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What is time-ordered

  1. Jul 24, 2014 #1
    Definition/Summary

    Time-order means in order of time. A sequence or product is in time-order if "earlier" items are placed to the right of "later" ones.

    For example, if [itex]t_1\ t_2\ \cdots t_n\text{ are times}[/itex][itex]\text{, and if }t_1>t_2>\cdots t_n[/itex], they are in time-order. And so is the product [itex]V(t_1)\ V(t_2)\ \cdots V(t_n)[/itex], where [itex]V[/itex] is an operator depending on time.

    And if [itex](x_1,y_1,z_1,t_1)\ (x_2,y_2,z_2,t_2)\ \cdots[/itex][itex](x_n,y_n,z_n,t_n)\text{ are position-time 4-vectors}[/itex][itex]\text{, and if }t_1>t_2>\cdots t_n[/itex], they are in time-order. And so is the product [itex]\mathcal{H}(x_1,y_1,z_1,t_1)\ \mathcal{H}(x_2,y_2,z_2,t_2)\ \cdots \mathcal{H}(x_n,y_n,z_n,t_n)[/itex], where [itex]\mathcal{H}[/itex] is an operator depending on position and time.

    Time-ordered integrals, and time-ordered products, are used in perturbation theory in quantum field theory: a time-ordered integral is either the integral of an ordinary product with time-ordered limits, or the integral of a time-ordered product with ordinary limits (and one can be converted to the other by using the time-ordering symbol [itex]T[/itex]).

    Equations

    EXAMPLE OF INTEGRAL OF PRODUCT WITH TIME-ORDERED LIMITS:

    [tex]S\ =\ \sum_{N\ =\ 1}^{\infty} (-i)^N\int_{-\infty}^{\infty}\int_{-\infty}^{t_1}\int_{-\infty}^{t_2}\cdots\int_{-\infty}^{t_{N-1}}V(t_1)\cdots V(t_N)\,dt_1\cdots dt_N[/tex]

    THE SAME INTEGRAL, WRITTEN AS AN INTEGRAL OF TIME-ORDERED PRODUCT WITH ORDINARY LIMITS:

    [tex]S\ =\ \sum_{N\ =\ 1}^{\infty}\frac{(-i)^N}{N!}\ \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}T\{V(t_1)\cdots V(t_N)\}\,dt_1\cdots dt_N[/tex]

    Extended explanation

    Time-ordered product:

    The time-ordered product of any items is the ordinary product of the same items, but with the items first rearranged into time-order.

    If the items depend on a 4-vector variable, [itex](x,y,z,t)[/itex], then the rearrangement is in order of the time-components, [itex]t[/itex], only.

    The [itex]T[/itex] symbol:

    The symbol [itex]T[/itex] placed before a product indicates that the items in the product are to be re-arranged into time-order before multiplying them:

    For example:

    [itex]T\,\{\mathcal{H}(\boldsymbol{a},3)\mathcal{H}( \boldsymbol{b},5.5)\mathcal{H}(\boldsymbol{c},7)\}\ =[/itex][itex]\ \mathcal{H}(\boldsymbol{a},3)\mathcal{H}( \boldsymbol{b},5.5)\mathcal{H}(\boldsymbol{c},7)[/itex]

    [itex]T\,\{\mathcal{H}(\boldsymbol{a},3)\mathcal{H}( \boldsymbol{b},7)\mathcal{H}(\boldsymbol{c},5.5)\}\ =[/itex][itex]\ \mathcal{H}(\boldsymbol{a},3)\mathcal{H}( \boldsymbol{c},5.5)\mathcal{H}(\boldsymbol{b},7)[/itex]

    [itex]T\,\{\mathcal{H}(\boldsymbol{a},7)\mathcal{H}( \boldsymbol{b},5.5)\mathcal{H}(\boldsymbol{c},3)\}\ =[/itex][itex]\ \mathcal{H}(\boldsymbol{c},3)\mathcal{H}( \boldsymbol{b},5.5)\mathcal{H}(\boldsymbol{a},7)[/itex]

    etc :wink:

    Perturbation theory:

    The S-matrix (in quantum field theory) is the limit as [itex]\tau_0\rightarrow -\infty\text{ and }\tau\rightarrow \infty[/itex] of an operator [itex]U(\tau,\tau_0)[/itex] satisfying:

    [tex]U(\tau,\tau_0)\ =\ 1 - i\int_{\tau_0}^{\tau}\,V(t)\,U(t,\tau_0)\,dt[/tex]

    and by repeated integration we obtain the Dyson series:

    [tex]S\ =\ \lim_{\tau_0\rightarrow -\infty,\,\tau\rightarrow \infty}\, U(\tau,\tau_0)\ =\ \sum_{N\ =\ 1}^{\infty} (-i)^N\int_{-\infty}^{\infty}\int_{-\infty}^{t_1}\int_{-\infty}^{t_2}\cdots\int_{-\infty}^{t_{N-1}}V(t_1)\cdots V(t_N)\,dt_1\cdots dt_N[/tex]

    The limits of integration are time-ordered, which is awkward to calculate :yuck:, so we change to the following integral, which has the same value, but has easy limits of integration:

    [tex]S\ =\ \sum_{N\ =\ 1}^{\infty}\frac{(-i)^N}{N!}\ \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}T\{V(t_1)\cdots V(t_N)\}\,dt_1\cdots dt_N[/tex]

    For the advantage of Lorentz covariance, we further change from integrals over the whole of time to integrals over the whole of space-time, and use a (scalar) Hamiltonian density [itex]\mathcal{H}(x)\ =\ \mathcal{H}(\boldsymbol{x},t)\text{ with }V(t)\ =\ \int\int\int\,d^3\boldsymbol{x}\,\mathcal{H}( \boldsymbol{x},t)[/itex], to obtain:

    [tex]S\ =\ \sum_{N\ =\ 1}^{\infty}\frac{(-i)^N}{N!}\ \int\,T\{\mathcal{H}(x_1)\cdots \mathcal{H}(x_N)\}\,d^4x_1\cdots d^4x_N[/tex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
  2. jcsd
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