Commutator of square angular momentum operator and position operator

elmp
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
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
 
Hm, I'm interested in this now as well, but in the OP's post I don't see how quite to arrive at:

elmp said:

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]

?
 
Nevermind my last post, if you do everything in vector notation (or matrix notation, things work out peachy keen.
 
sgd37 said:
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
 
elmp said:
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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top