# Ordered exponential

(Redirected from Path-ordered exponential)

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.

## Definition

Let A be an algebra over a real or complex field K, and a(t) be a parameterized element of A,

${\displaystyle a:K\to A.\,}$

The parameter t in a(t) is often referred to as the time parameter in this context.

The ordered exponential of a is denoted

{\displaystyle {\begin{aligned}\operatorname {OE} [a](t)\equiv {\mathcal {T}}\left\{e^{\int _{0}^{t}a(t')\,dt'}\right\}&\equiv \sum _{n=0}^{\infty }{\frac {1}{n!}}\int _{0}^{t}\cdots \int _{0}^{t}{\mathcal {T}}\left\{a(t'_{1})\cdots a(t'_{n})\right\}\,dt'_{1}\cdots dt'_{n}\\&\equiv \sum _{n=0}^{\infty }\int _{0}^{t}\int _{0}^{t'_{n}}\int _{0}^{t'_{n-1}}\cdots \int _{0}^{t'_{2}}a(t'_{n})\cdots a(t'_{1})\,dt'_{1}\cdots dt'_{n-2}dt'_{n-1}dt'_{n}\end{aligned}}}

where the term n = 0 is equal to 1 and where ${\displaystyle {\mathcal {T}}}$  is a higher-order operation that ensures the exponential is time-ordered: any product of a(t) that occurs in the expansion of the exponential must be ordered such that the value of t is increasing from right to left of the product; a schematic example:

${\displaystyle {\mathcal {T}}\left\{a(1.2)a(9.5)a(4.1)\right\}=a(9.5)a(4.1)a(1.2).}$

This restriction is necessary as products in the algebra are not necessarily commutative.

The operation maps a parameterized element onto another parameterized element, or symbolically,

${\displaystyle \operatorname {OE} {\mathrel {:}}\left(K\to A\right)\to \left(K\to A\right).}$

There are various ways to define this integral more rigorously.

### Product of exponentials

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:

${\displaystyle \operatorname {OE} [a](t)=\prod _{0}^{t}e^{a(t')\,dt'}\equiv \lim _{N\rightarrow \infty }\left(e^{a(t_{N})\,\Delta t}e^{a(t_{N-1})\,\Delta t}\cdots e^{a(t_{1})\,\Delta t}e^{a(t_{0})\,\Delta t}\right)}$

where the time moments {t0, …, tN} are defined as tii Δt for i = 0, …, N, and Δtt / N.

The ordered exponential is in fact a geometric integral.[1][2] [3]

### Solution to a differential equation

The ordered exponential is unique solution of the initial value problem:

{\displaystyle {\begin{aligned}{\frac {d}{dt}}\operatorname {OE} [a](t)&=a(t)\operatorname {OE} [a](t),\\[5pt]\operatorname {OE} [a](0)&=1.\end{aligned}}}

### Solution to an integral equation

The ordered exponential is the solution to the integral equation:

${\displaystyle \operatorname {OE} [a](t)=1+\int _{0}^{t}a(t')\operatorname {OE} [a](t')\,dt'.}$

This equation is equivalent to the previous initial value problem.

### Infinite series expansion

The ordered exponential can be defined as an infinite sum,

${\displaystyle \operatorname {OE} [a](t)=1+\int _{0}^{t}a(t_{1})\,dt_{1}+\int _{0}^{t}\int _{0}^{t_{1}}a(t_{1})a(t_{2})\,dt_{2}\,dt_{1}+\cdots .}$

This can be derived by recursively substituting the integral equation into itself.

## Example

Given a manifold ${\displaystyle M}$  where for a ${\displaystyle e\in TM}$  with group transformation ${\displaystyle g:e\mapsto ge}$  it holds at a point ${\displaystyle x\in M}$ :

${\displaystyle de(x)+\operatorname {J} (x)e(x)=0.}$

Here, ${\displaystyle d}$  denotes exterior differentiation and ${\displaystyle \operatorname {J} (x)}$  is the connection operator (1-form field) acting on ${\displaystyle e(x)}$ . When integrating above equation it holds (now, ${\displaystyle \operatorname {J} (x)}$  is the connection operator expressed in a coordinate basis)

${\displaystyle e(y)=\operatorname {P} \exp \left(-\int _{x}^{y}J(\gamma (t))\gamma '(t)\,dt\right)e(x)}$

with the path-ordering operator ${\displaystyle \operatorname {P} }$  that orders factors in order of the path ${\displaystyle \gamma (t)\in M}$ . For the special case that ${\displaystyle \operatorname {J} (x)}$  is an antisymmetric operator and ${\displaystyle \gamma }$  is an infinitesimal rectangle with edge lengths ${\displaystyle |u|,|v|}$  and corners at points ${\displaystyle x,x+u,x+u+v,x+v,}$  above expression simplifies as follows :

{\displaystyle {\begin{aligned}&\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{aligned}}}

Hence, it holds the group transformation identity ${\displaystyle \operatorname {OE} [-\operatorname {J} ]\mapsto g\operatorname {OE} [\operatorname {J} ]g^{-1}}$ . If ${\displaystyle -\operatorname {J} (x)}$  is a smooth connection, expanding above quantity to second order in infinitesimal quantities ${\displaystyle |u|,|v|}$  one obtains for the ordered exponential the identity with a correction term that is proportional to the curvature tensor.