Timeline for Perturbative solution to an Eigenvalue Problem with a continuous spectrum
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2011 at 11:06 | answer | added | Matthias Ludewig | timeline score: 0 | |
| Jul 29, 2011 at 22:06 | answer | added | Helge | timeline score: 1 | |
| Jul 29, 2011 at 21:32 | comment | added | psyduck | Yes, I believe when I was learning about scattering in quantum mechanics we found the green's function by adding a small imaginary component to the resolvant and took a limit $g^{\pm}(E) = \lim_{\epsilon \to 0}\frac{1}{H - E \pm i \epsilon}$ I did try that in the above scenario and still ended up with a solution that blew up. But, to be honest, I didn't do things as carefully as I should have. I'll try this again. | |
| Jul 29, 2011 at 21:16 | comment | added | paul garrett | If there is reason to believe that there should be a coherent outcome, in effect justifying "creative" heuristics, ... perturbing $p^2-q^2$ to $p^2-q^2+\epsilon^2$ is the/an obvious thing to (try to) justify. If this computation is in the middle of something else, the chances of justification need not be too bad... :) | |
| Jul 29, 2011 at 20:45 | comment | added | psyduck | By "blow up" I mean that if I try to evaluate an integral of the form $\int \frac{e^{ipx} h(p) }{p^2 - q^2} \delta'(p-q) dp = - \int \delta(p-q) \frac{d}{dp} ( \frac{e^{ipx} h(p) }{p^2 - q^2} ) dp$ the result is infinity. | |
| Jul 29, 2011 at 20:23 | comment | added | paul garrett | When you say that the one attempt "blows up"... in what sense, exactly? One reason for asking is that even when a series expression doesn't literally "converge", it still may have some validity as an asymptotic expansion, in various regimes. Poincare and others already found examples in various scenarios in celestial mechanics, where "approximate" solutions were eventually shown to not be parts of convergent infinite series... but, nevertheless, were perfectly fine asymptotic expansions (which is how they'd been used all along, in reality). Just a thought... | |
| Jul 29, 2011 at 20:21 | comment | added | psyduck | sorry. $\psi_p^0$ and $\psi_0^p$ were supposed to read $\phi_p^0$ above. | |
| Jul 29, 2011 at 20:19 | comment | added | psyduck | Until I read you post, I had not heard of the "trace class" before. But, after a google search, if my understanding is correct, unless I can show $\int < |(L^0 + i)^{-1} L^1 | \psi_p^0, \psi^p_0 > dp < \infty$ Then my search for a solution is hopeless. | |
| Jul 29, 2011 at 19:42 | comment | added | Helge | It's not possible to do this. The perturbation you consider is too singular. If you had that $(L^0 + i)^{-1} L^1$ was trace class, you could use wave operators .... but that's about to where what you want is possible. | |
| Jul 29, 2011 at 19:19 | history | asked | psyduck | CC BY-SA 3.0 |