$\def\ada{\text{ad}_A}$Suppose $A,B$ are operators in Quantum Mechanics, the adjoint endomorphism $\text{ad}_A$ can be expressed by $$ \text{ad}_A B=[A,B]=AB-BA $$ and $$\text{ad}_A^0B=B, \quad \ada^n B= [\underbrace{A,\cdots,[A}_{n\text{ times}},B]].\tag{1}$$
The BCH formula writes $$ \mbox{e}^{-A}B\mbox{e}^{A} = \mbox{e}^{\ada}B =\sum_{n=0}^1 \frac 1{n!} {\text{ad}_A^n}B,\tag{2} $$ whose proof can be found in wiki - BCH forumal .
In a special case where $$ [A,[A,B]]=[B,[B,A]]=0\quad \text{or}\quad \text{ad}_A^2B=\text{ad}_B^2A=0, \tag{3} $$ Glauber's formula holds, $$ \mbox{e}^{A}\mbox{e}^B=\mbox{e}^{A+B+[A,B]/2}.\tag{4} $$
These two proofs use the same trick. I'll repeat it for Eq.(4).
Let $C(\alpha)=\mbox{e}^{\alpha A}\mbox{e}^{\alpha B}$, take the derivative $$ \frac{\mbox{d} C}{\mbox{d}\alpha}=AC+\mbox{e}^{\alpha A}B\mbox{e}^{-\alpha A} C= ({A+\mbox{e}^{\alpha\ada} B})C,\tag{5} $$ where Eq.(1) and $\text{ad}_{\alpha A}=\alpha \text{ad}_A$ is used. When Eq(3) holds, Eq.(2) can be truncated as $$ \frac{\mbox d C}{\mbox dt}=({A+B+\alpha [A,B]})C.\tag{6} $$ Because $A,B$ are independent of $\alpha$, take the integral and setting $\alpha=1$ will finish the proof of eq. (4).
My question is: can the integral of Eq.(6) be taken if Eq.(3) does not hold? The difficulty is the integral of $\exp(\alpha\text{ad}_A)B$. By analogy with the integral of usual function $$ \int\mbox d x \exp({k x })=\frac 1k\exp({kx}), $$ I need someing like $\text{ad}_A^{-1}$, the "inverse" of $\text{ad}_A$. Since $\text{ad}_A$ is the derivative of the adjoint representation $$ \text{ad}_A =\mbox{d}(\text{Ad})_I (A), $$ the inverse may be someting like an integral. What's the expression of it, and can it be expressed in the form of operator products?
