For non-commuting vector operators we have in general $$\mathbf{A}\times \mathbf{B}= - \mathbf{B}\times \mathbf{A} +\sum_{ijk} \epsilon_{ijk} \big[ A_i,\,B_j\big]\hat{\mathbf{k}} $$

For the case $\mathbf{A}=\mathbf{B}$ this becomes $$ \mathbf{A}\times \mathbf{A}= - \mathbf{A}\times \mathbf{A} +\sum_{ijk} \epsilon_{ijk} \big[ A_i,\,A_j\big]\hat{\mathbf{k}} $$ This would imply $$2\left(\mathbf{A}\times \mathbf{A}\right)_k = \sum_{ij} \epsilon_{ijk}\big[ A_i,\,A_j\big] $$

Sources give the angular momentum commuator relations as $$\mathbf{J}\times \mathbf{J} = i\mathbf{J}$$ But we also know them as $$[J_i,J_j] = i\sum_{k}\epsilon_{ijk}J_k $$ I am curious about the connection these two equations. Can we derive the 2nd from the 1st? There is some discussion of it here on page 8. The 1st comes easily from the 2nd. I am not sure about the discussion of reversibility. What does $\mathbf{J}\times \mathbf{J} = i\mathbf{J}$ imply about $[J_i,J_j]$?


1 Answer 1


$$ \big( {\bf J} \times {\bf J} \big)_i = \epsilon_{ijk} J_j J_k = \frac{1}{2} \epsilon_{ijk} [ J_j , J_k ] = i J_i~. $$ Contract both sides with $\epsilon_{ipq}$. Then, $$ i \epsilon_{ipq} J_i = \frac{1}{2} \epsilon_{ijk} \epsilon_{ipq} [ J_j , J_k ] = \frac{1}{2} ( \delta_{jp} \delta_{kq} - \delta_{jq} \delta_{kp} ) [ J_j , J_k ] = [ J_p , J_q ]~. $$ Thus, we conclude $$ [ J_i , J_j ] = i \epsilon_{ijk} J_k ~. $$


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