I'm trying to model a system as being inside an infinite potential well with $V(x)=-ax^v$ where $a$ and $v$ are some positive real numbers.

However I'm a bit confused: if I take the - sign inside another constant, say $b=-a$, then $p(x)=\sqrt{2m(E-bx^v)}$ and if I do this integral I get quite a different answer than by having $p(x)=\sqrt{2m(E+ax^v)}$. $E>V(x)$ inside the well. What is the way to proceed here so that I get a correct answer? I know I could just "shift" my potential so that it becomes positive but I'm trying to be consistent with some previous work.

  • $\begingroup$ Check the integration again, you should get the same result. $\endgroup$ – acast00 May 21 '18 at 10:26
  • $\begingroup$ I don't think so... The result is some expression containing the hypergeometric function and it has different parameters for the 2 cases... $\endgroup$ – physicss May 21 '18 at 22:14

In order to have a potential well, the potential must have the form $V(x)=a|x|^v$ with a positive $a$ and $v$. Otherwise you risk to have a reflecting potential like $x^3$ or $-x^2$ (a parabola upside down).

  • $\begingroup$ Could you please elaborate on the "reflecting potential" part? i.e. what could go "wrong" in a $-x^2$ potential as compared to a $x^2$ well? $\endgroup$ – physicss May 21 '18 at 9:49
  • 1
    $\begingroup$ It is not a well, it is a barrier. $\endgroup$ – Vladimir Kalitvianski May 21 '18 at 9:52

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