I have the following $SU(2)$ matrix $$U(\mu,\theta,\phi)=\left( \begin{array}{cc} e^{i \mu } (\cos (\mu )-i \cos (\theta ) \sin (\mu )) & e^{i \varphi } \sin (\theta ) \sin (\mu ) \\ -e^{-i \varphi } \sin (\theta ) \sin (\mu ) & e^{-i \mu } (\cos (\mu )+i \cos (\theta ) \sin (\mu )) \\ \end{array} \right). \tag{1}$$ And I am thinking how to transform this matrix into the exponential form $$U=\exp(-i(a_1 J_1+a_2 J_2+a_3 J_3)), \tag{2}$$ where $a_1,a_2,a_3$ are free real parameters, $J_1,J_2,J_3$ are $SU(2)$ generators. In our case, in fundamental representation, $J_i=\sigma_i/2$, so we can expand this as $$ \begin{aligned} &U=\exp(-i(a_{1} {\color{red} J_{1}}+a_{2} {\color{red} J_{2}}+a_{3} {\color{red} J_{3}})) \\ &=\left( \begin{array}{cc} \cos \left(\frac{1}{2} a\right)-i x_{3} \sin \left(\frac{1}{2} a\right) & (-x_{2}+(-i) x_{1}) \sin \left(\frac{1}{2} a\right) \\ (x_{2}-i x_{1}) \sin \left(\frac{1}{2} a\right)& \cos \left(\frac{1}{2} a\right)+i x_{3} \sin \left(\frac{1}{2} a\right) \\ \end{array} \right). \end{aligned}\tag{3} $$ where we have defined $\vec{a}=a \vec{x}$, $\vec{x}$ is a unit vector and $a=||\vec{a}||$. (According to comment 1)
Since $SU(2)$ matrix is connected and compact, we can always able to write it into eq.(2). But I find it's really hard to transform eq.(1) into (2). Can anyone give me some advices?
I've tried that if the $e^{i\mu}$ and $e^{-i\mu}$ are disappeared in (1), the conversation from (1) to (2) is straightforward, $a_{1}\to -2 \mu \sin (\theta ) \sin (\varphi ),a_{2}\to -2 \mu \sin (\theta ) \cos (\varphi ),a_{3}\to 2 \mu \cos (\theta )$. However, I am puzzled how to obtain $a_1,a_2,a_3$ from original (1).
This might help identifying (1) with (3).
By taking the sum and the difference of the diagonal entries, and, separately, of the off-diagonal ones, you have the four equations, $$ \cos{a\over2}= 1+\sin^2\!\mu~(\cos\theta -1),\\ x_3 \sin{a\over2}=\sin \mu \cos\mu ~ (\cos\theta -1 ),\\ x_1 \sin{a\over2}= -\sin\phi \sin\theta\sin\mu,\\ x_2 \sin{a\over2}=-\cos\phi \sin\theta \sin\mu. $$
Note, crucially, squaring both sides and adding you get 1 on both sides!
So you have a from the first equation, $$ a= 2 \arccos \left [ 1+\sin^2\!\mu~(\cos\theta -1) \right ], $$ whence its sine,
$$ \sin{a\over2}= \sqrt{1-\cos^2{a\over2}}=... $$ whence $\hat x$ from the other three...