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Path-ordering
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In theoretical physics, path-ordering is the procedure (or a \mathcal P) that orders a product of operators according to the value of a chosen :

\mathcal P \left\{O_1(\sigma_1) O_2(\sigma_2) \cdots O_N(\sigma_N)\right\}
   \equiv O_{p_1}(\sigma_{p_1}) O_{p_2}(\sigma_{p_2}) \cdots O_{p_N}(\sigma_{p_N}).
     

Here p is a that orders the parameters by value:

p : \{1, 2, \dots, N\} \to \{1, 2, \dots, N\}
\sigma_{p_1} \leq \sigma_{p_2} \leq \cdots \leq \sigma_{p_N}.

For example:

\mathcal P \left\{ O_1(4) O_2(2) O_3(3) O_4(1) \right\} = O_4(1) O_2(2) O_3(3) O_1(4) .

In many fields of physics, the most common type of path-ordering is time-ordering, which is discussed in detail below.


Examples
If an operator is not simply expressed as a product, but as a function of another operator, we must first perform a of this function. This is the case of the , which is defined as a path-ordered exponential to guarantee that the Wilson loop encodes the of the . The parameter σ that determines the ordering is a parameter describing the contour, and because the contour is closed, the Wilson loop must be defined as a trace in order to be .


Time ordering
In quantum field theory it is useful to take the time-ordered product of operators. This operation is denoted by \mathcal T. (Although \mathcal T is often called the "time-ordering operator", strictly speaking it is neither an on states nor a on operators.)

For two operators A( x) and B( y) that depend on spacetime locations x and y we define:

\mathcal T \left\{A(x) B(y)\right\} := \begin{cases} A(x) B(y) & \text{if } \tau_x > \tau_y, \\ \pm B(y)A(x) & \text{if } \tau_x < \tau_y. \end{cases}
Here \tau_x and \tau_y denote the invariant scalar time-coordinates of the points x and y., The Quantum Theory of Fields, Vol. 3, Cambridge University Press, 1995, , p. 143.

Explicitly we have

\mathcal T \left\{A(x) B(y)\right\} := \theta (\tau_x - \tau_y) A(x) B(y) \pm \theta (\tau_y - \tau_x) B(y) A(x),
where \theta denotes the Heaviside step function and the \pm depends on if the operators are or in nature. If bosonic, then the + sign is always chosen, if fermionic then the sign will depend on the number of operator interchanges necessary to achieve the proper time ordering. Note that the statistical factors do not enter here.

Since the operators depend on their location in spacetime (i.e. not just time) this time-ordering operation is only coordinate independent if operators at separated points . This is why it is necessary to use \tau rather than t_0, since t_0 usually indicates the coordinate dependent time-like index of the spacetime point. Note that the time-ordering is usually written with the time argument increasing from right to left.

In general, for the product of n field operators the time-ordered product of operators are defined as follows:

\begin{align} \mathcal T \{ A_1(t_1) A_2(t_2) \cdots A_n(t_n) \} &= \sum_p \theta(t_{p_1} > t_{p_2} > \cdots > t_{p_n}) \varepsilon(p)
   A_{p_1}(t_{p_1}) A_{p_2}(t_{p_2}) \cdots A_{p_n}(t_{p_n}) \\
     
&= \sum_p \left( \prod_{j=1}^{n-1} \theta(t_{p_j} - t_{p_{j+1}}) \right) \varepsilon(p) A_{p_1}(t_{p_1}) A_{p_2}(t_{p_2}) \cdots A_{p_n}(t_{p_n}) \end{align}

where the sum runs all over p's and over the of n degree permutations and

   \varepsilon(p) \equiv \begin{cases}
     1 & \text{for bosonic operators,} \\
     \text{sign of the permutation} & \text{for fermionic operators.}
   \end{cases}
 
     

The in quantum field theory is an example of a time-ordered product. The S-matrix, transforming the state at to a state at , can also be thought of as a kind of "", analogous to the . We obtain a time-ordered expression because of the following reason:

We start with this simple formula for the exponential

\exp h = \lim_{N\to\infty} \left(1 + \frac{h}{N}\right)^N.

Now consider the discretized evolution operator

S = \cdots (1+h_{+3})(1+h_{+2})(1+h_{+1})(1+h_0)(1+h_{-1})(1+h_{-2})\cdots

where 1+h_{j} is the evolution operator over an infinitesimal time interval j\varepsilon,(j+1)\varepsilon. The higher order terms can be neglected in the limit \varepsilon\to 0. The operator h_j is defined by

h_j =\frac{1}{i\hbar} \int_{j\varepsilon}^{(j+1)\varepsilon} \, dt \int d^3 x \, H(\vec x,t).

Note that the evolution operators over the "past" time intervals appears on the right side of the product. We see that the formula is analogous to the identity above satisfied by the exponential, and we may write

S = {\mathcal T} \exp \left(\sum_{j=-\infty}^\infty h_j\right) = \mathcal T \exp \left(\int dt\, d^3 x \, \frac{H(\vec x,t)}{i\hbar}\right).

The only subtlety we had to include was the time-ordering operator \mathcal T because the factors in the product defining S above were time-ordered, too (and operators do not commute in general) and the operator \mathcal T ensures that this ordering will be preserved.


See also

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