How to efficiently compute the commutator $[\hat{r},\nabla^2]$?

Question

Given a system with Hamiltonian $ \hat{H} = \frac {\hat{p} ^2}{2m} + \hat{V}(r)$ in a certain state $|\psi \rangle$, I want to find if $\langle r \rangle$ varies with time.

From

$$ i \hbar\frac {d \langle r \rangle} {dt} = \langle [\hat{r},\hat{H}] \rangle$$

Since $[\hat{r}, \hat{V}(r)] = 0$, we have

$$[\hat{r},\hat{H}] = -\frac {\hbar ^2}{2m}[\hat{r},\nabla^2]$$

• What is the most efficient way to compute $[\hat{r},\nabla^2]$ ?

My approach:

Write $\nabla ^2$ in spherical coordinates, and since $\hat{r}$ commutes with the angular part what remains is (omitting the hats)

$$ \left[r, \frac {1}{r^2} \frac{\partial}{\partial r}\left( r^2\frac{\partial}{\partial r}\right)\right]$$

to be computed with the help of a test function.

Answer 1

Alright, so you're looking for the commutator $$\left[r,\frac{1}{r^2}\partial_r r^2\partial_r\right]$$ Where I let $\partial/\partial r=\partial_r$. Expanding, you get \begin{align} &\frac{1}{r}\partial_r r^2\partial_r - \frac{1}{r^2}\partial_r r^2\partial_rr \\ &= \frac{1}{r}\partial_r r^2\partial_r - \frac{1}{r^2}\partial_r r^3\partial_r-\frac{1}{r^2}\partial_r r^2 \\&=\frac{1}{r}\partial_r r^2\partial_r - \frac{1}{r^2}\partial_r r^3\partial_r-\partial_r - \frac{2}{r} \\&=\frac{1}{r}\left(r^2\partial^2_r+2r\partial_r \right)-\frac{1}{r^2}\left( r^3\partial^2_r+3r^2\partial_r\right)-\partial_r-\frac{2}{r} \\ &=r\partial^2_r+2\partial_r-r\partial^2_r-3\partial_r-\partial_r-\frac{2}{r} \\ &=-2\left(\partial_r+\frac{1}{r}\right) \end{align} So your equation simplifies to $$m\frac{d\langle r \rangle}{dt}=-i\hbar\nabla-\frac{i\hbar}{r} $$ Or $$m\frac{d\langle r \rangle}{dt}=\hat{p}-\frac{i\hbar}{r}$$