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

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.

• The Laplacian is not the momentum operator: it is the representation of the momentum operator onto the position basis (as such your initial equation makes no sense, on the left hand side you have an operator, on the right hand side you have a representation). – gented Apr 23 at 10:27

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 $$m\frac{d\langle r \rangle}{dt}=\hat{p}-\frac{i\hbar}{r}$$