# Non-commuting matrix exponentials

Reading this book, I came across the following formula

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

where $$A$$ and $$B$$ are two matrices and $$[A,B] := AB-BA$$.

I tried to find a proof, without success. It is impossible to use the binomial theorem since $$A$$ and $$B$$ do not commute. I've thought about developping the product in a power series, but I'm not sure the Cauchy product is allowed when $$[A,B] \ne 0$$.

I thought that $$[A,B^k]$$ could be useful so I searched the general expression

$$[A,B^n] = \sum_{i=0}^{n-1} B^i [A,B] B^{n-i-1}$$

Does anyone know a proof of the formula above? If so, is it possible to provide some hints? Or a full proof?

The author probably means an approximation rather than a true equality, as $(A+B)+\frac12[A,B]$ are only the first two terms in the BCH formula Zassenhaus formula.

• How can one be sure that the terms after can be neglected ? – ChocoPouce Dec 2 '13 at 17:24
• @ChocoPouce The trailing terms are not always negligible. You may search the internet for convergence results. For example, see sec. 3 and fig. 1 of Casas et al., Efficient computation of the Zassenhaus formula. – user1551 Dec 2 '13 at 18:50

The equality is true when $A,B$ are quasi-commutative, that is, when $[A,B]$ commutes with both $A$ and $B$.

The book you referenced was "Quantum Physics in One Dimension". The operators in quantum mechanics have to obey particular commutation relations. For instance the operators $x$ and $p$ obey the canonical commutation relation,

$$\left[ x, p \right] = i I.$$

This allows us to simplify the Zassenhaus formula. The operator $[x,p]$ commutes with every operator because it is proportional to the identity; if you consider the higher order terms in the Zassenhaus formula you will see they involve commutators which must be zero.

Meaning that,

$$\exp(x+p)= \exp(x)\exp(p)\exp(-[x,p]/2),$$

is an exact relationship for the operators $x$ and $p$.

Knowledge of specific commutation relations can allow you to simplify the Zassenhaus formula significantly.