Timeline for The eigenvalue of Schrodinger opeartor
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 4, 2020 at 21:01 | comment | added | Jochen Glueck | @Student: The question which unbounded types of potentials are admissible depends heavily on the sign of the potential. If you are interested in $V$ that are bounded above (but not necessarily below), things are easier than if you are interested in $V$ that might be unbounded above. | |
| May 4, 2020 at 20:39 | comment | added | Giorgio Metafune | I would recommend the book by B. Davies: spectral theory and differential operators. | |
| May 4, 2020 at 20:25 | comment | added | STUDENT | @WillieWong Yes, that's what I meant. Can we have weaker conditions like $V\in L^p(\Omega)$ such that the operator has a discrete increasing spectrum? However, I am more interested about the second one. | |
| May 4, 2020 at 20:18 | comment | added | Willie Wong | One possibility (after taking a quick look at G+T) of a reasonable question: what are the conditions on $V$ that guarantees the spectrum is discrete. Perhaps this is what the OP meant by question 1. | |
| May 4, 2020 at 20:04 | comment | added | Willie Wong | @JochenGlueck: I deleted my previous comment because my interpretation was not correct. Turns out that particular section of G+T is talking about Rayleigh quotients, and bounded coefficients are convenient. // And also, never noticed this before, but there is a typo in the 2001 edition of G+T: the operator and quadratic form on the bottom of page 212 disagree by a sign (which further muddies my second question to the OP). | |
| May 4, 2020 at 19:33 | comment | added | Jochen Glueck | Welcome to MathOverflow! What precisely do you mean by "the spectrum exists"? Every linear operator has a spectrum by definition. | |
| May 4, 2020 at 19:32 | comment | added | Willie Wong | What do you mean by weakest? (for question 1) // For your second question, as you wrote it, your operator $L$ is formally a negative operator when $V \equiv 0$. I just want to double check that you really meant to ask about the first eigenvalue being positive, or if rather you want the opposite direction? In any case, certainly the answer to the second question depends lots on $\Omega$ and details of $V$, if only just by thinking about the family $\lambda V$ of potentials and noting for all sufficiently small $\lambda$ the spectrum should be a small perturbation of that of $\Delta$. | |
| May 4, 2020 at 16:50 | review | First posts | |||
| May 4, 2020 at 17:00 | |||||
| May 4, 2020 at 16:44 | history | asked | STUDENT | CC BY-SA 4.0 |