d/dr is hermitian). But on higher dimensions, e.g. 3, one need to think of inner product with $r^2$ weight function, so it cancels the the boundary term. Of course this gives another term when doing integration by parts, and that is why $d/dr$ alone if not enough for hermiticity.

Of course this corresponds also to the fact that "plane wave" in radial coordinate has some function of $r$ in the denominator, because of energy conservation, e.g. $e^{ikr}/r$ in 3D.

Also see this nice paper, which describes the failure of this opproach in 2D.