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there are a few answered questions regarding the commutator of any two 3D angular momentum operator components $[L_i, L_j]$ , however, I am trying to go through fully using index notation so that I can arrive at the generic expression $$L_k=i \hbar \epsilon_{kij}x_ip_j$$ where $x$ and $p$ are the position and momentum operators in 3D.

Using $$[AB,C] = A[B,C] + [A,C]B$$ and cancelling out self-commutators, as well as using the canonical commutation relations I come to this line (skipped some steps):

$$[L_i, L_j] = \epsilon_{iab}\epsilon_{jcd}[x_ap_b,x_cp_d] = ... = \epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_bp_c\delta _{ad}) $$

Now if I try to contract the levi-civitas for each of the two terms in the brackets, I just get more kronecker deltas, with which I don't know what to do. If someone is feeling really generous today, could you go through the final steps to the solution so that I can understand what is happening (all the answers I have found just skip this assuming the reader will understand). Thank you!

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  • $\begingroup$ I believe your question is answered here: physics.stackexchange.com/questions/65947/… $\endgroup$
    – user83548
    Oct 30, 2015 at 18:09
  • $\begingroup$ It kind of is, except for the part where the person answering says "I suggest you work out the missing parts to understand how this levi-civita business works." Which is where I am stuck $\endgroup$ Oct 30, 2015 at 18:20

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You have mixed indices in the end of your line. Correctly: $$-\epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_cp_b\delta _{ad}).$$ So further, $-\epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_cp_b\delta _{ad})=\epsilon_{iab}\epsilon_{jcd}x_cp_b\delta _{ad}-\epsilon_{iab}\epsilon_{jcd}x_ap_d\delta _{bc}=\\=\epsilon_{idb}\epsilon_{jcd}x_cp_b - \epsilon_{iac}\epsilon_{jcd}x_ap_d=\epsilon_{iac}\epsilon_{jdc}x_ap_d-\epsilon_{ibd}\epsilon_{jcd}x_cp_b=\\=(\delta _{ij}\delta _{ad}-\delta _{id}\delta _{aj})x_ap_d-(\delta _{ij}\delta _{bc}-\delta _{ic}\delta _{bj})x_cp_b=\\=\delta _{ij}x_ap_a-x_jp_i-\delta _{ij}x_bp_b+x_ip_j=x_ip_j-x_jp_i=\epsilon_{ijk}L_k.$

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  • $\begingroup$ Thank you so much! One final question - why does the epsilon appear at the end? I understand it's role but I don't understand the jump. Edit: so there is no epsilon there? $\endgroup$ Oct 30, 2015 at 18:46
  • $\begingroup$ It is obtained by comparing the expression vector product and our record in index form. For example, for a specific index values $(i=1, j=2)$ we obtain: $x_1p_2-x_2p_1=\epsilon_{123}x_1p_2+\epsilon_{213}x_2p_1=[\mathbf{x} \times \mathbf{p}]_3=L_3$ $\endgroup$
    – kw_artem
    Oct 30, 2015 at 19:15
  • $\begingroup$ It can be seen that the first two of the levi-civitas index are permuted in the same manner as the indices of the $x$ and $p$ operators. $\endgroup$
    – kw_artem
    Oct 30, 2015 at 19:22

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