Consider the Baker-Hausdorff formula for two operators $A$ and $B$: $$e^BA e^{-B} =A+[B,A]+\frac{1}{2!}[B,[B,A]]+\frac{1}{3!}[B,[B,[B,A]]]+....,$$ where $[A,B]=AB-BA$. In the case of my problem, $[B,a_1]=\alpha\, a_1+\beta \,a_2+\gamma,$ and $[B,a_2]=\alpha\, a_2+\beta \,a_1,$ where the parameters are either imaginary $(\alpha, \beta)$ or complex $(\gamma)$. $a_1$ and $a_2$ are operators. So, I need to find the following: $$ a_1\,+(\alpha a_1+\beta a_2+\gamma)+\frac{1}{2!}\Big\{\alpha\Big(\alpha a_1+\beta a_2+\gamma\Big)+\beta\Big(\alpha\, a_2+\beta \,a_1\Big)\Big\}+\frac{1}{3!}\Big\{\alpha\Big(\alpha (\alpha\, a_1+\beta \,a_2+\gamma)+\beta (\alpha\, a_2+\beta \,a_1)+\gamma\Big)+\beta\Big(\alpha\, (\alpha\, a_2+\beta \,a_1)+\beta \,(\alpha\, a_1+\beta \,a_2+\gamma)\Big)\Big\}+....$$
In other words, in each new term, one should replace $a_1$ by $(\alpha\, a_2+\beta \,a_1+\gamma)$ and $a_2$ with $(\alpha\, a_2+\beta \,a_1)$, and repeat this procedure $N$ times, and finally calculate the limit when $N\rightarrow \infty$. Since I'm not familiar with Mathematica programming, I tried to find the "pattern" of successive terms and thereby find the limit of n'th term as $N\rightarrow\infty$. I tried
f0 = a1;
f1 = Collect[α a1 + β a2 + γ /. {a1 -> α a1 + β a2 + γ, a2 -> α a2 + β a1}, {a1, a2}] // Simplify
f2 = Collect[% /. {a1 -> α a1 + β a2 + γ, a2 -> α a2 + β a1}, {a1, a2}] // Simplify
f3 = Collect[% /. {a1 -> α a1 + β a2 + γ, a2 -> α a2 + β a1}, {a1, a2}] // Simplify
f4 = Collect[% /. {a1 -> α a1 + β a2 + γ, a2 -> α a2 + β a1}, {a1, a2}] // Simplify
Collect[f0 + f1 + 1/2 f2 + 1/6 f3 + 1/4! f4, {a1, a2}]
But, unfortunately, the pattern is not so clear to me. So I think it would be better to find it systematically. The answer is probably either exponential or trigonometric.
In fact, I'm trying to find equation 6 of the following reference: arxiv.org/pdf/1609.00075.pdf
