Quaternion Conjugate

The conjugate of a quaternion reverses the sign of its vector part while leaving its scalar part unchanged. It is the quaternion analogue of complex conjugation and is one of the basic operations used to define the quaternion norm, multiplicative inverse, and unit-quaternion rotation formulas.

For

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

(1)

where

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

the conjugate is

$\displaystyle \boxed{q^*=q_w-\mathbf q.}$

(2)

In components,

$\displaystyle \boxed{ (q_w,q_x,q_y,q_z)^*=(q_w,-q_x,-q_y,-q_z). }$

(3)

Connection with complex conjugation

The complex numbers form a two-dimensional subalgebra of the quaternions. If

$\displaystyle q=a+b\mathbf i,$

then the quaternion conjugate reduces to

$\displaystyle q^*=a-b\mathbf i,$

which is exactly the ordinary complex conjugate. The quaternion definition is therefore a direct extension of a familiar operation.

Conjugation is an involution

Applying conjugation twice returns the original quaternion:

$\displaystyle \boxed{(q^*)^*=q.}$

(4)

This follows immediately because the vector part changes sign twice.

Conjugation also preserves real scalars. If $a\in\mathbb{R}$ , then

$\displaystyle a^*=a.$

(5)

For real scalars $a,b$

and quaternions $p,q$

, conjugation is real-linear:

$\displaystyle \boxed{(ap+bq)^*=ap^*+bq^*.}$

(6)

Recovering scalar and vector parts

The conjugate provides a compact way to extract the two parts of a quaternion. Adding $q$ and $q^*$ cancels the vector part:

$\displaystyle q+q^*=2q_w.$

(7)

Therefore

$\displaystyle \boxed{\operatorname{Sc}(q)=\frac{q+q^*}{2}=q_w.}$

(8)

Subtracting instead cancels the scalar part:

$\displaystyle q-q^*=2\mathbf q,$

(9)

$\displaystyle \boxed{\operatorname{Vec}(q)=\frac{q-q^*}{2}=\mathbf q.}$

(10)

These identities are useful because they characterize scalar and pure-vector quaternions algebraically. A quaternion is purely real if and only if

$\displaystyle q=q^*,$

and is pure-vector if and only if

$\displaystyle q^*=-q.$

The crucial order-reversal rule

Quaternion multiplication is noncommutative, so conjugation of a product must be handled carefully. For Hamilton quaternions,

$\displaystyle \boxed{(pq)^*=q^*p^*.}$

(11)

The order reverses.

To see why, write

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

The scalar–vector Hamilton product is

$\begin{equation*}\begin{aligned} pq={}&\left(p_wq_w-\mathbf p\cdot\mathbf q\righ... ...hbf q+q_w\mathbf p+\mathbf p\times\mathbf q\right). \end{aligned}\end{equation*}$

Taking the conjugate leaves the scalar part fixed and reverses the vector part:

$\begin{equation*}\begin{aligned} (pq)^*={}&\left(p_wq_w-\mathbf p\cdot\mathbf q\... ...hbf q+q_w\mathbf p+\mathbf p\times\mathbf q\right). \end{aligned}\end{equation*}$

Now compute $q^*p^*=(q_w-\mathbf q)(p_w-\mathbf p)$ . The dot product is symmetric while the cross product changes sign under reversal,

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

The result is exactly equation (13).

For a product of several factors,

$\displaystyle \boxed{(q_1q_2\cdots q_n)^*=q_n^*\cdots q_2^*q_1^*.}$

(14)

This property makes quaternion conjugation an order-reversing involution, often called an anti-automorphism of the quaternion algebra.

Worked examples

Example 1: direct conjugation

Let

$\displaystyle q=2+\mathbf i-2\mathbf j+3\mathbf k.$

Then

$\displaystyle \boxed{q^*=2-\mathbf i+2\mathbf j-3\mathbf k.}$

A second conjugation returns $q$

Example 2: scalar and vector extraction

For

$\displaystyle q=4-2\mathbf i+\mathbf j+5\mathbf k,$

we have

$\displaystyle q^*=4+2\mathbf i-\mathbf j-5\mathbf k.$

Hence

$\displaystyle \frac{q+q^*}{2}=4,$

and

$\displaystyle \frac{q-q^*}{2}=-2\mathbf i+\mathbf j+5\mathbf k.$

Example 3: product order really reverses

Take

$\displaystyle p=\mathbf i, \qquad q=\mathbf j.$

Since $pq=\mathbf k$ ,

$\displaystyle (pq)^*=\mathbf k^*=-\mathbf k.$

But

$\displaystyle p^*q^*=(-\mathbf i)(-\mathbf j)=\mathbf k,$

which is not the answer. Reversing the order gives

$\displaystyle q^*p^*=(-\mathbf j)(-\mathbf i)=\mathbf j\mathbf i=-\mathbf k,$

as required.

Common pitfalls

Confusing conjugate with negative. For $q=q_w+\mathbf q$ ,

$\displaystyle q^*=q_w-\mathbf q, \qquad -q=-q_w-\mathbf q.$
They are generally different.
Forgetting product order reversal. The correct identity is

$\displaystyle (pq)^*=q^*p^*,$
not .
Treating scalar-first versus scalar-last as an algebraic change. Component storage order does not change the abstract operation $q\mapsto q^*$ .
Confusing conjugate with inverse. For a general quaternion, is not $q^{-1}$ . They coincide only for unit quaternions, which is derived in the quaternion inverse article.

Exercises

Find the conjugate of

$\displaystyle q=3-2\mathbf i+\mathbf j+4\mathbf k.$
Prove directly that .
Starting from $q^*=q_w-\mathbf q$ , prove

$\displaystyle \operatorname{Sc}(q)=\frac{q+q^*}{2}, \qquad \operatorname{Vec}(q)=\frac{q-q^*}{2}.$
Let $p=1+\mathbf i$ and $q=1+\mathbf j$ . Compute and and verify that they are equal.
Use the scalar–vector Hamilton product to prove

$\displaystyle (pq)^*=q^*p^*.$
Prove

$\displaystyle (pqr)^*=r^*q^*p^*.$
Show that a quaternion is pure-vector if and only if .
Show that a quaternion is real if and only if .

Solutions

1. Direct conjugate

$\displaystyle \boxed{q^*=3+2\mathbf i-\mathbf j-4\mathbf k.}$

2. Double conjugation

If $q=q_w+\mathbf q$ , then

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

$\displaystyle (q^*)^*=q_w+\mathbf q=q.$

3. Scalar and vector parts

Adding gives

$\displaystyle q+q^*=2q_w,$

while subtracting gives

$\displaystyle q-q^*=2\mathbf q.$

Therefore

$\displaystyle \boxed{\operatorname{Sc}(q)=\frac{q+q^*}{2}}, \qquad \boxed{\operatorname{Vec}(q)=\frac{q-q^*}{2}}.$

4. Product example

$\displaystyle pq=(1+\mathbf i)(1+\mathbf j)=1+\mathbf i+\mathbf j+\mathbf k.$

Thus

$\displaystyle (pq)^*=1-\mathbf i-\mathbf j-\mathbf k.$

Also

$\displaystyle q^*p^*=(1-\mathbf j)(1-\mathbf i) =1-\mathbf i-\mathbf j-\mathbf k.$

5. General product rule

The scalar part of $pq$

$\displaystyle p_wq_w-\mathbf p\cdot\mathbf q,$

and its vector part is

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

Conjugation changes the sign of the vector part. Expanding $q^*p^*$

produces the same scalar part and the negative of the same vector part because

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

Hence

$\displaystyle \boxed{(pq)^*=q^*p^*.}$

6. Three factors

Apply the two-factor rule twice:

$\displaystyle (pqr)^*=(p(qr))^*=(qr)^*p^*=r^*q^*p^*.$

7. Pure-vector condition

If $q=q_w+\mathbf q$ and $q^*=-q$

, then

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

. Conversely, if $q_w=0$

, then

8. Real condition

, then

$\displaystyle q_w+\mathbf q=q_w-\mathbf q,$

so $\mathbf q=0$ . Conversely, a real quaternion has zero vector part and is unchanged by conjugation.

What comes next

The conjugate allows the product $qq^*$ to collapse to a nonnegative real number. The next PhysicsLibrary article uses that fact to define and derive the quaternion norm.

Sources and historical notes

Hamilton's Elements of Quaternions, Tait's Elementary Treatise, Joly's Manual of Quaternions, and Hathaway's Primer of Quaternions all develop the operation historically denoted by $Kq$ , now usually written $q^*$ or $\bar q$ . Hathaway states explicitly that conjugation reverses the order of factors in a product. The notation, derivations, worked examples, exercises, and solutions here are newly written in modern form.

Bibliography

1: W. R. Hamilton, Elements of Quaternions, 2nd ed., Vol. I, edited by C. J. Joly, Longmans, Green, and Co., London, 1899. Public-domain historical source. Internet Archive copy
2: P. G. Tait, An Elementary Treatise on Quaternions, 3rd ed., Cambridge University Press, Cambridge, 1890. Public-domain historical source.
3: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public-domain historical source.
4: A. S. Hathaway, A Primer of Quaternions, 1896. Public-domain text; Project Gutenberg edition and L^ATEX source available. Project Gutenberg edition

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