Proof of the Baker-Campbell-Hausdorff formula

The Baker-Campbell-Hausdorff (BCH) formula appears in stochastic analysis and in quantum mechanics. In the context of stochastic analysis, the BCH formula provides a method to calculate the log-signature of the concatenation of two rough paths. In the context of quantum mechanics, the BCH formula allows to compute products of general operators in Hilbert spaces.

The log-signature of the concatenation of two paths in $\mathbb{R}^d$ expresses as a sum of Lie brackets of formal power series in the tensor algebra of $\mathbb{R}^d$. To learn how the BCH formula is used for computing log-signatures, the reader is referred to section 2.2.4, p. 37, of the book “Differential equations driven by rough paths” by Terry Lyons, Michael Caruan and Thierry Lévy, section 4, p. 4-5, of the notes “Calculation of iterated-integral signatures and log signatures” by Jeremy Reizenstein, and example 9, p. 16, of the notes “A primer on the signature method in machine learning” by Ilya Chevyrev and Andrey Kormilitzin.

This blog post provides a proof of the BCH formula from a quantum perspective, as stated in lemma 34, p. 56, of the lecture notes “Quantum mechanics” by Martin Plenio.

Let $\mathcal{H}$ be a finite Hilbert space over $\mathbb{C}$. Consider two operators $A:\mathcal{H}\rightarrow\mathcal{H}$ and $B:\mathcal{H}\rightarrow\mathcal{H}$. For the operators $A,B$, define the so-called commutator $[A,B]=AB-BA$. The BCH formula in this setting takes the form

$e^{B}Ae^{-B}=A+[B,A]+\frac{1}{2}[B,[B,A]]+\dots$

As a side note, the commutator $[A,B]$ corresponds to a Lie bracket in the rough path case.

To prove the BCH formula, introduce the auxiliary function $f(x)=e^{xB}Ae^{-xB}$, where $x\in\mathbb{R}$. Take the Taylor series expansion of $f(x)$ around $0$, i.e. consider $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}$. Setting $x=1$ leads to

$e^{B}Ae^{-B}=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}=A+f^{(1)}(0)+\frac{f^{(2)}(0)}{2}+\dots$

By application of definition 27, p. 51, of Plenio’s notes, it holds that $e^{xB}=\sum_{n=0}^{\infty}\frac{x^n B^n}{n!}$. Differentiating $e^{xB}$ gives $\frac{de^{xB}}{dx}=\sum_{n=1}^{\infty}\frac{x^{n-1} B^n}{(n-1)!}=B\sum_{k=0}^{\infty}\frac{x^k B^k}{k!}=Be^{xB}$. Similarly, $\frac{de^{-xB}}{dx}=-Be^{-xB}$. The chain rule yields

$\frac{df(x)}{dx}=\frac{de^{xB}}{dx}Ae^{-xB}+e^{xB}A\frac{de^{-xB}}{dx}$.

So, $\frac{df(x)}{dx}=Be^{xB}Ae^{-xB}-e^{xB}ABe^{-xB}$. For $x=0$, it is obvious that $f^{(1)}(0)=BA-AB=[B,A]$.

Working in a similar manner, the second-order derivative of $f$ is found to be

$\frac{d^{(2)}f(x)}{dx^2}=B^2e^{xB}Ae^{-xB}-2Be^{xB}ABe^{-xB}+e^{xB}AB^2e^{-xB}$.

For $x=0$, it can be seen that $f^{(2)}(0)=B^2A-2BAB+AB^2=[B,[B,A]]$.

The rest of the proof follows inductively.