Product of Pauli-matrix exponentials Given Pauli matrices $\sigma_x = \begin{pmatrix}  0 & 1 \\ 1 & 0 \end{pmatrix}$ and $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, can one write $e^{\alpha \sigma_z} e^{\beta \sigma_x}$ in terms of $e^{\beta \sigma_x} e^{\alpha \sigma_z}$ ($\alpha$, $\beta$ are some complex numbers). For example, is it possible to find some $M$ such that
\begin{equation}
e^{\alpha \sigma_z} e^{\beta \sigma_x} = e^{\beta \sigma_x} e^{\alpha \sigma_z}M?
\end{equation}
 A: Use the standard formula,
$$
e^{\alpha \sigma_z}= I\cosh \alpha + \sigma_z  \sinh\alpha = , \\
e^{\beta \sigma_x}= I\cosh \beta + \sigma_x  \sinh\beta ,
$$
to compute the quadruple product suggested in the comments,
$$
M=e^{-\alpha \sigma_z}e^{-\beta \sigma_x}e^{\alpha \sigma_z}e^{\beta \sigma_x}\\ = 
e^{-ia\sigma_z}e^{-ib\sigma_x}e^{ia \sigma_z} e^{ib\sigma_x}\\
= \Biggl (I\cos a\cos b+i\left (-\sigma_y\sin a \sin b  -\sigma_z \sin a \cos b-\sigma_x\sin b \cos a  \right ) \Biggr)\\ \times \Biggl ( I\cos a\cos b   +i\left (-\sigma_y \sin a \sin b  + \sigma_z \sin a \cos b   + \sigma_x  \sin b \cos a  \right ) \Biggr)\\
= I(1-2\sin^2a ~\sin^2 b)\\ +i\Bigl (\sigma_x 2\sin^2 a\sin b\cos b-\sigma_y 2\sin a \cos a\sin b\cos b -\sigma_z 2\sin^2b\sin a \cos a \Bigr ).
$$
Can you take it from here? You might be amused taking $a=\pi$, and seeing it geometrically, or else $b=\pi$ .
Note the all-important unitarity constraint/check: the sum of the squares of the coefficients of I and the σ s, respectively amounts to 1, as it should!
From the coefficient of the identity matrix, you find its  arccos, θ, and you just confirmed that the coefficient of the $i\hat n\cdot \vec \sigma$, for normalized unit $\hat n$,  messier  to write down, must be $\sin\theta$! So,
$$
M=e^{i\theta \hat n\cdot \vec \sigma}.
$$
For imaginary hyperbolic angles, so real trigonometric angles, this amounts to a quadruple group element product for SU(2), dubbed the group (not algebra) commutator.
