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: $$[\hat{r},\hat{p}]$$

  • $\begingroup$ 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

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).

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

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