The rotational analog of Ehrenfest's Theorem

[Bobbo Snap]

Homework Statement

Show $\frac{d}{dt}\langle\bf{L}\rangle = \langle \bf{N} \rangle$ where $\bf{N} = \bf{r}\times(-\nabla V)$

2. Homework Equations .
$$\frac{d}{dt}\langle A \rangle = \frac{i}{\hbar} \langle [H, A] \rangle$$

The Attempt at a Solution

I get to this point: $\frac{i}{\hbar}\langle [H,L] \rangle = \frac{i}{\hbar}\langle [H, r \times p] \rangle$ and then I'm stuck. I know the next step is supposed to use something like $[H, r \times p] = [H,r] \times p + r \times [H,p]$ but I don't see how to get this. I don't understand how operators distribute over the cross-product. Can anyone help or point me to some online resource where I can study how to work with operators/commutators/ etc.? I've been searching the internet all day with little to show for it.

 

[PhysicsGente]

Hey Bobbo Snap,

$[H,L] = HL - LH = H(r\times p) - (r\times p)H = (Hr)\times p - (rH)\times p = [H,r]\times p$

 

[Bobbo Snap]

Thanks PhysicsGente, your expression is exactly what I keep getting. That is,
$$[H, r \times p] = [H, r]\times p.$$

Maybe my mistake is trying to follow a solution I found which includes an extra term. It begins with the commutator reversed though, like this:
$$[r \times p, H] = r \times [p, H] + [r, H] \times p$$
I don't get how they come up with this.

 

[TSny]

I find it least confusing to just write it out in Cartesian components. For the x-component of the identity that you want to prove, use L_x = (r x p)_x = (yp_z - zp_y). Then simplify [H, L_x]. The y- and z-components of the identity then follow by "cyclic permutation" of x,y,z. Not very elegant, but it gets the job done.

 

[PhysicsGente]

Thanks PhysicsGente, your expression is exactly what I keep getting. That is,
$$[H, r \times p] = [H, r]\times p.$$

Maybe my mistake is trying to follow a solution I found which includes an extra term. It begins with the commutator reversed though, like this:
$$[r \times p, H] = r \times [p, H] + [r, H] \times p$$
I don't get how they come up with this.

You can also write it as $$[H, r \times p] = r\times [H, p].$$ I don't see why you'd need the second term though. Just calculate the commutator $[H, p]$ and you should be done.