Timeline for In what sense is a change of boundary conditions a finite rank perturbation?
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6 events
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| Oct 29 at 22:26 | comment | added | Christian Remling | It can probably also be seen in lazier style. Let's say we are on a half line $[0,\infty)$. Then for $u$ in the domain of the minimal operator (so $u(0)=u'(0)=0$) and $f=-u''+(V+i)u$, the resolvent $(T_{\alpha}+i)^{-1}f=u$ does not depend on the bc $\alpha$ (since $u$ satisfies all bc's), so the discrepancy takes place on a finite-dimensional space by the von Neumann theory. | |
| Oct 29 at 22:20 | comment | added | Christian Remling | It's not completely obvious, but it can be checked basically by writing it down since the resolvent is an integral operator, with kernel built from solutions via the variation of constants formula. | |
| Oct 29 at 22:17 | comment | added | George Coote | @ChristianRemling that makes a whole lot of sense, thanks! is it obvious why this is the case or are you aware of a source that elaborates a bit more? | |
| Oct 29 at 22:14 | comment | added | Christian Remling | It's a finite rank perturbation of the resolvent (and this is the intended meaning, but it's common to say it in this abbreviated style that however can not be taken at face value, as you observed). | |
| S Oct 29 at 22:02 | review | First questions | |||
| Oct 29 at 22:55 | |||||
| S Oct 29 at 22:02 | history | asked | George Coote | CC BY-SA 4.0 |