Timeline for Discreteness of set of energy eigenvalues
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 30, 2013 at 10:15 | comment | added | smiley06 | @guru Last and first lines of page 99-100 of 'Modern Quantum Mechanics' by J.J Sakurai. And $ \psi = 0 $ is called a trivial solution. | |
| Jul 30, 2013 at 9:09 | comment | added | guru | @smiley06: Can you quote exactly where this is mentioned in Sakurai? Page no. and line no.? Also you need to clarify what you mean by 'non-trivial'. | |
| Jun 15, 2013 at 6:15 | vote | accept | smiley06 | ||
| Jun 14, 2013 at 20:02 | answer | added | user24999 | timeline score: 11 | |
| May 27, 2013 at 11:07 | history | tweeted | twitter.com/#!/StackPhysics/status/338974767031939073 | ||
| May 23, 2013 at 19:10 | comment | added | Vibert | There might be some fancy proofs, but the simplest way to see this is through Sturm-Liouville theory, en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory. Schrödinger's equation is a special case of Sturm-Liouville (plug in $p(x) = 1$). Then assuming some conditions on $V(x)$ and the boundary conditions, there is a theorem (see Wiki) that tells you about the spectrum and the eigenfunctions. | |
| May 23, 2013 at 18:50 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 30 characters in body; edited tags
|
| May 23, 2013 at 17:09 | comment | added | jjcale | T is not a bounded operator and therefore not compact. But it's resolvent might be compact. You will find the answer certainly in one of the Reed-Simon books. | |
| May 23, 2013 at 16:37 | review | First posts | |||
| May 23, 2013 at 17:31 | |||||
| May 23, 2013 at 16:17 | history | asked | smiley06 | CC BY-SA 3.0 |