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Nov 28, 2022 at 22:26 vote accept ColourConfined
Nov 28, 2022 at 21:19 comment added ColourConfined Ok, I thought so. But now I am confused about the convergence argument. Taking the case of the finite well again, expanding around this hamiltonian, we would expect perturbative treatment of eignestate corrections not to converge (as we start with normalizable states but now with the linear potential they are all unnormalizable) but this doesn't necessary need be the case for the corrections of the energies.
Nov 28, 2022 at 20:05 comment added Hyperon Correct! You can easily find the non-normalizable eigenfunctions in the $p$-representation of the Hamiltonian: $H = p^2/2m +m g i \hbar \partial/ \partial p$.
Nov 28, 2022 at 19:52 comment added ColourConfined For 4. the spectrum is $E \in ( -\infty, \infty)$ but has no normalizable eigenstates?
Nov 28, 2022 at 13:30 history answered Hyperon CC BY-SA 4.0