Baker-Campbell-Hausdorff/Zassenhaus formula to first order in one matrix

Question

Is there a closed-form expression for the term of $e^{t(c \hat{X} + d \hat{Y})}$ that is first-order in $d$, where $t$, $c$, and $d$ are scalars and $\hat{X}$ and $\hat{Y}$ are finite-dimensional linear operators? I guess that to calculate this, you would use the Zassenhaus formula to expand the exponential, then collect all the terms proportional to $d^1$. Unfortunately, so far as I can find the Zassenhaus formula is usually presented organized in powers of $t$, and each power of $t$ has a term proportional to $d^1$, so you would need to consider every term in the $t$ expansion just to get the first term in the $d$ expansion.

(A general expression for the Zassenhaus formula reorganized in powers of $d$ would be even better, but that's probably too much to ask for!)

Answer 1

I don't know about using the Zassenhaus formula, but I did obtain the following expression for the terms that are first-order in $d$: $$\sum_{n\geq1}\frac{c^{n-1}t^n}{n!}\sum\limits_{r,s\geq0,\ \ r+s=n-1}\hat{X}^r\hat{Y}\hat{X}^s. $$

Basically, expand $e^{t(c \hat{X} + d \hat{Y})}$ as a power series: $$e^{t(c \hat{X} + d \hat{Y})}=\sum_{n=0}^\infty \frac{t^n}{n!}(c \hat{X} + d \hat{Y})^n. $$ Then, for each term $(c \hat{X} + d \hat{Y})^n$, we get a single $d$ whenever we obtain one copy of $d\hat{Y}$ and $n-1$ copies of $c\hat{X}$. For example, when $n=2$ we get $$cd(\hat{X}\hat{Y}+\hat{Y}\hat{X}),$$ and for $n=3$ we get $$ c^2d(\hat{X}\hat{X}\hat{Y}+\hat{X}\hat{Y}\hat{X}+\hat{Y}\hat{X}\hat{X}). $$ Summing over all $n$, we obtain the result.

Answer 2

1. The Baker-Campbell-Hausdorff (BCH) formulas to linear order in one of the operators $X$ and $Y$ are given by $$ Z ~=~ \ln (e^Xe^Y) ~=~\left\{\begin{array}{l} X+B(-{\rm ad}_X)Y +{\cal O}(Y^2), \cr\cr Y+B({\rm ad}_Y)X +{\cal O}(X^2), \end{array} \right. \tag{1}$$ where $$ B(x)~:=~\frac{x}{e^x-1}~=~1 -\frac{x}{2} +\frac{x^2}{12}-\frac{x^4}{720} +{\cal O}(x^6)\tag{2}$$ is the generating function of Bernoulli numbers.

2. The first order Zassenhaus formulas are given by $$ e^{-X}e^{X+Y} ~=~ \exp\left\{ E(-{\rm ad}_X)Y +{\cal O}(Y^2)\right\} ~=~ 1+ E(-{\rm ad}_X)Y +{\cal O}(Y^2)$$ $$~=~ \prod_{n=0}^{\infty}\exp\left\{ \frac{(-{\rm ad}_X)^nY}{(n+1)!}\right\} +{\cal O}(Y^2) , \tag{3}$$
   $$ e^{X+Y}e^{-Y}~=~\exp\left\{ E({\rm ad}_Y)X+{\cal O}(X^2)\right\} ~=~1+ E({\rm ad}_Y)X+{\cal O}(X^2)$$ $$~=~ \prod_{n=0}^{\infty}\exp\left\{ \frac{({\rm ad}_Y)^nX}{(n+1)!}\right\} +{\cal O}(X^2) , \tag{4}$$ where we have defined the reciprocal function $$ E(x)~=~\frac{1}{B(x)}~=~\frac{e^x-1}{x}~=~\sum_{n=0}^{\infty}\frac{x^n}{(n+1)!} .\tag{5}$$ Note that the operator ordering of the infinite product $\prod_{n=0}^{\infty}$ on the rhs. of eqs. (3) and (4) only matters beyond linear order.

Answer 3

Let $f:d\rightarrow e^{tcX+tdY}$. Then $f(d)=f(0)+Df_0(d)+O(d^2)$ when $t,c,X,Y$ are fixed. We can write the result in two different forms:

1. Let $ad_U:Z\rightarrow UZ-ZU$. $f(d)=e^{tcX}+e^{tcX}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(k+1)!}(ad_{tcX})^k(tYd)+O(d^2)$

or $f(d)=e^{tcX}(I+d\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(k+1)!}t^{k+1}c^k(ad_{X})^k(Y))+O(d^2)$.

2. $f(d)=e^{tcX}+dt\int_0^1 e^{\alpha tcX}Ye^{(1-\alpha)tcX}d\alpha+O(d^2)$.