Example conjugation operation

A rotation of 120° around the first diagonal permutes i, j, and k cyclically

Conjugating p by q refers to the operation p ↦ qpq⁻¹.

Consider the rotation f around the axis ${\displaystyle {\vec {v}}=\mathbf {i} +\mathbf {j} +\mathbf {k} }$, with a rotation angle of 120°, or 2π/3 radians.

${\displaystyle \alpha ={\frac {2\pi }{3}}}$

p ↦ q p for q = 1 + i + j + k/2 on the unit 3-sphere. Note this one-sided (namely, left) multiplication yields a 60° rotation of quaternions

The length of ${\displaystyle {\vec {v}}}$ is √3, the half angle is π/3 (60°) with cosine 1/2, (cos  60° = 0.5) and sine √3/2, (sin  60° ≈ 0.866). We are therefore dealing with a conjugation by the unit quaternion

${\displaystyle {\begin{aligned}u&=\cos {\frac {\alpha }{2}}+\sin {\frac {\alpha }{2}}\cdot {\frac {1}{\|{\vec {v}}\|}}{\vec {v}}\\&=\cos {\frac {\pi }{3}}+\sin {\frac {\pi }{3}}\cdot {\frac {1}{\sqrt {3}}}{\vec {v}}\\&={\frac {1}{2}}+{\frac {\sqrt {3}}{2}}\cdot {\frac {1}{\sqrt {3}}}{\vec {v}}\\&={\frac {1}{2}}+{\frac {\sqrt {3}}{2}}\cdot {\frac {\mathbf {i} +\mathbf {j} +\mathbf {k} }{\sqrt {3}}}\\&={\frac {1+\mathbf {i} +\mathbf {j} +\mathbf {k} }{2}}\end{aligned}}}$

If f is the rotation function,

${\displaystyle f(a\mathbf {i} +b\mathbf {j} +c\mathbf {k} )=u(a\mathbf {i} +b\mathbf {j} +c\mathbf {k} )u^{-1}}$

It can be proven that the inverse of a unit quaternion is obtained simply by changing the sign of its imaginary components. As a consequence,

${\displaystyle u^{-1}={\dfrac {1-\mathbf {i} -\mathbf {j} -\mathbf {k} }{2}}}$

and

${\displaystyle f(a\mathbf {i} +b\mathbf {j} +c\mathbf {k} )={\dfrac {1+\mathbf {i} +\mathbf {j} +\mathbf {k} }{2}}(a\mathbf {i} +b\mathbf {j} +c\mathbf {k} ){\dfrac {1-\mathbf {i} -\mathbf {j} -\mathbf {k} }{2}}}$

This can be simplified, using the ordinary rules for quaternion arithmetic, to

${\displaystyle f(a\mathbf {i} +b\mathbf {j} +c\mathbf {k} )=c\mathbf {i} +a\mathbf {j} +b\mathbf {k} }$

As expected, the rotation corresponds to keeping a cube held fixed at one point, and rotating it 120° about the long diagonal through the fixed point (observe how the three axes are permuted cyclically).

Quaternion-derived rotation matrix

A quaternion rotation ${\displaystyle \mathbf {p'} =\mathbf {q} \mathbf {p} \mathbf {q} ^{-1}}$ (with ${\displaystyle \mathbf {q} =q_{r}+q_{i}\mathbf {i} +q_{j}\mathbf {j} +q_{k}\mathbf {k} }$) can be algebraically manipulated into a matrix rotation ${\displaystyle \mathbf {p'} =\mathbf {Rp} }$, where ${\displaystyle \mathbf {R} }$ is the rotation matrix given by:^[10]

${\displaystyle \mathbf {R} ={\begin{bmatrix}1-2s(q_{j}^{2}+q_{k}^{2})&2s(q_{i}q_{j}-q_{k}q_{r})&2s(q_{i}q_{k}+q_{j}q_{r})\\2s(q_{i}q_{j}+q_{k}q_{r})&1-2s(q_{i}^{2}+q_{k}^{2})&2s(q_{j}q_{k}-q_{i}q_{r})\\2s(q_{i}q_{k}-q_{j}q_{r})&2s(q_{j}q_{k}+q_{i}q_{r})&1-2s(q_{i}^{2}+q_{j}^{2})\end{bmatrix}}}$

Here ${\displaystyle s=\|q\|^{-2}}$ and if q is a unit quaternion, ${\displaystyle s=1^{-2}=1}$.

This can be obtained by using vector calculus and linear algebra if we express ${\displaystyle \mathbf {p} }$ and ${\displaystyle \mathbf {q} }$ as scalar and vector parts and use the formula for the multiplication operation in the equation ${\displaystyle \mathbf {p'} =\mathbf {q} \mathbf {p} \mathbf {q} ^{-1}}$. If we write ${\displaystyle \mathbf {p} }$ as ${\displaystyle \left(0,\ \mathbf {p} \right)}$, ${\displaystyle \mathbf {p} '}$ as ${\displaystyle \left(0,\ \mathbf {p} '\right)}$ and ${\displaystyle \mathbf {q} }$ as ${\displaystyle \left(q_{r},\ \mathbf {v} \right)}$, where ${\displaystyle \mathbf {v} =\left(q_{i},q_{j},q_{k}\right)}$, our equation turns into ${\displaystyle \left(0,\ \mathbf {p} '\right)=\left(q_{r},\ \mathbf {v} \right)\left(0,\ \mathbf {p} \right)s\left(q_{r},\ -\mathbf {v} \right)}$. By using the formula for multiplication of two quaternions that are expressed as scalar and vector parts,

${\displaystyle \left(r_{1},\ {\vec {v}}_{1}\right)\left(r_{2},\ {\vec {v}}_{2}\right)=\left(r_{1}r_{2}-{\vec {v}}_{1}\cdot {\vec {v}}_{2},\ r_{1}{\vec {v}}_{2}+r_{2}{\vec {v}}_{1}+{\vec {v}}_{1}\times {\vec {v}}_{2}\right),}$

this equation can be rewritten as

${\displaystyle {\begin{aligned}(0,\ \mathbf {p} ')=&((q_{r},\ \mathbf {v} )(0,\ \mathbf {p} ))s(q_{r},\ -\mathbf {v} )\\=&(q_{r}0-\mathbf {v} \cdot \mathbf {p} ,\ q_{r}\mathbf {p} +0\mathbf {v} +\mathbf {v} \times \mathbf {p} )s(q_{r},\ -\mathbf {v} )\\=&s(-\mathbf {v} \cdot \mathbf {p} ,\ q_{r}\mathbf {p} +\mathbf {v} \times \mathbf {p} )(q_{r},\ -\mathbf {v} )\\=&s(-\mathbf {v} \cdot \mathbf {p} q_{r}-(q_{r}\mathbf {p} +\mathbf {v} \times \mathbf {p} )\cdot (-\mathbf {v} ),\ (-\mathbf {v} \cdot \mathbf {p} )(-\mathbf {v} )+q_{r}(q_{r}\mathbf {p} +\mathbf {v} \times \mathbf {p} )+(q_{r}\mathbf {p} +\mathbf {v} \times \mathbf {p} )\times (-\mathbf {v} ))\\=&s\left(-\mathbf {v} \cdot \mathbf {p} q_{r}+q_{r}\mathbf {v} \cdot \mathbf {p} ,\ \mathbf {v} \left(\mathbf {v} \cdot \mathbf {p} \right)+q_{r}^{2}\mathbf {p} +q_{r}\mathbf {v} \times \mathbf {p} +\mathbf {v} \times \left(q_{r}\mathbf {p} +\mathbf {v} \times \mathbf {p} \right)\right)\\=&\left(0,\ s\left(\mathbf {v} \otimes \mathbf {v} +q_{r}^{2}\mathbf {I} +2q_{r}[\mathbf {v} ]_{\times }+[\mathbf {v} ]_{\times }^{2}\right)\mathbf {p} \right),\end{aligned}}}$

where ${\displaystyle \otimes }$ denotes the outer product, ${\displaystyle \mathbf {I} }$ is the identity matrix and ${\displaystyle [\mathbf {v} ]_{\times }}$ is the transformation matrix that when multiplied from the right with a vector ${\displaystyle \mathbf {u} }$ gives the cross product ${\displaystyle \mathbf {v} \times \mathbf {u} }$.

Since ${\displaystyle \mathbf {p} '=\mathbf {R} \mathbf {p} }$, we can identify ${\displaystyle \mathbf {R} }$ as ${\displaystyle s\left(\mathbf {v} \otimes \mathbf {v} +q_{r}^{2}\mathbf {I} +2q_{r}[\mathbf {v} ]_{\times }+[\mathbf {v} ]_{\times }^{2}\right)}$, which upon expansion should result in the expression written in matrix form above.

Recovering the axis-angle representation

The expression ${\displaystyle \mathbf {q} \mathbf {p} \mathbf {q} ^{-1}}$ rotates any vector quaternion ${\displaystyle \mathbf {p} }$ around an axis given by the vector ${\displaystyle \mathbf {a} }$ by the angle ${\displaystyle \theta }$, where ${\displaystyle \mathbf {a} }$ and ${\displaystyle \theta }$ depends on the quaternion ${\displaystyle \mathbf {q} =q_{r}+q_{i}\mathbf {i} +q_{j}\mathbf {j} +q_{k}\mathbf {k} }$.

${\displaystyle \mathbf {a} }$ and ${\displaystyle \theta }$ can be found from the following equations:

${\displaystyle {\begin{aligned}(a_{x},a_{y},a_{z})={}&{\frac {(q_{i},q_{j},q_{k})}{\sqrt {q_{i}^{2}+q_{j}^{2}+q_{k}^{2}}}}\\[2pt]\theta =2\operatorname {atan2} &\left({\sqrt {q_{i}^{2}+q_{j}^{2}+q_{k}^{2}}},\,q_{r}\right),\end{aligned}}}$

where ${\displaystyle \operatorname {atan2} }$ is the two-argument arctangent.

Care should be taken when the quaternion approaches a scalar, since due to degeneracy the axis of an identity rotation is not well-defined.

The composition of spatial rotations

A benefit of the quaternion formulation of the composition of two rotations R_B and R_A is that it yields directly the rotation axis and angle of the composite rotation R_C = R_BR_A.

Let the quaternion associated with a spatial rotation R be constructed from its rotation axis S with the rotation angle ${\displaystyle \varphi }$ around this axis. The associated quaternion is given by

$${\displaystyle S=\cos {\frac {\varphi }{2}}+\mathbf {S} \sin {\frac {\varphi }{2}}.}$$

Then the composition of the rotation R_B with R_A is the rotation R_C = R_BR_A with rotation axis and angle defined by the product of the quaternions

$${\displaystyle A=\cos {\frac {\alpha }{2}}+\mathbf {A} \sin {\frac {\alpha }{2}}\quad {\text{and}}\quad B=\cos {\frac {\beta }{2}}+\mathbf {B} \sin {\frac {\beta }{2}},}$$

that is

$${\displaystyle C=\cos {\frac {\gamma }{2}}+\mathbf {C} \sin {\frac {\gamma }{2}}=\left(\cos {\frac {\beta }{2}}+\mathbf {B} \sin {\frac {\beta }{2}}\right)\left(\cos {\frac {\alpha }{2}}+\mathbf {A} \sin {\frac {\alpha }{2}}\right).}$$

Expand this product to obtain

$${\displaystyle \cos {\frac {\gamma }{2}}+\mathbf {C} \sin {\frac {\gamma }{2}}=\left(\cos {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}-\mathbf {B} \cdot \mathbf {A} \sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\right)+\left(\mathbf {B} \sin {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}+\mathbf {A} \sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}+\mathbf {B} \times \mathbf {A} \sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\right).}$$

Divide both sides of this equation by the identity, which is the law of cosines on a sphere,

$${\displaystyle \cos {\frac {\gamma }{2}}=\cos {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}-\mathbf {B} \cdot \mathbf {A} \sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}},}$$

and compute

$${\displaystyle \mathbf {C} \tan {\frac {\gamma }{2}}={\frac {\mathbf {B} \tan {\frac {\beta }{2}}+\mathbf {A} \tan {\frac {\alpha }{2}}+\mathbf {B} \times \mathbf {A} \tan {\frac {\beta }{2}}\tan {\frac {\alpha }{2}}}{1-\mathbf {B} \cdot \mathbf {A} \tan {\frac {\beta }{2}}\tan {\frac {\alpha }{2}}}}.}$$

This is Rodrigues' formula for the axis of a composite rotation defined in terms of the axes of the two rotations. He derived this formula in 1840 (see page 408).^[11]

The three rotation axes A, B, and C form a spherical triangle and the dihedral angles between the planes formed by the sides of this triangle are defined by the rotation angles. Hamilton^[12] presented the component form of these equations showing that the quaternion product computes the third vertex of a spherical triangle from two given vertices and their associated arc-lengths, which is also defines an algebra for points in Elliptic geometry.

Axis–angle composition

The normalized rotation axis, removing the ${\textstyle \cos {\frac {\gamma }{2}}}$ from the expanded product, leaves the vector which is the rotation axis, times some constant. Care should be taken normalizing the axis vector when ${\displaystyle \gamma }$ is ${\displaystyle 0}$ or ${\displaystyle k2\pi }$ where the vector is near ${\displaystyle 0}$; which is identity, or 0 rotation around any axis.

$${\displaystyle {\begin{aligned}\gamma &=2\cos ^{-1}\left(\cos {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}-\mathbf {B} \cdot \mathbf {A} \sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\right)\\\mathbf {D} &=\mathbf {B} \sin {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}+\mathbf {A} \sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}+\mathbf {B} \times \mathbf {A} \sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\end{aligned}}}$$

Or with angle addition trigonometric substitutions...

$${\displaystyle {\begin{aligned}\gamma &=2\cos ^{-1}\left(\left(1-\mathbf {A} \cdot \mathbf {B} \right)\cos {\frac {\beta -\alpha }{2}}+\left(1+\mathbf {A} \cdot \mathbf {B} \right)\cos {\frac {\beta +\alpha }{2}}\right)\\\mathbf {D} &=\left(\sin {\frac {\beta +\alpha }{2}}+\sin {\frac {\beta -\alpha }{2}}\right)\mathbf {A} +\left(\sin {\frac {\beta +\alpha }{2}}-\sin {\frac {\beta -\alpha }{2}}\right)\mathbf {B} +\left(\cos {\frac {\beta -\alpha }{2}}-\cos {\frac {\beta +\alpha }{2}}\right)\mathbf {B} \times \mathbf {A} \end{aligned}}}$$

finally normalizing the rotation axis: ${\textstyle {\frac {\mathbf {D} }{2\sin {\frac {1}{2}}\gamma }}}$ or ${\textstyle {\frac {\mathbf {D} }{\|\mathbf {D} \|}}}$.

Differentiation with respect to the rotation quaternion

The rotated quaternion p' = q p q⁻¹ needs to be differentiated with respect to the rotating quaternion q, when the rotation is estimated from numerical optimization. The estimation of rotation angle is an essential procedure in 3D object registration or camera calibration. For unitary q and pure imaginary p, that is for a rotation in 3D space, the derivatives of the rotated quaternion can be represented using matrix calculus notation as

${\displaystyle {\begin{aligned}{\frac {\partial \mathbf {p'} }{\partial \mathbf {q} }}\equiv \left[{\frac {\partial \mathbf {p'} }{\partial q_{0}}},{\frac {\partial \mathbf {p'} }{\partial q_{x}}},{\frac {\partial \mathbf {p'} }{\partial q_{y}}},{\frac {\partial \mathbf {p'} }{\partial q_{z}}}\right]=\left[\mathbf {pq} -(\mathbf {pq} )^{*},(\mathbf {pqi} )^{*}-\mathbf {pqi} ,(\mathbf {pqj} )^{*}-\mathbf {pqj} ,(\mathbf {pqk} )^{*}-\mathbf {pqk} \right].\end{aligned}}}$

A derivation can be found in.^[13]

Quaternions

The complex numbers can be defined by introducing an abstract symbol i which satisfies the usual rules of algebra and additionally the rule i² = −1. This is sufficient to reproduce all of the rules of complex number arithmetic: for example:

${\displaystyle (a+b\mathbf {i} )(c+d\mathbf {i} )=ac+ad\mathbf {i} +b\mathbf {i} c+b\mathbf {i} d\mathbf {i} =ac+ad\mathbf {i} +bc\mathbf {i} +bd\mathbf {i} ^{2}=(ac-bd)+(bc+ad)\mathbf {i} .}$

In the same way the quaternions can be defined by introducing abstract symbols i, j, k which satisfy the rules i² = j² = k² = i j k = −1 and the usual algebraic rules except the commutative law of multiplication (a familiar example of such a noncommutative multiplication is matrix multiplication). From this all of the rules of quaternion arithmetic follow, such as the rules on multiplication of quaternion basis elements. Using these rules, one can show that:

${\displaystyle {\begin{aligned}&(a+b\mathbf {i} +c\mathbf {j} +d\mathbf {k} )(e+f\mathbf {i} +g\mathbf {j} +h\mathbf {k} )=\\&(ae-bf-cg-dh)+(af+be+ch-dg)\mathbf {i} +(ag-bh+ce+df)\mathbf {j} +(ah+bg-cf+de)\mathbf {k} .\end{aligned}}}$

The imaginary part ${\displaystyle b\mathbf {i} +c\mathbf {j} +d\mathbf {k} }$ of a quaternion behaves like a vector ${\displaystyle {\vec {v}}=(b,c,d)}$ in three-dimensional vector space, and the real part a behaves like a scalar in R. When quaternions are used in geometry, it is more convenient to define them as a scalar plus a vector:

${\displaystyle a+b\mathbf {i} +c\mathbf {j} +d\mathbf {k} =a+{\vec {v}}.}$

Some might find it strange to add a number to a vector, as they are objects of very different natures, or to multiply two vectors together, as this operation is usually undefined. However, if one remembers that it is a mere notation for the real and imaginary parts of a quaternion, it becomes more legitimate. In other words, the correct reasoning is the addition of two quaternions, one with zero vector/imaginary part, and another one with zero scalar/real part:

${\displaystyle q_{1}=s+{\vec {v}}=\left(s,{\vec {0}}\right)+\left(0,{\vec {v}}\right).}$

We can express quaternion multiplication in the modern language of vector cross and dot products (which were actually inspired by the quaternions in the first place^[14]). When multiplying the vector/imaginary parts, in place of the rules i² = j² = k² = ijk = −1 we have the quaternion multiplication rule:

${\displaystyle {\vec {v}}{\vec {w}}=-{\vec {v}}\cdot {\vec {w}}+{\vec {v}}\times {\vec {w}},}$

where:

• ${\displaystyle {\vec {v}}{\vec {w}}}$ is the resulting quaternion,
• ${\displaystyle {\vec {v}}\times {\vec {w}}}$ is vector cross product (a vector),
• ${\displaystyle {\vec {v}}\cdot {\vec {w}}}$ is vector scalar product (a scalar).

Quaternion multiplication is noncommutative (because of the cross product, which anti-commutes), while scalar–scalar and scalar–vector multiplications commute. From these rules it follows immediately that (see details):

${\displaystyle q_{1}q_{2}=\left(s+{\vec {v}}\right)\left(t+{\vec {w}}\right)=\left(st-{\vec {v}}\cdot {\vec {w}}\right)+\left(s{\vec {w}}+t{\vec {v}}+{\vec {v}}\times {\vec {w}}\right).}$

The (left and right) multiplicative inverse or reciprocal of a nonzero quaternion is given by the conjugate-to-norm ratio (see details):

${\displaystyle q_{1}^{-1}=\left(s+{\vec {v}}\right)^{-1}={\frac {\left(s+{\vec {v}}\right)^{*}}{\lVert s+{\vec {v}}\rVert ^{2}}}={\frac {s-{\vec {v}}}{s^{2}+\lVert {\vec {v}}\rVert ^{2}}},}$

as can be verified by direct calculation (note the similarity to the multiplicative inverse of complex numbers).

Rotation identity

Let ${\displaystyle {\vec {u}}}$ be a unit vector (the rotation axis) and let ${\displaystyle q=\cos {\frac {\alpha }{2}}+{\vec {u}}\sin {\frac {\alpha }{2}}}$. Our goal is to show that

${\displaystyle {\vec {v}}'=q{\vec {v}}q^{-1}=\left(\cos {\frac {\alpha }{2}}+{\vec {u}}\sin {\frac {\alpha }{2}}\right)\,{\vec {v}}\,\left(\cos {\frac {\alpha }{2}}-{\vec {u}}\sin {\frac {\alpha }{2}}\right)}$

yields the vector ${\displaystyle {\vec {v}}}$ rotated by an angle ${\displaystyle \alpha }$ around the axis ${\displaystyle {\vec {u}}}$. Expanding out (and bearing in mind that ${\displaystyle {\vec {u}}{\vec {v}}={\vec {u}}\times {\vec {v}}-{\vec {u}}\cdot {\vec {v}}}$), we have

${\displaystyle {\begin{aligned}{\vec {v}}'&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+\left({\vec {u}}{\vec {v}}-{\vec {v}}{\vec {u}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-{\vec {u}}{\vec {v}}{\vec {u}}\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-\left(\left({\vec {u}}\times {\vec {v}}\right)-\left({\vec {u}}\cdot {\vec {v}}\right)\right){\vec {u}}\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-\left(\left({\vec {u}}\times {\vec {v}}\right){\vec {u}}-\left({\vec {u}}\cdot {\vec {v}}\right){\vec {u}}\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-\left(\left(\left({\vec {u}}\times {\vec {v}}\right)\times {\vec {u}}-\left({\vec {u}}\times {\vec {v}}\right)\cdot {\vec {u}}\right)-\left({\vec {u}}\cdot {\vec {v}}\right){\vec {u}}\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-\left(\left({\vec {v}}-\left({\vec {u}}\cdot {\vec {v}}\right){\vec {u}}\right)-0-\left({\vec {u}}\cdot {\vec {v}}\right){\vec {u}}\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-\left({\vec {v}}-2{\vec {u}}\left({\vec {u}}\cdot {\vec {v}}\right)\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}+\left(2{\vec {u}}\left({\vec {u}}\cdot {\vec {v}}\right)-{\vec {v}}\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]\end{aligned}}}$

If we let ${\displaystyle {\vec {v}}_{\bot }}$ and ${\displaystyle {\vec {v}}_{\|}}$ equal the components of ${\displaystyle {\vec {v}}}$ perpendicular and parallel to ${\displaystyle {\vec {u}}}$ respectively, then ${\displaystyle {\vec {v}}={\vec {v}}_{\bot }+{\vec {v}}_{\|}}$ and ${\displaystyle {\vec {u}}\left({\vec {u}}\cdot {\vec {v}}\right)={\vec {v}}_{\|}}$, leading to

${\displaystyle {\begin{aligned}{\vec {v}}'&={\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}+\left(2{\vec {u}}\left({\vec {u}}\cdot {\vec {v}}\right)-{\vec {v}}\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&=\left({\vec {v}}_{\|}+{\vec {v}}_{\bot }\right)\cos ^{2}{\frac {\alpha }{2}}+2\left({\vec {u}}\times {\vec {v}}\right)\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}+\left({\vec {v}}_{\|}-{\vec {v}}_{\bot }\right)\sin ^{2}{\frac {\alpha }{2}}\\[6pt]&={\vec {v}}_{\|}\left(\cos ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\alpha }{2}}\right)+\left({\vec {u}}\times {\vec {v}}\right)\left(2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}\right)+{\vec {v}}_{\bot }\left(\cos ^{2}{\frac {\alpha }{2}}-\sin ^{2}{\frac {\alpha }{2}}\right)\\[6pt]\end{aligned}}}$

Using the trigonometric pythagorean and double-angle identities, we then have

${\displaystyle {\begin{aligned}{\vec {v}}'&={\vec {v}}_{\|}\left(\cos ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\alpha }{2}}\right)+\left({\vec {u}}\times {\vec {v}}\right)\left(2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}\right)+{\vec {v}}_{\bot }\left(\cos ^{2}{\frac {\alpha }{2}}-\sin ^{2}{\frac {\alpha }{2}}\right)\\[6pt]&={\vec {v}}_{\|}+\left({\vec {u}}\times {\vec {v}}\right)\sin \alpha +{\vec {v}}_{\bot }\cos \alpha \end{aligned}}}$

This is the formula of a rotation by ${\displaystyle \alpha }$ around the u→ axis.

The hypersphere of rotations

Visualizing the space of rotations

Unit quaternions represent the group of Euclidean rotations in three dimensions in a very straightforward way. The correspondence between rotations and quaternions can be understood by first visualizing the space of rotations itself.

Two separate rotations, differing by both angle and axis, in the space of rotations. Here, the length of each axis vector is relative to the respective magnitude of the rotation about that axis.

In order to visualize the space of rotations, it helps to consider a simpler case. Any rotation in three dimensions can be described by a rotation by some angle about some axis; for our purposes, we will use an axis vector to establish handedness for our angle. Consider the special case in which the axis of rotation lies in the xy plane. We can then specify the axis of one of these rotations by a point on a circle through which the vector crosses, and we can select the radius of the circle to denote the angle of rotation.

Similarly, a rotation whose axis of rotation lies in the xy plane can be described as a point on a sphere of fixed radius in three dimensions. Beginning at the north pole of a sphere in three-dimensional space, we specify the point at the north pole to be the identity rotation (a zero angle rotation). Just as in the case of the identity rotation, no axis of rotation is defined, and the angle of rotation (zero) is irrelevant. A rotation having a very small rotation angle can be specified by a slice through the sphere parallel to the xy plane and very near the north pole. The circle defined by this slice will be very small, corresponding to the small angle of the rotation. As the rotation angles become larger, the slice moves in the negative z direction, and the circles become larger until the equator of the sphere is reached, which will correspond to a rotation angle of 180 degrees. Continuing southward, the radii of the circles now become smaller (corresponding to the absolute value of the angle of the rotation considered as a negative number). Finally, as the south pole is reached, the circles shrink once more to the identity rotation, which is also specified as the point at the south pole.

Notice that a number of characteristics of such rotations and their representations can be seen by this visualization. The space of rotations is continuous, each rotation has a neighborhood of rotations which are nearly the same, and this neighborhood becomes flat as the neighborhood shrinks. Also, each rotation is actually represented by two antipodal points on the sphere, which are at opposite ends of a line through the center of the sphere. This reflects the fact that each rotation can be represented as a rotation about some axis, or, equivalently, as a negative rotation about an axis pointing in the opposite direction (a so-called double cover). The "latitude" of a circle representing a particular rotation angle will be half of the angle represented by that rotation, since as the point is moved from the north to south pole, the latitude ranges from zero to 180 degrees, while the angle of rotation ranges from 0 to 360 degrees. (the "longitude" of a point then represents a particular axis of rotation.) Note however that this set of rotations is not closed under composition. Two successive rotations with axes in the xy plane will not necessarily give a rotation whose axis lies in the xy plane, and thus cannot be represented as a point on the sphere. This will not be the case with a general rotation in 3-space, in which rotations do form a closed set under composition.

The sphere of rotations for the rotations that have a "horizontal" axis (in the xy plane).

This visualization can be extended to a general rotation in 3-dimensional space. The identity rotation is a point, and a small angle of rotation about some axis can be represented as a point on a sphere with a small radius. As the angle of rotation grows, the sphere grows, until the angle of rotation reaches 180 degrees, at which point the sphere begins to shrink, becoming a point as the angle approaches 360 degrees (or zero degrees from the negative direction). This set of expanding and contracting spheres represents a hypersphere in four dimensional space (a 3-sphere). Just as in the simpler example above, each rotation represented as a point on the hypersphere is matched by its antipodal point on that hypersphere. The "latitude" on the hypersphere will be half of the corresponding angle of rotation, and the neighborhood of any point will become "flatter" (i.e. be represented by a 3-D Euclidean space of points) as the neighborhood shrinks. This behavior is matched by the set of unit quaternions: A general quaternion represents a point in a four dimensional space, but constraining it to have unit magnitude yields a three-dimensional space equivalent to the surface of a hypersphere. The magnitude of the unit quaternion will be unity, corresponding to a hypersphere of unit radius. The vector part of a unit quaternion represents the radius of the 2-sphere corresponding to the axis of rotation, and its magnitude is the sine of half the angle of rotation. Each rotation is represented by two unit quaternions of opposite sign, and, as in the space of rotations in three dimensions, the quaternion product of two unit quaternions will yield a unit quaternion. Also, the space of unit quaternions is "flat" in any infinitesimal neighborhood of a given unit quaternion.

Non-commutativity

Non-commutative rotation of dice

The multiplication of quaternions is non-commutative. This fact explains how the p ↦ q p q⁻¹ formula can work at all, having q q⁻¹ = 1 by definition. Since the multiplication of unit quaternions corresponds to the composition of three-dimensional rotations, this property can be made intuitive by showing that three-dimensional rotations are not commutative in general.

The figure to the right illustrates this with dice. Use your right hand to create a pair of 90 degree rotations. Both dice are initially configured as shown in the upper left-hand corner (with 1 dot on the top face.) Path A begins with a rotation about the –y axis (using the right-hand rule.), followed by a rotation about the +z axis, resulting in the configuration shown in the lower left corner (5 dots on the top face.) Path B reverses the order of operations, resulting with 3 dots on top.

If you don't have dice, set two books next to each other. Rotate one of them 90 degrees clockwise around the z axis, then flip it 180 degrees around the x axis. Take the other book, flip it 180° around x axis first, and 90° clockwise around z later. The two books do not end up parallel. This shows that, in general, the composition of two different rotations around two distinct spatial axes will not commute.

Orientation

The vector cross product, used to define the axis–angle representation, does confer an orientation ("handedness") to space: in a three-dimensional vector space, the three vectors in the equation a × b = c will always form a right-handed set (or a left-handed set, depending on how the cross product is defined), thus fixing an orientation in the vector space. Alternatively, the dependence on orientation is expressed in referring to such ${\displaystyle {\vec {u}}}$ that specifies a rotation as to axial vectors. In quaternionic formalism the choice of an orientation of the space corresponds to order of multiplication: ij = k but ji = −k. If one reverses the orientation, then the formula above becomes p ↦ q⁻¹ p q, i.e., a unit q is replaced with the conjugate quaternion – the same behaviour as of axial vectors.

Software applications by convention used

The table below groups applications by their adherence to either quaternion convention:^[16]

While use of either convention does not impact the capability or correctness of applications thus created, the authors of ^[16] argued that the Shuster convention should be abandoned because it departs from the much older quaternion multiplication convention by Hamilton and may never be adopted by the mathematical or theoretical physics areas.

Advantages of quaternions

The representation of a rotation as a quaternion (4 numbers) is more compact than the representation as an orthogonal matrix (9 numbers). Furthermore, for a given axis and angle, one can easily construct the corresponding quaternion, and conversely, for a given quaternion one can easily read off the axis and the angle. Both of these are much harder with matrices or Euler angles.

In video games and other applications, one is often interested in "smooth rotations", meaning that the scene should slowly rotate and not in a single step. This can be accomplished by choosing a curve such as the spherical linear interpolation in the quaternions, with one endpoint being the identity transformation 1 (or some other initial rotation) and the other being the intended final rotation. This is more problematic with other representations of rotations.

When composing several rotations on a computer, rounding errors necessarily accumulate. A quaternion that is slightly off still represents a rotation after being normalized: a matrix that is slightly off may not be orthogonal any more and is harder to convert back to a proper orthogonal matrix.

Quaternions also avoid a phenomenon called gimbal lock which can result when, for example in pitch/yaw/roll rotational systems, the pitch is rotated 90° up or down, so that yaw and roll then correspond to the same motion, and a degree of freedom of rotation is lost. In a gimbal-based aerospace inertial navigation system, for instance, this could have disastrous results if the aircraft is in a steep dive or ascent.

Conversion to and from the matrix representation