How to Simplify Commutators Using Levi-Civita Symbol?

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KostasV
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Handwritten derivation of the commutator [li, xj] using the Levi-Civita symbol, ending with the result [li, xj] = i h epsilon_ijk x_k.

Homework Statement


The problem statement can be seen in the picture i uploaded.

Homework Equations


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The Attempt at a Solution


The attempt to the solution can be seen in the picture i uploaded.
I reached to the A and i don't know how to proceed to the solution that is given below. How does the minus and δkj disappear?
If i do the double summation on k and j I think that every term gets zero either because of εijk (levi-cevita) or δkj (kronecker)
 
You used the index j twice: once for the index on x and then again as a dummy summation index when writing out the angular momentum in terms of the levi-civita symbol.
 
TSny said:
You used the index j twice: once for the index on x and then again as a dummy summation index when writing out the angular momentum in terms of the levi-civita symbol.
I can't see why this is wrong ... :/
 
TSny said:
You used the index j twice: once for the index on x and then again as a dummy summation index when writing out the angular momentum in terms of the levi-civita symbol.
Moreover , if i use , let's say the index m on x (not on x that comes from angular momentum , yes on x that is alone) , then i still have the minus on ih bar ... My solutions say that it should not be there ...
 
Handwritten derivation on lined paper of the commutator [Li, xj] using the Levi-Civita symbol, summing over k and m with a Kronecker delta, ending in -ε_ijm x_m and then i ħ ε_ijk x_k.
Ok i think i understand why i can't have the same index on these two .
Moreover i think i found how i get rid of this minus ...
I must use the fact that εijk=-εikj wright ?
Is now the solution correct ? (Uploaded photo)
 
Yes. That looks very good.
 
TSny said:
Yes. That looks very good.
Thank you very much for your help :D