THE BAKER-CAMPBELL-HAUSDORFF FORMULA

If $A \in {\cal L(X)}$ is a bounded linear operator on a Banach space X, the operator e^A can be defined by the usual power series. If also $B \in {\cal L(X)}$, and B commutes with A, it is easy to see from the power series that e^A e^B = e^A+B. The BCH formula tells how to compute e^A e^B in general. We begin the proof with a simple lemma on differential equations.

Lemma 1 Let Y be a Banach space, and suppose $A \in {\cal L(Y)}$, $w : (-a,a) \to Y$ is a continuous function, and $u_0 \in Y$. Then the initial-value problem

$$\frac{du}{dt} = Au + w(t) , \qquad u(0) = u_0$$

for a function $u : (-a,a) \to Y$ has a unique solution, given by

$$u(t) = e^{tA}u_0 + \int_0^t e^{(t-s)A} w(s)\,ds .$$

In particular, if w is constant,

u(t) = e^tAu₀ + f(t,A) w,

where

$$f(t,z) = \frac{e^{tz} - 1}{z} .$$

The proof is identical to that of the special case $X = {{\rm I}\kern-.18em{\rm R}}$; note however that f(t,z) is an entire function (for every t), so that f(t,A) is defined.

We now take $Y = {\cal L(X)}$. Each $A \in {\cal L(X)}$ defines a linear operator $\mbox{ad$\,A$} \in {\cal L(Y)}$ by the formula

$$\mbox{ad$\,A$} (B) = [A,B] := AB - BA \qquad \mbox{for all $B \in {\cal L(X)}$}.$$

(It is easy to see that $\mbox{ad$\,A$}$ is bounded; in fact $\Vert \mbox{ad$\,A$}\Vert \leq 2 \Vert A\Vert$.) The next lemma suggests that $\mbox{ad$\,A$}$ might be a natural object to study.

Lemma 2 Let $A, B \in {\cal L(X)}$; then

$$e^A B e^{-A} = e^{\mbox{ad$\,A$}} B .$$

Proof For $t \in {{\rm I}\kern-.18em{\rm R}}$, define $\rho_t : {\cal L(X)} \to {\cal L(X)}$ by

$$\rho_t(B) = e^{tA} B e^{-tA}.$$

Then

$$\frac{d}{dt} (\rho_t(B)) = [ A, \rho_t(B) ] = \mbox{ad$\,A$} (\rho_t(B) .$$

Noting that $\rho_0 (B) = B$, we see from Lemma 1 that

$$\rho_t (B) = e^{t \mbox{ad$\,A$}} (B) ;$$

putting t = 1 gives the result.

Lemma 3 Let $A : (-a,a) \to {\cal L(X)}$ be continuously differentiable. Then

$$e^{A(t)} \frac{d}{dt} e^{-A(t)} = -g(\mbox{ad$\,A(t)$}) \frac{dA}{dt},$$

where g(z) = f(1,z).

Proof For $s \in {{\rm I}\kern-.18em{\rm R}}$, $t \in (-a,a)$, define

$$B(s,t) = e^{sA(t)} \frac{\partial}{\partial t} \left( e^{-sA(t)} \right) .$$

Then

while B(0,t) = I. Applying Lemma 1, we find

$$B(s,t) = - f(s, \mbox{ad$\,A(t)$}) \frac{dA}{dt} ;$$

putting s = 1 gives the result.

Theorem 1 (Baker-Campbell-Hausdorff) Let A and B belong to ${\cal L(X)}$, with $\Vert A\Vert$, $\Vert B\Vert$ sufficiently small. Then the operater $\ln (e^A e^B) \in {\cal L(X)}$ is uniquely defined, and

$$\ln (e^A e^B) = B \int_0^1 h \left( e^{t \mbox{ad$\,A$}} e^{\mbox{ad$\,B$}} \right) A\,dt ,$$

where, for $\vert z-1\vert < 1$,

$$h(z) = \frac{\ln z}{z-1} = 1 - \frac{z-1}{2} + \frac{(z-1)^2}{3} - \cdots .$$

Proof The series for h shows that the integral is defined provided $\Vert I - e^{t \mbox{ad$\,A$}} e^{\mbox{ad$\,B$}}\Vert < 1$, for all $t \in [0,1]$. Now, for $t \in [0,1]$ define

$$C(t) = \ln (e^{tA} e^B) .$$

(The standard power series for e^z and $\ln (1+z)$ show that C(t) is well-defined on [0,1] if $\Vert A\Vert$ and $\Vert B\Vert$ are sufficiently small.) Then

$$e^{C(t)} \frac{d}{dt} e^{-C(t)} = e^{tA} e^B \frac{d}{dt} \left( e^{-B}e^{-tA} \right) = - A ,$$

and we deduce from Lemma 3 that  

$$A = g( \mbox{ad$\,C(t)$} ) \frac{dC}{dt} .$$ (1)

Lemma 2 implies that $e^{\mbox{ad$\,C(t)$}} = e^{t\mbox{ad$\,A$}} e^{\mbox{ad$\,B$}}$, so that

$$\mbox{ad$\,C(t)$} = \ln \left( e^{t \mbox{ad$\,A$}} e^{\mbox{ad$\,B$}} \right) .$$

Now $g(\ln z) h(z) = 1$ if $\vert z-1\vert < 1$, so

$$g(\mbox{ad$\,C(t)$}) = g \left( \ln \left( e^{t \mbox{ad$\,A... ...\left( e^{t \mbox{ad$\,A$}} e^{\mbox{ad$\,B$}} \right) ^{-1},$$

and (1) gives

$$\frac{dC}{dt} = h \left( e^{t \mbox{ad$\,A$}} e^{\mbox{ad$\,B$}} \right) A;$$

the BCH formula follows by integration.

The integrand in the BCH formula can be expanded as a power series in $\mbox{ad$\,A$}$ and $\mbox{ad$\,B$}$ applied to A, and integrated term-by-term. (There are fewer terms than one might expect at first, because $\mbox{ad$\,A$} (A) = 0$.) Applying the exponential function to both sides of the formula, the result can be written

$$e^A e^B = \exp ( A + B + {1\over 2}[A,B] + \frac{1}{12}([A,[A,B]] - [B,[A,B]]) + \cdots ) .$$

When A and B commute, we recover the result that e^A e^B = e^A+B. A less trivial case, which is at least of some formal relevance to quantum mechanics, is when the commutator [A,B] commutes with both A and B; in that case we find $e^A e^B = e^{A + B + {1\over 2}[A,B]}$. (In quantum mechanics, we are interested in operators such that [A,B] is a multiple of the identity. Unfortunately, it can be easily shown that such a pair of operators cannot both be bounded, so the proof of BCH breaks down.)