3
$\begingroup$

It is well known that

$$[\hat{x},\hat{p_x}] = [\hat{y}, \hat{p_y}] = [\hat{z}, \hat{p_z}] = i\hbar$$

But what if instead we wanted to know the commutator of the net displacement $\hat{r} = \sqrt{\hat{x}^2+\hat{y}^2+\hat{z}^2}$ and the net momentum $\hat{p}= \sqrt{\hat{p_x}^2+\hat{p_y}^2+\hat{p_z}^2}$ ?

That is, what is the following: $$[\hat{r},\hat{p}]$$

$\endgroup$
2
  • $\begingroup$ Use $[A^2, B^2] = [A^2, B]B + B[A^2,B] = A[A,B]B+[A,B]AB +BA[A,B] +B[A,B]A$. $\endgroup$
    – DanielC
    Apr 20 '20 at 9:06
  • 1
    $\begingroup$ that is a very implicit equation for $[A,B]$ $\endgroup$
    – MystMan
    Apr 20 '20 at 9:19
0
$\begingroup$

It's not trivial to calculate this commutation relation using the fundamental canonical commutation relations, as the square-root is not an analytic function and the standard approach of expanding the functions as a power-series about zero and then taking the commutation relations of the different powers will not work here.

What one can do, is to calculate it explicitly

$$ \langle {\bf{r}}' | \left[\hat{r}, \hat{p}\right] | {\bf{r}} \rangle = (r'-r) \langle {\bf{r}}' | \hat{p} | {\bf{r}} \rangle = (r'-r)\int\!\frac{d^3p}{(2\pi\hbar)^3}||{\bf{p}}|| e^{-\frac{i}{\hbar}{{\bf{p}\cdot({{\bf{r}'-{\bf{r}}})}}}}$$

we are free to choose, in the integration the angle between ${\bf{p}}$ and ${\bf{r}}-{\bf{r}'}$ to be $\theta$, and then we get $$ \langle {\bf{r}}' | \left[\hat{r}, \hat{p}\right] | {\bf{r}} \rangle = \frac{r'-r}{\hbar}\int_0^{\infty} \frac{p^3 dp}{(2\pi\hbar)^2} \int_0^{\pi}\sin\theta d\theta e^{-\frac{i}{\hbar}p(r'-r)\cos\theta} = \frac{i}{(2\pi\hbar)^2}\int_0^{\infty}p^2 \left[e^{-ip(r'-r)/\hbar}-e^{ip(r'-r)/\hbar}\right]dp$$

now we can substitute $p^2 e^{\pm i px/\hbar} = -\hbar^2\partial^2_{x}e^{\pm ipx/\hbar}$ and we get $$ \langle {\bf{r}}' | \left[\hat{r}, \hat{p}\right] | {\bf{r}} \rangle = -\frac{i}{(2\pi)^2}\partial^{2}_{r'-r}\int_0^{\infty}\left[e^{-ip(r'-r)/\hbar}-e^{ip(r'-r)/\hbar}\right]dp$$

This integral does not converge formally, but we can add $-\eta$ to the exponent and take the limit $\eta\to 0$ at the end (anyhow we expect something that will have delta-functions, so we are not intimidated by this non-convergence). Then we get $$ \langle {\bf{r}}' | \left[\hat{r}, \hat{p}\right] | {\bf{r}} \rangle = -\frac{\hbar}{(2\pi)^2}\partial^{2}_{r'-r} \left[\frac{1}{r'-r-i\eta}-\frac{1}{r'-r+i\eta}\right] = \frac{i\hbar}{2\pi}\partial^{2}_{r'-r}\delta(r'-r)$$ where I used the identity $\lim_{\eta\to 0} \eta/(x^2+\eta^2) = \pi\delta(x)$ at the end.

I am probably wrong with the prefactors for up to a sign, and $2\pi$, but at least the dimensions are correct, so I think I got the general correct result. This is not a trivial operator, and certainly not proportional to the identity as $[\hat{x}, \hat{p}]$, since $ \langle {\bf{r}}' | {\bf{r}} \rangle = \delta({\bf{r}}'-{\bf{r}})$ which is not proportional to $\delta(r'-r)$ (there are angle dependencies).

$\endgroup$
1
  • $\begingroup$ why the surprise? most commutators are $\endgroup$
    – user245141
    Apr 20 '20 at 10:24

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.