$\left[A, e^{\lambda B}\right]=\lambda e^{\lambda B}[A, B]$ proof?

Question

In a derivation, my professor used the following property: $$\left[A, e^{\lambda B}\right]=\lambda e^{\lambda B}[A, B]$$ Where $A$ and $B$ are operators that commute with their commutator $[B,[A,B]]=[A,[A,B]]=0$. How can this be proven?

Answer 1

It is easy to prove that, in general, the adjoint action is but $$ e^{-\lambda B} A e^{-\lambda B}= A -\lambda [B,A]+ {\lambda^2\over 2!} [B,[B,A]]+ ... $$ But, in your case, the series truncates after the leading two terms.

It is then evident that $$ [A,e^{\lambda B}] = e^{\lambda B} ( e^{-\lambda B} A e^{\lambda B} -A)=- e^{\lambda B} \lambda [B,A]. $$