Proving commutator [${L}_i, f(r)]=0$

Question

I am trying to prove that the angular momentum component operator ${L}_i$ commutes with any function of ${r}\equiv \sqrt{{\textbf{x}}\cdot{\textbf{x}}}$, i.e.

$$[L_i, f( r)]=0.$$

I first worked out that $L_i$ commutes with $\textbf{x}\cdot\textbf{x}$, i.e. $$[L_i,\textbf{x}^2]=0.$$ I then expanded $f(r)$ in terms of a power series: $$f(r)=\sum_n a_nr^n=\sum_na_n(\textbf{x}\cdot\textbf{x})^{n\over 2}.$$

Next, I want to show that $L_i$ commutes with $(\textbf{x}\cdot\textbf{x})^{n\over 2}$, where $n$ is a positive integer, i.e. $$[L_i,(\textbf{x}\cdot\textbf{x})^{n\over 2}]=0.$$

What commutator rule can I use to evaluate this?

Answer 1

The answers given by Philip and Alexander are correct, but let me add another snippet of information: one could calculate the commutation in the polar coordinates! Then: $$\hat{L}_i = \frac{\partial}{\partial \phi_i},$$ whereas $f(r)$ is obviously independent on angle $\phi_i$.

Answer 2

There are multiple ways of proving this, as other answers show. Here is another one. There is a general identity that states that if $[x, y]=0$, then $[x,f(y)]=0$, where $f$ is some function. The result you are after follows immediately from this, with $f(y)=\sqrt y$.

To prove this identity, let us first show that $[x,y]=0$ implies $[x, y^n]$, where $n$ is any non-negative integer. We can do this by induction. Suppose that $[x,y^m]=0$ for all $m<n$. Then, $$\begin{aligned} \left[x, y^n\right] & = xy^n-y^nx \\ &= xy^{n-1}y -y^{n-1}yx \\ &= y^{n-1}xy-y^{n-1}yx\\ &=y^{n-1}[x,y] \\ &= 0. \end{aligned}$$ Then, since $f(y)$ can be expanded into a Taylor series involving powers of $y$, it follows from $[x,y]=0$ that $[x,f(y)]=0$.

Edit: As has been brought up in comments, the square root function is not actually analytic on $\mathbb{C}$, although it can be chosen to be analytic on $\mathbb{C}-\{0\}$. This would mean the proof may not be valid for this function. I'm not exactly sure that the function being non-analytic at $0$ invalidates this approach though, so I will keep this answer so others can comment on this, and also because $[x,f(y)]=0$ (with $f$ analytic) is a useful identity.

Answer 3

You could use the following rule that states that if you have two operators $A$ and $B$ such that $[B,[A,B]]=0$, then

$$[A, f(B)] = f'(B)[A,B].$$

Answer 4

Hint 1:

There is a basic identity (derived upon several restrictions) that in your case reads as - $$[L_i,f(r)]=[L_i,r]f'(r)$$ Applying the same relation you can show that if operator commutes with $r$, it commutes with any function of $r$ You can also use the following for $L_i$ to relax the expression - $$[AB,C]=A[B,C]+[A,C]B$$

@Philips answer also useful, though in general case (when the operator doesn't commute with the commutator) the order (which side to land the $f'$) matters. In this case you should be able to use the identities to break the expression up to $[x,p]$ commutator, which is constant, thus commutes with everything.

Hint 2:

The following reference seems very helpful for the most general case

http://hdl.lib.byu.edu/1877/1263

General answer:

Essentially, angular momentum operators operate on angle variables (real space representation), while the $f(r)$ function is independent of angular coordinates. Thus you can switch freely the order of operators.

The following reference has solution to your question

http://www.physics.ucc.ie/apeer/PY3102/Angular_momentum.pdf

Answer 5

Encountered the same question in 2023. Though using spherical coordinates (as pointed out by @Roger Vadim) is for sure one way to prove $[L_i, f(r)]=0$, if one is sticking to the Cartesian ones, one may try, for example:

$$ \begin{split} [L_z, f(r)]g(r) &=L_zf(r)g(r)-f(r)L_zg(r)\\ &=(xp_y-yp_x)f(r)g(r)-f(r)(xp_y-yp_x)g(r)\\ &=(-i\hbar x\frac{\partial}{\partial y}+i\hbar y\frac{\partial}{\partial x})[f(r)g(r)]-f(r)(-i\hbar x\frac{\partial}{\partial y}+i\hbar y\frac{\partial}{\partial x})g(r)\\ &=i\hbar [-xg(r)\frac{\partial f(r)}{\partial y}+yg(r)\frac{\partial f(r)}{\partial x}]\\ &=i\hbar[-xg(r)\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+yg(r)\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}]\\ &=i\hbar[-xg(r)\frac{\partial f}{\partial r}\frac{y}{r}+yg(r)\frac{\partial f}{\partial r}\frac{x}{r}]\\ &=0 \end{split}$$

where g(r) is just a test function and $\frac{\partial r}{\partial y}=\frac{\partial \sqrt{x^2+y^2+z^2}}{\partial y}=\frac{y}{r}$ and $\frac{\partial r}{\partial x}=\frac{x}{r}$. Thus, $[L_z,f(r)]=0$. The same goes for $[L_x,f(r)]$ and $[L_y,f(r)]$.