Quaternion product — Hamilton product: Question bank

Before we start, one anchor so no symbol is unearned:

Definition The moving parts (keep this open)

A quaternion is $q = (w, v)$ : a plain number $w$ (the scalar part) glued to an arrow $v = (x, y, z)$ (the vector part). Symbols you will meet below:

Basis units $i, j, k$ are the pure quaternions $i = (0, (1, 0, 0))$ , $j = (0, (0, 1, 0))$ , $k = (0, (0, 0, 1))$ — each is a zero scalar attached to one coordinate axis.
Conjugate $\overset{q}{ˉ}$ ("q-bar") flips the sign of the vector part: $\overset{q}{ˉ} = (w, - v)$ . It reverses the twist while keeping the scalar.
Norm (length) $∣ q ∣ = w^{2} + x^{2} + y^{2} + z^{2}$ — the ordinary 4D length of the four numbers. A unit quaternion has $∣ q ∣ = 1$ .
Hat $\overset{n}{^}$ means a unit axis: a direction arrow of length $1$ (so $∣ \overset{n}{^} ∣ = 1$ ), used to name the axis you spin about.

The Hamilton product of $p = (w_{1}, v_{1})$ and $q = (w_{2}, v_{2})$ is $pq = (scalar w_{1} w_{2} - v_{1} \cdot v_{2}, vector w_{1} v_{2} + w_{2} v_{1} + v_{1} \times v_{2})$ Here $v_{1} \cdot v_{2}$ (dot product) is a number measuring "how aligned," and $v_{1} \times v_{2}$ (cross product, see Cross Product & Right-Hand Rule) is an arrow pointing sideways by the right-hand rule.

True or false — justify

ij = k

and

j i = k

both hold because

i, j, k

are just fancy imaginary units.

False.

ij = k

but

j i = - k

; the cross-product piece flips sign when you swap order, so the two products differ.

The Hamilton product is commutative because the dot product inside it is commutative.

False. The dot term is symmetric, but the cross term

v_{1} \times v_{2}

is anti-symmetric, and that single term makes the whole product non-commutative.

The product of two unit quaternions is always another unit quaternion.

True. Norms multiply,

∣ pq ∣ = ∣ p ∣ ∣ q ∣

, so

1 \times 1 = 1

— this is exactly why chaining rotations never rescales orientation.

q_{A} q_{B}

means "do rotation

A

first, then

B

," reading left to right like English.

False. Hamilton convention applies right-to-left:

q_{A} q_{B}

means

B

first, then

A

. The leftmost factor acts last.

Quaternion multiplication being non-commutative is a defect of the notation we could fix with a better convention.

False. It mirrors physical reality — 3D rotations genuinely don't commute (roll-then-pitch

\neq =

pitch-then-roll), so any faithful rule must be non-commutative.

For a pure quaternion

u = (0, u)

, the product

q u

already gives the rotated vector.

False. A single product leaks scalar and changes length; you need the sandwich

q u q^{- 1}

to get a clean rotated pure quaternion.

The half-angle

θ /2

inside

q = (cos \frac{θ}{2}, sin \frac{θ}{2} \overset{n}{^})

is an arbitrary convention.

False. The rotation is applied twice in the sandwich

q u q^{- 1}

, so each factor must carry half the angle for the total to come out as

θ

Since

ij k = - 1

and

i^{2} = - 1

, quaternions are just complex numbers with extra letters.

False. In

C

ab = ba

; quaternions are non-commutative. The extra structure is precisely what encodes 3D orientation.

Spot the error

"I computed the vector part as just

v_{1} \times v_{2}

Missing the two scaling terms

w_{1} v_{2} + w_{2} v_{1}

; dropping them discards all the pure-spin (scalar-weighted) information.

"The scalar part is

w_{1} w_{2} + v_{1} \cdot v_{2}

Wrong sign — it is a minus:

w_{1} w_{2} - v_{1} \cdot v_{2}

, inherited from

i^{2} = j^{2} = k^{2} = - 1

"To combine rotate-about-

z

then rotate-about-

x

, I write

q_{z} q_{x}

Backwards. "

x

after

z

" means

x

acts last, so the last factor goes leftmost:

q_{x} q_{z}

"Products of unit quaternions stay unit, so I never renormalize in my filter."

True in exact math, false in floating point — rounding accumulates over thousands of gyro updates, so you must periodically set

q \leftarrow q /∣ q ∣

v_{1} \times v_{2}

is symmetric, so I can compute it in either order."

The cross product is anti-symmetric:

v_{2} \times v_{1} = - v_{1} \times v_{2}

. Order flips the sign.

q^{- 1} = \overset{q}{ˉ}

always, so I can skip the norm."

Only for unit

q

. In general

q^{- 1} = \overset{q}{ˉ} /∣ q ∣^{2}

where

\overset{q}{ˉ} = (w, - v)

; forgetting the

∣ q ∣^{2}

misrotates and rescales any non-unit quaternion.

"The rotation quaternion for

θ

about

\overset{n}{^}

(θ, \overset{n}{^})

Wrong entirely — it is

(cos \frac{θ}{2}, sin \frac{θ}{2} \overset{n}{^})

; the raw angle never appears, only its half-angle cosine and sine.

Why questions

Why does the cross product, and not the dot product, cause non-commutativity?

Dot is symmetric so it is identical either way; only the cross term changes sign under swapping, so it is the sole asymmetric contribution.

Why must we rotate a vector with a sandwich

q u q^{- 1}

instead of one product?

The sandwich cancels the scalar "leakage" of each product and preserves length, leaving a pure quaternion of the same magnitude — a genuine rotation.

Why do gyro-integration filters prefer Hamilton products over stacking rotation matrices?

Only 4 numbers, no trig each step, and the unit-norm property auto-preserves a valid orientation — cheaper and drift-resistant, see [[Quaternion Kinematics —

\overset{q}{˙} = \frac{1}{2} q ω

]].

Why does the half-angle appear inside a rotation quaternion?

Because the vector is rotated by conjugation, applying

q

effectively twice; splitting the angle in half makes the two half-turns sum to the full angle.

Why do

ij = k

j k = i

k i = j

follow a cyclic (right-handed) pattern?

The units behave like right-handed coordinate axes under the cross product (

\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}

), so their products inherit the same cyclic order, see Cross Product & Right-Hand Rule.

Why is the Hamilton product equivalent to multiplying rotation matrices?

Both encode "do one rotation, then another" as a single combined rotation; the quaternion is just a 4-number coordinate for the same element of Rotation Matrices — SO(3).

Why does an attitude filter like the Multiplicative EKF (MEKF) chain products rather than add corrections?

Rotations compose multiplicatively, not additively; adding quaternions leaves the unit sphere and no longer represents a rotation, so corrections must be applied by Hamilton multiplication.

Edge cases

What is

q \overset{q}{ˉ}

for a unit quaternion, and why?

It equals

(1, 0)

, the identity — because for unit

q

the conjugate

\overset{q}{ˉ} = (w, - v)

is its inverse, and a rotation followed by its opposite does nothing.

Both

q

and

- q

describe the same rotation — true or false?

True. Negating all four numbers flips the half-angle by

36 0^{\circ}

, i.e. a full extra turn, which is physically identical; this is the "double cover" of rotations.

What does multiplying by the scalar quaternion

(1, 0)

do?

Nothing — it is the multiplicative identity, leaving any quaternion unchanged, just as multiplying a rotation matrix by

I

does.

What rotation does

q = (1, 0)

represent?

The zero rotation (

θ = 0

), since

cos 0 = 1

and

sin 0 \overset{n}{^} = 0

— no turn at all.

What happens if you feed a zero quaternion

(0, 0)

as a rotation?

It is invalid: its norm

∣ q ∣ = 0

, so it has no inverse and cannot represent a rotation; a valid rotation quaternion must be unit-norm.

v_{1}

and

v_{2}

are parallel, what happens to the cross term?

It vanishes (

v_{1} \times v_{2} = 0

), so the product is commutative in that special case — parallel-axis rotations do commute.

v_{1} ⊥ v_{2}

(like

i

and

j

), what does the dot term contribute to the scalar part?

Nothing — the dot is

0

, so the scalar part reduces to

w_{1} w_{2}

, and all the action lives in the cross-product vector term.

What is the product

q q

(a rotation composed with itself)?

A rotation by twice the angle about the same axis, since the half-angles add to

2 \cdot \frac{θ}{2} = θ

per factor, giving

2 θ

total — same axis, doubled turn.

Recall

Recall One-line self-test

Sole source of non-commutativity? ::: the cross-product term $v_{1} \times v_{2}$ .
Sign in the scalar part? ::: minus: $w_{1} w_{2} - v_{1} \cdot v_{2}$ .
Meaning of $q_{A} q_{B}$ ? ::: do $B$ first, then $A$ (right-to-left).
What is $\overset{q}{ˉ}$ ? ::: the conjugate $(w, - v)$ — flips the vector part's sign.
How is $∣ q ∣$ computed? ::: $w^{2} + x^{2} + y^{2} + z^{2}$ , the 4D length.

Connections

Quaternion product — Hamilton product — parent topic these traps target.
Cross Product & Right-Hand Rule — root of the cyclic units and non-commutativity.
Rotation Matrices — SO(3) — the matrix equivalent of the product.
Axis-Angle & Euler Rodrigues — where the half-angle comes from.
Quaternion Kinematics — $\dot q = \tfrac12 q\,\omega$ — chaining products in gyro integration.
Multiplicative EKF (MEKF) — multiplicative corrections in filters.
Gimbal Lock & Euler Angles — the failure quaternions avoid.

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