I was looking into quaternion mulitplication, but I'm having a hard time understanding how it works when the exponent of, say $j$, is not an integer.
The basic algebraic definitions of quaternions are clear to me. I know that $i\cdot j=k$ and $j \cdot i = -k$ and so on. But it is not so clear to me what $i \cdot j^{\frac 1 2}$ would evaluate to. Does it even equal one specific value? Or does it behave like, say, complex logarithms with infinite values?
Could anyone with a good grasp of quaternion multiplication help me out? Thanks.
Well, the best we could say a priori without any other knowledge is that $j^{\frac12}$ must be some quaternion $q$ (if any such exists) such that $q^2=j.$ Let's see what we can determine about such a beast!
For any quaternion $q,$ we know $q=a+bi+cj+dk$ for some unique real numbers $a,b,c,d.$
Next, note that $$\begin{eqnarray}q^2 & = &(a+bi+cj+dk)q\\ & = & aq+biq+cjq+dkq\\ & = & a^2+abi+acj+adk+biq+cjq+dkq.\end{eqnarray}$$
We also have the following: $$biq=abi+-b^2+bck-bdj=-b^2+abi-bdj+bck\\cjq=acj-bck-c^2+cdi=-c^2+cdi+acj-bck\\dkq=adk+bdj-cdi-d^2=-d^2-cdi+bdj+adk.$$
Thus, in general, $$q^2=a^2-b^2-c^2-d^2+2abi+2acj+2adk.$$
That looks promising! Since we want $q^2=j,$ then it gives us a system of equations to solve, namely:
$$\begin{array}{}0=a^2-b^2-c^2-d^2\\0=2ab\\1=2ac\\0=2ad.\end{array}$$ By solving it, we find that (as is typical for square roots of nonzero numbers) there are two distinct solutions!