Commutation relation using Levi-Civita symbol

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The discussion focuses on computing the commutation relations between the position operator R and angular momentum L, specifically addressing the second part of the question regarding R^2 and L. A participant expresses confusion over why certain terms cancel out in their calculations, expecting them to add instead. It is clarified that both terms in question are equal and should be shown to separately equal zero. The conversation also explores specific cases, particularly when j = 1, to analyze the implications for the terms involving the Levi-Civita symbol. The resolution confirms that the nonzero possibilities of the epsilon symbol lead to a correct understanding of the commutation relations.
davon806
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Homework Statement


Hi,I have got a question as follow:
Compute the commutation relations of the position operator R and the angular momentum L.Deduce the commutation relations of R^2 with the angular momentum L

Homework Equations

The Attempt at a Solution


In fact I have got the solutions to this problem.I am having trouble for the 2nd part of the question.
In the red box of the image,why did the terms cancel out?Since position vector is commutative I expect they should add up?

Thanks!
 

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You are correct that the two terms in the first line of the red box are equal. Try to show that each term separately equals zero.
 
TSny said:
You are correct that the two terms in the first line of the red box are equal. Try to show that each term separately equals zero.
But if k =/= l ,then the first term becomes i(h bar) (x1x2 + x2x1+x3x2+x2x3 + x3x1 + x1x3)?
 
Note that ##j## is some fixed value.

Take the specific case where ##j = 1##. Then what does the first term become when you sum over ##k## and ##l##?
 
TSny said:
Note that ##j## is some fixed value.

Take the specific case where ##j = 1##. Then what does the first term become when you sum over ##k## and ##l##?
So if j =1,then we can either have k = 2,l = 3 or k = 3,l = 2.Hence the nonzero possibility of epsilon becomes e(132) or e(123)
i(h bar)(x2x3 - x3x2) =0 ?
 
Yes, that's correct.
 
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