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.


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

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.

  • $\begingroup$ 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). $\endgroup$
    – gented
    Apr 23, 2019 at 10:27

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

  • $\begingroup$ This seems correct, however it is the same approach I propose in the question. I am interested in faster ways to do it if there are any. Also watch out for that third entry in the commutator! $\endgroup$
    – TeneT
    Apr 23, 2019 at 13:25
  • $\begingroup$ Sorry, I didn't you were looking for another approach; you may want to make that more clear in your question. Thank you for pointing that out! I just edited it. $\endgroup$
    – Zachary
    Apr 23, 2019 at 13:31

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