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Homework Help: Determine the potential V(x)

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A particel is moving along the x-axis in a potential, V(x). The ground state of the wavefunction is:

    [tex]\[\psi \left( x \right)={{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}},\][/tex]
    where a is a constant of dimension length. We also know that V(a) = 0.

    Determine V(x).

    2. Relevant equations

    The stationary Schrödinger equation:

    [tex]\[-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+V\psi =E\psi \][/tex]

    3. The attempt at a solution

    Well, first I derive the wavefunction two times and get:

    & \frac{d}{dx}\psi \left( x \right)=-\frac{2}{{{a}^{4}}}{{x}^{3}}{{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}} \\
    & \frac{d}{dx}\psi '\left( x \right)=\left( \frac{4{{x}^{6}}}{{{a}^{8}}}-\frac{6{{x}^{2}}}{{{a}^{4}}} \right){{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}} \\

    Then I insert it in the Schrödinger equation.
    And here is where I get confused. It's been a while since I've had QM, so I can't really remember what's next.
    I figure that the potential is NOT just what I get when I isolate V in the Schrödinger equation. So what's is my next step ?

  2. jcsd
  3. Feb 25, 2010 #2


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    Why not?...Keep in mind that [itex]E[/itex], being an eigenvalue, must be a constant (no x-dependence).
  4. Feb 25, 2010 #3
    So the potential V(x) is just:

    [tex]\[V\left( x \right)=\frac{2\hbar {{x}^{6}}-3{{a}^{4}}\hbar {{x}^{2}}+E{{a}^{8}}m}{{{a}^{8}}m}\][/tex] ?
  5. Feb 25, 2010 #4


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    You still need to evaluate what E equals. E is an eigenvalue. Its value changes depending on the eigenstate of the system.
  6. Feb 25, 2010 #5
    Ahhh yes...
    But what is En for the stationary state ? Can't find it on the internet, and I haven't got my QM book in ages. If I recall correctly it's different for the infinite well, harmonic oscillator, finite well and free particle. So which one should I use ?
  7. Feb 25, 2010 #6


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    You tell us....you are told that V(a)=0, so what must E be for that to be true?
  8. Feb 25, 2010 #7


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    You can use [tex]E=<\psi|H|\psi>[/tex] since your wave function is an eigenstate.

    EDIT: Wait, disregard that...
  9. Feb 26, 2010 #8
    I think I got it now...
    V(a) = 0, which removes the potential part, replace the x's with a's and isolate E.
    And then after that I use that E in the same Schrödinger equation and isolate V.
  10. Feb 26, 2010 #9


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    Another way to look at it is that you've already derived an expression for [itex]V(x)[/itex] that includes an unknown constant, [itex]E[/itex] (in post #3), and you are given an initial value, [itex]V(a)=0[/itex]. So, you can determine the value of that constant by plugging [itex]x=a[/itex] into your expression, setting it equal to zero, and solving for [itex]E[/itex].
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