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Setup: the time-independent Schrödinger equation (eigenvalue problem):

$(-\frac{\hbar^2}{2m}\Delta +V)\psi = E\psi$

(On any 'reasonable' domain $D \subset \mathbb{R^3}$ with 'reasonable' boundary conditions ($\psi \in L^2(D)$ atleast) with $V: D \rightarrow \mathbb{R}$ a real valued potential function, again can roughly assume it decays at large distances, but can have isolated singularities; can reduce down to a $1$D problem to simplify it as well)

When dealing with computing bound states/bound state energies of say an electron in complicated potentials, one trick I have seen informally is that we can replace the given complicated potential with something that has 'similar behaviour' but has known eigenstates/eigenvalues to 'estimate' the eigenstates/eigenvalues. What I wish to know is, under what conditions is something like this scheme faithful/what conditions need to be imposed to achieve this kind of approximation? Roughly speaking, If $V$ is the given potential, and $V_{\epsilon}$ is another potential whose eigenvalues and eigenstates are known, such that $\|V_\epsilon-V\|<\epsilon$ (in some norm) then if $E_{\epsilon}, E$ denote the ground state energies (the lowest energy eigenvalues) in $V_{\epsilon}, V$ respectively, then what is an estimate on $|E_{\epsilon}-E|$? Is it of the order $\epsilon$?

I do not know what this class of problems is called so I could not look for it by just googling for example. Any references on it are requested.

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  • $\begingroup$ isn't this just what we try to achieve with perturbation theory? assuming the ground state is not degenerate, the correction $\delta E=E-E_\epsilon$ equals $\int (V-V_\epsilon)|\Psi_\epsilon|^2 d^3 r$ plus terms of order $\epsilon^2$ (with $E_\epsilon$ and $\Psi_\epsilon$ the ground state eigenvalue and eigenfunction in the potential $V_\epsilon$) $\endgroup$
    – Carlo Beenakker
    Commented Mar 18, 2023 at 11:08
  • $\begingroup$ Thank you for your comment; could you also provide a source for such an analysis? With respect to how the error terms could be controlled $\endgroup$
    – Rohan Didmishe
    Commented Mar 18, 2023 at 11:11
  • $\begingroup$ Well in this case there is no parameter valued expansion I am assuming a-priori; is that the only case in which some scheme like this can be justified? @CarloBeenakker $\endgroup$
    – Rohan Didmishe
    Commented Mar 18, 2023 at 11:13
  • $\begingroup$ Denoting the unperturbed potential by $V_{\epsilon } $ is maximally confusing here. Write $V=V_0 + \epsilon V' $ with the $V_0 $ problem solvable and develop the perturbation series in $\epsilon $. Unless there are degeneracies in the spectrum of the $V_0 $ problem, the spectrum of the $V$ problem can indeed be written as a power series in $\epsilon $. $\endgroup$
    – Michael Engelhardt
    Commented Mar 18, 2023 at 13:40
  • $\begingroup$ I see, thank you for your inputs. $\endgroup$
    – Rohan Didmishe
    Commented Mar 18, 2023 at 14:11

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