Timeline for What's the reasons to use time-ordering operator?
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22 events
| when toggle format | what | by | license | comment | |
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| S Dec 9, 2016 at 2:10 | history | bounty ended | CommunityBot | ||
| S Dec 9, 2016 at 2:10 | history | notice removed | CommunityBot | ||
| Dec 8, 2016 at 1:23 | vote | accept | Jack | ||
| Dec 6, 2016 at 17:17 | comment | added | Jahan Claes | In the context of condensed matter, I would say the time ordering is a "trick" in the following sense: You can calculate time ordered green's functions using path integrals/feynmann diagrams, but you need the retarded green's functions to get observables. Then there's some formulas you can use to relate the time ordered green's functions to the retarded green's functions. So the time ordering is convenient, and then there are "tricks" to get the actual green's functions you want. | |
| Dec 6, 2016 at 15:57 | comment | added | Qmechanic♦ | Possible duplicates: physics.stackexchange.com/q/103503/2451 and links therein. | |
| Dec 6, 2016 at 14:34 | answer | added | ZeroTheHero | timeline score: 4 | |
| Dec 5, 2016 at 4:25 | comment | added | QuantumDot | @tparker you mean deriving the equivalent path integral from different operator Hamiltonians? | |
| Dec 5, 2016 at 3:58 | comment | added | tparker | @QuantumDot It largely comes down to definitions. Yes, I agree that omitting the $\mathcal{T}$ in the equations in which it appears would render them incorrect, but most of those quantities can be defined in a different way that does not require the time-ordering symbol. For example, one could in principle start with many different operator Hamiltonians and rederive the equivalent Lagrangian actions for each one independently, but using $\mathcal{T}$ gives a convenient shortcut rule that works for any Hamiltonian. If you don't like the word "trick," maybe "algorithm" would be better. | |
| Dec 5, 2016 at 2:22 | comment | added | QuantumDot | @tparker That can't be right (unless I misunderstand the question). A trick is something you don't have to use, but you do because it is convenient. I'm pretty sure that omitting the time ordering is simply wrong. For example, the time evolution operator wouldn't satisfy its defining equation without it.... | |
| Dec 2, 2016 at 21:12 | comment | added | tparker | Prof. Mark Srednicki (of the QFT textbook) once told me that in the context of QFT, he mostly views it as a technical trick that facilitates translating between the Hamiltonian operator and Lagrangian path-integral formalisms, without much physical significance. | |
| Dec 1, 2016 at 1:17 | answer | added | Sean E. Lake | timeline score: 16 | |
| Dec 1, 2016 at 0:48 | comment | added | Elliot Yu | Here are some related questions and answers: physics.stackexchange.com/q/173603, physics.stackexchange.com/a/56217/55689 | |
| S Dec 1, 2016 at 0:12 | history | bounty started | Jack | ||
| S Dec 1, 2016 at 0:12 | history | notice added | Jack | Canonical answer required | |
| Nov 30, 2016 at 10:49 | history | edited | Jack | CC BY-SA 3.0 |
How to understand time-ordering operator from the physical viewpoint ?
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| Nov 23, 2016 at 8:44 | history | edited | Jack | CC BY-SA 3.0 |
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| Nov 22, 2016 at 12:48 | history | edited | Jack | CC BY-SA 3.0 |
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| Nov 16, 2016 at 1:20 | comment | added | Jack | Hi, Unnikrishnan.Thanks for your nice implications,I will learn and improve my skills.I am also glad to join this interesting community. | |
| Nov 16, 2016 at 1:12 | history | edited | Jack | CC BY-SA 3.0 |
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| Nov 10, 2016 at 12:15 | comment | added | UKH | Hi Jack. Your title should emphasize on what you are asking. You have provisions to add tags on the subject of your queston. | |
| Nov 10, 2016 at 7:29 | review | First posts | |||
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| Nov 10, 2016 at 7:19 | history | asked | Jack | CC BY-SA 3.0 |