Quaternion product — Hamilton product: Worked examples

Intuition What this page is

The parent note built the rule. Here we drill it. Every quaternion product you will ever meet in GNC (Guidance, Navigation & Control — the on-board maths that keeps a spacecraft or drone pointed the right way) falls into one of a handful of case classes — parallel axes, perpendicular axes, identity, pure vectors, non-commuting pairs, the sandwich rotation, and the limiting "tiny angle" case that gyro integrators live in. We work one fully-solved example per class so that when your filter throws you a case at 3 a.m., you have already seen its shape.

Notation reminder (nothing new assumed): a quaternion is $q = (w, v)$ where $w$ is one plain number (the scalar part) and $v = (x, y, z)$ is an arrow in 3D (the vector part). Two ways of writing the same quaternion appear below — always in the same component order:

paired form $q = (w, (x, y, z))$ — scalar first, then the arrow's three components in brackets;
flat form $q = (w, x, y, z)$ — the identical four numbers with the inner brackets dropped.

These are the same object; $(w, x, y, z)$ just means $(w, (x, y, z))$ . Scalar is always the first slot, and the vector components are always in $x, y, z$ order. The hat on a letter, as in $\overset{n}{^}$ , means "this arrow has length exactly $1$ " — a pure direction with no size; the coordinate directions $\overset{x}{^} = (1, 0, 0)$ , $\overset{y}{^} = (0, 1, 0)$ , $\overset{z}{^} = (0, 0, 1)$ are the standard unit arrows along the axes. The product rule, which we USE constantly below, is $pq = (scalar w_{1} w_{2} - v_{1} \cdot v_{2}, vector w_{1} v_{2} + w_{2} v_{1} + v_{1} \times v_{2})$

The scenario matrix

Every Hamilton-product situation is one row of this table. The examples below hit each cell. ( $\overset{q}{ˉ}$ = the conjugate $(w, - v)$ defined in the box just below.)

#	Case class	What makes it special	Example
A	Identity element	one factor is $1 = (1, 0)$	Ex 1
B	Pure imaginary × pure imaginary, perpendicular	both scalars $0$ , axes $⊥$ → dot $= 0$	Ex 2
C	Pure imaginary, parallel/equal axes	both scalars $0$ , axes parallel → cross $= 0$	Ex 3
D	Order swap (non-commutativity)	compute $pq$ and $q p$ , compare	Ex 4
E	Two real rotations, general axes	full formula, all three terms nonzero	Ex 5
F	Sandwich (rotate a vector) $q u \overset{q}{ˉ}$	conjugate $\overset{q}{ˉ}$ from the box below; two products, length preserved	Ex 6
G	Sign / negative-scalar (half-angle $> 18 0^{\circ}$ )	$cos \frac{θ}{2} < 0$ ; double-cover $q \equiv - q$	Ex 7
H	Limiting case: infinitesimal rotation	$θ \to 0$ , links to gyro integration	Ex 8
I	Exam twist: solve for the unknown factor	given $q p = r$ , find $p$ via inverse	Ex 9

Two building blocks we reuse:

Case A — the identity element

Worked example Ex 1 — multiply by

$1$ (on both sides) Compute $q \cdot 1$ and $1 \cdot q$ where $q = (w, v)$ is arbitrary and $1 = (1, (0, 0, 0))$ .

Forecast: guess — should multiplying by "the number one" change anything, and does the side matter?

$q \cdot 1$ scalar part $= w \cdot 1 - v \cdot 0 = w$ . Why this step? Dotting any arrow with the zero arrow gives $0$ , so no scalar leaks in.
$q \cdot 1$ vector part $= w 0 + 1 v + v \times 0 = v$ . Why this step? Cross with the zero arrow is zero; the only survivor is $1 \cdot v$ . So $q \cdot 1 = (w, v) = q$ .
$1 \cdot q$ scalar part $= 1 \cdot w - 0 \cdot v = w$ . Why this step? Now $1$ is the first factor; its vector part is $0$ , so the dot term is still $0$ . Order of the dot doesn't matter (dot is symmetric).
$1 \cdot q$ vector part $= 1 v + w 0 + 0 \times v = v$ . Why this step? The cross term is $0 \times v = 0$ — this is the only term that could break symmetry, and it vanishes. So $1 \cdot q = (w, v) = q$ as well.

Verify: $q \cdot 1 = 1 \cdot q = q$ . The identity $1$ is the one factor that commutes with everything — not because non-commutativity switched off, but because its cross term is always zero. Every other pair keeps the order-dependence we exploit in Ex 4.

Case B — perpendicular pure imaginaries

Worked example Ex 2 —

$i j$ from the formula $p = i = (0, (1, 0, 0))$ , $q = j = (0, (0, 1, 0))$ .

Forecast: the parent claims $ij = k$ . Which term of the formula produces it — dot or cross?

Scalar $= 0 \cdot 0 - (1, 0, 0) \cdot (0, 1, 0) = 0 - 0 = 0$ . Why this step? The two axes are at $9 0^{\circ}$ ; perpendicular arrows have zero dot product, so no scalar appears.
Vector $= 0 \cdot (0, 1, 0) + 0 \cdot (1, 0, 0) + (1, 0, 0) \times (0, 1, 0)$ . Why this step? Both scalars are $0$ , so the two weighting terms vanish; only the cross survives.
$(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ by the right-hand rule ( $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ ).
Result $= (0, (0, 0, 1)) = k$ . ✅

Figure — what you are looking at: the black arrow along $x$ is $i$ , the black arrow along $y$ is $j$ . Where they meet is a small right-angle mark labelled " $9 0^{\circ}$ (dot $= 0$ )". The red arrow is the result $k = i \times j$ , pointing out of the $x$ – $y$ plane by the right-hand rule. Red = the answer we built.

Verify: matches the multiplication table $ij = k$ . Length: $∣ k ∣ = 1$ , and $∣ i ∣∣ j ∣ = 1$ , so length is preserved — a healthy sign.

Case C — parallel / equal axes

Case D — order matters (non-commutativity)

Worked example Ex 4 — compare

$ij$ and $j i$ Compute both products and subtract.

Forecast: will $ij$ equal $j i$ ? If not, where does the difference come from?

$ij = (0, (0, 0, 1)) = k$ (from Ex 2).
$j i$ : scalar $= 0 - (0, 1, 0) \cdot (1, 0, 0) = 0$ (still perpendicular). Why this step? Dot is symmetric, so it is unchanged by swapping order — the scalar part can never cause disagreement.
Vector of $j i = (0, 1, 0) \times (1, 0, 0) = (0, 0, - 1)$ . Why this step? Cross product is anti-symmetric: $b \times a = - (a \times b)$ . Swapping flips the sign.
$j i = (0, (0, 0, - 1)) = - k$ .
$ij - j i = k - (- k) = 2 k \neq = 0$ . Non-commutative. ✅

Figure — what you are looking at: black arrows $a$ (along $x$ ) and $b$ (along $y$ ) are the two factors. The red arrow pointing up is $ij = + k$ ; the red arrow pointing down is $j i = - k$ . Same two black inputs, opposite red outputs — the red arrow literally reverses when you swap the order. Red = the (order-dependent) answer.

Verify: $ij = - j i$ , exactly matching the parent. The cross term is the sole culprit — the dot term was identical both ways. This is not algebraic pedantry: it says "yaw-then-roll ≠ roll-then-yaw," a physical fact about 3D orientation.

Case E — two real rotations, general axes

Worked example Ex 5 —

$9 0^{\circ}$ about $z$ , then $9 0^{\circ}$ about $x$ Unit quaternion for angle $θ$ about unit axis $\overset{n}{^}$ (the hat means $∣ \overset{n}{^} ∣ = 1$ ): $q = (cos \frac{θ}{2}, sin \frac{θ}{2} \overset{n}{^})$ . $q_{z} = (\frac{1}{2}, 0, 0, \frac{1}{2})$ , $q_{x} = (\frac{1}{2}, \frac{1}{2}, 0, 0)$ (flat form: scalar, then $x, y, z$ ). Compute $q_{x} q_{z}$ (" $z$ first, then $x$ ").

Forecast: all three formula terms are nonzero here — expect a genuinely "mixed" quaternion.

Scalar $= \frac{1}{2} \cdot \frac{1}{2} - (\frac{1}{2}, 0, 0) \cdot (0, 0, \frac{1}{2}) = \frac{1}{2} - 0 = \frac{1}{2}$ . Why this step? The two axes $\overset{x}{^}$ and $\overset{z}{^}$ are perpendicular → their dot is $0$ , so only the $w_{1} w_{2}$ half-angle product remains.
Vector $= w_{1} v_{2} \frac{1}{2} (0, 0, \frac{1}{2}) + w_{2} v_{1} \frac{1}{2} (\frac{1}{2}, 0, 0) + v_{1} \times v_{2} (\frac{1}{2}, 0, 0) \times (0, 0, \frac{1}{2})$ . Why this step? Both scalars are nonzero now, so the two weighting terms survive alongside the cross.
$= (0, 0, \frac{1}{2}) + (\frac{1}{2}, 0, 0) + (0, - \frac{1}{2}, 0)$ — note $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ .
Sum $= (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ . So $q_{x} q_{z} = (\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ in flat form.

Verify: norm $= 4 \cdot \frac{1}{4} = 1$ ✅ — the product of unit quaternions is a unit quaternion, so it is a valid rotation. This is the workhorse case a filter runs every millisecond.

Case F — the sandwich: rotate a real vector

Worked example Ex 6 — rotate

$u = (1, 0, 0)$ by $9 0^{\circ}$ about $z$ Embed $u = (0, (1, 0, 0))$ . Use $q_{z} = (\frac{1}{2}, 0, 0, \frac{1}{2})$ and its conjugate $\overset{q}{ˉ}_{z} = (\frac{1}{2}, 0, 0, - \frac{1}{2})$ (from the Conjugate box: same scalar, negated vector). Compute $u^{'} = q_{z} u \overset{q}{ˉ}_{z}$ .

Forecast: rotating the $x$ -axis by $9 0^{\circ}$ about $z$ should land on the $y$ -axis. Watch the length stay $1$ .

First product $a = q_{z} u$ . Scalar $= \frac{1}{2} \cdot 0 - (0, 0, \frac{1}{2}) \cdot (1, 0, 0) = 0$ . Why this step? We rotate in two half-steps; $q u$ is the first squeeze. Axis $\overset{z}{^} ⊥ \overset{x}{^}$ so dot $= 0$ .
Vector of $a = \frac{1}{2} (1, 0, 0) + 0 + (0, 0, \frac{1}{2}) \times (1, 0, 0) = (\frac{1}{2}, 0, 0) + (0, \frac{1}{2}, 0) = (\frac{1}{2}, \frac{1}{2}, 0)$ . So $a = (0, (\frac{1}{2}, \frac{1}{2}, 0))$ .
Second product $u^{'} = a \overset{q}{ˉ}_{z}$ . Scalar $= 0 \cdot \frac{1}{2} - (\frac{1}{2}, \frac{1}{2}, 0) \cdot (0, 0, - \frac{1}{2}) = 0$ . Why this step? The scalar of the result must be $0$ for a pure vector — a built-in sanity check.
Vector of $u^{'} = 0 + \frac{1}{2} (\frac{1}{2}, \frac{1}{2}, 0) + (\frac{1}{2}, \frac{1}{2}, 0) \times (0, 0, - \frac{1}{2})$ . Why this step? This is the second half-turn: $w_{2} v_{1}$ scales $a$ 's arrow by $\overset{q}{ˉ}_{z}$ 's scalar, and the cross term adds the sideways twist $\overset{q}{ˉ}_{z}$ contributes. Together the two half-turns compose into the full $9 0^{\circ}$ . Cross $= (\frac{1}{2} \cdot (- \frac{1}{2}) - 0, 0 - \frac{1}{2} \cdot (- \frac{1}{2}), 0) = (- \frac{1}{2}, \frac{1}{2}, 0)$ . Sum $= (\frac{1}{2}, \frac{1}{2}, 0) + (- \frac{1}{2}, \frac{1}{2}, 0) = (0, 1, 0)$ .
$u^{'} = (0, (0, 1, 0)) \Rightarrow u^{'} = (0, 1, 0) = \overset{y}{^}$ . ✅

Figure — what you are looking at: the black arrow along $x$ is the input $u = (1, 0, 0)$ ; the dashed black arc is the $+ 9 0^{\circ}$ sweep about $z$ ; the red arrow along $y$ is the output $u^{'} = (0, 1, 0)$ . Same length, quarter turn — red = where the vector lands.

Verify: $∣ u^{'} ∣ = 1 = ∣ u ∣$ (length preserved), and $\overset{x}{^}$ went to $\overset{y}{^}$ exactly as a $+ 9 0^{\circ}$ turn about $z$ should. The half-angle inside $q$ is what makes two half-turns compose into one full rotation.

Case G — negative scalar & the double cover

Worked example Ex 7 — a

$27 0^{\circ}$ rotation about $z$ $θ = 27 0^{\circ} \Rightarrow \frac{θ}{2} = 13 5^{\circ}$ , $cos 13 5^{\circ} = - \frac{1}{2}$ , $sin 13 5^{\circ} = + \frac{1}{2}$ . So $q = (- \frac{1}{2}, 0, 0, \frac{1}{2})$ (flat form).

Forecast: a $27 0^{\circ}$ turn about $z$ is the same physical orientation as $- 9 0^{\circ}$ about $z$ . Will the two quaternions look equal — or opposite?

Quaternion for $- 9 0^{\circ}$ about $z$ : $\frac{θ}{2} = - 4 5^{\circ}$ , giving $q^{'} = (\frac{1}{2}, 0, 0, - \frac{1}{2})$ . Why this step? We compare two ways of naming the same physical turn.
Compare: $q = (- \frac{1}{2}, 0, 0, \frac{1}{2})$ and $- q^{'} = (- \frac{1}{2}, 0, 0, \frac{1}{2})$ . They are equal, i.e. $q = - q^{'}$ . Why this step? Every rotation has two quaternions, $q$ and $- q$ — this is the double cover. A negative scalar just means the half-angle passed $9 0^{\circ}$ .
Prove they act identically by sandwiching $\overset{x}{^} = (1, 0, 0)$ with $q$ . First $a = q x$ where $x = (0, (1, 0, 0))$ : scalar $= 0 - (0, 0, \frac{1}{2}) \cdot (1, 0, 0) = 0$ ; vector $= - \frac{1}{2} (1, 0, 0) + 0 + (0, 0, \frac{1}{2}) \times (1, 0, 0) = (- \frac{1}{2}, \frac{1}{2}, 0)$ , so $a = (0, (- \frac{1}{2}, \frac{1}{2}, 0))$ . Why this step? The only honest test of "same rotation" is: do they move a basis vector the same way?
Then $x^{'} = a \overset{q}{ˉ}$ with $\overset{q}{ˉ} = (- \frac{1}{2}, 0, 0, - \frac{1}{2})$ : scalar $= 0 - (- \frac{1}{2}, \frac{1}{2}, 0) \cdot (0, 0, - \frac{1}{2}) = 0$ ; vector $= - \frac{1}{2} (- \frac{1}{2}, \frac{1}{2}, 0) + (- \frac{1}{2}, \frac{1}{2}, 0) \times (0, 0, - \frac{1}{2}) = (\frac{1}{2}, - \frac{1}{2}, 0) + (- \frac{1}{2}, - \frac{1}{2}, 0) = (0, - 1, 0)$ . So $\overset{x}{^} \mapsto (0, - 1, 0) = - \overset{y}{^}$ . A $27 0^{\circ}$ (equivalently $- 9 0^{\circ}$ ) turn about $z$ sends $\overset{x}{^}$ to $- \overset{y}{^}$ . ✅

Verify: $∣ q ∣ = \frac{1}{2} + \frac{1}{2} = 1$ ✅, and both $q$ and $q^{'} = - q$ send $\overset{x}{^} \to - \overset{y}{^}$ (you get the identical vector because $(- q) u \overline{(- q)} = q u q^{- 1}$ — the two minus signs cancel). A negative $w$ is never "wrong"; it is one of two valid labels. Filters flip $q \to - q$ when $w < 0$ to keep the shortest-path representation.

Case H — the limiting case (tiny rotation)

Worked example Ex 8 — infinitesimal update,

$θ \to 0$ A gyro reports angular velocity $ω$ over a tiny time $Δ t$ , giving a small rotation of angle $θ = ∣ ω ∣Δ t$ about axis $\overset{n}{^}$ . Take $θ = 0.01$ rad about $\overset{n}{^} = (0, 0, 1)$ and multiply it onto $q_{0} = 1 = (1, 0)$ .

Forecast: for a tiny angle, what happens to $cos \frac{θ}{2}$ and $sin \frac{θ}{2}$ ? Guess the approximate quaternion.

Taylor-expand: $cos \frac{θ}{2} = 1 - \frac{1}{2} (\frac{θ}{2})^{2} + O (θ^{4})$ and $sin \frac{θ}{2} = \frac{θ}{2} - \frac{1}{6} (\frac{θ}{2})^{3} + O (θ^{5})$ . We keep only the leading terms: $cos \frac{θ}{2} \approx 1$ (error $\sim θ^{2} /8$ ) and $sin \frac{θ}{2} \approx \frac{θ}{2}$ (error $\sim θ^{3} /48$ ). Why this step? The neglected terms are $O (θ^{2})$ and smaller; for $θ ≲ 0.1$ rad ( $\approx 6^{\circ}$ , a typical gyro step) that relative error is under $\frac{0. 1 ^{2}}{8} \approx 0.13%$ , safely below sensor noise — so the linearization is trustworthy in exactly the regime it is used.
Build the small quaternion. With $\overset{n}{^} = (0, 0, 1)$ and $\frac{θ}{2} = 0.005$ : $δ q = (cos \frac{θ}{2}, sin \frac{θ}{2} \overset{n}{^}) \approx (1, (0, 0, 0.005)), δ q = (1, 0, 0, 0.005) (flat form).$ Numerically $cos 0.005 \approx 0.99998750$ and $sin 0.005 \approx 0.00499998$ , both rounding to the kept values. Why this step? This is the linear approximation gyro integrators actually store: scalar $\approx 1$ , vector $\approx \frac{1}{2} ω Δ t$ .
Apply the update onto the current attitude $q_{0} = 1$ : $q_{1} = δ q q_{0} = δ q \cdot 1 = δ q = (1, 0, 0, 0.005)$ . Why this step? By Case A (Ex 1), multiplying by the identity $1$ leaves $δ q$ unchanged — so the new attitude is just $δ q$ itself. Starting from "no rotation," one tiny step is the tiny rotation.
Renormalize: $∣ δ q ∣ = 1^{2} + 0.00 5^{2} = 1.000025 \approx 1.0000125$ , so $q_{1} \leftarrow q_{1} /∣ q_{1} ∣ \approx (0.99998750, 0, 0, 0.00499994)$ . Why this step? Dropping the $O (θ^{2})$ term left $δ q$ slightly off-unit (norm $> 1$ by $\sim 1.25 \times 1 0^{- 5}$ ); dividing by the norm restores a valid rotation (the parent's "renormalize" warning in action).

Verify: the stored vector part $\frac{1}{2} θ = 0.005$ matches $\frac{1}{2} ∣ ω ∣Δ t$ exactly — this recovers the kinematics $\dot q=\tfrac12 q\,\omega$. As $θ \to 0$ : $cos \frac{θ}{2} \to 1$ , $sin \frac{θ}{2} \to 0$ , so $δ q \to (1, 0) = 1$ — no rotation, no change. Smooth limiting behaviour ✅.

Case I — exam twist: solve for the unknown factor

Worked example Ex 9 — given

$q p = r$ , find $p$ You know the total rotation $r = k = (0, (0, 0, 1))$ and the last-applied factor $q = i = (0, (1, 0, 0))$ , while the first-applied factor $p$ is unknown. Find $p$ .

Forecast: to "undo" a left multiplication by $q$ , what do we multiply by, and on which side?

From $q p = r$ , left-multiply both sides by $q^{- 1}$ : $q^{- 1} q p = q^{- 1} r \Rightarrow p = q^{- 1} r$ . Why this step? Multiplication is non-commutative, so we must cancel $q$ on the same side it sits — the left. Order discipline is the whole point.
$q = i$ is unit, so $q^{- 1} = \overset{q}{ˉ} = (0, (- 1, 0, 0)) = - i$ (conjugate: same scalar $0$ , negated vector). Why this step? For unit quaternions inverse $=$ conjugate; flip the vector part.
$p = (- i) (k)$ . Scalar $= 0 - (- 1, 0, 0) \cdot (0, 0, 1) = 0$ . Vector $= 0 + 0 + (- 1, 0, 0) \times (0, 0, 1) = (0, 1, 0) = \overset{y}{^}$ . So $p = (0, (0, 1, 0)) = j$ . Why this step? Used the Hamilton formula once more; $(- \overset{x}{^}) \times \overset{z}{^} = + \overset{y}{^}$ .
Answer: $p = j$ .

Verify: check $q p = i j = k = r$ ✅ (from Ex 2). The unknown factor is recovered exactly — this is precisely how a MEKF backs out an error quaternion from measured vs. predicted attitude. See Multiplicative EKF (MEKF).

Recall

Recall Which term dies in each degenerate case?

Perpendicular axes ::: dot term $= 0$ (Ex 2). Parallel/equal axes ::: cross term $= 0$ (Ex 3). One factor is $1$ ::: everything but $v$ dies; identity, commutes both sides (Ex 1). Tiny angle $θ \to 0$ ::: quaternion $\to (1, 0)$ , no change (Ex 8).

Recall Test yourself

$ij - j i = ?$ ::: $2 k$ — the doubled cross term (Ex 4). How to undo a left factor in $q p = r$ ? ::: $p = q^{- 1} r$ , multiply on the left (Ex 9). Two quaternions for one rotation? ::: $q$ and $- q$ (double cover, Ex 7). Small-angle vector part of $δ q$ ? ::: $\frac{1}{2} ω Δ t$ (Ex 8).

Connections

Quaternion product — Hamilton product — the rule these examples drill.
Cross Product & Right-Hand Rule — the source of every sign flip in Ex 2, 4, 5.
Axis-Angle & Euler Rodrigues — where the half-angle $\frac{θ}{2}$ in Ex 5–8 comes from.
Quaternion Kinematics — $\dot q = \tfrac12 q\,\omega$ — the limit Ex 8 recovers.
Multiplicative EKF (MEKF) — the error-quaternion trick of Ex 9.
Rotation Matrices — SO(3) — cross-check Ex 6 against a rotation matrix.

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