Quaternions, rotations, spherical coordinates

A unit quaternion (or "rotor") \(\mathbf{R}\) can rotate a vector \(\vec{v}\) into a new vector \(\vec{v}'\) according to the formula \begin{equation} \vec{v}' = \mathbf{R}\, \vec{v}\, \mathbf{R}^{-1}. \end{equation} In principle, a unit quaternion obeys \(\bar{\mathbf{R}} = \mathbf{R}^{-1}\). In practice, however, there are cases where the system is (slightly slower, but) more stable numerically if the explicit inversion is used. And since the inversion is such a simple operation, we just use it.

The Wigner \(\mathfrak{D}\) matrices can be derived directly in terms of components of the unit rotor corresponding to the rotation of which \(\mathfrak{D}\) is a representation. That is, we don't need to use Euler angles or rotation matrices at all. In fact, the resulting expressions are at least as simple as those using Euler angles, but are faster and more accurate to compute. The derivation is shown on this page.

When necessary to make contact with previous literature, we will use the standard spherical coordinates \((\vartheta, \varphi)\), in the physics convention. To explain this, we first establish our right-handed basis vectors: \((\vec{x}, \vec{y}, \vec{z})\), which remain fixed at all times. Thus, \(\vartheta\) measures the angle from the positive \(\vec{z}\) axis and \(\varphi\) measures the angle of the projection into the \(\vec{x}\)-\(\vec{y}\) plane from the positive \(\vec{x}\) axis, as shown in the figure.

Now, if \(\hat{n}\) is the unit vector in that direction, we construct a related rotor \begin{equation} R_{(\vartheta, \varphi)} = e^{\varphi \vec{z}/2}\, e^{\vartheta \vec{y}/2}. \end{equation} This can be obtained as a quaternionic.array object in python as

>>> import quaternionic
>>> vartheta, varphi = 0.1, 0.2
>>> R_tp = quaternionic.array.from_spherical_coordinates(vartheta, varphi)

Here, rotations are given assuming the right-hand screw rule, so that this corresponds to an initial rotation through the angle \(\vartheta\) about \(\vec{y}\), followed by a rotation through \(\varphi\) about \(\vec{z}\). (The factors of \(1/2\) are present because \(R_{(\vartheta, \varphi)}\) is essentially the square-root of the rotation; the effect of this rotor produces the full rotations through \(\vartheta\) and \(\varphi\).) This rotor is particularly useful, because \(\hat{n}\) and the two standard tangent vectors at that point are all given by rotations of the basis vectors \((\vec{x}, \vec{y}, \vec{z})\). Specifically, we have \begin{align} \hat{\vartheta} &= R_{(\vartheta, \varphi)}\,\vec{x}\,R_{(\vartheta, \varphi)}^{-1}, \\\\ \hat{\varphi} &= R_{(\vartheta, \varphi)}\, \vec{y}\, R_{(\vartheta, \varphi)}^{-1}, \\\\ \hat{n} &= R_{(\vartheta, \varphi)}\, \vec{z}\, R_{(\vartheta, \varphi)}^{-1}. \end{align} Note that we interpret the rotations in the active sense, where a point moves, while our coordinate system and basis vectors are left fixed. Hopefully, this won't cause too much confusion if we express the original point with the vector \(\vec{z}\), for example, and the rotated point is therefore given by some \(\mathbf{R}\, \vec{z}\, \bar{\mathbf{R}}\).