From the definition, we understand $i,j,k$ are quaternion units.
Quaternions are generally represented in the form:
$a + bi + cj + dk$ where a, b, c, and d are real numbers, and i, j, and k are the fundamental Quaternion Units.
https://en.wikipedia.org/wiki/Quaternion
We also know that
$i^2 = j^2 = k^2 = ijk = -1$
So, when we apply rotation with Q quaternion(as a unit quaternion like i, j or k) to vector we expect to get reverse of it. $$Q = (w:0.707,x:0.707,y:0,z:0) (or Axis: [ x:1, y:0, z:0 ], angle:90)$$
$$v = (w:0,x:0,y:1,z:0) $$
then from the rotation formula $$ q V q^-1 $$ we get result $$r=(w:0,x:0,y:0,z:1)$$
And when we rotate it again with same $Q$ we get result $$(w:0,x:0,y:-1,z:1)$$ which is $-v$.
Thus, could we say that Quaternion units $i,j,k$ are always 90 degree rotations!?
Which seems intiutively correct to me, rotating 90 degrees twice you would get the reverse by 180 degrees.
Also, while asking this question, i re-checked the definition of unit quaternion(versor) and determine the unit quaternion as
$$ Unit Quaternion: ( w:0, x:0.577, y:0.577, z:0.577 ) (or Axis:[ 0.577, 0.577, 0.577 ], angle:180)$$
And applied the rotation formula twice and find out that it works, it reverses the vector.
Edit1: I checked it again but applying rotation by the "Unit Quaternion" twice gets the same vector, not reverse!
So, are those i,j and k unit quaternions have this value : $$ ( w:0, x:0.577, y:0.577, z:0.577 ) $$
Or what is the value of unit quaternions(or i,j,k) actually?
Edit2: My false assumption is that multiplying vector with $i^2=-1$ is identical to applying $i$ quaternion twice to vector with the rotation formula but multiplication is non-commutative and it will not work i think.
Edit3: Finally i tested to directly multiply vector by $i^2$(without rotation formula) and i get the reverse of vector as expected from $i^2=-1$. So, $$ ( 0, 0.577, 0.577, 0.577 ) $$ is the unit quaternion i think but i am not sure which one is this i,j or k and how i,j and k differs in this case?