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Eigenvalue for harmonic oscillator

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data
    The Hamiltonian for a particle in a harmonic potential is given by
    [itex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{Kx^2}{2}[/itex], where K is the spring constant. Start with the trial wave function [itex]\psi(x)=exp(\frac{-x^2}{2a^2})[/itex]
    and solve the energy eigenvalue equation [itex]\hat{H}\psi(x)=E\psi(x)[/itex]. You must find the value of the constant, a, which will make applying the Hamiltonian to the function return a constant time[sic, I assume he meant "times"] the function. Then find the energy eigenvalue.

    2. Relevant equations
    [itex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{Kx^2}{2}[/itex]
    [itex]\psi(x)=exp(\frac{-x^2}{2a^2})[/itex]
    [itex]\hat{H}\psi(x)=E\psi(x)[/itex]

    3. The attempt at a solution
    Application of the Hamiltonian gave me:
    [itex]\hat{H}\psi(x)=[\frac{h^2}{2m}(\frac{1}{a^2}+\frac{x^2}{a^2})+\frac{1}{2}Kx^2]\psi(x)=E\psi(x)[/itex]

    If I understand the problem correctly, since E and a must be constant, I must come up with an a, devoid of any x's or functions of x's so taking the Hamiltonian will yield something equally devoid of x's and functions of x's.

    I mean, I get that I should have [itex][\frac{h^2}{2m}(\frac{1}{a^2}+\frac{x^2}{a^2})+\frac{1}{2}Kx^2]=E[/itex] for some constants E, a. The problem is, I have no clue how to go about solving that and getting a to not involve any x's, nor am I convinced that it's possible. Using the quadratic formula on it gives a big mess (involving x), so I'm pretty sure that's the wrong route.

    Thanks for the help.
     
  2. jcsd
  3. Oct 1, 2013 #2

    ehild

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    You have a sign error: How is the operator [itex]\hat{p}[/itex] defined?

    To remove x from the equation, collect the terms with x2. You have to choose the parameter a properly, so as the two terms cancel each other.


    ehild
     
  4. Oct 1, 2013 #3

    TSny

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    Also check to see if you have enough factors of ##a## in the denominator of your ##x^2/a^2## term.
     
  5. Oct 1, 2013 #4
    Thanks, I found the sign error and got it worked out. As for the factors of a, you're right, I actually typed it in wrong.
     
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