From Kostrikin, A. I. (1982). Introduction to Algebra. Springer-Verlag,
$$\Gamma: \operatorname{SP}(1)\subset\mathbb{H} \to \operatorname{SU}(2)$$ $$a+bi+cj+dk \mapsto \left(\begin{matrix} a+bi & c+di \\ -(c-di) & a-bi \\ \end{matrix}\right),$$ is an isomorphism. And respectively, they can induce rotations in $\operatorname{SO}(3)$ by $\Phi$ and $\psi$:
$$\Phi_g=H^{-1}\circ\varphi_g\circ H: \mathbb R^3 \to \mathbb R^3, g \in \operatorname{SU}(2);$$ $$\psi_p: \mathbb H^- \to \mathbb H^-, p \in \operatorname{SP}(1),$$ $$p \mapsto pqp^{-1}$$ where $$ H: \mathbb R^3 \to M_2^+$$ $$ (x,y,z) \mapsto \left( \begin{matrix} z & x+iy \\ x-iy & -z\\ \end{matrix} \right);$$ $$ \varphi_g: M \mapsto gMg^{-1}$$ and $\mathbb H^-$ is the pure quaternion subspace of $\mathbb H$, which can be seen equal to $\mathbb R^3$.
My question arises here:
According to my calculation, for example, if we pick $q = \sin(\dfrac{\theta}{2}) + k \cos(\dfrac{\theta}{2}) \in \mathbb H$, then $$\Phi_q = \left(\begin{matrix} \cos(\theta) & -\sin(\theta)& 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{matrix} \right)$$ but $$\Gamma(q) = \left(\begin{matrix} \cos(\dfrac{\theta}{2}) & i\sin(\dfrac{\theta}{2}) \\ i\sin(\dfrac{\theta}{2}) & \cos(\dfrac{\theta}{2}) \\ \end{matrix} \right)$$ thus $$\psi_{\Gamma(q)} = \left(\begin{matrix} 1 & 0& 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{matrix} \right) \neq \Phi_q.$$
Is my calculation wrong or are the $\Phi$ and $\psi$ not good inductions or is $\Gamma$ not a good isomorphism so that $q$ and $\Gamma(q)$ is the same rotation in $\operatorname{SO}(3)$?
Thank you.
--update--
I find that $$ \Gamma (q): \left( \begin{matrix} a+di & c+bi \\ -c+bi & a-di \end{matrix} \right) $$ should do the right isomorphic.