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

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SUMMARY

The commutation relations of the squared position operator \( \hat{R}^2 \) with angular momentum \( \hat{L} \) are derived using the fundamental commutation relations of quantum mechanics. Specifically, the relation is established as \([\hat{R}^2, \hat{L}_j] = i\hbar \epsilon_{jkl} x_k x_i + i\hbar x_i \epsilon_{jkl} x_k\). The use of the Levi-Civita symbol \( \epsilon_{ijk} \) is crucial in the derivation, particularly when applying the properties of commutators. The discussion highlights the importance of correctly incorporating the factor of \( i \) in the Levi-Civita symbol during calculations.

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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
 
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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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