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


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

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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

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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