Say the quaternions real and imaginary part are written as $(q_1, \vec q)$. One useful multiplication property is $qr=(q_1r_1 - \langle\vec q, \vec r\rangle, q_1\vec r + r_1\vec q + \vec q \times \vec r)$. I am interested in why for unit quaternion $q=\left(cos\left(\frac{\phi}{2}\right), \vec v \sin\left(\frac{\phi}{2}\right)\right), |v|=1$ the formula $x'=qxq^{-1}=qxq^*$ is a rotation of the point $x$ (represented as imaginary quaternion) by an angle $\phi$ around rotation axis $\vec v$

I know a proof similar to the wikipedia proof, it justs plugs in $q$ and $q^*$, uses the multiplication property, then simplifies and the key step is to recognize that what comes out is a rotation formula .

I am looking for something more intuitive. Like for complex numbers it is easy to see that multiplying $z=re^{i\phi}$ and $e^{i\theta}$ gives you a rotation with $re^{i(\phi+\theta)}$

In this post one has for imaginary quaternion $\mathbf v$ that $e^\mathbf{v}= \cos|v|+ \mathbf{v}\;\dfrac{\sin |v|}{|v|}$ so it looks somewhat similar to $q$ from above, but I don't see how it helps me.

  • $\begingroup$ Is it really completely obvious that multiplication by $e^{i\theta}$ is a rotation by $\theta$? I suspect it only seems so because one uses it so much that it becomes almost second nature. Most of us don't get nearly that much exercise with quaternions. $\endgroup$
    – David K
    Aug 10, 2015 at 2:46

1 Answer 1


Fun fact: the square roots of $-1$ in $\Bbb H$ are precisely the purely imaginary unit quaternions. For any such quaternion $\bf u$, the subspace $\Bbb R\oplus\Bbb R{\bf u}\subset \Bbb H$ is a real subalgebra isomorphic to $\Bbb C$. Indeed all of the numbers $a+b{\bf u}$ would just act among themselves like complex numbers $a+bi$. Let's denote this subspace by $A({\bf u})$, and its orthogonal complement by $B({\bf u})$. (Use the standard Euclidean inner product on $\Bbb H$ inducing the standard Euclidean norm $\|\cdot\|$.)

Write $B({\bf u})=\langle {\bf v},{\bf w}\rangle$ where $\{{\bf u},{\bf v},{\bf w}\}$ is an ordered orthonormal basis for the purely imaginary quaternions with the same orientation as $\{{\bf i},{\bf j},{\bf k}\}$. Then in particular, $\{1,{\bf u},{\bf v},{\bf w}\}$ has the same multiplication table as $\{1,{\bf i},{\bf j},{\bf k}\}$ (this follows from the rule ${\bf ab}=-{\bf a}\cdot{\bf b}+{\bf a}\times{\bf b}$).

Denote ${\bf q}=\exp(\theta{\bf u})=\cos\theta+{\bf u}\sin\theta$ and let $L_{\bf q}({\bf x})={\bf q}{\bf x}$ and $R_{\bf q}({\bf x})={\bf xq}$. Observe $L_{\bf q}$ and $R_{\bf q}$ are both counterclockwise rotations in the $\langle 1,{\bf u}\rangle$-plane $A({\bf u})$ by an angle $\theta$, and $L_{\bf q}$ is a clockwise rotation whereas $R_{\bf q}$ is a counterclockwise rotation in the $\langle {\bf v},{\bf w}\rangle$-plane $B({\bf u})$. Composing the two maps would double up their effect in the first plane and cancel their effects in the second; if instead we compose the maps $L_{\bf q}$ and $R_{\bf q^{-1}}$ then the effects in the first plane will cancel and the effects in the second plane will double up.

Therefore, ${\bf x}\mapsto {\bf qxq}^{-1}$ where ${\bf q}=\exp(\theta{\bf u})$ will rotate purely imaginary quaternions around the axis $\Bbb R\bf u$ by the angle $2\theta$ according to the right-hand rule.

  • $\begingroup$ Hi, thanks for the answer sorry for looking at it so late. Helped out a lot. $\endgroup$
    – myelf
    Oct 8, 2015 at 16:52
  • $\begingroup$ I've been looking for an intuitive explanation of this, and this exactly fits the bill. Nicely done, great answer! $\endgroup$
    – RghtHndSd
    Sep 26, 2016 at 15:14

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