Can Hadamard's formula be used for fermionic operators?

Question

Can I use this special case of Hadamard's formula $$e^\hat B \hat A e^{-\hat B}= A + [B,A]+\frac{1}{2!}[B, [B,A]] + \dots$$ for fermionic operators?

Suppose I have fermionic operators that obey anticommutation relations $\{a,a^{\dagger}\}=1$ and $\{a,a\}=\{a^{\dagger},a^{\dagger}\}=0$. The commutator for fermions $[a,a^{\dagger}]=1-2a^{\dagger}a$.

Then, if $A=a^{\dagger}$ and $B=a$, I can get

$e^\hat a \hat a^{\dagger} e^{-\hat a}=a^{\dagger}+[a,a^{\dagger}]+\frac{1}{2!}[a, [a,a^{\dagger}]]+ \dots = a^{\dagger}+(1-2a^{\dagger}a)+\frac{1}{2!}[a, (1-2a^{\dagger}a)]+\dots$

Is this formula universal for fermionic and bosonic operators?

Answer 1

Hadamard's formula $$ e^XYe^{-X}~=~e^{[X,\cdot]_C}Y \tag{1}$$ also works if one or both operators $X$ and $Y$ are Grassmann-odd (or even don't carry definite Grassmann-parity). Here it is important that $[\cdot,\cdot]_C$ in eq. (1) is the commutator; not the supercommutator nor the anticommutator. The proof is very similar to the Grassmann-even case.

NB: Be aware that a Grassmann-odd operator $X$ does not need to square to zero, cf. e.g. SUSY charge operators.