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

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: $$D_{m'm}^{j=1/2}(\varphi ,\theta,\phi=2\pi ) = \begin{pmatrix} -e^{-i\varphi/2} \cos \frac{\theta}{2} & -e^{-i\varphi/2} \sin \frac{\theta}{2}\\ e^{i\varphi/2} \sin \frac{\theta}{2} & -e^{i\varphi/2} \cos \frac{\theta}{2} \end{pmatrix} \neq - \mathbb{I}$$

• Are you conflicted about the definition? You may compose the three Euler angles to an $\phi \hat n$. Oct 6, 2020 at 13:33
• @CosmasZachos - I understand the definition but I'm not sure about the exact relationship between the Euler angles and $\hat{n}$. The proposed angles $\alpha=\varphi,\beta=\theta,\gamma=\phi = 2\pi$ (which intuitively make sense) don't actually yield $D_{m'm}^{j}(\alpha ,\beta ,\gamma )=-\mathbb{I}_{2\times 2}$ as one would expect in the aforementioned example. Note that I work in the standard $zyz$ convention. Oct 6, 2020 at 13:39
• @CosmasZachos - I updated my question and added the Wigner-D matrix calculation for $j=1/2$ and the proposed angles. Clearly, the proposed angles are wrong and I would like to understand why. Oct 6, 2020 at 13:56
• I already proposed one such construction: first, we need to rotate the system such that one of its axes (say $z$) points in the direction of $\hat{n}$. This can be achieved by taking $\alpha = \varphi$ (rotate the system around $z$ by the azimuthal angle of $\hat{n}$, such that $\hat{n}$ now lies in the $xz$ plane of the rotated system) and $\beta=\theta$ (rotate the new system through the polar angle of $\hat{n}$ about the previously rotated $y$ axis). Now, when $\hat{n}$ and $z$ coincide, rotate the system around $z$ by an amount of $\phi=2\pi$. But this construction seems to be wrong. Oct 6, 2020 at 14:19
• Well, draw the picture. You effectively rotated by π around n in pictorial terms. This does not amount to a reflection in the fixed axis system, rightly so. Oct 6, 2020 at 14:43

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 $$\hat{\boldsymbol{n}}$$.)
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 $$R_z(\Phi)R_y(\Theta)R_z(-\Phi)$$.
• Thank you for the great reference. There seems to be some ambiguity in the parameters, because the convenient explicit form $U_{MM^{\prime}}^{J}\left(\omega;\Theta,\Phi\right)=i^{M-M^{\prime}}e^{-i\left(M-M^{\prime}\right)\Phi}\left(\frac{1-i\tan\frac{\omega}{2}\cos\Theta}{\sqrt{1+\tan^{2}\frac{\omega}{2}\cos^{2}\Theta}}\right)^{M+M^{\prime}}d_{MM^{\prime}}^{J}\left(\xi\right)$ yields $\delta_{MM^{\prime}}$ as opposed to $-\delta_{MM^{\prime}}$ for $J=1/2,\omega=2\pi,\xi=0$. On the other hand, if $\xi=2\pi$ then the result is correct. Perhaps $\omega$ is restricted to $[0,2\pi)$? Oct 6, 2020 at 18:34
• Yes I seem to remember that $\omega$ is restricted because the correspondence between the $\Theta,\Phi$ and the Euler angles is geometrical but I can't find the paper where I read this. Or maybe there's a $\pm$ in the square root that one chooses with a geometrical argument. Oct 6, 2020 at 18:37
• In §4.5.4, which is dedicated to orthogonality and completeness they state that the parameters are defined in the domain $0\leq\Theta\leq\pi,0\leq\Phi<2\pi,0\leq\omega<2\pi$. But beyond that, since $\xi$ is determined by $\sin\frac{\xi}{2}=\sin\frac{\omega}{2}\sin\Theta$ there seems to be an inherent ambiguity in $d_{MM^{\prime}}^{1/2}(\xi)$ which can be $\pm\delta_{MM^{\prime}}$ depending on $\xi$. In light of this, what would be the proper way to handle the aforementioned situation of a spin-$1/2$ system with $\omega=2\pi$ (or even $\omega=4\pi$, which brings back the original spinor)? Oct 6, 2020 at 18:50