Timeline for What is the standard notation for a multiplicative integral?
Current License: CC BY-SA 4.0
8 events
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| Aug 30, 2022 at 17:16 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
fixed the dead link
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| Jul 22, 2010 at 2:14 | comment | added | Theo Johnson-Freyd |
@George Lowther: That's correct. Feynman's idea works for any analytic function of a noncommuting (but time-indexed) variable, provided that terms at the same time don't contribute. A large part of the value is that the user can manipulate time-ordered mathematics as in regular mathematics. For example, let $x(t)$ be a $\operatorname{Lie}(G)$-valued function. Then $\exp(\int_a^b x(t)dt)\exp(\int_b^c x(t)dt) \neq \exp(\int_a^c x(t)dt)$ in general, whereas the corresponding time-ordered equation is true.
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| Jul 21, 2010 at 12:46 | comment | added | Steve Huntsman | I've seen both TO and PO used as well. | |
| Jul 21, 2010 at 2:56 | comment | added | George Lowther | Yes, but you have even numbers of anticommuting terms in the products at each point. So grouped together they will commute. (I think) | |
| Jul 21, 2010 at 2:48 | comment | added | José Figueroa-O'Farrill | "terms associated to space-like separated points commute"... or anticommute! Remember there's a spin-statistics theorem, at least in 4-dimensional relativistic quantum field theory. | |
| Jul 21, 2010 at 2:43 | comment | added | George Lowther | As far as I can make out, physicists have a habit of simply slapping a capital T in front of an expression and declaring it to be "time-ordered". Not just exponentials, but whatever takes their fancy really. I guess, this assumes that everything can be expanded out into a sum of products of terms associated with a time index which are then ordered. Or, into fields associated with space-time points where it is assumed that terms associated to space-like separated points commute. | |
| Jul 21, 2010 at 2:15 | comment | added | Theo Johnson-Freyd | Now I see that George Lowther has given the same answer above. | |
| Jul 21, 2010 at 2:14 | history | answered | Theo Johnson-Freyd | CC BY-SA 2.5 |