# Quaternions multiplication

1. Jun 27, 2012

### amiras

1. The problem statement, all variables and given/known data
Its not really homework problem, and you may find it silly because its only multiplication problem, but I cannot get the right answer by multiplying quaternions.

Basically this is what i want to show:

exp(iψ/2)exp(kθ/2)exp(iф/2) = cos(θ/2)exp(i[ψ+ф]/2) + ksin(θ/2)exp(i[ψ-ф]/2)

2. Relevant equations

3. The attempt at a solution

I begin writing:
exp(kθ/2) = cos(θ/2) + ksin(θ/2)

Then multiplying:
exp(iψ/2)*[cos(θ/2) + ksin(θ/2)] = exp(iψ/2)cos(θ/2) + exp(iψ/2)*ksin(θ/2) =
= cos(θ/2)exp(iψ/2) + sin(θ/2) exp(iψ/2)*k

Since scalar terms can are commutative in quaternions algebra.

Finally multiplying answer above with the final exp(iф/2)

[cos(θ/2)exp(iψ/2) + sin(θ/2) exp(iψ/2)*k] * exp(iф/2) =
= cos(θ/2)exp(iψ/2)exp(iф/2) + sin(θ/2)exp(iψ/2)*k*exp(iф/2) =
= cos(θ/2)exp(i(ψ+ф)/2) + sin(θ/2)*k*exp(-iψ/2)*exp(iф/2) =
= cos(θ/2)exp(i(ψ+ф)/2) + k*sin(θ/2)*exp(i(ф-ψ)/2)

Here I used that exp(iψ/2)*k = k*exp(-iψ/2)

And no matter how I do it I always get the same answer, with the last exponential term having ф-ψ, and the paper says it should be ψ-ф

2. Jun 27, 2012

### tiny-tim

hi amiras!

yours looks correct to me …

exp(iψ/2)exp(kθ/2)exp(iф/2)

= exp(iψ/2)cos(θ/2)exp(iф/2) + exp(iψ/2)ksin(θ/2)exp(iф/2)

= cos(θ/2)exp(i(ψ+ф)/2) + kexp(-iψ/2)sin(θ/2)exp(iф/2)

3. Jun 27, 2012

### amiras

Thanks for confirming me, from now on I'l stop blindly following and start thinking by myself. :)