1. Not finding help here? Sign up for a free 30min 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!

Variational Principle

  1. Mar 12, 2008 #1

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
  2. jcsd
  3. Mar 12, 2008 #2

    pam

    User Avatar

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

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    kdv

    User Avatar

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Variational Principle
  1. Variational Principle (Replies: 3)

  2. Variational Principle (Replies: 7)

Loading...