Equivalence notation Euler angles-angular momentum in Wigner $D$ matrix

Question

In Wigner little-$d$ function the convention that I found in wikipedia https://en.wikipedia.org/wiki/Wigner_D-matrix is z-y-z as shown here.

A 3-dimensional rotation operator can be written as $$R(\alpha,\beta,\gamma)=e^{-i\alpha J_z}e^{-i\beta J_y}e^{-i\gamma J_z},$$

I was wondering how can we choose $\alpha,\beta,\gamma$, such that the effective $R(\alpha,\beta,\gamma)$ as $e^{-i\delta J_x}$, that is, how to choose the set of angles $(\alpha,\beta,\gamma)$ to generate a rotation along the $x$-axis?

Answer 1

You can build a rotation around the $x$-axis by

• first rotating $90°$ around the $z$-axis,
• then rotating some angle around the $y$-axis,
• and finally rotating back $90°$ around the $z$-axis.

Representing rotations by 3$\times$3 matrices, and especially using the basic 3D rotations $$ e^{-i\theta J_x}=R_x(\theta)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix} \\ e^{-i\theta J_y}=R_y(\theta)=\begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix} \\ e^{-i\theta J_z}=R_z(\theta)=\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ you can easily verify it.

$$\begin{align} R(-90°,\delta,90°) &=R_z(-90°)R_y(\delta)R_z(90°) \\ &=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix} \begin{pmatrix}\cos\delta&0&\sin\delta\\0&1&0\\-\sin\delta&0&\cos\delta\end{pmatrix} \begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix} \\ &=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix} \begin{pmatrix}0&-\cos\delta&\sin\delta\\1&0&0\\0&\sin\delta&\cos\delta\end{pmatrix} \\ &=\begin{pmatrix}1&0&0\\0&\cos\delta&-\sin\delta\\0&\sin\delta&\cos\delta\end{pmatrix} \\ &=R_x(\delta) \end{align}$$