Representing a rotation around an arbitrary axis using Wigner $D$-matrix

Question

It is known that an arbitrary rotation can be expressed in terms of three consecutive rotations called the Euler rotations. So instead of expressing the rotation operator as $\hat{R}(\hat{n},\phi) = \exp\left(-\frac{i\phi}{\hbar} \hat{n}\cdot\vec{J}\right )$ one can write $\hat{R}(\alpha,\beta,\gamma) = \hat{R}_z(\alpha)\hat{R}_y(\beta)\hat{R}_z(\gamma)$ where $(\alpha,\beta,\gamma)$ are the so-called Euler angles. My question is fairly simple: what is the relationship between a given $\hat{n}$ and $(\alpha,\beta,\gamma)$?

Let me be more specific. Suppose we have a spin-$1/2$ system and some spinor $|\chi\rangle$ associated with it. Now, suppose I want to rotate this spinor through an angle $\phi = 2\pi$ around some arbitrary axis $\hat{n}=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)$, where $\theta,\varphi$ are the usual polar and azimuthal angles in the original spherical coordinate system. Obviously, we can use the following identity $$\hat{R}(\hat{n},\phi) = \mathbb{I}\cos \frac{\phi}{2} - i(\hat{n}\cdot\vec{\sigma}) \sin\frac{\phi}{2}$$ and conclude that $\hat{R}(\hat{n},\phi=2\pi)=-\mathbb{I}$ for any $\hat{n}$. But then I wanted to see if the same result can be obtained using the Wigner D-matrices (which are tied to Euler rotations). Evidently, one must rotate the original coordinate system first such that one of its axes aligns with $\hat{n}$ and then rotate $|\chi\rangle$ around that axis. But how exactly can this be done in just three steps (angles)? Initially I thought that the correct sequence should be $\alpha=\varphi,\beta=\theta,\gamma=\phi$, however for the aforementioned example it yields: $$D_{m'm}^{j=1/2}(\varphi ,\theta,\phi=2\pi ) = \begin{pmatrix} -e^{-i\varphi/2} \cos \frac{\theta}{2} & -e^{-i\varphi/2} \sin \frac{\theta}{2}\\ e^{i\varphi/2} \sin \frac{\theta}{2} & -e^{i\varphi/2} \cos \frac{\theta}{2} \end{pmatrix} \neq - \mathbb{I}$$

Answer 1

I suspect what you want is something called the $U^J_{MM'}$ rotation matrices: \begin{align} U^{J}_{MM'}(\omega;\Theta,\Phi)\equiv \langle JM\vert e^{-i\omega \hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{J}} } \vert JM'\rangle\, , \end{align} where $\Theta,\Phi$ determine the rotation axis (i.e. the direction of $\hat{\boldsymbol{n}}$.)

The source for this is section 4.5 of "the bible"

Varshalovich, D.A., Moskalev, A.N. and Khersonskii, V.K.M., 1988. Quantum theory of angular momentum.

In short, $U^{J}_{MM'}(\omega;\Theta,\Phi)$ can be expanded in terms of the "usual" $D$-functions \begin{align} U^{J}_{MM'}(\omega;\Theta,\Phi) =\sum_{M''} D^J_{MM''}(\Phi,\Theta,-\Phi) e^{-i M'' \omega } D^J_{M''M}(\Phi,-\Theta,-\Phi) \, . \end{align} The interpretation is clear: $D^J_{MM''}(\Phi,\Theta,-\Phi)$ is a rotation by $\Theta$ about an axis $\hat y'$ in the $xy$ plane that has been rotated by $\Phi$ about $\hat z$, and $D^J_{M''M}(\Phi,-\Theta,-\Phi) $ is the inverse rotation. Thus, the result is a rotation about $z'$ that has been rotated by $R_z(\Phi)R_y(\Theta)R_z(-\Phi)$.