Closed forms of the Zassenhaus formula

TL;DR

This paper derives closed-form expressions for the Zassenhaus formula when operators have a specific linear commutator relation, simplifying the disentanglement process in algebraic and physical applications.

Contribution

It provides explicit closed-form formulas for the Zassenhaus expansion under a particular commutator condition, extending the understanding of operator exponential decompositions.

Findings

Closed-form formulas for specific operator commutators.

Equivalent versions of the exponential decomposition.

Explicit functions g_r, g_c, g_ell for operator disentanglement.

Abstract

The Zassenhaus formula finds many applications in theoretical physics or mathematics, from fluid dynamics to differential geometry. The non-commutativity of the elements of the algebra implies that the exponential of a sum of operators cannot be expressed as the product of exponentials of operators. The exponential of the sum can then be decomposed as the product of the exponentials multiplied by a supplementary term which takes generally the form of an infinite product of exponentials. Such a procedure is often referred to as ``disentanglement''. However, for some special commutators, closed forms can be found. In this work, we propose a closed form for the Zassenhaus formula when the commutator of operators $\hat{X}$ and $\hat{Y}$ satisfy the relation $[\hat{X},\hat{Y}]=u\hat{X}+v\hat{Y}+c\mathcal{I}$. Such an expression boils down to three equivalent versions, a left-sided, a centered and a right-sided formula: \begin{equation*} e^{\hat{X}+\hat{Y}}=e^{\hat{X}}e^{\hat{Y}}e^{g_{r}(u,v)[\hat{X},\hat{Y}]}=e^{\hat{X}}e^{g_{c}(u,v)[\hat{X},\hat{Y}]}e^{\hat{Y}}=e^{g_{\ell}(u,v)[\hat{X},\hat{Y}]}e^{\hat{X}}e^{\hat{Y}}, \end{equation*} with respective arguments, \begin{eqnarray*} g_{r}(u,v)&=&g_{c}(v,u)e^{u}=g_{\ell}(v,u)=\frac{u\left(e^{u-v}-e^{u}\right)+v\left(e^{u}-1\right)}{vu(u-v)} \end{eqnarray*} for $u\ne v$ and \begin{eqnarray*} g_{r}(u,u)=\frac{u+1-e^u}{u^2}\;\;\;\;\mathrm{with}\;\;\;\; g_r(0,0)=-1/2. \end{eqnarray*} With additional special case \begin{eqnarray*} g_{r}(0,v)= -\frac{e^{-v}-1+v}{v^{2}}, \quad & g_{r}(u,0)=\frac{e^{u}(1-u)-1}{u^{2}}. \end{eqnarray*}