# Angular Momentum crossed with itself

I need some help with proving identity:

$$L × L = iℏL.$$

I am using a triple scalar product however I reach dead-end (perhaps I am doing it wrong ?)

Using this identity

d = a×(b×c)

letting a = L and b and c using definition of angular momentum I have b = r; c = p

I get:

= $$\epsilon_{mni}L_n\epsilon_{ijk}r_jp_k$$ = $$r_mL_kp_k - p_mL_jr_j$$

how do I proceed after ?

I don't recognize the identity $$\mathbf d=\mathbf a\times(\mathbf b\times\mathbf c)$$ - could you define the variables there? Anyway, you shouldn't need that to prove the relation you are trying to prove. Remember, the operator $$\mathbf L$$ is a vector matrix equal to $$\begin{pmatrix}\mathbf{L_x}\\\mathbf{L_y}\\\mathbf{L_z}\end{pmatrix}$$, which means that the cross product of $$\mathbf L$$ and itself can be evaluated as$$\begin{vmatrix}\hat i&\hat j&\hat k\\\mathbf{L_x}&\mathbf{L_y}&\mathbf{L_z}\\\mathbf{L_x}&\mathbf{L_y}&\mathbf{L_z}\end{vmatrix}=\begin{pmatrix}\mathbf{L_yL_z}-\mathbf{L_zL_y}\\\mathbf{L_zL_x}-\mathbf{L_xL_z}\\\mathbf{L_xL_y}-\mathbf{L_yL_z}\end{pmatrix}=\begin{pmatrix}[\mathbf{L_y},\mathbf{L_z}]\\ [\mathbf{L_z},\mathbf{L_x}]\\ [\mathbf{L_x},\mathbf{L_y}]\end{pmatrix},$$where we are using the commutation relation $$[\mathbf A,\mathbf B]=\mathbf{AB}-\mathbf{BA}$$. The identies we need here are\begin{align*}[\mathbf{L_x},\mathbf{L_y}]&=i\hbar\mathbf{L_z}\\ [\mathbf{L_y},\mathbf{L_z}]&=i\hbar\mathbf{L_x}\\ [\mathbf{L_z},\mathbf{L_x}]&=i\hbar\mathbf{L_y}.\end{align*}
Hopefully, these identities have already been reviewed in your course - if not, you should go back and try to prove them from the definition of the angular momentum operators! From here, it becomes pretty clear that$$\begin{pmatrix}[\mathbf{L_y},\mathbf{L_z}]\\ [\mathbf{L_z},\mathbf{L_x}]\\ [\mathbf{L_x},\mathbf{L_y}]\end{pmatrix}=i\hbar\mathbf L,$$so we have shown that $$\mathbf L\times\mathbf L=i\hbar\mathbf L$$ as desired.