I know that, if we consider $$e^{i \alpha Z} e^{i \beta X} e^{i \gamma Z}$$
(where $X, Z$ are the Pauli matrices) then we can get any element of $SU(2)$ (the so-called $ZXZ$ decomposition). If I write down an element of $SU(2)$ as $$e^{i (aX + bY + cZ)}$$
is there a relationship between $(a,b,c)$ and $(\alpha, \beta, \gamma)$? I tried to set up the equation but it seems highly non-trivial to solve it.
(Partial answer)
Of course, you may use the BCH / Zassenhaus formula, but the computations promise to become quickly untractable. Instead, it is more judicious to make good use of the following well-known formula : $$ e^{i\theta\,\hat{n}\cdot\vec{\sigma}} = \cos\theta + i\,\hat{n}\cdot\vec{\sigma}\sin\theta, $$ where $\hat{n}$ is a unit vector. In consequence, the second expression corresponds to $\theta = \sqrt{a^2+b^2+c^2}$ and $\hat{n} = (\frac{a}{\theta},\frac{b}{\theta},\frac{c}{\theta})$, when the first expression is decomposed as $$ e^{i\alpha\sigma_z}e^{i\beta\sigma_x}e^{i\gamma\sigma_z} = (\cos\alpha + i\sigma_z\sin\alpha)(\cos\beta + i\sigma_x\sin\beta)(\cos\gamma + i\sigma_z\sin\gamma) $$ thanks to the same formula. I let you expand this expression with the help of the additional relation $\sigma_\mu\sigma_\nu = \delta_{\mu\nu} + i\varepsilon_{\mu\nu\rho}\sigma_\rho$, where $\delta_{\mu\nu}$ is the Kronecker delta and $\varepsilon_{\mu\nu\rho}$ the Levi-Civita symbol, and equate the different terms.