I'm currently writing a numerical code to solve the quantum system (in 3D) with an attractive ($v_0 < 0$) Gaussian potential, $$ V(r) = v_0 \exp(-r^2/r_0^2). $$ It would be nice to have some reference values (at least for the energies) to compare to and to test my accuracy. Does anyone know of any article/textbook where such data is provided?
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2$\begingroup$ Did you perform a literature search? I am finding quite a few articles about the problem. There are even exact solutions for a problem with a pseudo-Gaussian potential. $\endgroup$– FlatterMannCommented Apr 11, 2023 at 5:53
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$\begingroup$ Is $v_0$ positive? $\endgroup$– Roger V.Commented Apr 11, 2023 at 6:50
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$\begingroup$ For 1D potential see: iopscience.iop.org/article/10.1088/0305-4470/17/3/002, researchgate.net/publication/…, demonstrations.wolfram.com/… I think it could be good, if you set the problem for 3D in your question and indicate which resources you have studied and how you tried to approach this problem. (I do not vote for closing, since it could be a useful reference question, if properly formulated and answered.) $\endgroup$– Roger V.Commented Apr 11, 2023 at 6:55
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$\begingroup$ Thank you for your comments already. FlatterMann: I did some literature search before, but apparently not for the right search terms. I tried some more and now found at least the following article: arxiv.org/abs/1805.00006, even though I did not find which value of the parameter "A" (corresponding to v0 in my case) they chose. However, within there are further references linked, I will check them out. RogerVadim: v0 is negative in my case, to represent an attractive potential. I will edit the question text to add it. Also I will edit the question itself to include 3D. $\endgroup$– physhCommented Apr 11, 2023 at 7:41
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$\begingroup$ @FlatterMann and I have made some suggestions on modifying your question - if you expand it accordingly and make it more precise, it might be reopened. I also suggest adding links to the papers you found to the text itself (e.g., as references) - since comments risk to be deleted. $\endgroup$– Roger V.Commented Apr 12, 2023 at 7:55
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