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

  • $\begingroup$ Are you conflicted about the definition? You may compose the three Euler angles to an $\phi \hat n$. $\endgroup$ Oct 6, 2020 at 13:33
  • $\begingroup$ @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. $\endgroup$
    – grjj3
    Oct 6, 2020 at 13:39
  • $\begingroup$ @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. $\endgroup$
    – grjj3
    Oct 6, 2020 at 13:56
  • $\begingroup$ 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. $\endgroup$
    – grjj3
    Oct 6, 2020 at 14:19
  • $\begingroup$ 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. $\endgroup$ Oct 6, 2020 at 14:43

1 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 $R_z(\Phi)R_y(\Theta)R_z(-\Phi)$.

  • $\begingroup$ 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 $[0,2\pi)$? $\endgroup$
    – grjj3
    Oct 6, 2020 at 18:34
  • $\begingroup$ 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. $\endgroup$ Oct 6, 2020 at 18:37
  • 1
    $\begingroup$ 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)? $\endgroup$
    – grjj3
    Oct 6, 2020 at 18:50

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