We will consider the expression \(\mathrm{e}^{\hat{A}}\hat{B}\mathrm{e}^{-\hat{A}}\) with operators \(\hat{A}\) and \(\hat{B}\). In order to derive a formula we introduce an auxiliary operator \(\hat{F}(t)=\mathrm{e}^{\hat{A}t}\hat{B}\mathrm{e}^{-\hat{A}t}\) that depends on the parameter \(t\). The operator \(\hat{F}(t)\) obeys the differential equation \[ \frac{\mathrm{d}}{\mathrm{d}t}\hat{F}(t)=[\hat{A},\hat{F}(t)] \] with the initial condition \(\hat{F}(0)=\hat{B}\). From this differential equation we obtain an integral equation \[ \hat{F}(t)=\hat{B}+\int_{0}^{t}[\hat{A},\hat{F}(t_{1})]\,\mathrm{d}t_{1}\,. \] The solution can be written as \(\hat{F}=\hat{F}_0 + \hat{F}_1 + \hat{F}_2 + \cdots\), where \(\hat{F}_0 = \hat{B}\) and \[ \hat{F}_{n+1}(t)=\int_0^t [\hat{A},\hat{F}_n (t_1 )]\,\mathrm{d}t_1\,. \] The integral in the term \(\hat{F}_n\) is of the form \[ \int_0^t\mathrm{d}t_1\cdots\int_0^{t_{n-1}}\mathrm{d}t_n = \frac{1}{n!}\int_0^t\mathrm{d}t_1\cdots\int_0^t\mathrm{d}t_n = \frac{1}{n!}t^n\,. \] We obtain \[ \hat{F}(t)=\hat{B}+t[\hat{A},\hat{B}]+\frac{1}{2!}t^{2}[\hat{A},[\hat{A},\hat{B}]]+\cdots\,. \] Substituting \(t=1\) we get the equation \[ \mathrm{e}^{\hat{A}}\hat{B}\mathrm{e}^{-\hat{A}}=\hat{B}+[\hat{A},\hat{B}]+\frac{1}{2!}[\hat{A},[\hat{A},\hat{B}]]+\cdots\,. \]
In partucular, if the commutator \([\hat{A},\hat{B}]\) commutes with both operators \(\hat{A}\) and \(\hat{B}\), \([[\hat{A},\hat{B}],\hat{A}]=[[\hat{A},\hat{B}],\hat{B}]=0\), the equation takes the form \[ \exp(\hat{A}t)\hat{B}\exp(-\hat{A}t)=\hat{B}+t[\hat{A},\hat{B}]\,. \]
We will derive the formula for the operator \(\mathrm{e}^{\hat{A}+\hat{B}}\) when the commutator \([\hat{A},\hat{B}]\) commutes with both operators \(\hat{A}\) and \(\hat{B}\). For this purpose we introduce an auxiliary operator \(\hat{F}(t)=\mathrm{e}^{(\hat{A}+\hat{B})t}\) that depends on the parameter \(t\). This operator obeys the diferential equation \[ \frac{\mathrm{d}\hat{F}}{\mathrm{d}t}=(\hat{A}+\hat{B})\hat{F} \] with the initial condition \(\hat{F}(0)=1\). We search the solution in the form \[ \hat{F}(t)=\mathrm{e}^{\hat{A}t}\mathrm{e}^{\hat{B}t}\hat{X}(t)\,, \] where \([\hat{X}(t),\hat{A}]=0\) and \([\hat{X}(t),\hat{B}]=0\). The derivative with respoect to \(t\) is \[ \begin{aligned} \frac{\mathrm{d}\hat{F}}{\mathrm{d}t} & = \mathrm{e}^{\hat{A}t}\mathrm{e}^{\hat{B}t}\frac{\mathrm{d}\hat{X}(t)}{\mathrm{d}t} +\hat{A}\mathrm{e}^{\hat{A}t}\mathrm{e}^{\hat{B}t}\hat{X}(t) +\mathrm{e}^{\hat{A}t}\hat{B}\mathrm{e}^{\hat{B}t}\hat{X}(t)\\ & = \frac{\mathrm{d}\hat{X}(t)}{\mathrm{d}t}\mathrm{e}^{\hat{A}t}\mathrm{e}^{\hat{B}t} +\hat{A}\hat{F}(t)+(\hat{B}+t[\hat{A},\hat{B}])\hat{F}(t)\,. \end{aligned} \] Taking into account the differential equation for the operator \(\hat{F}(t)\) we get \[ \frac{\mathrm{d}\hat{X}(t)}{\mathrm{d}t}=-t[\hat{A},\hat{B}]\hat{X}(t)\,. \] The solution of this equation is \(\hat{X}(t)=\exp\left(-\frac{t^{2}}{2}[\hat{A},\hat{B}]\right)\). Taking \(t=1\) we obtain the formula \[ \mathrm{e}^{\hat{A}+\hat{B}}=\mathrm{e}^{\hat{A}t}\mathrm{e}^{\hat{B}t}\mathrm{e}^{-\frac{1}{2}[\hat{A},\hat{B}]}\,. \]