Find $\vec{n}$ such that $e^{-i\alpha S_z} e^{-i\beta S_y} e^{-i\gamma S_z} = e^{-i \vec{n} \cdot \vec{S}}$ Any idea how to effciently find vector $\vec{n}$ in the formula
$$e^{-i\alpha S_z} e^{-i\beta S_y} e^{-i\gamma S_z} = e^{-i \vec{n} \cdot \vec{S}},$$
where $\alpha,\beta,\gamma$ are given and $\vec{S} = (S_x, S_y, S_z)$ and operators $S_i$ satisfy commutation relations $[S_i, S_j] = \epsilon_{ijk}S_k$? I know how to do it graphically with rotation on a sphere, but I am looking for a different method, solely algebraical, which I could apply to larger groups e.g. SU(3).
 A: You start from the remarkably simple "Gibbs formula" generic group composition law for SU(2 in terms of Pauli matrices, the defining (so, faithful) representation of SU(2). As a group identity, the result will then hold for all representations! Note however, you have chosen your S antihermitian, and I'm sure you don't mean that, as they do not give you unitary exponentials. With an i on the r.h.s. of their Lie algebra, they'd be hermitian generators of SU(2), so then half the corresponding Pauli matrices. I'll just let you clean up your normalizations to your satisfaction, and multiply the Pauli-half-angle analog of your two exponentials, leaving the second, messier, multiplication for you. 
$$
e^{ia \sigma_z} e^{ib \sigma_y}  \\
= \exp \left ( i \frac{\arccos (\cos a ~ \cos b)}{\sqrt{1-\cos^2 a \cos^2 b } } ~ (\sigma_z \sin a \cos b + \sigma_y \sin b \cos a + \sigma_x \sin a \sin b) \right ),
$$
basically the simplest formula of spherical trigonometry.
You will not have such luck for SU(3), of course. With judicious maneuvering, you might be able to use this rep fact.
