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

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?

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}