How do disentangling and reordering of exponential operators work? I have seen in several sources that by invoking Lie groups, $$e^{\alpha_1 g_1+\alpha_2 g_2 + \dots} = e^{\beta_1 g_1}e^{\beta_2 g_2}\dots $$ where $g_i$ are elements of a Lie algbera.
For example, take the two-mode squeeze operator in quantum optics:
$$e^{-\xi\hat{a}\hat{b}+\xi^*\hat{a}^\dagger\hat{b}^\dagger} = e^{-\frac{\xi^*}{|\xi|}\tanh|\xi|\hat{a}^\dagger\hat{b}^\dagger} e^{-\ln\cosh|\xi| \left(\hat{a}^\dagger\hat{a}+\hat{b}^\dagger\hat{b}+1\right)} e^{\frac{\xi}{|\xi|}\tanh|\xi| \hat{a}\hat{b}}.$$
A few other examples can be the displacement and the single-mode squeeze operators.
My question is what are the conditions under which we can disentangle the operators like this and also reorder them?
 A: Section III of this classic illustrates the method. I'll bypass the subtle math and cut to the chase for your specific example, taking the trivial case of ξ real... you do general things to your satisfaction, yourself, or check the ref in the comment of @ZeroTheHero above.
This is an identity between exponentials of operators. In Lie group theory, composition of such exponentials (group elements) amounts to a single group element: an exponential of a linear combination of nested commutators of these operators ("the Lie algebra" of your l.h.s.). All commutators, even an infinity of them, ultimately close into a finite number of operators, a finite-dimensional Lie algebra. (There are also infinite-dimensional Lie algebras, but let's not go there...)
So what is the Lie algebra in your example? It's su(1,1), but don't worry about it. I'll map it to the Pauli matrices, so you only need to recall their commutation relations, not even knowing names and such of the relevant Lie algebras; you only need to know these matrices are a faithful representation of the algebra: they reproduce all commutation relations of it exactly.
So,  define
$$
\sigma^+\equiv i a^\dagger b^\dagger, \qquad \sigma^-\equiv i a b, 
\qquad  \sigma_3\equiv 1+ a^\dagger a+ b^\dagger b, 
$$
and confirm that these obey this Lie algebra,
$$
[\sigma_3,\sigma^{\pm}]= \pm \sigma^{\pm}, \qquad [\sigma^+,\sigma^-]= \sigma_3. 
$$

*

*Now you know Pauli matrices obey this Lie algebra as well, so, if it held for them that
$$
e^{i\xi(\sigma^-- \sigma^+)}  = e^{i \tanh \xi ~\sigma^+ } e^{-\ln \cosh \xi ~ \sigma_3} e^{-i \tanh \xi ~\sigma^-} ,
$$
then the CBH combinatorics would be identical for your operators as well, and your identity would hold.

Indeed, the l.h.s is but
$$
e^{\xi \sigma_2}= \cosh \xi ~ 1\!\!1 +\sinh \xi ~ \sigma_2~. 
$$
The r.h.s., by dint of the two nilpotent exponents and the diagonal middle one, is
$$
(1\!\!1 + i \tanh \xi ~\sigma^+ ) ~~\operatorname{diag}(1/\cosh \xi , \cosh \xi) ~~(1\!\!1 - i \tanh \xi ~\sigma^- )\\ 
=\cosh \xi ~ 1\!\!1 -\sinh \xi ~ \sigma_2~,
$$
the complex conjugate of the above. Hmmmm...
I believe your stated identity has flakey signs on the left side, as seen by taking small ξ  and comparing the expanded exponentials!
In any case, you get the drift...
Check Prob 5 here to see the versatility of the method.
