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In Sakurai's Modern Quantum Mechanics p.198-199, he states that for the matrix $$J_y^{(j=1)} = \frac{J_+-J_-}{2i} = \frac{\hbar}{2} \begin{pmatrix} 0 & -\sqrt{2}i & 0 \\ \sqrt{2}i & 0 & -\sqrt{2}i \\ 0 & \sqrt{2}i & 0 \end{pmatrix}$$ we have the relationship: $$\left(\frac{J_y^{(j=1)}}{\hbar}\right)^3=\frac{J_y^{(j=1)}}{\hbar}$$

However, when I use this in a problem it yields undesirable results. So I verify it in Mathematica. It turns out that $$-2\left(\frac{J_y^{(j=1)}}{\hbar}\right)^3=\frac{J_y^{(j=1)}}{\hbar}$$ with a factor of -2.

As a result, I can't figure out the final result using Taylor expansion of $\exp(iJ_y\beta/\hbar)$: $$\exp(iJ_y\beta/\hbar) \to 1-\left(\frac{J_y}{\hbar}\right)^2(1-\cos\beta)-i\left(\frac{J_y}{\hbar}\right)\sin\beta$$

Wigner d-matrix for j=1 Mathematica

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closed as unclear what you're asking by Brandon Enright, Kyle Kanos, ACuriousMind, Martin, Chris Mueller Apr 20 '15 at 21:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ 1. You haven't stated an explicit question. 2. I suspect your explicit question would be off-topic as homework-like. $\endgroup$ – ACuriousMind Apr 20 '15 at 14:43
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you are using the wrong command. Use MatrixPower[J/h,3] to calculate cube of J/h.

In[7]:= MatrixPower[J/h, 3]

Out[7]= {{0, -(i/Sqrt[2]), 0}, {i/Sqrt[2], 0, -(i/Sqrt[2])}, {0, i/ Sqrt[2], 0}}

Clearly, this is equal to J/h.

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