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There is something that is not clear to me

Here is what I know:

  • Pauli matrices are $\sigma_1 = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, $\sigma_2 = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$, $\sigma_3 = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$
  • Each matrix in $A=SU(2)$ can be represented as $\begin{pmatrix}\alpha & -\bar \beta \\ \beta & \bar\alpha\end{pmatrix},\, |\alpha|^2 + |\beta|^2 = 1$ and decomposed as $A=wI-ix\sigma_1-iy\sigma_2-iz\sigma_3$ with $x^2+y^2+z^2+w^2=1$. That is $I,-i\sigma_1,-i\sigma_2, -i\sigma_3$ are sort of a basis for $SU(2)$, but values have to be normalized
  • Each matrix in $A=SU(2)$ can be obtained as $A=\exp (H)$ where $H$ is skew Hermitian. Now a basis for skew Hermitian matrices is also $-i\sigma_1,-i\sigma_2, -i\sigma_3$ and $H=-\alpha i\sigma_1-\beta i\sigma_2 -i\gamma \sigma_3$

My question is: is there a simple relation between the parametrization $w,x,y,z\in S^3$ for $$A\in SU(2), A=wI-ix\sigma_1-iy\sigma_2-iz\sigma_3$$ and the parametrization $\alpha,\beta,\gamma\in \mathbb{R}^3$ for $$A\in SU(2), A=\exp (H),H=-\alpha i\sigma_1-\beta i\sigma_2 -i\gamma \sigma_3$$

The question arises because I see often the first parametrization in use (for example as a way to represent unit quaternions) but never the second one

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The exponentiating a Pauli vector is straightforward, $$A\in SU(2),~~ A=\exp (H), \qquad H=-\alpha i\sigma_1-\beta i\sigma_2 -i\gamma \sigma_3 .$$ Define $$ \theta ^2\equiv \alpha^2+\beta^2+\gamma^2 , \qquad n_1=\alpha /\theta;\qquad n_2=\beta /\theta;\qquad n_3=\gamma /\theta , $$ so that the antihermitean logarithm of the group element is but $$ H= -i\theta~ \vec n \cdot \vec \sigma , ~~ \Longrightarrow ~\exp (H)= \cos \theta -i\sin\theta ~~\vec n \cdot \vec \sigma ~. $$

That is to say $$ w=\cos\theta; \qquad x= \frac{\sin\theta}{\theta} \alpha ; \qquad y= \frac{\sin\theta}{\theta} \beta ; \qquad z= \frac{\sin\theta}{\theta} \gamma ~ $$ a euclidean unit four-vector, alright.

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