I am trying to determine the ground state energy and wave function for a particle of mass m confined to a one dimensional potential of the form $\lambda^2 / x^2 - \lambda/x$ . I found a paper that explains how the potential $1/x^2$ can be solved for the needed quantities, but i am confused how the first potential is different from the second. one thing is the first potential have ground state, while the second one has no ground state.

how is the solution to my question different from the solution to the 1/x^2 potential.

  • $\begingroup$ highenergy.phys.ttu.edu/~akchurin/PHYS5302ProjectPapers/… when i read this paper, this is what i understood. If i am wrong please change them to the correct tags. $\endgroup$ Commented Jan 12, 2017 at 13:50
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    $\begingroup$ It's not clear to me what your precise question is. Do you want someone to do an analysis similar to the paper you link of your modified potential for you? (That would be off-topic as homework-like in my eyes) $\endgroup$
    – ACuriousMind
    Commented Jan 12, 2017 at 14:24

1 Answer 1


if you look at the graphs 1/x^2,1/x^2-1/x, the second one has a minimum at a finite x while the first one has no minimum. This will explain the ground state too.

  • $\begingroup$ Yes, i see that, i am just not sure how to proceed with the solution... to paraphrase my question.. how is the solution to my problem different from the solutuion to the 1/x^2 potential. $\endgroup$ Commented Jan 12, 2017 at 13:04
  • $\begingroup$ can you give me the link to the solution you found ? $\endgroup$
    – Ismasou
    Commented Jan 12, 2017 at 13:10
  • $\begingroup$ highenergy.phys.ttu.edu/~akchurin/PHYS5302ProjectPapers/… $\endgroup$ Commented Jan 12, 2017 at 13:10
  • $\begingroup$ The best way I see for solving this diff equation is to use the series method, you can find an explanation here tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx $\endgroup$
    – Ismasou
    Commented Jan 12, 2017 at 13:23
  • $\begingroup$ I will try. and inform you with the results. $\endgroup$ Commented Jan 12, 2017 at 13:29

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