When studying symmetries in quantum mechanics, one often has to calculate $UBU^\dagger$ where $B$ is a self-adjoint operator and $U$ is a unitary operator. More often than not $U$ has an exponential form $U=e^{-A}$ with $A$ self-adjoint operator, so we have to calculate $e^{-A}Be^A$.

I've stumbled across this formula when studying the angular momentum, while proving that $\frac{L_j}{i\hbar}$ are the representations of the generators of the Lie algebra of rotations.

In the proof $e^{-A}Be^A$ was expanded into a sum of nested commutators, using what has been called "Hadamard formula".

$e^{-A}Be^A=\sum_{n=0}^{\infty}\frac{1}{n!}[...[[[B,A],A],A]...]$ with $n$ commutators

I didn't find any reference for such a formula and I don't understand how we can get this result.

When studying symmetries in quantum mechanics, one often has to calculate $UBU^\dagger$ where $B$ is a self-adjoint operator and $U$ is a unitary operator. More often than not $U$ has an exponential form $U=e^{-A}$ with $A$ self-adjoint operator, so we have to calculate $e^{-A}Be^A$.

I've stumbled across this formula when studying the angular momentum, while proving that $\frac{L_j}{i\hbar}$ are the representations of the generators of the Lie algebra of rotations.

In the proof $e^{-A}Be^A$ was expanded into a sum of nested commutators, using what has been called "Hadamard formula" $$ e^{-A}Be^A=\sum_{n=0}^{\infty}\frac{1}{n!}[...[[[B,A],A],A]...] $$ with $n$ commutators.

I didn't find any reference for such a formula and I don't understand how we can get this result.