Rotations using Quaternions

Section 2.2 Rotations using Quaternions

(The material in this section was originally published in [1].)

It is important to realize that \(\pm i\text{,}\) \(\pm j\text{,}\) and \(\pm k\) are not the only square roots of \(-1\text{.}\) Rather, any imaginary quaternion squares to a negative number, so it is only necessary to choose its norm to be 1 in order to get a square root of \(-1\text{.}\) The imaginary quaternions of norm 1 form a sphere; in the above notation, this is the set of points

\begin{equation} q_2^2 + q_3^2 + q_4^2 = 1\tag{2.2.1} \end{equation}

(and \(q_1=0\)). Any such unit imaginary quaternion \(u\) can be used to construct a complex subalgebra of \(\HH\text{,}\) which we will also denote by \(\CC\text{,}\) namely

\begin{equation} \CC = \{a + b\,u\}\tag{2.2.2} \end{equation}

with \(a,b\in\RR\text{.}\) Furthermore, we can use the identity~(\ref{cis}) to write

\begin{equation} e^{u\theta} = \cos\theta + u\sin\theta\tag{2.2.3} \end{equation}

This means that any quaternion can be written in the form

\begin{equation} q = r e^{u\theta}\tag{2.2.4} \end{equation}

where

\begin{equation} r=|q|\tag{2.2.5} \end{equation}

and where \(u\) denotes the direction of the imaginary part of \(q\text{.}\)

We define (algebraic) conjugation of one quaternion \(q\) by another quaternion \(p\) by \(pqp^{-1}\text{.}\) ¹ The norm of \(p\) is irrelevant here, so we might as well assume that \(|p|=1\text{,}\) in which case \(p^{-1}=\bar{p}\)

What is the result of conjugating a quaternion by \(i\text{?}\)

A useful strategy for solving problems involving the quaternions is to break up the quaternions into a pair of complex numbers. Write \(q\) in terms of a pair of complex numbers via

\begin{equation} q = q_1 + q_2 j\tag{2.2.6} \end{equation}

Then \(i\) commutes with the complex numbers \(q_1\) and \(q_2\text{,}\) but anticommutes with \(j\text{.}\) Thus,

\begin{equation} iq\bar\imath = i q_1 \bar\imath + i q_2 j \bar\imath = i q_1 \bar\imath - i q_2 \bar\imath j = q_1 - q_2 j\tag{2.2.7} \end{equation}

Conjugation by \(i\) therefore leaves the complex plane untouched, but yields a rotation by \(\pi\) in the \(jk\)-plane. Analogous results would hold for conjugation by any other imaginary quaternionic unit, such as \(u\text{.}\)

What is the result of conjugating a quaternion by \(e^{i\theta}\text{?}\)

Interchanging the roles of \(i\) and \(j\) in the previous discussion, conjugation by \(j\) yields a rotation by \(\pi\) in the \(ki\)-plane, so that

\begin{equation} j e^{-i\theta} \bar\jmath = e^{i\theta}\tag{2.2.8} \end{equation}

Multiplying both of these equations on the right by \(j\) yields the important relation

\begin{equation} j e^{-i\theta} = e^{i\theta} j\tag{2.2.9} \end{equation}

Thus,

\begin{align} e^{i\theta} q e^{-i\theta} \amp= e^{i\theta} q_1 e^{-i\theta} + e^{i\theta} q_2 j e^{-i\theta}\notag\\ \amp= e^{i\theta} q_1 e^{-i\theta} + e^{i\theta} q_2 e^{i\theta} j\notag\\ \amp= q_1 + q_2 e^{2i\theta} j\tag{2.2.10} \end{align}

corresponding to a rotation by \(2\theta\) in the \(jk\)-plane.

The name is unfortunate; algebraic conjugation has nothing to do with complex conjugation.