# Angular Momentum Operator

I'm looking at a review question I was given and it quite frankly has me stumped.

"Using matrix representations find $L^{3}_{x},L^{3}_{y},L^{3}_{z}$ and from these show that $L_{x}, L_{y},L_{z}$ satisfy the same algebraic equations."

What has me stumped is the $L^{3}_{x},L^{3}_{y},L^{3}_{z}$, I'm not even sure how it makes sense to talk of the cube of L? Any pointers in the right direction here would be incredibly helpful!

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In units of $\hbar$, $$L_x=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0\end{bmatrix},L_y=\begin{bmatrix}0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0\end{bmatrix},L_z=\begin{bmatrix}0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ Note that, to compute these, I took the derivative of the rotation matrices (multiplied by $i$) at $\theta =0$. From here, it is a simple matter of computation. For example, $$L_x^3=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0\end{bmatrix}\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0\end{bmatrix}^2=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0\end{bmatrix}\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0\end{bmatrix}=L_x$$ Similarly, $L_y^3=L_y$ and $L_z^3=L_z$. Thus, of course the operators $L_x^3,L_y^3,L_z^3$ obey the same commutation relations as the operators $L_x,L_y,L_z$: they're the exact same operators!
Thanks! Looking at it now in terms of the algebra you've laid out makes sense (I should have guessed $LL^{2}=L^{3}$). I'm left wondering why I would want to use $L^{3}$ instead of $L$; for computational expediency in select situations? – atomicpedals Nov 22 '11 at 3:27
@atomicpedals ". . . why would I want to use $L^3$ instead of $L$;" First of all, be careful when you write $L^3$. $\mathbf{L}$ is a vector, so if you write $L^3$, I'm going to think you mean $(L_x^2+L_y^2+L_z^2)^{3/2}$ (because it doesn't make sense to cube a vector, although it does make sense to cube the magnitude of a vector), which is not your meaning. As for why would you work with $L_i^3$ over $L_i$: you wouldn't. I'm pretty sure this exercise is just for the sake of practice. – Jonathan Gleason Nov 22 '11 at 3:38