From the Kinematics of Precession Motion to Generalized Rabi Cycles

Received Sep 29, 2017; Accepted Jan 16, 2018

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction: Quaternions and Vector-Parameters

The Clifford algebra of quaternions $\mathbb{H}\cong C{\mathcal{l}}_{\mathrm{0,2}}(\mathbb{R})\cong C{\mathcal{l}}_{\mathrm{3}}^{\mathrm{0}}(\mathbb{R})$ is known to be intimately related to the rotation group in ${\mathbb{R}}^{\mathrm{3}}$ and one way to see it is via the standard spin covering map $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(\mathrm{3})\cong \mathrm{S}\mathrm{U}(\mathrm{2})\stackrel{{\mathbb{Z}}_{\mathrm{2}}}{\to }\mathrm{S}\mathrm{O}(\mathrm{3})$ that is topologically a projection from the unit sphere ${\mathbb{S}}^{\mathrm{3}}\stackrel{{\mathbb{Z}}_{\mathrm{2}}}{\to }\mathbb{R}{\mathbb{P}}^{\mathrm{3}}$. The basis of bivectors in $C{\mathcal{l}}_{\mathrm{3}}$ is spanned by three units satisfying$$\begin{array}{}\text{(1)}& {\mathbf{i}}^{\mathrm{2}}={\mathbf{j}}^{\mathrm{2}}={\mathbf{k}}^{\mathrm{2}}=\mathbf{i}\mathbf{j}\mathbf{k}=-\mathbf{1}\mathbf{,}\end{array}$$where $\mathbf{1}$ denotes the identity element, so one can express each quaternion $\zeta \in \mathbb{H}$ in the form$$\begin{array}{}\text{(2)}& \zeta =\left({\zeta }_{\mathrm{0}},\mathit{\zeta }\right)={\zeta }_{\mathrm{0}}\mathbf{1}+{\zeta }_{\mathrm{1}}\mathbf{i}+{\zeta }_{\mathrm{2}}\mathbf{j}+{\zeta }_{\mathrm{3}}\mathbf{k}.\end{array}$$Hence, as a linear space, $\mathbb{H}$ may obviously be identified with ${\mathbb{R}}^{\mathrm{4}}\cong \mathbb{R}\times {\mathbb{R}}^{\mathrm{3}}$ by introducing coordinates as shown above. At the same time, quaternions have a specific Clifford multiplication rule that can be expressed in the above notation as$$\begin{array}{}\text{(3)}& \left({\zeta }_{\mathrm{0}},\mathit{\zeta }\right)\left({\tilde{\zeta }}_{\mathrm{0}},\tilde{\mathit{\zeta }}\right)=\left({\zeta }_{\mathrm{0}}{\tilde{\zeta }}_{\mathrm{0}}-\mathit{\zeta }·\tilde{\mathit{\zeta }},{\zeta }_{\mathrm{0}}\tilde{\mathit{\zeta }}+{\tilde{\zeta }}_{\mathrm{0}}\mathit{\zeta }+\mathit{\zeta }\times \tilde{\mathit{\zeta }}\right),\end{array}$$where $·$ and $\times$ stand for the dot and cross products in ${\mathbb{R}}^{\mathrm{3}}$, respectively. Just like in the case of complex numbers, Clifford conjugation in $\mathbb{H}$ is given by sign inversion of the imaginary (vector) part $\mathit{\zeta }$: that is, $\zeta =({\zeta }_{\mathrm{0}},\mathit{\zeta })\to {\zeta }^{\ast }=({\zeta }_{\mathrm{0}},-\mathit{\zeta })$, and since there are no zero divisors, the quaternion norm defined as ${‖\zeta ‖}^{\mathrm{2}}\mathbf{1}={\zeta }^{\ast }\zeta$ yields the inverse of every nonzero element ${\zeta }^{-\mathrm{1}}$

$$\begin{array}{}\hspace{1em}\end{array}$$$$\begin{array}{}\hspace{1em}\end{array}$$$$\begin{array}{}\end{array}$$$$\begin{array}{}\hspace{1em}\end{array}$$$$\begin{array}{}\left(\sqrt{\hspace{0.17em}}\right)\end{array}$$$$\begin{array}{}\left(\begin{array}{cc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\end{array}\right)\end{array}$$

$\hspace{0.17em}\hspace{0.17em}$$$\begin{array}{}\left(\hspace{0.17em}\right)\end{array}$$$$\begin{array}{}\hspace{0.17em}\end{array}$$$$\begin{array}{}\hspace{0.17em}\end{array}$$

$$\begin{array}{}\end{array}$$

$$\begin{array}{}\end{array}$$$$\begin{array}{}\left({\hspace{0.17em}}^{}\right)\hspace{0.17em}{\hspace{0.17em}}^{}\end{array}$$

$$\begin{array}{}\hspace{1em}\end{array}$$$$\begin{array}{c}\hspace{0.17em}\end{array}$$

$$\begin{array}{}\end{array}$$$$\begin{array}{}\begin{array}{c}\end{array}\end{array}$$$$\begin{array}{}\left(\hspace{0.17em}\hspace{0.17em}\right)\end{array}$$

$$\begin{array}{}\hspace{1em}\end{array}$$$$\begin{array}{}\left(\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\right)\end{array}$$

$$\begin{array}{}\end{array}$$