Commutator of square angular momentum operator and position operator

[elmp]

can someone please help me with this. it's killing me.

Homework Statement

to show $$\left[\vec{L}^{2}\left[\vec{L}^{2},\vec{r}\right]\right]=2\hbar^{2}(\vec{r}\vec{L}^{2}+\vec{L}^{2}\vec{r})$$

Homework Equations

I have already established a result (from the hint of the question) that
$$\left[\vec{L}^{2},\vec{r}\right]=2\imath\hbar\vec{r}\times\vec{L}+2\hbar^{2}\vec{r}$$

but where to go from that?

The Attempt at a Solution

basically I will need to show either one of (4.55) from http://www.eng.fsu.edu/~dommelen/quantum/style_a/commute.html, but I am running out of clues.

 

[sgd37]

if you know vector component notation (look it up) then this is a pretty simple problem

$$\vec{r} \times \vec{L} = \epsilon_{ijk} r_j L_k$$ where $$\epsilon_{ijk}$$ is the levi-civita tensor (look it up)

if you don't you could use the triple product expansion

$$\vec{r} \times (\vec {r} \times \vec{p}) = \vec{r} (\vec{r} \cdot \vec{p}) - \vec{p}( r^2)$$

and then apply all the other commutation relations you know, but that is tedious

 

[rynlee]

Hm, I'm interested in this now as well, but in the OP's post I don't see how quite to arrive at:

Homework Equations

I have already established a result (from the hint of the question) that
$$\left[\vec{L}^{2},\vec{r}\right]=2\imath\hbar\vec{r}\times\vec{L}+2\hbar^{2}\vec{r}$$

I can find that
[L^2,z]=2i\hbar(xL_y-yL_x-i\hbar z)

and the equivalents for x and y, but how do you relate that to the commutator

$$\left[\vec{L}^{2},\vec{r}\right]$$

?

 

[rynlee]

Nevermind my last post, if you do everything in vector notation (or matrix notation, things work out peachy keen.

 

[elmp]

if you know vector component notation (look it up) then this is a pretty simple problem

$$\vec{r} \times \vec{L} = \epsilon_{ijk} r_j L_k$$ where $$\epsilon_{ijk}$$ is the levi-civita tensor (look it up)

if you don't you could use the triple product expansion

$$\vec{r} \times (\vec {r} \times \vec{p}) = \vec{r} (\vec{r} \cdot \vec{p}) - \vec{p}( r^2)$$

and then apply all the other commutation relations you know, but that is tedious

hmmm, i must be really stupid, but i still don't see how to proceed

 

[grey_earl]

can someone please help me with this. it's killing me.

Homework Statement

to show $$\left[\vec{L}^{2}\left[\vec{L}^{2},\vec{r}\right]\right]=2\hbar^{2}(\vec{r}\vec{L}^{2}+\vec{L}^{2}\vec{r})$$

Homework Equations

I have already established a result (from the hint of the question) that
$$\left[\vec{L}^{2},\vec{r}\right]=2\imath\hbar\vec{r}\times\vec{L}+2\hbar^{2}\vec{r}$$

but where to go from that?

Use that fact that the commutator is linear and its product rule, so you get
$$\left[\vec{L}^{2}, \left[\vec{L}^{2},\vec{r}\right]\right] = \left[\vec{L}^{2}, 2\mathrm{i}\hbar\vec{r}\times\vec{L}+2\hbar^{2}\vec{r} \right] = 2\mathrm{i}\hbar \left[\vec{L}^{2}, \vec{r}\times\vec{L} \right] + 2\hbar^{2} \left[\vec{L}^{2}, \vec{r} \right] = 2\mathrm{i}\hbar \vec{r}\times \left[\vec{L}^{2}, \vec{L} \right] + 2\mathrm{i}\hbar \left[\vec{L}^{2}, \vec{r} \right] \times\vec{L} + 2\hbar^{2} \left[\vec{L}^{2}, \vec{r} \right]$$

Then you are almost done, to have a nice result you need the triple product expansion given before.