The ordered exponential, also called the path-ordered exponential, is a mathematical operation defined in non-commutative algebras, equivalent to the exponential of the integral in the commutative algebras. In practice the ordered exponential is used in matrix and operator algebras. It is a kind of product integral, or Volterra integral.
Let be an algebra over a field, and be an element of parameterized by the real numbers,
a:\R\toA.
The parameter in is often referred to as the time parameter in this context.
The ordered exponential of is denoted
\begin{align} \operatorname{OE}[a](t)\equiv
| ||||||||||
l{T}\left\{e |
\right\}&\equiv
infty | |
\sum | |
n=0 |
1 | |
n! |
t | |
\int | |
0 |
dt'1 …
t | |
\int | |
0 |
dt'n l{T}\left\{a(t'1) … a(t'n)\right\}\\ &=
infty | |
\sum | |
n=0 |
t | |
\int | |
0 |
dt'1
t'1 | |
\int | |
0 |
dt'2
t'2 | |
\int | |
0 |
dt'3 …
t'n-1 | |
\int | |
0 |
dt'n a(t'n) … a(t'1) \end{align}
where the term is equal to 1 and where
l{T}
l{T}\left\{a(1.2)a(9.5)a(4.1)\right\}=a(9.5)a(4.1)a(1.2).
The operation maps a parameterized element onto another parameterized element, or symbolically,
\operatorname{OE}l{:}(\R\toA)\to(\R\toA).
There are various ways to define this integral more rigorously.
The ordered exponential can be defined as the left product integral of the infinitesimal exponentials, or equivalently, as an ordered product of exponentials in the limit as the number of terms grows to infinity:
\operatorname{OE}[a](t)=
t | |
\prod | |
0 |
ea(t')\equiv \limN\left(
a(tN)\Deltat | |
e |
a(tN-1)\Deltat | |
e |
…
a(t1)\Deltat | |
e |
a(t0)\Deltat | |
e |
\right)
where the time moments are defined as for, and .
The ordered exponential is in fact a geometric integral.[1] [2] [3]
The ordered exponential is unique solution of the initial value problem:
\begin{align}
d | |
dt |
\operatorname{OE}[a](t)&=a(t)\operatorname{OE}[a](t),\\[5pt] \operatorname{OE}[a](0)&=1. \end{align}
The ordered exponential is the solution to the integral equation:
\operatorname{OE}[a](t)=1+
t | |
\int | |
0 |
a(t')\operatorname{OE}[a](t')dt'.
This equation is equivalent to the previous initial value problem.
The ordered exponential can be defined as an infinite sum,
\operatorname{OE}[a](t)=1+
t | |
\int | |
0 |
a(t1)dt1+
t | |
\int | |
0 |
dt1
t1 | |
\int | |
0 |
dt2 a(t1)a(t2)+ … .
This can be derived by recursively substituting the integral equation into itself.
Given a manifold
M
e\inTM
g:e\mapstoge
x\inM
de(x)+\operatorname{J}(x)e(x)=0.
Here,
d
\operatorname{J}(x)
e(x)
\operatorname{J}(x)
e(y)=\operatorname{P}\exp\left(-
y | |
\int | |
x |
J(\gamma(t))\gamma'(t)dt\right)e(x)
with the path-ordering operator
\operatorname{P}
\gamma(t)\inM
\operatorname{J}(x)
\gamma
|u|,|v|
x,x+u,x+u+v,x+v,
\begin{align} &\operatorname{OE}[-\operatorname{J}]e(x)\\[5pt] ={}&\exp[-\operatorname{J}(x+v)(-v)]\exp[-\operatorname{J}(x+u+v)(-u)]\exp[-\operatorname{J}(x+u)v]\exp[-\operatorname{J}(x)u]e(x)\\[5pt] ={}&[1-\operatorname{J}(x+v)(-v)][1-\operatorname{J}(x+u+v)(-u)][1-\operatorname{J}(x+u)v][1-\operatorname{J}(x)u]e(x). \end{align}
Hence, it holds the group transformation identity
\operatorname{OE}[-\operatorname{J}]\mapstog\operatorname{OE}[\operatorname{J}]g-1
-\operatorname{J}(x)
|u|,|v|