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

| cite | improve this answer | |
  • $\begingroup$ 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 '19 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 '19 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.