Explicit Description of the Zassenhaus Formula

TL;DR

This paper provides an explicit infinite series expansion of the Zassenhaus formula for the exponential of a sum of operators, offering a new perspective on the Baker-Campbell-Hausdorff formula.

Contribution

It introduces a novel explicit description of the Zassenhaus formula and offers an alternative expression for the Baker-Campbell-Hausdorff formula.

Findings

Explicit infinite sum expansion of e^{A+B}

New expression for the Baker-Campbell-Hausdorff formula

Enhanced understanding of operator exponential decompositions

Abstract

We explicitly describe an expansion of $e^{A+B}$ as an infinite sum of the products of $B$ multiplied by the exponential function of $A$. This is the explicit description of the Zassenhaus formula. We also express the Baker-Campbell-Hausdorff formula in a different manner.

Equations

$$\displaystyle{\rm e}^{t(A+B)}\$$

$$\displaystyle Z_{n}\$$

$$\displaystyle{\rm e}^{\beta(A+B)}\$$

$$\displaystyle\mathscr{Z}_{n}\$$

$$\displaystyle(A+B)^{n}\$$

$$\displaystyle{\rm e}^{A+B}\$$

$$\displaystyle X_{m+1}\$$

$$\displaystyle X_{m}\$$

$$\displaystyle X_{m+1,1}\$$

$$\displaystyle X_{m+1,m+1}\$$

$$\displaystyle X_{m+1,p}\$$

$$\displaystyle X_{m,1}\$$

$$\displaystyle X_{m,m}\$$

$$\displaystyle X_{m,p}\$$

$$\times {\mathcal{B}}_{m-(k_1+\dots +k_{p-1})}{\mathcal{B}}_{k_{p-1}}\cdots {\mathcal{B}}_{k_2}{\mathcal{B}}_{k_1}.$$

$$\displaystyle{\rm e}^{A+B}\$$

$$\displaystyle=\ \Big{\{}1+\sum_{m=1}^{\infty}\mathscr{B}_{m}+\sum_{m=1}^{\infty}\sum_{k_{1}=1}^{m-1}\frac{k_{1}}{m}\,\mathscr{B}_{m-k_{1}}\mathscr{B}_{k_{1}}+\sum_{m=1}^{\infty}\sum_{k_{1}=1}^{m-2}\sum_{k_{2}=1}^{m-k_{1}-1}\frac{k_{1}k_{2}}{m(m-k_{1})}\,\mathscr{B}_{m-k_{1}-k_{2}}\mathscr{B}_{k_{2}}\mathscr{B}_{k_{1}}$$

$$+\sum _{m=1}^{\infty }\sum _{k_1=1}^{m-3}\sum _{k_2=1}^{m-k_1-2}\sum _{k_3=1}^{m-k_1-k_2-1}\frac{k_1{k}_2{k}_3}{m(m-k_1)(m-k_1-k_2)}{\mathcal{B}}_{m-k_1-k_2-k_3}{\mathcal{B}}_{k_3}{\mathcal{B}}_{k_2}{\mathcal{B}}_{k_1}$$

$$\displaystyle\ \ \ \ \ \ \ \ \ \ +\ldots\Big{\}}\,{\rm e}^{A}\,.$$

$$\displaystyle{\rm e}^{A+B}\$$

$$\displaystyle{\rm e}^{Z}\ =\ {\rm e}^{X}{\rm e}^{Y}\$$

$$\displaystyle{\rm e}^{X}{\rm e}^{Y}\$$

$$\displaystyle\mathscr{X}_{n}\$$

$$\displaystyle{\rm e}^{X}{\rm e}^{Y}\$$

$$\displaystyle\mathscr{Y}_{n}\$$

$$\displaystyle\frac{1}{2}\big{\{}$$

$$\displaystyle=\ \frac{1}{2}\Big{\{}1+\Big{(}\mathscr{X}_{1}-\mathscr{X}_{2}+\mathscr{X}_{3}+\ldots\Big{)}+\Big{(}\frac{1}{2}\mathscr{X}_{1}\mathscr{X}_{1}-\frac{1}{3}\mathscr{X}_{2}\mathscr{X}_{1}-\frac{2}{3}\mathscr{X}_{1}\mathscr{X}_{2}+\ldots\Big{)}+\Big{(}\frac{1}{6}\mathscr{X}_{1}\mathscr{X}_{1}\mathscr{X}_{1}+\ldots\Big{)}\Big{\}}$$

$$\displaystyle\ \ \ \ +\frac{1}{2}\Big{\{}1+\Big{(}\mathscr{Y}_{1}+\mathscr{Y}_{2}+\mathscr{Y}_{3}+\ldots\Big{)}+\Big{(}\frac{1}{2}\mathscr{Y}_{1}\mathscr{Y}_{1}+\frac{2}{3}\mathscr{Y}_{2}\mathscr{Y}_{1}+\frac{1}{3}\mathscr{Y}_{1}\mathscr{Y}_{2}+\ldots\Big{)}+\Big{(}\frac{1}{6}\mathscr{Y}_{1}\mathscr{Y}_{1}\mathscr{Y}_{1}+\ldots\Big{)}\Big{\}}$$

$$\displaystyle=\ 1+(X+Y)-\frac{1}{4}\Big{(}[Y,X]-[X,Y]\Big{)}+\frac{1}{12}\Big{(}[Y,[Y,X]]+[X,[X,Y]]\Big{)}$$

$$\displaystyle\ \ \ \ +\frac{1}{2}(X+Y)^{2}-\frac{1}{12}\Big{(}[Y,X](X+Y)-(X+Y)[X,Y]\Big{)}-\frac{1}{6}\Big{(}(X+Y)[Y,X]-[X,Y](X+Y)\Big{)}$$

$$+\frac{1}{6}(X+Y{)}^3+\dots$$

$$\displaystyle=\ 1+(X+Y)+\Big{\{}\frac{1}{2}[X,Y]+\frac{1}{2}(X+Y)^{2}\Big{\}}$$

$$\displaystyle\ \ \ \ +\Big{\{}\frac{1}{12}\Big{(}[Y,[Y,X]]+[X,[X,Y]]\Big{)}+\frac{1}{4}\Big{(}[X,Y](X+Y)+(X+Y)[X,Y]\Big{)}+\frac{1}{3!}(X+Y)^{3}\Big{\}}+\ldots\,.$$

$$\displaystyle X_{2}\$$

$$\displaystyle X_{3,1}\$$

$$\displaystyle X_{4,1}\$$

$$\displaystyle X_{4,2}\$$

$$\displaystyle X_{4,3}\$$

$$\displaystyle X_{4,4}\$$

Full text

Introduction

In various topics in physics and mathematics, we often have to expand the exponential function of two operators $A$ and $B$ such as ${\rm e}^{A+B}$ in a certain situation (for instance, [1] and [2]). An expansion is described as the Zassenhaus formula ([3] and references therein):

[TABLE]

Its transposed version is also given as

[TABLE]

Unfortunately, however, the above two expressions are rather complicated because we sequentially obtain the explicit expression of higher order terms in the operators $A$ and $B$. In this paper, we will obtain a new description of the Zassenhaus formula in which all of the higher order terms are explicitly expressed.

Derivation

First of all, we expand $(A+B)^{n}$ and move all the operator $A$ to the right in each term, and define the following expression:

[TABLE]

where $X_{m}$ are polynomials involving $B^{l}$, commutators $[A,[A,\cdots[A,B]]]$, and their multiplications. By using $X_{m}$, we obtain the exponential function of $A+B$ as the following form:

[TABLE]

There exists a recursion relation among $X_{m}$ in such a way that

[TABLE]

where $\mathscr{L}_{A}{\cal O}$ is the commutator between $A$ and a certain operator ${\cal O}$ such as $\mathscr{L}_{\mathscr{A}}{\cal O}=[A,{\cal O}]$. It is easy to derive (5) when we compute $(A+B)^{n+1}$ as the product of $(A+B)$ and $(A+B)^{n}$ in terms of (3). Let us evaluate the relation (5). It is convenient to express $X_{m}$ as the sum of new polynomials $X_{m,p}$:

[TABLE]

where $p$ denotes the power of $B$ in $X_{m,p}$, whose examples can be seen in the appendix. Substituting the expression (6) into the relation (5), we find three recursion relations

[TABLE]

We immediately obtain the solutions of these relations (the proof is exhibited in the appendix):

[TABLE]

Here we introduced the terminology $\mathscr{B}^{\prime}_{m}$ defined the above. The solution (8c) can be described in an explicit way if we iteratively use (8c) until we reach $X_{m-(k_{1}+\ldots+k_{p-1}),1}=\mathscr{B}^{\prime}_{m-(k_{1}+\ldots+k_{p-1})}$ given by (8a). Hence we obtain

[TABLE]

For simplicity, we further introduced the description $\mathscr{B}_{m}\equiv\frac{1}{m!}\mathscr{B}^{\prime}_{m}$. Applying (9) to (3) and (6), we obtain the explicit expansion of ${\rm e}^{A+B}$ in terms of the products of $\mathscr{B}_{m}$:

[TABLE]

Relabeling $k_{i}$ and $m-(k_{1}+\ldots+k_{p-1})$ to $n_{i}$ and $n_{p}$ respectively, we obtain the final form

[TABLE]

We have a comment that each $k_{i}$, as well as the new label $n_{i}$, is unbounded from above because $m$ goes to infinity. It turns out that (11) is the the explicit description of the Zassenhaus formula (2) without using the functions $\mathscr{Z}_{n}$. We understand that the exponential form ${\rm e}^{B}$ in the right-hand side of (2) can be obtained from (11) when we extract the terms of the products only of $\mathscr{B}_{1}=B$. However, it is hard to extract ${\rm e}^{\mathscr{Z}_{m}}$ of arbitrary $m$ from (11). Because of the iterative definition of $\mathscr{Z}_{m}$ in (2), we have to obtain the explicit expression of all $\mathscr{Z}_{l}$ where $l\leq m-1$ beforehand to determine $\mathscr{Z}_{m}$.

Furthermore, if we transpose (11) and rename $A^{{\rm T}}$ and $B^{{\rm T}}$ to $A$ and $B$, we obtain

[TABLE]

This is the explicit description of (1) without using the functions $Z_{n}$. We should notice that the ordering of the operators $\mathscr{B}_{n_{i}}$ is different from that of (11).

The descriptions we obtained are quite useful if the product of the operator $\mathscr{B}_{n_{i}}$ is truncated at a certain level such as $\mathscr{B}_{n_{k}}\mathscr{B}_{n_{k-1}}\cdots\mathscr{B}_{n_{1}}=0$, which originates from the nilpotency of the operator $B$ of degree $k$, i.e., $B^{k}=0$.

Baker-Campbell-Hausdorff formula

We can also discuss the Baker-Campbell-Hausdorff (BCH, for short) formula

[TABLE]

by using the descriptions (11) and (12), though the general form (13) has already been well-known (see, for instance, [4]). We would like to use the operator ${\rm e}^{Z}$ rather than $Z$, because we often encounter the exponential form such as ${\rm e}^{X}{\rm e}^{Y}$ in quantum mechanics. Multiplying (11) by ${\rm e}^{-A}$ from the right and replacing $A+B$ and $-A$ with $X$ and $Y$ respectively, we obtain

[TABLE]

On the other hand, multiplying (12) by ${\rm e}^{-A}$ from the left and replacing $-A$ and $A+B$ with $X$ and $Y$ respectively, we find

[TABLE]

The original BCH formula $Z=\log({\rm e}^{X}{\rm e}^{Y})$ consists only of $X$, $Y$ and their commutators. On the other hand, we immediately find that powers of $X$ and $Y$ directly contribute to the expansion in both (14) and (15). These two seem to be different feature. However, the powers of $X$ and $Y$ in (14) and (15) originate from the corresponding power of $X+Y$ in the Taylor expansion of (13).

We recognize that the operators $X$ and $Y$ in (14) and (15) do not appear on equal footing with each other. In order to describe an expression on equal footing, we simply sum up (14) and (15), and divide it by two. For instance, we evaluate this up to cubic powers of the operators $X$ and $Y$ in such a way that

[TABLE]

This coincides with the Taylor expansion of ${\rm e}^{X}{\rm e}^{Y}$ in the form of (13). Indeed, we can find the coincidence in any powers of the operators. Once we establish the above new descriptions, it would be interesting to apply it to a generalization of the BCH formula such as ${\rm e}^{X}{\rm e}^{Y}{\rm e}^{Z}$ developed in [5].

Acknowledgments

I would like to thank Reona Arai, Tetsutaro Higaki, Hideaki Iida, Hiroyasu Miyazaki, Toshifumi Noumi, Noriaki Ogawa, Masato Taki, Akinori Tanaka and Masahide Yamaguchi for helpful discussions. I would also like to thank Katsushi Ito, Marco Matone and Hector Moya-Cessa for valuable correspondence. I am supported by the Iwanami-Fujukai Foundation. I am also supported in part by the MEXT-Supported Program for the Strategic Research Foundation at Private Universities “Topological Science” (Grant No. S1511006) and by the MEXT Grant-in-Aid for Scientific Research on Innovative Areas “Nuclear Matter in Neutron Stars Investigated by Experiments and Astronomical Observations” (No. 15H00841 by Muneto Nitta).

Appendix

Here we explicitly exhibit a series of $X_{m,p}$ defined in (6). When we consider $(A+B)^{2}$ as the form (3), we obtain $X_{2}$ and $X_{2,p}$ as follows:

[TABLE]

In the case of $(A+B)^{3}$, the components $X_{3,p}$ are

[TABLE]

In the same way, the explicit forms of $X_{4,p}$ and $X_{5,p}$ are given as

[TABLE]

Here we prove (8c) as the solution of the recursion relation (7c) by mathematical induction. Let us assume that each $X_{k,l}$ with $1\leq l\leq k\leq m$ satisfies the expression (8c). We compute $\mathscr{L}_{A}X_{m,p}+BX_{m,p-1}$:

[TABLE]

The final form is nothing but $X_{m+1,p}$. Hence we proved that (8c) is the solution of the relation (7c).