quaternion conjugate QuaternionConjugate 2026-08-23 19:12:04 2026-08-23 19:12:04 Definition Converted to passive rotation convention quaternion conjugate Hamilton product scalar part % vector part anti-automorphism noncommutative algebra % this is the default PhysicsLibrary preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \section*{Quaternion Conjugate} The \emph{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 \begin{equation} q=q_w+q_x\mathbf i+q_y\mathbf j+q_z\mathbf k =q_w+\mathbf q, \end{equation} where \[ \mathbf q=q_x\mathbf i+q_y\mathbf j+q_z\mathbf k, \] the conjugate is \begin{equation} \boxed{q^*=q_w-\mathbf q.} \end{equation} In components, \begin{equation} \boxed{ (q_w,q_x,q_y,q_z)^*=(q_w,-q_x,-q_y,-q_z). } \end{equation} \section{Connection with complex conjugation} The complex numbers form a two-dimensional subalgebra of the quaternions. If \[ q=a+b\mathbf i, \] then the quaternion conjugate reduces to \[ q^*=a-b\mathbf i, \] which is exactly the ordinary complex conjugate. The quaternion definition is therefore a direct extension of a familiar operation. \section{Conjugation is an involution} Applying conjugation twice returns the original quaternion: \begin{equation} \boxed{(q^*)^*=q.} \end{equation} This follows immediately because the vector part changes sign twice. Conjugation also preserves real scalars. If $a\in\mathbb R$, then \begin{equation} a^*=a. \end{equation} For real scalars $a,b$ and quaternions $p,q$, conjugation is real-linear: \begin{equation} \boxed{(ap+bq)^*=ap^*+bq^*.} \end{equation} \section{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: \begin{equation} q+q^*=2q_w. \end{equation} Therefore \begin{equation} \boxed{\operatorname{Sc}(q)=\frac{q+q^*}{2}=q_w.} \end{equation} Subtracting instead cancels the scalar part: \begin{equation} q-q^*=2\mathbf q, \end{equation} so \begin{equation} \boxed{\operatorname{Vec}(q)=\frac{q-q^*}{2}=\mathbf q.} \end{equation} These identities are useful because they characterize scalar and pure-vector quaternions algebraically. A quaternion is purely real if and only if \[ q=q^*, \] and is pure-vector if and only if \[ q^*=-q. \] \section{The crucial order-reversal rule} Quaternion multiplication is noncommutative, so conjugation of a product must be handled carefully. For Hamilton quaternions, \begin{equation} \boxed{(pq)^*=q^*p^*.} \end{equation} The order reverses. To see why, write \[ p=p_w+\mathbf p, \qquad q=q_w+\mathbf q. \] The scalar--vector Hamilton product is \begin{equation} \begin{aligned} pq={}&\left(p_wq_w-\mathbf p\cdot\mathbf q\right)\\ &+\left(p_w\mathbf 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\right)\\ &-\left(p_w\mathbf 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, \[ \mathbf q\times\mathbf p=-\mathbf p\times\mathbf q. \] The result is exactly equation (13). For a product of several factors, \begin{equation} \boxed{(q_1q_2\cdots q_n)^*=q_n^*\cdots q_2^*q_1^*.} \end{equation} This property makes quaternion conjugation an order-reversing involution, often called an \emph{anti-automorphism} of the quaternion algebra. \section{Worked examples} \subsection*{Example 1: direct conjugation} Let \[ q=2+\mathbf i-2\mathbf j+3\mathbf k. \] Then \[ \boxed{q^*=2-\mathbf i+2\mathbf j-3\mathbf k.} \] A second conjugation returns $q$. \subsection*{Example 2: scalar and vector extraction} For \[ q=4-2\mathbf i+\mathbf j+5\mathbf k, \] we have \[ q^*=4+2\mathbf i-\mathbf j-5\mathbf k. \] Hence \[ \frac{q+q^*}{2}=4, \] and \[ \frac{q-q^*}{2}=-2\mathbf i+\mathbf j+5\mathbf k. \] \subsection*{Example 3: product order really reverses} Take \[ p=\mathbf i, \qquad q=\mathbf j. \] Since $pq=\mathbf k$, \[ (pq)^*=\mathbf k^*=-\mathbf k. \] But \[ p^*q^*=(-\mathbf i)(-\mathbf j)=\mathbf k, \] which is not the answer. Reversing the order gives \[ q^*p^*=(-\mathbf j)(-\mathbf i)=\mathbf j\mathbf i=-\mathbf k, \] as required. \section{Common pitfalls} \begin{enumerate} \item \textbf{Confusing conjugate with negative.} For $q=q_w+\mathbf q$, \[ q^*=q_w-\mathbf q, \qquad -q=-q_w-\mathbf q. \] They are generally different. \item \textbf{Forgetting product order reversal.} The correct identity is \[ (pq)^*=q^*p^*, \] not $p^*q^*$. \item \textbf{Treating scalar-first versus scalar-last as an algebraic change.} Component storage order does not change the abstract operation $q\mapsto q^*$. \item \textbf{Confusing conjugate with inverse.} For a general quaternion, $q^*$ is not $q^{-1}$. They coincide only for unit quaternions, which is derived in the quaternion inverse article. \end{enumerate} \section{Exercises} \begin{enumerate} \item Find the conjugate of \[ q=3-2\mathbf i+\mathbf j+4\mathbf k. \] \item Prove directly that $(q^*)^*=q$. \item Starting from $q^*=q_w-\mathbf q$, prove \[ \operatorname{Sc}(q)=\frac{q+q^*}{2}, \qquad \operatorname{Vec}(q)=\frac{q-q^*}{2}. \] \item Let $p=1+\mathbf i$ and $q=1+\mathbf j$. Compute $(pq)^*$ and $q^*p^*$ and verify that they are equal. \item Use the scalar--vector Hamilton product to prove \[ (pq)^*=q^*p^*. \] \item Prove \[ (pqr)^*=r^*q^*p^*. \] \item Show that a quaternion is pure-vector if and only if $q^*=-q$. \item Show that a quaternion is real if and only if $q^*=q$. \end{enumerate} \section{Solutions} \subsection*{1. Direct conjugate} \[ \boxed{q^*=3+2\mathbf i-\mathbf j-4\mathbf k.} \] \subsection*{2. Double conjugation} If $q=q_w+\mathbf q$, then \[ q^*=q_w-\mathbf q, \] so \[ (q^*)^*=q_w+\mathbf q=q. \] \subsection*{3. Scalar and vector parts} Adding gives \[ q+q^*=2q_w, \] while subtracting gives \[ q-q^*=2\mathbf q. \] Therefore \[ \boxed{\operatorname{Sc}(q)=\frac{q+q^*}{2}}, \qquad \boxed{\operatorname{Vec}(q)=\frac{q-q^*}{2}}. \] \subsection*{4. Product example} \[ pq=(1+\mathbf i)(1+\mathbf j)=1+\mathbf i+\mathbf j+\mathbf k. \] Thus \[ (pq)^*=1-\mathbf i-\mathbf j-\mathbf k. \] Also \[ q^*p^*=(1-\mathbf j)(1-\mathbf i) =1-\mathbf i-\mathbf j-\mathbf k. \] \subsection*{5. General product rule} The scalar part of $pq$ is \[ p_wq_w-\mathbf p\cdot\mathbf q, \] and its vector part is \[ p_w\mathbf q+q_w\mathbf p+\mathbf p\times\mathbf q. \] 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 \[ \mathbf q\times\mathbf p=-\mathbf p\times\mathbf q. \] Hence \[ \boxed{(pq)^*=q^*p^*.} \] \subsection*{6. Three factors} Apply the two-factor rule twice: \[ (pqr)^*=(p(qr))^*=(qr)^*p^*=r^*q^*p^*. \] \subsection*{7. Pure-vector condition} If $q=q_w+\mathbf q$ and $q^*=-q$, then \[ q_w-\mathbf q=-q_w-\mathbf q, \] so $q_w=0$. Conversely, if $q_w=0$, then $q^*=-q$. \subsection*{8. Real condition} If $q=q^*$, then \[ q_w+\mathbf q=q_w-\mathbf q, \] so $\mathbf q=0$. Conversely, a real quaternion has zero vector part and is unchanged by conjugation. \section{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. \section{Sources and historical notes} Hamilton's \emph{Elements of Quaternions}, Tait's \emph{Elementary Treatise}, Joly's \emph{Manual of Quaternions}, and Hathaway's \emph{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. \begin{thebibliography}{9} \bibitem{Hamilton1899} W. R. Hamilton, \emph{Elements of Quaternions}, 2nd ed., Vol. I, edited by C. J. Joly, Longmans, Green, and Co., London, 1899. Public-domain historical source. \PMlinkexternal{Internet Archive copy}{https://archive.org/details/elementsofquater01hamiuoft} \bibitem{Tait1890} P. G. Tait, \emph{An Elementary Treatise on Quaternions}, 3rd ed., Cambridge University Press, Cambridge, 1890. Public-domain historical source. \bibitem{Joly1905} C. J. Joly, \emph{A Manual of Quaternions}, Macmillan and Co., London, 1905. Public-domain historical source. \bibitem{Hathaway1896} A. S. Hathaway, \emph{A Primer of Quaternions}, 1896. Public-domain text; Project Gutenberg edition and \LaTeX{} source available. \PMlinkexternal{Project Gutenberg edition}{https://www.gutenberg.org/ebooks/9934} \end{thebibliography}