Physical intro
In quantum mechanics, spin-½ systems are described using two-dimensional vectors. In the physical experiment, we can measure the spin along three orthogonal axes X, Y, Z. Each of these axes has an associated orthonormal basis.
Let's say our first basis, associated with the axis X, is: $$B = \big\{(1,0) ;\; (0,1)\big\}$$ If a particle in general orientation $v=(a,b)$ flies inside our measurement system. The probability we measure “X+” is: $$ P = |B_1 \cdot v|^2 $$ Similarly, if we wanted to know the probability of measuring “X−”, it would be: $$ P = |B_2 \cdot v|^2 $$
From the experiments we know that if a particle has spin either X+ or X− and we measure its spin along an orthogonal axis (eg. Y), the probability of measuring Y+ is 50 %, the same applies for Y−. That means that there is no statistical correlation between sping along X and Y.
Mathematical intro
Motivated by the physics, we define:
Let $B$, $C$ be two orthonormal bases. We call them uncorrelated if $$ |b \cdot c|^2 = \frac{1}{2} $$ for all $b \in B, \; c \in C$.
It can be shown that if we work in $\mathbb{R}^2$ and start with the basis $B = \big\{(1,0) ;\; (0,1)\big\}$, there is only one basis that is uncorrelated to it: $$ C = \Big\{ \tfrac{1}{\sqrt{2}} \, (1, 1) ;\; \tfrac{1}{\sqrt{2}} \, (1, -1) \Big\}. $$ This is as far as we can get with $\mathbb{R}^2$. In a sense, this is a complete set of uncorrelated bases in $\mathbb{R}^2$.
If we let ourselves work in $\mathbb{C}^2$, it turns out there is a third basis, uncorrelated to both $B$ and $C$: $$ D = \Big\{ \tfrac{1}{\sqrt{2}} \, (1, \mathrm{i}) ;\; \tfrac{1}{\sqrt{2}} \, (1, -\mathrm{i}) \Big\}. $$ Furthermore we can “turn” any of the bases by a factor of $e^{\mathrm{i}\varphi}$ without changing anything (ie. the bases will be still uncorrelated). However, this phase factor won't help us find a new basis uncorellated to $B, C, D$, so we're stuck with three uncorellated bases.
This is basically what we need in quantum mechanics, as our physical space has three spacial dimensions, so three uncorellated bases is just enough. However, what I'm interested in is, if we hypothetically needed a fourth basis, would quaternions help?
The question
If we work in $\mathbb{H}^2$, a two-dimensional quaternionic Hilbert space, is there a set of four pairwise uncorellated bases? Specifically, is there an orthonormal basis $E$ such that it is uncorellated to $B, C, D$ defined earlier?