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Homework Help: Variational Principle

  1. Mar 12, 2008 #1


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    1. The problem statement, all variables and given/known data

    Find the best bound state on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form

    [tex] \psi(x) = \frac{A}{x^2 + b^2} [/tex]

    where A is determined by normalization and b is an adjustable parameter.

    2. Relevant equations

    The variational principle

    [tex] \langle \psi |H| \psi \rangle \geq E_{gs} [/tex]

    3. The attempt at a solution

    This problem has some really nasty integrals. E.g., just to normalize the wavefunction, you need to calculate:

    [tex] \int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2} [/tex]

    I was able to find most of them using residue theory. It was a last resort...the only method I could think of! I had to dig through my old complex analysis notes to remind myself of the technique using a semicircular contour of radius r and then letting r go to infinity. The integral of f(z) over the arc is zero, leaving the integral on the real axis, which is the part I want to evaluate. Anyway, I'm actually pretty sure of my answers, which are as follows

    The best value of b is:

    [tex] b^2 = \frac{\hbar}{m\omega \sqrt{2}} [/tex]

    [tex] \langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs} [/tex]

    My question is, in a subsequent problem, Griffiths asks us to generalize this to a trial wavefunction of the form

    [tex] \psi(x) = \frac{A}{(x^2 + b^2)^n} [/tex]

    Huh? Did I miss something?! I had enough trouble with the integrals the first time around!! I don't think contour integration is going to help me. Am I missing some more obvious method for evaluating these integrals?
    Last edited: Mar 12, 2008
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  3. Mar 12, 2008 #2


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    That first integral is more easily done with the substitution x=tan y, and some trig.
  4. Mar 12, 2008 #3


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    hmm...you're right that does work rather nicely. Not too sure about the general case though.
  5. Mar 12, 2008 #4


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    note that

    [tex] \frac{1}{(x^2+b^2)^3} = -\frac{1}{2} \frac{d}{db^2} ( \frac{1}{(x^2+b^2)^2}) [/tex]

    You can generalize to an arbitrary n. Therefore, if you know how to integrate with 1/(b^2+x^2), you can integrate any power n. Your result will contain the nth derivative which you can probably find explicitly (it depends, what is the result of the integral with 1/(x^2+b^2) ?)
  6. Nov 7, 2009 #5
    Not to grave dig here too bad, but I don't think the OP came to the right solution and I didn't want someone else to be confused.

    The ground state energy of a harmonic oscillator is known to be

    [tex] E_{gs}=\frac{1}{2} \hbar \omega >
    \frac{1}{\sqrt{2}} \hbar \omega [/tex]

    The variational principle can only find a lower bound on the energy, meaning that the approximation must be greater than or equal to the actual energy of the system.

    Working out the problem myself, I got a little different value of b. I believe this is where the OP went wrong.
    Last edited: Nov 7, 2009
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