Maths - Powers of Quaternions

We want to calculate an expression for:

(a + i b + j c + k d)n

where n is an integer.

When we calculated the powers of complex numbers we used the binomial theorem:

(a + i b)n=
(n-k)! k!
(-i)k a n-k bk

However there are a couple of differences in this case: these are 4 terms instead of 2 (so we would have to use a quadnomial instead of a binomial) but more importantly the terms don't all commute so we must use all permutations rather than integer multiplications of combinations which the binomial theorem gives.

see combinatorics


So lets try the simplest case where: (a + i b + j c + k d)²

Taking all the permutations gives:

permutation dimension  
a a 1 commutes
a b i commutes
a c j commutes
a d k commutes
b a i commutes
b b -1 commutes
b c ij = k anti-commutes
b d ik -j anti-commutes
c a j commutes
c b ji =-k anti-commutes
c c -1 commutes
c d jk = i anti-commutes
d a k commutes
d b ki = j commutes
d c kj = -i anti-commutes
d d -1 anti-commutes

The anti-commuting values cancel out so taking the other terms gives:

(a + i b + j c + k d)² = a²-b²-c²-d² + i 2ab + j 2ac + k 2ad

Going on to higher powers gives:

n (a + i b + j c + k d)n w x y z
1 (a + i b + j c + k d)1 a b c d
2 (a + i b + j c + k d)2 a²-b²-c²-d² 2ab 2ac 2ad
3 (a + i b + j c + k d)3 a3-3b²a-3c²a -3d²a 3a²b-b3-3c²a-3d²a 3a²c-c3-3b²c-3d²c 3a²d-d3-3b²d-3c²d
4 (a + i b + j c + k d)4        

metadata block
see also:


Correspondence about this page

Book Shop - Further reading.

Where I can, I have put links to Amazon for books that are relevant to the subject, click on the appropriate country flag to get more details of the book or to buy it from them.


cover us uk de jp fr caVisualizing Quaternions by Andrew J. Hanson

cover us uk de jp fr ca Quaternions and Rotation Sequences.

Terminology and Notation

Specific to this page here:


This site may have errors. Don't use for critical systems.

Copyright (c) 1998-2023 Martin John Baker - All rights reserved - privacy policy.