Timeline for Using Fredholm theory to evaluate a ratio of operator determinants
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2017 at 16:35 | history | edited | Shayne | CC BY-SA 3.0 |
edited body
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| Jul 27, 2017 at 23:14 | answer | added | Christian Remling | timeline score: 2 | |
| Jul 27, 2017 at 23:05 | comment | added | Shayne | @Christian I was advised to post here because Coleman is a graduate level text, but I think you are right. I'll try there, thank you! | |
| Jul 27, 2017 at 22:54 | comment | added | Christian Remling | No, the operator is not bounded. By the way, this site is for research level mathematics. For more basic questions, math stackexchange tends to work much better. | |
| Jul 27, 2017 at 22:53 | comment | added | Shayne | @ChristianRemling Yes, the operator is bounded and on a finite interval $[-T/2,T/2]$. | |
| Jul 27, 2017 at 22:51 | comment | added | Shayne | @MichaelRenardy yes, sorry, I forgot to mention that the operator is indeed bounded. | |
| Jul 27, 2017 at 22:49 | history | edited | Shayne | CC BY-SA 3.0 |
Added extra, relevant information
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| Jul 27, 2017 at 22:49 | comment | added | Christian Remling | @MichaelRenardy: In fact, $\lambda_j\to-\infty$. The OP is certainly suppressing assumptions, a general Schrodinger operator need not have any eigenvalues at all. | |
| Jul 27, 2017 at 22:47 | comment | added | Christian Remling | Also, of course your "definition" of $\det (d^2/dt^2-W)$ is a very badly divergent product; only the ratio can possibly make sense. | |
| Jul 27, 2017 at 22:46 | history | edited | Shayne | CC BY-SA 3.0 |
added 6 characters in body
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| Jul 27, 2017 at 22:43 | comment | added | Christian Remling | The claim is that $\prod (\lambda + \lambda^{(1)}_j)/(\lambda + \lambda^{(2)}_j)\to 1$, and for this one needs to know the asymptotics of $\lambda_j$. I assume you consider operators on a bounded interval, and then this topic has been well studied. It has nothing to do with Fredholm theory. | |
| Jul 27, 2017 at 22:42 | comment | added | Michael Renardy | This cannot be right without putting some constraint on how $\lambda$ goes to infinity, since the $\lambda_n^{(1)}$ and $\lambda_n^{(2)}$ presumably go to infinity as well (the operator is unbounded!). | |
| Jul 27, 2017 at 22:38 | review | First posts | |||
| Jul 27, 2017 at 22:41 | |||||
| Jul 27, 2017 at 22:34 | history | asked | Shayne | CC BY-SA 3.0 |