Quaternion Product in Scalar Vector Form

The component formula for multiplying two quaternions is correct but can look like a collection of unrelated signs. A more revealing form separates each quaternion into a scalar part and a three dimensional vector part. In that notation the Hamilton product becomes a compact combination of the ordinary dot product and cross product.

Let

$\displaystyle p=p_w+\mathbf p, \qquad q=q_w+\mathbf q,$

where $p_w$ and $q_w$ are real scalars and $\mathbf p$ and $\mathbf q$ are pure quaternions identified with ordinary three dimensional vectors. The Hamilton product is

$\displaystyle pq = \left( p_wq_w-\mathbf p\cdot\mathbf q \right) + \left( p_w\mathbf q +q_w\mathbf p +\mathbf p\times\mathbf q \right).$

(1)

Equivalently, writing a quaternion as the ordered pair $(q_w,\mathbf q)$ ,

$\displaystyle (p_w,\mathbf p)(q_w,\mathbf q) = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p+\mathbf p\times\mathbf q \right).$

(2)

Equation (2) is one of the most useful quaternion formulas in physics and engineering. It exposes the geometric content of Hamilton multiplication and makes the sign convention immediately visible: the PhysicsLibrary Hamilton product contains the term $+\mathbf p\times\mathbf q$ .

The later PhysicsLibrary attitude articles use passive coordinate transformations. That application convention does not change equation (2). Hamilton algebra and active or passive attitude interpretation are separate choices.

Convention declaration

PhysicsLibrary uses Hamilton multiplication,

$\displaystyle \mathbf i\mathbf j=\mathbf k, \qquad \mathbf j\mathbf k=\mathbf i, \qquad \mathbf k\mathbf i=\mathbf j,$

with reversed products changing sign. Quaternions are displayed scalar first:

$\displaystyle q = q_w+\mathbf q \quad\longleftrightarrow\quad \begin{bmatrix} q_w\ q_x\ q_y\ q_z \end{bmatrix},$

(3)

where

$\displaystyle \mathbf q=q_x\mathbf i+q_y\mathbf j+q_z\mathbf k.$

The product convention is determined by the algebra, not by whether software stores the scalar component first or last.

For comparison, a flipped multiplication used in some engineering literature reverses the Hamilton factors. Its scalar vector formula contains $-\mathbf p\times\mathbf q$ instead of the plus sign in equation (2). PhysicsLibrary does not use that multiplication unless an article explicitly states otherwise.

Product of two pure quaternions

The key result behind equation (2) is the product of two pure quaternions.

Let

$\displaystyle \mathbf a = a_x\mathbf i+a_y\mathbf j+a_z\mathbf k$

and

$\displaystyle \mathbf b = b_x\mathbf i+b_y\mathbf j+b_z\mathbf k.$

Expand without changing the factor order:

$\begin{equation*} \begin{aligned} \mathbf a\mathbf b ={}& a_xb_x\mathbf i^2 +a_x... ...hbf i +a_zb_y\mathbf k\mathbf j +a_zb_z\mathbf k^2. \end{aligned}\end{equation*}$

Using the Hamilton basis products gives the scalar part

$\displaystyle -\left( a_xb_x+a_yb_y+a_zb_z \right) = -\mathbf a\cdot\mathbf b,$

and the vector part

$\begin{equation*} \begin{aligned} &(a_yb_z-a_zb_y)\mathbf i\ &+(a_zb_x-a_xb_z)\mathbf j\ &+(a_xb_y-a_yb_x)\mathbf k, \end{aligned}\end{equation*}$

which is exactly $\mathbf a\times\mathbf b$ .

Therefore

$\displaystyle \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b + \mathbf a\times\mathbf b.$

(4)

Equation (4) is the central bridge between Hamilton quaternion multiplication and ordinary vector analysis.

Deriving the general scalar vector product

Write

$\displaystyle p=p_w+\mathbf p, \qquad q=q_w+\mathbf q.$

Distributivity gives

$\displaystyle pq = p_wq_w +p_w\mathbf q +\mathbf p q_w +\mathbf p\mathbf q.$

Real scalars commute with quaternion factors, so

$\displaystyle \mathbf p q_w=q_w\mathbf p.$

Using equation (4),

$\displaystyle \mathbf p\mathbf q = -\mathbf p\cdot\mathbf q +\mathbf p\times\mathbf q.$

Therefore

$\begin{equation*} \begin{aligned} pq ={}& p_wq_w -\mathbf p\cdot\mathbf q\ &+ p_w\mathbf q +q_w\mathbf p +\mathbf p\times\mathbf q. \end{aligned}\end{equation*}$

Separating scalar and vector parts gives

$\displaystyle \operatorname{Sc}(pq) = p_wq_w-\mathbf p\cdot\mathbf q,$

(5)

and

$\displaystyle \operatorname{Vec}(pq) = p_w\mathbf q +q_w\mathbf p +\mathbf p\times\mathbf q.$

(6)

Together, equations (5) and (6) reproduce equation (2).

Connection to the component formula

Let

$\displaystyle \mathbf p= \begin{bmatrix} p_x\\ p_y\\ p_z \end{bmatrix}, \qquad \mathbf q= \begin{bmatrix} q_x\\ q_y\\ q_z \end{bmatrix}.$

The dot product is

$\displaystyle \mathbf p\cdot\mathbf q = p_xq_x+p_yq_y+p_zq_z,$

and the cross product is

$\displaystyle \mathbf p\times\mathbf q = \begin{bmatrix} p_yq_z-p_zq_y\ p_zq_x-p_xq_z\ p_xq_y-p_yq_x \end{bmatrix}.$

Substitution into equations (5) and (6) gives

$\displaystyle [pq]_{\mathrm{PL}} = \begin{bmatrix} p_wq_w-p_xq_x-p_yq_y-p_zq_z\... ..._y\ p_wq_y-p_xq_z+p_yq_w+p_zq_x\ p_wq_z+p_xq_y-p_yq_x+p_zq_w \end{bmatrix}.$

(7)

Thus the compact scalar vector formula and the four component formula are exactly equivalent.

Geometric meaning of the pure quaternion product

Equation (4),

$\displaystyle \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b +\mathbf a\times\mathbf b,$

contains both the metric and oriented geometry of three dimensional Euclidean space.

If $\gamma$ is the angle from $\mathbf a$ to $\mathbf b$ , then

$\displaystyle \mathbf a\cdot\mathbf b = \lVert\mathbf a\rVert \lVert\mathbf b\rVert \cos\gamma$

and

$\displaystyle \lVert\mathbf a\times\mathbf b\rVert = \lVert\mathbf a\rVert \lVert\mathbf b\rVert \sin\gamma.$

Therefore the scalar part of $\mathbf a\mathbf b$ contains the negative cosine information, while the vector part contains the oriented sine information.

This is why quaternion multiplication is naturally sensitive to order: the dot product is symmetric,

$\displaystyle \mathbf a\cdot\mathbf b = \mathbf b\cdot\mathbf a,$

but the cross product is antisymmetric,

$\displaystyle \mathbf b\times\mathbf a = -\mathbf a\times\mathbf b.$

Reversing the product

From equation (2),

$\displaystyle pq = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p+\mathbf p\times\mathbf q \right).$

Reversing the order gives

$\displaystyle qp = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p-\mathbf p\times\mathbf q \right).$

(8)

The scalar parts are identical. The difference lies entirely in the sign of the cross product.

Subtracting gives the commutator

$\displaystyle [p,q] = pq-qp = 2\mathbf p\times\mathbf q.$

(9)

The commutator is therefore always a pure quaternion.

Adding the two products gives

$\displaystyle pq+qp = 2\left( p_wq_w-\mathbf p\cdot\mathbf q \right) + 2\left( p_w\mathbf q+q_w\mathbf p \right).$

(10)

The cross product cancels from the symmetric combination.

When do two quaternions commute?

Equation (9) gives an immediate condition:

$\displaystyle pq=qp \quad\Longleftrightarrow\quad \mathbf p\times\mathbf q=0.$

Thus two quaternions commute exactly when their vector parts are parallel, antiparallel, or one vector part is zero.

For example, every pair of quaternions of the form

$\displaystyle a+b\mathbf u, \qquad c+d\mathbf u,$

with the same fixed unit pure quaternion $\mathbf u$ , commute with each other. They form a two dimensional subalgebra isomorphic to the complex numbers.

Important special cases

Real scalar times a quaternion

If $p=(a,\mathbf 0)$ , then

$\displaystyle pq = (aq_w,\;a\mathbf q).$

(11)

This is ordinary real scalar multiplication.

Two pure quaternions

If $p=(0,\mathbf p)$ and $q=(0,\mathbf q)$ , then

$\displaystyle pq = \left( -\mathbf p\cdot\mathbf q,\; \mathbf p\times\mathbf q \right).$

(12)

A pure quaternion squared

Set $\mathbf q=\mathbf p$ . Because

$\displaystyle \mathbf p\times\mathbf p=\mathbf 0,$

equation (12) gives

$\displaystyle \mathbf p^2 = -\lVert\mathbf p\rVert^2.$

(13)

Therefore every unit pure quaternion satisfies

$\displaystyle \mathbf u^2=-1.$

This result is fundamental to the later axis angle exponential.

Parallel pure quaternions

If $\mathbf b=\lambda\mathbf a$ , then

$\displaystyle \mathbf a\times\mathbf b=\mathbf 0.$

Hence

$\displaystyle \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b,$

(14)

which is purely real.

Perpendicular pure quaternions

$\displaystyle \mathbf a\cdot\mathbf b=0,$

then

$\displaystyle \mathbf a\mathbf b = \mathbf a\times\mathbf b,$

(15)

which is pure.

For example,

$\displaystyle \mathbf i\mathbf j = \mathbf i\times\mathbf j = \mathbf k.$

Reversing the factors gives

$\displaystyle \mathbf j\mathbf i=-\mathbf k.$

Hamilton and flipped multiplication

The sign of the cross product is a direct convention diagnostic.

PhysicsLibrary Hamilton multiplication uses

$\displaystyle (p_w,\mathbf p)(q_w,\mathbf q) = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p+\mathbf p\times\mathbf q \right).$

(16)

If another source defines a flipped product by reversing Hamilton factors,

$\displaystyle p\otimes q\equiv qp,$

then its scalar vector form is

$\displaystyle (p_w,\mathbf p)\otimes(q_w,\mathbf q) = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p-\mathbf p\times\mathbf q \right).$

(17)

Therefore

$\displaystyle +\mathbf p\times\mathbf q$

identifies Hamilton multiplication and

$\displaystyle -\mathbf p\times\mathbf q$

identifies the reversed product when all other definitions are held fixed.

This algebraic distinction must not be confused with the later choice between active and passive attitude interpretation. PhysicsLibrary retains equation (16) while using passive coordinate transformations for attitude.

Worked example 1: product from scalar and vector parts

Let

$\displaystyle p=2+\mathbf i-2\mathbf j+\mathbf k, \qquad q=-1+3\mathbf i+2\mathbf k.$

Then

$\displaystyle p_w=2, \qquad \mathbf p= \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix},$

and

$\displaystyle q_w=-1, \qquad \mathbf q= \begin{bmatrix} 3\\ 0\\ 2 \end{bmatrix}.$

The dot product is

$\displaystyle \mathbf p\cdot\mathbf q = 1(3)+(-2)(0)+1(2) = 5.$

The cross product is

$\displaystyle \mathbf p\times\mathbf q = \begin{bmatrix} -4\ 1\ 6 \end{bmatrix}.$

The scalar part is

$\displaystyle p_wq_w-\mathbf p\cdot\mathbf q = 2(-1)-5 = -7.$

The vector part is

$\begin{equation*} \begin{aligned} p_w\mathbf q+q_w\mathbf p+\mathbf p\times\math... ...rix}\ &= \begin{bmatrix} 1\\ 3\\ 9 \end{bmatrix}. \end{aligned}\end{equation*}$

Therefore

$\displaystyle pq=-7+\mathbf i+3\mathbf j+9\mathbf k.$

(18)

This agrees with the component calculation in the preceding PhysicsLibrary article, quaternion definition and basic algebra.

Worked example 2: pure quaternion product

Let

$\displaystyle \mathbf a= \begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}, \qquad \mathbf b= \begin{bmatrix} 0\\ 1\\ 3 \end{bmatrix}.$

Their dot product is

$\displaystyle \mathbf a\cdot\mathbf b=2,$

and their cross product is

$\displaystyle \mathbf a\times\mathbf b = \begin{bmatrix} 6\\ -3\\ 1 \end{bmatrix}.$

Therefore

$\displaystyle \mathbf a\mathbf b = -2+6\mathbf i-3\mathbf j+\mathbf k.$

(19)

Reversing the factors keeps the scalar part and reverses the vector part:

$\displaystyle \mathbf b\mathbf a = -2-6\mathbf i+3\mathbf j-\mathbf k.$

(20)

Worked example 3: recovering the basis rule

Take

$\displaystyle \mathbf a=\mathbf i, \qquad \mathbf b=\mathbf j.$

Because the two unit vectors are perpendicular,

$\displaystyle \mathbf i\cdot\mathbf j=0,$

and

$\displaystyle \mathbf i\times\mathbf j=\mathbf k.$

Equation (4) gives

$\displaystyle \mathbf i\mathbf j=\mathbf k.$

(21)

Thus the familiar Hamilton basis rule is contained directly in the scalar vector product.

Common pitfalls

Using the wrong sign on the cross product. PhysicsLibrary uses Hamilton multiplication, so the vector part contains $+\mathbf p\times\mathbf q$ .
Forgetting the minus sign on the dot product. For pure quaternions, $\mathbf a\mathbf b=-\mathbf a\cdot\mathbf b+\mathbf a\times\mathbf b$ .
Treating quaternion multiplication as componentwise multiplication. The dot and cross products couple all three vector components.
Assuming the product of two pure quaternions is pure. Its scalar part is generally $-\mathbf a\cdot\mathbf b$ .
Assuming the product of two pure quaternions is real. Its vector part is generally $\mathbf a\times\mathbf b$ .
Interchanging factors. Reversing the factors reverses the cross product and therefore generally changes the quaternion.
Confusing storage order with multiplication convention. A scalar last array can still use Hamilton multiplication.
Changing the cross product sign because the attitude convention is passive. The passive PhysicsLibrary attitude convention does not alter Hamilton multiplication.

Exercises

The exercises are stated first so the article can be used for self study. Complete solutions follow afterward.

Derive the pure product. Starting from the Hamilton basis products, derive

$\displaystyle \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b +\mathbf a\times\mathbf b.$
Scalar vector product. Let

$\displaystyle p=(2,[1,-2,1]^T), \qquad q=(-1,[3,0,2]^T).$
Compute using equation (2).
Reverse the order. For the quaternions in Exercise 2, compute without repeating the complete calculation. Use the symmetry of the dot product and antisymmetry of the cross product.
Pure quaternion square. Prove from equation (4) that

$\displaystyle \mathbf a^2=-\lVert\mathbf a\rVert^2.$
Unit pure quaternion. If $\lVert\mathbf u\rVert=1$ , prove that $\mathbf u^2=-1$ .
Perpendicular vectors. Let $\mathbf a\cdot\mathbf b=0$ . Show that

$\displaystyle \mathbf a\mathbf b=\mathbf a\times\mathbf b$
and

$\displaystyle \mathbf b\mathbf a=-\mathbf a\times\mathbf b.$
Parallel vectors. If $\mathbf b=\lambda\mathbf a$ , show that $\mathbf a\mathbf b$ is purely real.
Commutator. Derive

$\displaystyle [p,q]=2\mathbf p\times\mathbf q.$
Commuting quaternions. Use Exercise 8 to prove that and commute if and only if their vector parts are parallel, antiparallel, or one is zero.
Convention diagnostic. A source states

$\displaystyle (p_w,\mathbf p)\star(q_w,\mathbf q) = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p-\mathbf p\times\mathbf q \right).$
Does $\star$ agree with PhysicsLibrary Hamilton multiplication? Explain.

Solutions

1. Derive the pure product

Expand

$\displaystyle \mathbf a = a_x\mathbf i+a_y\mathbf j+a_z\mathbf k$

and

$\displaystyle \mathbf b = b_x\mathbf i+b_y\mathbf j+b_z\mathbf k.$

The square terms contribute

$\displaystyle -a_xb_x-a_yb_y-a_zb_z = -\mathbf a\cdot\mathbf b.$

The mixed basis terms contribute

$\displaystyle (a_yb_z-a_zb_y)\mathbf i + (a_zb_x-a_xb_z)\mathbf j + (a_xb_y-a_yb_x)\mathbf k,$

which is $\mathbf a\times\mathbf b$ . Therefore

$\displaystyle \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b + \mathbf a\times\mathbf b.$

2. Scalar vector product

For

$\displaystyle p_w=2, \qquad \mathbf p= \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix},$

and

$\displaystyle q_w=-1, \qquad \mathbf q= \begin{bmatrix} 3\\ 0\\ 2 \end{bmatrix},$

we have

$\displaystyle \mathbf p\cdot\mathbf q=5$

and

$\displaystyle \mathbf p\times\mathbf q= \begin{bmatrix} -4\\ 1\\ 6 \end{bmatrix}.$

Thus the scalar part is $-7$ , and the vector part is

$\displaystyle 2\mathbf q-\mathbf p+\mathbf p\times\mathbf q = \begin{bmatrix} 1\\ 3\\ 9 \end{bmatrix}.$

Therefore

$\displaystyle pq=-7+\mathbf i+3\mathbf j+9\mathbf k.$

3. Reverse the order

The scalar part remains $-7$ . Reversing the factors changes

$\displaystyle \mathbf p\times\mathbf q$

$\displaystyle -\mathbf p\times\mathbf q.$

Therefore

$\displaystyle qp=-7+9\mathbf i+\mathbf j-3\mathbf k.$

4. Pure quaternion square

Set $\mathbf b=\mathbf a$ in equation (4):

$\displaystyle \mathbf a^2 = -\mathbf a\cdot\mathbf a + \mathbf a\times\mathbf a.$

Since

$\displaystyle \mathbf a\cdot\mathbf a=\lVert\mathbf a\rVert^2$

and

$\displaystyle \mathbf a\times\mathbf a=\mathbf 0,$

we obtain

$\displaystyle \mathbf a^2=-\lVert\mathbf a\rVert^2.$

5. Unit pure quaternion

If $\lVert\mathbf u\rVert=1$ , Exercise 4 gives immediately

$\displaystyle \mathbf u^2=-1.$

6. Perpendicular vectors

$\displaystyle \mathbf a\cdot\mathbf b=0,$

then

$\displaystyle \mathbf a\mathbf b=\mathbf a\times\mathbf b.$

Reversing the factors gives

$\displaystyle \mathbf b\mathbf a = \mathbf b\times\mathbf a = -\mathbf a\times\mathbf b.$

7. Parallel vectors

$\displaystyle \mathbf b=\lambda\mathbf a,$

then

$\displaystyle \mathbf a\times\mathbf b = \lambda\mathbf a\times\mathbf a = \mathbf 0.$

Therefore

$\displaystyle \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b,$

which is purely real.

8. Commutator

Use equations (2) and (8). The scalar parts cancel, as do the terms $p_w\mathbf q+q_w\mathbf p$ . The cross terms give

$\displaystyle \mathbf p\times\mathbf q - (-\mathbf p\times\mathbf q) = 2\mathbf p\times\mathbf q.$

Hence

$\displaystyle [p,q] = 2\mathbf p\times\mathbf q.$

9. Commuting quaternions

The quaternions commute exactly when

$\displaystyle [p,q]=0.$

By Exercise 8 this is equivalent to

$\displaystyle \mathbf p\times\mathbf q=0.$

A cross product vanishes exactly when the two vectors are parallel, antiparallel, or one is zero.

10. Convention diagnostic

The operation $\star$ does not agree with PhysicsLibrary Hamilton multiplication because its vector part contains

$\displaystyle -\mathbf p\times\mathbf q.$

PhysicsLibrary Hamilton multiplication contains

$\displaystyle +\mathbf p\times\mathbf q.$

The sign difference identifies a reversed or flipped multiplication when the remaining definitions are held fixed. It is not caused by scalar first or scalar last storage, and it is not caused by the later passive attitude interpretation.

What comes next

The scalar vector form makes several later quaternion identities almost automatic.

For the conjugate,

$\displaystyle q^*=q_w-\mathbf q,$

equation (2) immediately shows why

$\displaystyle qq^*=q^*q$

is purely real: the cross product of $\mathbf q$ with itself vanishes. That result leads directly to the quaternion norm and inverse.

The later orientation articles use the same Hamilton product to construct passive frame transformations. No algebraic sign in equation (2) is changed when the application changes from general quaternion algebra to passive attitude representation.

Sources and historical notes

Hamilton's quaternion algebra contains both scalar and vector parts of products. Later quaternion and vector analysis texts made the relationship to dot and cross products increasingly explicit. Joly's A Manual of Quaternions and Macfarlane's Vector Analysis and Quaternions are useful public domain historical references for this decomposition.

The modern PhysicsLibrary notation uses ordinary dot and cross product symbols and scalar first quaternion components. The exposition and worked examples in this entry are newly organized for physics and engineering use.

Bibliography

1: W. R. Hamilton, Elements of Quaternions, 2nd ed., edited by C. J. Joly, Longmans, Green, and Co., 1899. Public domain historical source. Internet Archive scan https://archive.org/details/elementsofquater01hamiuoft
2: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public domain historical source. Internet Archive scan https://archive.org/details/manualofquaterni00jolyrich
3: A. Macfarlane, Vector Analysis and Quaternions, John Wiley and Sons, New York, 1906. Public domain historical source. Project Gutenberg edition https://www.gutenberg.org/ebooks/13609
4: H. Sommer, I. Gilitschenski, M. Bloesch, S. Weiss, R. Siegwart, and J. Nieto, “Why and How to Avoid the Flipped Quaternion Multiplication,” Aerospace, vol. 5, no. 3, article 72, 2018. Published under CC BY 4.0. Publisher article https://www.mdpi.com/2226-4310/5/3/72

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