Straight to the point: what's the result of the commutator of the magnitude of the position and the momentum operators and how can I approach it, i.e., $[|\mathbf{\hat{x}}|,|\mathbf{\hat{p}}|]=$ ?

My efforts: (1) trying to use $|\mathbf{\hat{x}}|=\sqrt{\sum \hat{x}_i\hat{x}_i}$ doesn't seem to help because of the square root; (2) identifying $|\mathbf{\hat{x}}|$ as the radial position operator $\hat{r}$ and using the "radial momentum" operator $\hat{p}_r$ was of no help because $\hat{p}_r \neq |\mathbf{\hat{p}}|$ and $\hat{p}_r$ is the square root of a sum relating $|\mathbf{\hat{p}}|^2$ and the square of the angular momentum operator.

ADDENDUM (Oct 19th, 2020)

I see there may be a number of technical difficulties defining $|\mathbf{\hat{x}}|$ and $|\mathbf{\hat{p}}|$ because these are square roots of operators $\hat{x}_i$ and $\hat{p}_i$. To be clearer, I'm looking for a formal expression for $[|\mathbf{\hat{x}}|,|\mathbf{\hat{p}}|]$ that might (but not necessarily should) overlook technical issues regarding square root of operators. For instance, I wonder if that's the case of the well-known result $[x_i,F(p_x,p_y,p_z)] = i\hbar \frac{\partial F}{\partial p_i}$ if we take $F=|\mathbf{\hat{p}}|$ so that $[x_i,|\mathbf{\hat{p}}|] =i\hbar\frac{p_i}{|\mathbf{\hat{p}}|}$.

  • 4
    $\begingroup$ Out of curiosity, where did such an object come up? $\endgroup$ – jacob1729 Oct 15 '20 at 19:23
  • $\begingroup$ I've been studying some models proposing modifications to the canonical commutation relation for position and momentum and, among several different commutators appearing there, this particular commutator never shows up. I haven't found a physical reason for that and, if there's none indeed, I think I could envisage situations it would appear. $\endgroup$ – andrehgomes Oct 15 '20 at 19:32
  • 1
    $\begingroup$ I would say the most natural approach would be using spectral decompositions of x and p explicitly and change the signs for the part x<0 and p<0. $\endgroup$ – chau Oct 15 '20 at 19:46
  • 3
    $\begingroup$ I'm not sure how this question can have a well-defined answer unless you either a) define what you mean by $\lvert \hat{x}\rvert$ technically or b) define what goal you are pursuing. $\endgroup$ – ACuriousMind Oct 16 '20 at 16:45
  • 2
    $\begingroup$ @andrehgomes the square root of operators is not well defined in general. It is possible to define it in some cases (e.g. bounded operators, which is not the case here), which is why ACuriousMind refers to a "technical" definition. $\endgroup$ – fqq Oct 17 '20 at 14:15

Here are some incomplete untested ideas.

  1. Define operators $$\begin{align}J_-~:=&~\frac{1}{2}{\bf x}^2,\qquad J_+~:=~\frac{1}{2}{\bf p}^2,\cr J_z~:=&~\frac{1}{4}\sum_{j=1}^3 \{x^j, p_j\}_+.\end{align}\tag{1}$$ One may prove that they form an $sl(2,\mathbb{R})$ Lie algebra $$ [J_-,J_+]~=~2i\hbar J_z, \qquad [J_z,J_{\pm}]~=~i\hbar J_{\pm}.\tag{2}$$ In this language OP wants to calculate the commutator $2[\sqrt{J_-},\sqrt{J_+}]$.

  2. Define normalization $$\sigma_{\pm}~:=~\frac{J_{\pm}}{\sqrt{2}\hbar}, \qquad \sigma_z~:=~\frac{J_z}{i\hbar}.\tag{3}$$ Then $$ [\sigma_+,\sigma_-]~=~\sigma_z, \qquad [\sigma_z,\sigma_{\pm}]~=~\sigma_{\pm}.\tag{4}$$ The $2\times 2$ Pauli matrices satisfy the same Lie algebra (although the underlying associative algebra is different).

  3. It might be possible to define a notion of square root operators similar to this Phys.SE post.

  4. It might be possible to adapt a representation a la Holstein-Primakoff or Dyson-Maleev, cf. this Phys.SE post, or perhaps some squeezed state methods.


In quantum mechanics, there are different possibilities to express an operator. Let's consider a matrix of the operator $\hat{A} = [|\hat{\bf{x}}|, |\hat{\bf{p}}|]$ in coordinate reperesentation: $$ A(\bf{x},\bf{x'}) = \langle \bf{x} | \hat{A}|\bf{x'}\rangle = (|\bf{x}| - |\bf{x'}|) \langle \bf{x} | |\hat{\bf{p}}||\bf{x'}\rangle $$ According to the spectral representation of $\bf{p}$, we have $$ \langle \bf{x} | |\hat{\bf{p}}||\bf{x'}\rangle = {\rm \frac1{(2\pi\hbar)^3}} \int |\bf{p}| e^{\frac{i}\hbar\bf{p}(\bf{x}-\bf{x'})} \bf{d^3 p} {\rm\ \equiv \frac1{(2\pi\hbar)^3} f(}{\bf x - x'}{\rm )} $$ Further, I shall treat the function $$ f(\bf{R}) = \int |\bf{p}| e^{\frac{i}\hbar\bf{p}\bf{R}} \bf{d^3 p} $$ as a generalized function. Then we have $$ f(\bf{R}) = -\hbar^2\Delta_{\bf{R}} \int \frac1{|\bf{p}|} e^{\frac{i}\hbar\bf{p}\bf{R}} \bf{d^3 p} = {\rm -4\pi\hbar^4} \Delta_{\bf{R}}{\rm \frac1{{\bf R}^2}}\quad {\rm (1)} $$ It is well known that $$ -\Delta \frac1{|\bf{R}|} = {\rm 4\pi\delta(}\bf{R}{\rm )} $$ If analogous local representation exists for the function (1), then it might be possible to express $\hat{A}$ as a simple combination of $\hat{\bf x}$ and $\hat{\bf p}$ operators. Otherwise, I suppose there is no simple expression for $\hat{A}$.

  • $\begingroup$ That's very insightful! This and this may relate to your answer. I see I need extra study to better appreciate the meaning of these ideas. Thanks! $\endgroup$ – andrehgomes Oct 17 '20 at 14:19

I'll provide a tentative solution based on a result derived by M. K. Transtrum and J.-F. S. Van Huele, J. Math. Phys. 46, 063510 (2005). They derived a general expression for the commutator of functions $f(A,B)$ and $g(A,B)$ of noncommuting operators $A$ and $B$:

\begin{equation} \left[f(A,B),g(A,B)\right] = \sum_{k=1}^\infty \frac{(-c)^k}{k!} \left( \frac{\partial^k g}{\partial A^k} \frac{\partial^k f}{\partial B^k} - \frac{\partial^k f}{\partial A^k} \frac{\partial^k g}{\partial B^k}\right), \quad \text{where} \quad c=[A,B]. \end{equation}

Tentative solution

I consider the particular case $f=f(A)$ and $g=g(B)$:

\begin{equation} \left[f(A),g(B)\right] = - \sum_{k=1}^\infty \frac{(-c)^k}{k!} \frac{\partial^k f}{\partial A^k} \frac{\partial^k g}{\partial B^k} = \left[ - \sum_{k=1}^\infty \frac{(-c)^k}{k!} \frac{\partial^k }{\partial A^k} \frac{\partial^k }{\partial B^k} \right] f(A)g(B), \end{equation}

where I believe the last step is not problematic as long we understand its meaning: derivative $\partial_A\equiv\frac{\partial}{\partial_A}$ acts on $f(A)$ and derivative $\partial_B\equiv\frac{\partial}{\partial_B}$ acts on $g(B)$. Finally we simplify the result to

\begin{equation} [f(A),g(B)] = \left( 1-e^{-c\partial_A \partial_B} \right)f(A)g(B) \quad \text{or} \quad g(B)f(A)=e^{-c\partial_A \partial_B}f(A)g(B). \end{equation}

Heading to $[|\mathbf{\hat{x}}|,|\mathbf{\hat{p}}|]$, I'll now ommit the "hats" for simplicity and use the notation $\mathbf{x}=(x,y,z)$ and $\mathbf{p}=(p_x,p_y,p_z)$ for the position and momentum operators respectively. My approach is writing |\mathbf{x}| and |\mathbf{p}| as power series:

\begin{equation} |\mathbf{x}| = \sum_{abc}A_{abc} x^a y^b z^c \quad \text{and} \quad |\mathbf{p}| = \sum_{uvw}B_{uvw} p_x^u p_y^v p_z^w. \end{equation}


\begin{equation} [|\mathbf{x}|,|\mathbf{p}|] = \sum_{abc}\sum_{uvw} A_{abc}B_{uvw} (x^a y^b z^c p_x^u p_y^v p_z^w - p_x^u p_y^v p_z^w x^a y^b z^c). \end{equation}

The last term can be recast as

\begin{equation} p_x^u p_y^v p_z^w x^a y^b z^c = (p_x^u x^a) (p_y^v y^b) (p_z^w z^c) = (e^{-i\hbar\partial_x \partial_{p_x}} x^a p_x^u) (e^{-i\hbar\partial_y \partial_{p_y}} y^b p_y^v) (e^{-i\hbar\partial_z \partial_{p_z}} z^c p_z^w) = e^{-i\hbar(\partial_x \partial_{p_x} + \partial_y \partial_{p_y} + \partial_z \partial_{p_z})} x^a p_x^u y^b p_y^v z^c p_z^w = e^{-i\hbar\partial_\mathbf{x}\cdot\partial_\mathbf{p}} x^a p_x^u y^b p_y^v z^c p_z^w, \end{equation}


\begin{equation} \partial_\mathbf{x}\cdot\partial_\mathbf{p} \equiv \sum_i \frac{\partial}{\partial x_i} \frac{\partial}{\partial p_i}. \end{equation}


\begin{equation} [|\mathbf{x}|,|\mathbf{p}|] = \sum_{abc}\sum_{uvw} A_{abc}B_{uvw} (1-e^{-i\hbar\partial_\mathbf{x}\cdot\partial_\mathbf{p}})x^a y^b z^c p_x^u p_y^v p_z^w = (1-e^{-i\hbar\partial_\mathbf{x}\cdot\partial_\mathbf{p}})|\mathbf{x}| |\mathbf{p}| \end{equation}

or, explicitly,

\begin{equation} [|\mathbf{x}|,|\mathbf{p}|] = - \sum_{n=1}^\infty \frac{(-i\hbar)^n}{n!}(\partial_\mathbf{x} \cdot \partial_\mathbf{p})^n |\mathbf{x}| |\mathbf{p}|. \end{equation}

A few remarks:

  • The one-dimensional version of this equation recovers the expected expression given in the beginning of this answer.

  • The r.h.s. is rotationally invariant as expected considering the l.h.s.


  • Maybe a nice, closed form for the above result could be achieved writing $\partial_\mathbf{x} \cdot \partial_\mathbf{p}$ in spherical polar coordinates?

  • This derivation seems to work without modifications for any $[|\mathbf{x}|^n,|\mathbf{p}|^m]$. If the last remark is met with success, we can check if the proposed answer recovers some commutators that can be computed easily -- e.g., one with $n=m=2$.

I'm investigating these prospects and I will modify this answer accordingly.

  • 1
    $\begingroup$ Downvoters, please share your issues with this answer so I can try to improve it or understand what may be wrong with it. $\endgroup$ – andrehgomes Oct 17 '20 at 11:43

Here I'll overlook technical issues that may be relevant when defining operators $|\mathbf{x}|$ and $|\mathbf{p}|$ and head toward a formal expression for $[|\mathbf{x}|,|\mathbf{p}|]$. I'll omit the "hat" above operators for simplicity.

I start defining the symmetrized radial momentum operator,

$$p_r \equiv \frac{1}{2}\left( \frac{\mathbf{x}}{|\mathbf{x}|} \cdot \mathbf{p} + \mathbf{p} \cdot \frac{\mathbf{x}}{|\mathbf{x}|} \right).$$

This operator came into my attention on Alvarez & González, Am. J. Phys. 57, 923 (1989) and Liboff, Nebenzahl & Fleischmann, Am. J. Phys. 41, 976 (1973) but I learned it is standard content in QM textbooks. In position space represention $\mathbf{p}=-i\hbar\mathbf{\nabla}$ and spherical coordinates, straightforward calculation reveals

$$p_r = -i\hbar \left(\frac{\partial}{\partial r} + \frac{1}{r} \right),$$ where $r\equiv|\mathbf{x}|$. In this last form, the following result is easily verified:


From the relation $[A,f(B)]=[A,B]\frac{\partial f}{\partial B}$ valid if $[A,[A,B]]=0$ we conclude

$$[|\mathbf{x}|,f(p_r)]=i\hbar \frac{\partial f}{\partial p_r}.$$

Now I set $f(p_r) = |\mathbf{p}| = \sqrt{p_r^2 + \frac{L^2}{r^2}}$, with the last equality coming from the splitting of $|\mathbf{p}|$ into radial and angular parts, and with $L$ the angular momentum. From the previous commutator, we get

$$[|\mathbf{x}|,|\mathbf{p}|]=i\hbar \frac{1}{|\mathbf{p}|}\left[p_r + \frac{1}{2}\frac{\partial}{\partial p_r}\left(\frac{L^2}{r^2}\right) \right] = i\hbar \frac{p_r}{|\mathbf{p}|}.$$

The last derivative vanished, as I see it, because $L^2$ comprises only of angular pieces of $|\mathbf{p}|$ but no radial piece and because $\partial/\partial p_r$ doesn't act on $r$ as we can check applying it to $[|\mathbf{x}|,p_r]=i\hbar$. Finally, with the definition of $p_r$ above, we get

$$\boxed{ [|\mathbf{x}|,|\mathbf{p}|] = \frac{i\hbar}{2}\frac{1}{|\mathbf{p}|} \left( \frac{\mathbf{x}}{|\mathbf{x}|} \cdot \mathbf{p} + \mathbf{p} \cdot \frac{\mathbf{x}}{|\mathbf{x}|} \right) = i\hbar \left( \frac{\mathbf{p}}{|\mathbf{p}|} \cdot \frac{\mathbf{x}}{|\mathbf{x}|} + \frac{i\hbar}{|\mathbf{p}| |\mathbf{x}|} \right). }$$

A few remarks:

  • This result is rotationally invariant as expected.

  • For the cases I checked, the derivation presented here predicts results derived by other means. For instance, $[|\mathbf{x}|,|\mathbf{p}|^2]=[|\mathbf{x}|,\sum p_i p_i] = \sum\left( [|\mathbf{x}|,p_i] p_i + p_i [|\mathbf{x}|,p_i]\right) = i\hbar\left( \frac{\mathbf{x}}{|\mathbf{x}|}\cdot\mathbf{p} + \mathbf{p}\cdot\frac{\mathbf{x}}{|\mathbf{x}|}\right)$ and the derivation presented here gets the same, $[|\mathbf{x}|,|\mathbf{p}|^2]=i\hbar\frac{\partial |\mathbf{p}|^2}{\partial p_r} = 2i\hbar p_r = i\hbar \left( \frac{\mathbf{x}}{|\mathbf{x}|} \cdot \mathbf{p} + \mathbf{p} \cdot \frac{\mathbf{x}}{|\mathbf{x}|} \right).$


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