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

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spacetimedude
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Homework Statement


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

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 [tex][\hat R,\hat L_j]=i \hbar \epsilon _{ijk}x_k[/tex] after setting up the commutator as [tex]\epsilon _{jkl}[x_i,x_kP_l][/tex] where I did not include the i in the epsilon.

Now, I did the same with with [itex][\hat R^2,\hat L_j][/itex] and set it up as [tex][\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[/tex], in which I simplified using the commutator property, and which is then equal to [tex]i\hbar\epsilon_{jkl}x_kx_i+i\hbar x_i\epsilon_{jkl}x_k[/tex]. 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
 
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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?
 
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