How to show that $$\langle\vec{r}|[L_{j},f(\hat{r})]|\psi\rangle = 0~?$$

note that $\hat{r}$ is a operator and not a unit vector What I know so far is that the commutation relation of a normal radial function the angular moment $[L_{j}, f(r)] = 0$ as well as that radial function of a operator $\hat{r}$ by $$ f(\hat{r}) = \int d^{3} r' f(r') |\vec{r'}\rangle \langle\vec{r'}| $$ Based on this, one can imply the eigenvalue problem $f(\hat{r})|\vec{r}\rangle = f(r)|\vec{r}\rangle $ . Another thing I know is that $\langle\vec{r}|p_{k}|\psi\rangle = \frac{\hbar}{i}\frac{d}{dx_{k}} \langle\vec{r}|\psi\rangle$ and $L_{j} = \epsilon_{jkm}x_{k}p_{m}$. I tried expanding out the commutation in the middle and then split the expression into $\langle\hat{r}|L_{j}f(\hat{r})|\psi\rangle - \langle\vec{r}|f(\hat{r})L_{j}|\psi\rangle$ I was thinking next of moving the operators around and evaluating what they work on?

  • $\begingroup$ Comment to the post (v2): Note that the notation $\hat{r}$ often denotes the unit vector $\vec{r}/r$. Consider to explicitly define your notation for clarity. $\endgroup$
    – Qmechanic
    Sep 15, 2017 at 6:46
  • $\begingroup$ Sorry about that. I have fixed it by noting that $\hat{r}$ is a operator and not a unit vector $\endgroup$
    – Sam
    Sep 15, 2017 at 6:54
  • $\begingroup$ Is $f(\vec{r}) = f( | \vec{r}|)$? $\endgroup$ Sep 15, 2017 at 7:10
  • $\begingroup$ My bad there was a typo. In that second paragraph with the commutation split up, it should be $L_{j}f(\hat{r})$ not $f(\vec{r})$ $\endgroup$
    – Sam
    Sep 15, 2017 at 7:40
  • $\begingroup$ Fixed additional typo $\endgroup$
    – Sam
    Sep 15, 2017 at 7:41

1 Answer 1


There's a bunch of commutator stuff that's really standard and not worth going over, but the core of the result is the fact that $f(r)$ is purely a function of the radius. This means, therefore, that when you do \begin{align} \langle \vec r|\hat{p}_k f(\hat r)|\psi\rangle & = -i\hbar\frac{\partial}{\partial x_k} \langle \vec r|f(\hat r)|\psi\rangle \\& = -i\hbar\frac{\partial}{\partial x_k} \bigg[ f(r) \langle \vec r|\psi\rangle \bigg] \\& = -i\hbar \bigg[ \frac{\partial f(r) }{\partial x_k}\langle \vec r|\psi\rangle + f(r)\frac{\partial}{\partial x_k} \langle \vec r|\psi\rangle \bigg], \end{align} the second term reduces to $\langle \vec r|f(\hat r)\hat{p}_k|\psi\rangle$, but the first term gives you \begin{align} \frac{\partial f(r) }{\partial x_k} &= f'(r)\frac{\partial r}{\partial x_k} \\&= f'(r)\frac{\partial}{\partial x_k}\sqrt{\sum_j x_j^2} \\&= \frac{x_k}{r}f'(r), \end{align} and therefore \begin{align} -i\hbar \frac{\partial f(r) }{\partial x_k}\langle \vec r|\psi\rangle &= -i\hbar \frac{x_k}{r}f'(r)\langle \vec r|\psi\rangle \\ &= \left\langle \vec r\middle|-i\hbar \frac{\hat{x}_k}{\hat{r}}f'(\hat r)\middle|\psi\right\rangle . \end{align} If you now add the component $\hat{x}_j$ into the matrix element, together with the antisymmetric $\epsilon_{ijk}$ of the cross product, you will be left with a matrix element of an operator proportional to $$ \epsilon_{ijk}\hat x_j \hat x_k, $$ which vanishes identically.


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