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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.

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  • $\begingroup$ Check the integration again, you should get the same result. $\endgroup$
    – acast00
    Commented May 21, 2018 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
    Commented May 21, 2018 at 22:14

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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).

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  • $\begingroup$ 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
    Commented May 21, 2018 at 9:49
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    $\begingroup$ It is not a well, it is a barrier. $\endgroup$
    – Vladimir Kalitvianski
    Commented May 21, 2018 at 9:52

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