# 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.

• Well, for A self-adjoint, $e^{\pm A}$ is not unitary, so you may want to adjust that. Jun 18, 2021 at 21:09

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.

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.

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.