THE BAKER-CAMPBELL-HAUSDORFF FORMULA

TERENCE TAO

Abstract. The derivation of the Baker-Campbell-Hausdorff formula.

1. Introduction

Let us work purely formally; we discuss at the end of the note how to make the following computations rigorous. Let A, B, C, etc. lie in a formal algebra (e.g. n×n matrices). We let ad_{A} denote the adjoint operator ad_{A}X := [A, X] := AX − XA, and A ↦ e^{A} denote the exponential operator, thus in particular e^{0} = 1 and \frac{d}{dt}e^{tA} = Ae^{tA} = e^{tA}A.

Lemma 1.1. We have e^{A}Xe^{−A} = e^{ad^{A}}X.

Proof Let f(t) := e^{tA}Xe^{−tA}. Then f(0) = X and

\frac{d}{dt}f(t) = Ae^{tA}Xe^{−tA} − e^{tA}Xe^{−tA}A
= ad_{A}f(t).

The claim follows.

Corollary 1.2. If e^{A}e^{B} = e^{C}, then e^{ad^{A}}e^{ad^{B}} = e^{ad(C)}.

Lemma 1.3 (Duhamel formula). Let C(t) depend on a parameter t. Then

\frac{d}{dt}e^{C(t)} = ∫_{0}^{1} e^{sC(t)}(\frac{d}{dt}C(t)) exp((1 − s)C(t)) ds

= ([∫_{0}^{1} e^{sad_{C(t)}} ds]\frac{d}{dt}C(t))e^{C(t)}

= [\frac{e^{ad^{C(t)}} − 1}{ad_{C(t)}})\frac{d}{dt}C(t)]e^{C(t)}.

Proof Let f(s) := e^{−sC(t)} \frac{d}{dt}e^{sC(t)}. Then f(0) = 0 and

\frac{d}{ds}f(s) = −e^{−sC(t)}C(t)\frac{d}{dt}e^{sC(t)} + e^{−sC(t)} \frac{d}{dt}[C(t)e^{sC(t)}]

= e^{−sC(t)}[\frac{d}{dt}C(t)]e^{sC(t)}.

Applying the fundamental theorem of calculus and changing variables, we obtain the first identity. The second identity then follows from Lemma 1.1, and the third from formal integration.

1991 Mathematics Subject Classification. 42B15.

We can rearrange the above formula as

\frac{d}{dt}C(t) = \frac{ad_{C(t)}}{e^{ad^{C(t)}} − 1}[(\frac{d}{dt}e^{C(t)})e^{−C(t)}].

We apply this in particular to

C(t) := log(e^{tA}e^{B})

and conclude

\frac{d}{dt} log(e^{tA}e^{B}) = \frac{ad_{C(t)}}{e^{ad^{C(t)}} − 1}A.

From Corollary 1.2 we have

\frac{ad_{C(t)}}{e^{ad^{C(t)}} − 1} = f(e^{ad^{C(t)}}) = f(e^{tad_{A}}e^{ad^{B}})

where

f(x) := \frac{log x}{x − 1} = 1 − \frac{x − 1}{2} + \frac{(x − 1)^{2}}{3} − ….

From the fundamental theorem of calculus we thus have the Baker-Campbell-Hausdorff formula

log(e^{A}e^{B}) = B + ∫_{0}^{1} f(e^{tad_{A}}e^{ad^{B}})A dt.

One could expand this formula out as a power series if desired:

log(e^{A}e^{B}) = B + ∑_{n=0}^{∞} \frac{(−1)^{n}}{n} ∑_{r_{i}+s_{i}>0^{r_{1},s_{1},... ,rn,sn≥0}} \frac{(ad_{A})^{r_{1}}(ad_{B})^{s_{1}} … (ad_{A})^{r_{n}}(ad_{B})^{s_{n}}}{r_{1}!s_{1}! … r_{n}!s_{n}!(r_{1} + … + r_{n} + 1)} A.

This expresses log(e^{A}e^{B}) as a formal series in commutators of A and B. The above formulae can be easily justified in the case that A and B take values in a nilpotent Lie algebra (and e^{A} and e^{B} take values in the corresponding Lie group), as all series terminate in that case. Alternatively, one can justify these formulae for matrix-valued (or even operator-valued) A and B provided that the operator norms of A and B are sufficiently small.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

   E-mail address: tao@math.ucla.edu