Disentangling exponential of number operator and creation and annihilation operators

I think i have found a method using the answers to these two questions:

https://mathoverflow.net/questions/163172/lie-group-about-the-quantum-harmonic-oscillator

How do disentangling and reordering of exponential operators work?

We can map the following matrices to the ladder operators:

$a^daggerequiv A=left[matrix{0 & 1 & 0\0 &0 &0\0 &0 &0}right]$, $a equiv B=left[matrix{0 & 0 & 0\0 &0 &1\0 &0 &0}right]$, $Iequiv C=left[matrix{0 & 0 & -1\0 &0 &0\0 &0 &0}right]$, $Nequiv D= frac12left[matrix{1 & 0 & 0\0 &-1 &0\0 &0 &1}right]$

The matrices A,B,C,D satisfy the commutation relations of the ladder operators. Then evaluate left hand side and right hand side using these matrices and match coefficients. It seems to work, but I'd like some confirmation this is the right approach as i have no experience with lie algebras.

Not an answer, but an extended comment on your basically sound approach, since the comment format does not permit such extended comments. The group involved is the oscillator group, and the 3d rep you found is a faithful one, so any group relation for it will also hold for the abstract group in general, so, all representations! I will call your central element C of your answer Z, and it can filter out of all expressions, commuting with everything.

The generic statement supported by Lie's theorem is that the product of all group elements will close to an exponential of some linear combination of all generators in the Lie algebra, so, then, $$ ?^{?Z} ?^{??^†} ?^{??}?^{??}=?^{?'Z+??+??+??^†}. $$ However, since Z commutes with everything, we can invert the first factor of the l.h.s. to the right, and incorporate it into a new parameter $phi'-theta=phi$, so that $$ ?^{??^†} ?^{??}?^{??}=?^{?Z+??+??+??^†}, tag{*} $$ where the parameters $phi,alpha,beta, gamma$ are guaranteed to be functions of $G,A,B$.

Now, by the nilpotency of the first three generators, and the diagonally of the fourth, the l.h.side trivially evaluates to $$ e^{-A/2} begin{bmatrix}e^A & G & BG\0 &1 &B\0 &0 &e^Aend{bmatrix}, $$ with determinant $e^{A/2}$.

This must equal $$ exp begin{bmatrix} alpha/2 & gamma & -phi\0 &-alpha/2 &beta\0 &0 &alpha/2end{bmatrix}. $$ Its determinant is $e^{alpha/2}$ by the identity $e^{operatorname{Tr} M} = det e^M$.

Now, to second order in its parameters, it expands to $$ begin{bmatrix}1+ alpha/2 +alpha^2/8& gamma & -phi-phialpha/2+betagamma/2\0 &1-alpha/2 +alpha^2/8&beta\0 &0 &1+alpha/2+alpha^2/8end{bmatrix}. $$

Comparing with the above l.h.side dictates, to second order, $$A=alpha, qquad B=beta e^{alpha/2}, qquad G=gamma e^{alpha/2},$$ but then you realize the upper-rightmost entry is mismatched, and requires a non-vanishing $phi$, $$ BGe^{-A/2}= betagamma e^{alpha/2}= betagamma/2 -phi(1+alpha/2), $$ to pick up the slack. One had to go to second order to see this, as your need at least one commutation $[a,a^dagger]$ to produce the central element.

So, then, $phi$ is actually essential in your amended expression (*): this is not a degree of freedom that could be omitted. Apologies (with Pascal) for lacking the time to make the comment shorter.