Angular Moment Operator Vector Identity Question

[Chronum]

In my EM class, this vector identity for the angular momentum operator (without the $i$) was stated without proof. Is there anywhere I can look to to actually find a good example/proof on how this works? This is in spherical coordinates, and I can't seem to find this vector identity anywhere. I've tried Googling for hours now, and I've legitimately come up with no example with a good accompanying explanation, of how this particular identity is working. This is the identity in question. Any help would be appreciated.

$(\vec{r}\times\nabla)\psi = \nabla\times \vec{r}\psi - \psi(\nabla\times \vec{r})$

EDIT: I've also seen this identity on Wikipedia, which _may_ satisfy my requirement if I'm thinking of the curl operator right.​

$\nabla \times(\psi \vec{r}) = \psi(\nabla \times \vec{r}) + \nabla\psi\times\vec{r}$

For the first term on the RHS, we can use the anticommutativity of the cross product, and state​

$\psi(\nabla \times \vec{r}) = - \psi (\vec{r} \times \nabla)$

and then the identity becomes a trivial moving around of terms. Am I thinking about that in the right way?​

 

[blue_leaf77]

It shouldn't be that hard to show using product rule of derivative but for some reason I got the sign reversed.
$$
(\mathbf r \times \nabla)_i \psi = \epsilon_{ijk}r_j\partial_k \psi
$$
but $r_j\partial_k \psi = \partial_k (r_j \psi) - \psi \partial_k r_j$. Hence
$$
(\mathbf r \times \nabla)_i \psi = \epsilon_{ijk}(\partial_k (r_j \psi) - \psi \partial_k r_j) = -\epsilon_{ikj}\partial_k (r_j \psi) + \epsilon_{ikj}\psi \partial_k r_j = -(\nabla\times\mathbf r \psi)_i + \psi(\nabla \times \mathbf r)_i
$$
Still trying to figure out where I went wrong.
EDIT: The above is correct.

 

[Chronum]

It shouldn't be that hard to show using product rule of derivative but for some reason I got the sign reversed.
$$
(\mathbf r \times \nabla)_i \psi = \epsilon_{ijk}r_j\partial_k \psi
$$
but $r_j\partial_k \psi = \partial_k (r_j \psi) - \psi \partial_k r_j$. Hence
$$
(\mathbf r \times \nabla)_i \psi = \epsilon_{ijk}(\partial_k (r_j \psi) - \psi \partial_k r_j) = -\epsilon_{ikj}\partial_k (r_j \psi) + \epsilon_{ikj}\psi \partial_k r_j = -(\nabla\times\mathbf r \psi)_i + \psi(\nabla \times \mathbf r)_i
$$
Still trying to figure out where I went wrong.

Thank you so much. I've never really been introduced the inner nitty gritties of the vector calculus identites/their sources, so this tells me I should probably acquaint myself better with them.

 

[blue_leaf77]

It looks like you got this identity not correct:

$(\vec{r}\times\nabla)\psi = \nabla\times \vec{r}\psi - \psi(\nabla\times \vec{r})$​

whereas mine is, check https://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_2

For the first term on the RHS, we can use the anticommutativity of the cross product, and state​

ψ(∇×⃗r)=−ψ(⃗r×∇)\psi(\nabla \times \vec{r}) = - \psi (\vec{r} \times \nabla)

and then the identity becomes a trivial moving around of terms. Am I thinking about that in the right way?​

No, that's incorrect. The LHS is a vector but the RHS is a vector operator. While curl operation and vector cross product share many properties, their anticommutativity property is different.

 

[PeroK]

It shouldn't be that hard to show using product rule of derivative but for some reason I got the sign reversed.
$$
(\mathbf r \times \nabla)_i \psi = \epsilon_{ijk}r_j\partial_k \psi
$$
but $r_j\partial_k \psi = \partial_k (r_j \psi) - \psi \partial_k r_j$. Hence
$$
(\mathbf r \times \nabla)_i \psi = \epsilon_{ijk}(\partial_k (r_j \psi) - \psi \partial_k r_j) = -\epsilon_{ikj}\partial_k (r_j \psi) + \epsilon_{ikj}\psi \partial_k r_j = -(\nabla\times\mathbf r \psi)_i + \psi(\nabla \times \mathbf r)_i
$$
Still trying to figure out where I went wrong.

I think you're right and the OP is wrong.

 

[blue_leaf77]

I think you're right and the OP is wrong.

Yes, just confirmed it from the Wikipedia link I shared above.

 

[PeroK]

Yes, just confirmed it from the Wikipedia link I shared above.

Yes, just saw it.

@Chronum there are no shortcuts with this sort of identity. You just have to work through each component.

I tend to use the determinant form of the cross product and just do the first component. The others generally follow by the cyclic symmetry.

 

[Chronum]

Yes, just saw it.

@Chronum there are no shortcuts with this sort of identity. You just have to work through each component.

I tend to use the determinant form of the cross product and just do the first component. The others generally follow by the cyclic symmetry.

Maybe this is also complicated by the fact that I'm trying to do this in spherical coordinates, when I can potentially do it in cartesian. I'm uploading a snippet of the notes I'm looking at, just making sure I'm not misinterpreting anything. The given relation holds true without the extra term because we're dealing with a radial vector here.

 

[blue_leaf77]

Maybe this is also complicated by the fact that I'm trying to do this in spherical coordinates, when I can potentially do it in cartesian.

If $\mathbf r$ is a position vector (you call it radial vector) then yes the relation in your note is correct after multiplication with -1 in either sides of the equation. Have you checked the wiki link in post#4?
Why not trying what Perok suggested in post #7 by verifying the first component, e.g. the x-component?
Moreover, the first relation (the corrected version) you posted in post #1 holds for any vector field $\mathbf r$ and scalar field $\psi$ regardless of the coordinate system.