Clean proof of Baker-Campbell-Hausdorff Formula

Question

I am thinking of the cleanest way to prove the BCH formula and I have come up with this.

First, work out $e^{\lambda A}Be^{-\lambda A}$ by expanding the exponentials (sums go from $0$ to $\infty$): $$\left(\sum_{n}\frac{\lambda^n}{n!}A^n \right)B\left(\sum_{k}\frac{(-\lambda)^k}{k!}A^k \right).$$ This can be written as $$\sum_{n,k}\frac{(-1)^k\lambda^{n+k}}{n!k!}A^nBA^k.$$ We define $m=n+k$, and rewrite the previous expression as $$\sum_{m=0}^{\infty}\sum_{n=0}^m\frac{(-1)^{m-n}\lambda^{m}}{n!(m-n)!}A^nBA^{m-n}.$$ By dividing and multiplyling by $m!$ inside the sum we finally arrive to $$\sum_{m=0}^{\infty}\frac{\lambda^m}{m!} \sum_{n=0}^m(-1)^{m-n}\frac{m!}{n!(m-n)!}A^nBA^{m-n}.$$

The formula is usually presented as $$e^{\lambda A}Be^{-\lambda A}=B+\lambda[A,B]+\frac{\lambda^2}{2!}[A,[A,B]]+...$$

By comparing with what I got, proving BCH is reduced to proving $$\underbrace{[A,[A,[...,[A,B]...]}_{m}=\sum_{n=0}^m(-1)^{m-n}\frac{m!}{n!(m-n)!}A^nBA^{m-n}.$$

At this point I thought this could be easily proven by induction, but now I'm not sure it is that simple. The equation is true for $m=1$, and if we assume it is true for $m$, then we get $$\underbrace{[A,[A,[...,[A,B]...]}_{m+1}=A\underbrace{[A,[A,[...,[A,B]...]}_{m}-\underbrace{[A,[A,[...,[A,B]...]}_{m}A$$ $$=A\left(\sum_{n=0}^m(-1)^{m-n}\frac{m!}{n!(m-n)!}A^nBA^{m-n}\right)-\left(\sum_{n=0}^m(-1)^{m-n}\frac{m!}{n!(m-n)!}A^nBA^{m-n}\right)A$$ $$=\sum_{n=0}^m(-1)^{m-n}\frac{m!}{n!(m-n)!}(A^{n+1}BA^{m-n}-A^nBA^{m+1-n})$$ $$=\sum_{n=0}^m(-1)^{m+1-n}\frac{m!}{n!(m-n)!}(A^nBA^{m+1-n}-A^{n+1}BA^{m-n}).$$

I would like to see this is equal to $$\sum_{n=0}^{m+1}(-1)^{m+1-n}\frac{(m+1)!}{n!(m+1-n)!}A^nBA^{m+1-n},$$ since that would complete the proof. I've tried to work it by inserting commutators here and there, but the algebra becomes too involved. Any help would be truly appreciated.

• Maybe the last expression is more transparent if read as $$\sum_{n=0}^{m+1}(-1)^{m+1-n}\begin{pmatrix}m+1\\n \end{pmatrix}A^nBA^{m+1-n}$$

Answer 1

Start with $f(\lambda):=e^{\lambda A}Be^{-\lambda A}$ but take derivatives at $\lambda=0$: $$\begin{align} f(0) &= B \\ f'(0) &= \left( e^{\lambda A}ABe^{-\lambda A} + e^{\lambda A}B(-A)e^{-\lambda A} \right)_{\lambda=0} = \left. e^{\lambda A}[A,B]e^{-\lambda A}\right|_{\lambda=0} = [A,B] \\ f''(0) &= \left( e^{\lambda A}A[A,B]e^{-\lambda A} + e^{\lambda A}[A,B](-A)e^{-\lambda A} \right)_{\lambda=0} = \left. e^{\lambda A}[A,[A,B]]e^{-\lambda A}\right|_{\lambda=0} = [A,[A,B]] \\ \vdots\\ f^{(k)}(0) &= \underbrace{[A,[A,\cdots,[A,B]\cdots]]}_{[A,\cdot]\text{ applied $k$ times}} = [A,\cdot]^k B \end{align}$$ The last identity can be proved by induction.

This gives $$ f(\lambda) = \sum_{k=0}^{\infty} \frac{1}{k!}\lambda^k f^{(k)}(0) = \sum_{k=0}^{\infty} \frac{1}{k!}\lambda^k [A,\cdot]^k B = e^{\lambda[A,\cdot]} B . $$