Given two element of SU(2) given by $$A=e^{a_1 i+a_2 j +a_3 k}$$ and $$B=e^{b_1 i+b_2 j +b_3 k}$$
What is $$C=AB=e^{c_1 i + c_2 j+c_3 k}$$ ?
I thought it would be something like $\overrightarrow{c} =\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}$ due to the Baker–Campbell–Hausdorff formula but I can't prove it using the formula like:
$$e^{a_1 i+a_2 j +a_3 k} = \cos(|a|) + \frac{(a_1 i+a_2 j +a_3 k)}{|a|}\sin(|a|)$$
To get$$\left(\cos|a|+a_1\operatorname{sinc}|a|i+a_2\operatorname{sinc}|a|j+a_3\operatorname{sinc}|a|k\right)\left(\cos|b|+b_1\operatorname{sinc}|b|i+b_2\operatorname{sinc}|b|j+b_3\operatorname{sinc}|b|k\right)\\=\cos|a|\cos|b|-\vec{a}\cdot\vec{b}\operatorname{sinc}|a|\operatorname{sinc}|b|\\+[a_1\operatorname{sinc}|a|\cos|b|+b_1\operatorname{sinc}|b|\cos|a|+(a_2b_3-a_3b_2)\operatorname{sinc}|a|\operatorname{sinc}|b|]i\\+\cdots\\=\cos|c|+\frac{c_1i+c_2j+c_3k}{|c|}\sin|c|$$we need$$\cos|c|=\cos|a|\cos|b|-\vec{a}\cdot\vec{b}\operatorname{sinc}|a|\operatorname{sinc}|b|,\,\\\vec{c}\operatorname{sinc}|c|=\vec{a}\operatorname{sinc}|a|\cos|b|+\vec{b}\operatorname{sinc}|b|\cos|a|+\operatorname{sinc}|a|\operatorname{sinc}|b|\vec{a}\times \vec{b}.$$Or if we denote unit vectors with hats,$$\cos|c|=\cos|a|\cos|b|-\hat{a}\cdot\hat{b}\sin|a|\sin|b|,\,\\\hat{c}\sin|c|=\hat{a}\sin|a|\cos|b|+\hat{b}\sin|b|\cos|a|+\sin|a|\sin|b|\hat{a}\times\hat{b}.$$