1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [tex]\[\begin{align}
    & \frac{d}{dx}\psi \left( x \right)=-\frac{2}{{{a}^{4}}}{{x}^{3}}{{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}} \\
    & \frac{d}{dx}\psi '\left( x \right)=\left( \frac{4{{x}^{6}}}{{{a}^{8}}}-\frac{6{{x}^{2}}}{{{a}^{4}}} \right){{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}} \\
    \end{align}\]
    [/tex]

    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 ?


    Regards
     
  2. jcsd
  3. Feb 25, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook