Hadamard formula in quantum mechanics 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.
 A: The formula is closely related to the Baker-Campbell-Hausdorff formula, it can be found on the linked Wiki page as "An important lemma". The proof is relatively simple: define a matrix valued function $f(s)$ via
$$
f(s) = e^{sA} B e^{-sA} .
$$
By differentiating, we obtain a differential equation for $f(s)$:
$$
\frac{df}{ds} = A e^{sA} B e^{-sA} - e^{sA} B e^{-sA}A = \big[A, f(s) \big],
$$
together with the initial condition $f(0) = B$. If you think of $[A,\cdot]$ as a super-operator on the matrix $f(s)$, then it's easy to see that the solution to the above differential equation is
$$
f(s) = e^{s[A,\cdot]} B = \sum_{n = 0}^{\infty} \frac{s^n}{n!} \underbrace{\big[A, \big[A, \ldots \big[A}_{n \text{ commutators}},B \big] \ldots \big] \big]
$$
(Even if the expression $e^{s[A,\cdot]}$ does not make sense to you, you can see by inspecting the right-hand side that we indeed have a solution to the differential equation.) Setting $s=1$, we obtain the desired result.
A: One standard method to dealing with operator exponentials is to write them as their definition - that is, $$e^A = \sum_n\frac{A^n}{n!}.$$ The Hadamard lemma can then be derived with some algebra:
$$\begin{align*}
e^{-A}Be^A &= \left(1 - A + \frac{A^2}{2!} - \ldots\right)B\left(1 + A + \frac{A^2}{2!} + \ldots\right)\\
&= B + BA - AB + \ldots\\
&= B + [B,A] + \frac{1}{2}[[B,A],A] + \ldots
\end{align*}$$
as desired. You'll have to be careful to match up all the terms with the same degree of $A$, but you can prove that they're all accounted for by induction.
A: Making the answer of @onetoinfinity more explicit I proceed as follows.
First to group the terms (commutators) in the right way, one should know explicitly how $[A,[A,[...[A,B]...]]$ look like, which I like to denote $[A,\cdot\ ]^nB$. The answer is
$$
[A,\cdot\ ]^nB := \underbrace{[A,[A,[...[A}_{n},B]...]] = \sum_{k=0}^n \binom{n}{k}(-)^k A^{n-k}BA^k.
$$
The proof is by induction below. Now the expression on the LHS can be written as
$$
e^A B e^{-A} = \sum_{n=0}^\infty \sum_{k=0}^n (-)^k \frac{A^{n-k}BA^k}{(n-k)!k!}
= \sum_{n=0}^\infty \frac{1}{n!}[A,\cdot\ ]^nB = \left(e^{[A,\cdot\ ]}\right)B
$$
q.e.d.

The proof by induction (slightly different from this)

*

*For $n=1$ it is clear that $[A,\cdot\ ]^1B = AB - BA = [A,B]$

*To show that for some $n \ge 1$
$$
[A,\cdot\ ]^n B = \sum_{k=0}^n \binom{n}{k}(-)^k A^{n-k}BA^k \implies [A,\cdot\ ]^{n+1} B = \sum_{k=0}^{n+1} \binom{n+1}{k}(-)^k A^{n+1-k}BA^k
$$
we observe that
$$
[A,\cdot\ ]^{n+1}B = [A,[A,\cdot\ ]^{n}B] = A[A,\cdot\ ]^{n}B - ([A,\cdot\ ]^{n}B)A
 \tag{*}$$
and
$$
\binom{n+1}{k} = \frac{n+1}{n+1-k}\binom{n}{k} \\= \frac{n+1-k+k}{n+1-k}\binom{n}{k} =
\binom{n}{k} + \frac{k}{n+1-k}\binom{n}{k} = \binom{n}{k} + \binom{n}{k-1}.
$$
The hypothesised expression for $[A,\cdot\ ]^{n+1} B$ becomes

\begin{align}
\sum_{k=0}^{n+1} \binom{n+1}{k}&(-)^k A^{n+1-k}BA^k \\
&= \sum_{k=0}^{n+1}\left\{ \binom{n}{k}(-)^k A^{n+1-k}BA^k + \binom{n}{k-1}(-)^k A^{n+1-k}BA^k \right\} \\
&= \sum_{k=0}^{n+1}\binom{n}{k}(-)^k A^{n+1-k}BA^k - \sum_{k=0}^{n+1} \binom{n}{k-1}(-)^k A^{n+1-k}BA^k \\
&= A(\sum_{k=0}^{n}\binom{n}{k}(-)^k A^{n-k}BA^k) - (\sum_{k=1}^{n+1} \binom{n}{k-1}(-)^{k-1} A^{n-(k-1)}BA^{k-1})A \\
&= A\left(\sum_{k=0}^{n}\binom{n}{k}(-)^k A^{n-k}BA^k\right) - \left(\sum_{k=0}^{n} \binom{n}{k}(-)^k A^{n-(k)}BA^{k}\right)A \\[5pt]
&= A[A,\cdot\ ]^{n}B - ([A,\cdot\ ]^{n}B)A \\[8pt]
&= [A,[A,\cdot\ ]^nB]
\end{align}
where a shift in the index $k$ was done in the 4th line (equality), the assumption for $n$ was used in the 5th line and $(*)$ was used in the last line.
