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Mar 18, 2023 at 22:53 history edited Willie Wong
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Mar 18, 2023 at 14:11 comment added Rohan Didmishe I see, thank you for your inputs.
Mar 18, 2023 at 13:40 comment added Michael Engelhardt 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 $.
Mar 18, 2023 at 11:16 history edited YCor CC BY-SA 4.0
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Mar 18, 2023 at 11:13 comment added Rohan Didmishe 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
Mar 18, 2023 at 11:11 comment added Rohan Didmishe 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
Mar 18, 2023 at 11:08 comment added Carlo Beenakker 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 $V_\epsilon$)
S Mar 18, 2023 at 11:02 review First questions
Mar 18, 2023 at 11:42
S Mar 18, 2023 at 11:02 history asked Rohan Didmishe CC BY-SA 4.0