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Also asked on MSE (https://math.stackexchange.com/questions/4987654/in-what-sense-is-a-change-of-boundary-conditions-a-finite-rank-perturbation and https://math.stackexchange.com/questions/4875398/how-is-changing-the-boundary-conditions-a-finite-rank-perturbation?noredirect=1&lq=1) with no answers.

I am concerned with differential operators, let's say Schrodinger operators $T = -\Delta + V$ defined on $\mathcal D(T) = \{f \in H^2(\Omega) : V f \in L^2(\Omega)\} \cap \mathcal B(\Omega)$ for suitable $\Omega \subseteq \mathbb R$ and $\mathcal B(\Omega)$ is certain boundary conditions. Often we have results about spectra only for $\Omega = [0, \infty)$. Taking $\widetilde V(x) = V(-x)$ and reflecting your input functions similarly, you can quickly extend these results to $\Omega = (-\infty, 0]$, but it seems non-trivial to step to any conclusions about $\Omega = \mathbb R$. It seems you cannot just take the direct sum of these two operators. Sets like the singular continuous spectrum are notoriously unstable even under rank-one perturbations, so it becomes a reasonable question of how the singular continuous spectrum of the half-line problems will relate to the singular continuous spectrum of the whole line problem.

In the paper "Operators with Singular Continuous Spectrum V: Sparse Potentials" by Simon and Stolz, it is said at the bottom of page 5 that:

Whole-line problems differ from the direct sum of two half-line problems only by adding boundary conditions at 0, a finite rank perturbation

This does not immediately make sense to me, since a finite rank perturbation will be bounded and so shouldn't change the domain of the operator. In general this paper is a bit brief on the relation between the half-line and whole-line problems. Could someone help make this explicit (ie. explicitlyish what perturbation is being taken and why it is finite rank) or give a reference that does so?

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  • $\begingroup$ 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). $\endgroup$ Commented Oct 29 at 22:14
  • $\begingroup$ @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? $\endgroup$ Commented Oct 29 at 22:17
  • $\begingroup$ 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. $\endgroup$ Commented Oct 29 at 22:20
  • $\begingroup$ 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. $\endgroup$ Commented Oct 29 at 22:26

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