Timeline for Asymptotic question about time ordered exponentials
Current License: CC BY-SA 3.0
26 events
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| Nov 15, 2017 at 6:44 | comment | added | fedja | @DavidSpeyer It is a correct argument, but it has certain assumptions that have been completely swept under the rug. Perhaps, the quoted paper tells more about them, so you can see if they cover your situation or not. | |
| S Nov 14, 2017 at 23:00 | history | suggested | jeq | CC BY-SA 3.0 |
Copied image to imgur.com, as it was not being displayed because of broken link. Added link to original image source via Wayback Machine.
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| Nov 14, 2017 at 22:43 | review | Suggested edits | |||
| S Nov 14, 2017 at 23:00 | |||||
| Jul 2, 2013 at 16:05 | history | wiki removed | Robert Cartaino | ||
| May 2, 2012 at 10:56 | history | edited | Jon | CC BY-SA 3.0 |
Modified link to image
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| May 2, 2012 at 10:44 | history | edited | Jon | CC BY-SA 3.0 |
Modified link to image
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| May 2, 2012 at 10:31 | history | edited | Jon | CC BY-SA 3.0 |
Resized image
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| May 2, 2012 at 10:21 | history | edited | Jon | CC BY-SA 3.0 |
Trying to display an image
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| May 1, 2012 at 21:54 | comment | added | Jon | David, you are welcome. Ask as much as you like. | |
| May 1, 2012 at 20:26 | comment | added | David E Speyer | More questions in my "answer". Sorry about taking so long to get it. (I am also reading Fedoryuk's book, but you are a lot clearer than he is.) | |
| May 1, 2012 at 17:17 | comment | added | Jon | David, thank you very much. I was already in difficulty while I was writing down the text. | |
| May 1, 2012 at 16:24 | comment | added | David E Speyer | I took the liberty of cleaning up some of your TeX. There are some strange interactions between the Markdown syntax which MO uses and LaTeX. | |
| May 1, 2012 at 16:23 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 20 characters in body; added 2 characters in body
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| May 1, 2012 at 15:49 | history | edited | Jon | CC BY-SA 3.0 |
Minor edit
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| May 1, 2012 at 15:43 | history | edited | Jon | CC BY-SA 3.0 |
Minor edit
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| Apr 30, 2012 at 23:06 | comment | added | David E Speyer | I think I've unconfused myself again. See the note I left below. | |
| Apr 28, 2012 at 13:45 | comment | added | Jon | David, thank you very much for your appreciation of my work. Since I proved that theorem I have produced a lot of applications. One of the most successful paper is this arxiv.org/abs/quant-ph/9811037 but I have also devised applications to PDEs in arxiv.org/abs/hep-th/0508246 . This is all published material. You can find a lot more on arxiv as currently I applied these techniques to Yang-Mills equations and scalar field theory. In this area, at a classical level, I have proved a theorem with some helpful support from Terry Tao. | |
| Apr 27, 2012 at 20:01 | comment | added | David E Speyer | Actually, I am still confused. But I am seeing similar formulas in other sources, so presumably this is right. Hope I can figure out how! | |
| Apr 27, 2012 at 18:59 | vote | accept | David E Speyer | ||
| Apr 27, 2012 at 18:59 | comment | added | David E Speyer | I take it back. Your answer is right. It's just that I am going to need to get way inside those asymptotics to see the terms I care about. Thanks! | |
| Apr 26, 2012 at 22:09 | comment | added | David E Speyer | I just started thinking about the problem that motivated this again. I am suspicious of this answer. The leading term is a linear combination of the terms $| n;1 \rangle \langle n; -1|$, meaning that the leading term takes the eigenvectors of $A(-1)$ to the eigenvectors of $A(1)$, right? Independent of whatever happens in the middle of the "integral"? I am pretty sure this isn't true, and that my problem suggests a counterexample. I'll try to compute one in a few days. | |
| Dec 1, 2011 at 13:38 | comment | added | Jon | David, you are welcome. I have fixed the equation for $\tilde {\bf A}(t)$ as $r$ appears to multiply the eigenvalues of course. Then $\tilde {\bf B}(r)$ is a series, analogous to the Magnus one (but we physicists prefer to call it Dyson series), that is meaningful in the limit $r\rightarrow\infty$. | |
| Dec 1, 2011 at 13:36 | history | edited | Jon | CC BY-SA 3.0 |
Corrected an equation
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| Dec 1, 2011 at 13:22 | comment | added | David E Speyer | Thanks! This will take some time to digest, but I have the following immediate questions: In your equation for $\tilde{B}(r)$, there is no $r$ on the right hand side. Is that right? Also, you describe your answer as a series in inverse powers of $r$, but I don't see such a series appearing anywhere in your answer. Thanks again! | |
| Dec 1, 2011 at 9:55 | history | edited | Jon | CC BY-SA 3.0 |
Minor changes
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| Dec 1, 2011 at 9:40 | history | answered | Jon | CC BY-SA 3.0 |