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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! |
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