Expression for $\{B,e^A\}$ where $\{[B,A],A\}=0$

Question

I found this problem in a quantum field theory class, talking about fermionic modes. $B$ and $A$ are operators that obey anti-commutation relations. I have tried to solve it doing the following: \begin{align*} \{B,e^A\}=\{B,\sum_n \frac{A^n}{n!}\}=\sum_n\frac{1}{n!}\{B,A^n\} \end{align*} So the problem is about calculating $\{B,A^n\}$ under the condition $\{[B,A],A\}=0$.

Answer 1

A sketch:

We have $[B,\,A^2]=0$. Use $\{B,\,UV\}=[B,\,U]V+U\{B,\,V\}$, so $\{B,\,A^2V\}=A^2\{B,\,V\}$ and $\{B,\,A^{2n}V\}=A^{2n}\{B,\,V\}$, to show$$\{B,\,e^A\}=(\cosh A)\underbrace{2B}_{\{B,\,I\}}+(\operatorname{sinch}A)\{B,\,A\},$$provided we have the necessary convergence conditions to suitably manipulate infinite series.