What Are the Commutation Relations of \( \hat{R}^2 \) with \( \hat{L} \)?

spacetimedude
Messages
87
Reaction score
1

Homework Statement


Deduece the commutation relations of position operator (squared) \hat R^2 with angular momentum \hat L

Homework Equations


[xi,xj]=0, Lj= εijkxjPk, [xi, Pl]=ih, [xi,Lj]=iℏϵijkxk

The Attempt at a Solution


The previous question related R and L and the result was [\hat R,\hat L_j]=i \hbar \epsilon _{ijk}x_k after setting up the commutator as \epsilon _{jkl}[x_i,x_kP_l] where I did not include the i in the epsilon.

Now, I did the same with with [\hat R^2,\hat L_j] and set it up as [\hat R^2,\hat L_j]=[x_ix_i,L_j]=\epsilon_{jkl}[x_i,P_l]x_kx_i+x_i\epsilon_{jkl}[x_i,P_l]x_k, in which I simplified using the commutator property, and which is then equal to i\hbar\epsilon_{jkl}x_kx_i+i\hbar x_i\epsilon_{jkl}x_k. I don't think I can reduce it any further.
The solution has the i included in the epsilon in the setup and I don't know why that is.

Any help will be appreciated
 
Physics news on Phys.org
I can't follow your use of the epsilon symbol. Why not try calculating:

##[x^2, p_x]##

And from there:

##[R^2, L_x]##

Before you do the calculation, though, what do you think the answer will be?
 
Last edited:
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