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Homework Help: Perturbation of the simple harmonic oscillator

  1. Dec 3, 2007 #1
    [SOLVED] Perturbation of the simple harmonic oscillator

    1. The problem statement, all variables and given/known data
    An additional term V0e-ax2 is added to the potential of the simple harmonic oscillator (V and a are constants, V is small, a>0). Calculate the first-order correction of the ground state. How does the correction change when a gets bigger?

    2. Relevant equations

    3. The attempt at a solution
    [tex]\alpha=\frac{m\omega}{\hbar}, E_0^1=\int ^{\infty}_{-\infty} (\frac{\alpha}{\pi})^{1/4}e^{-\alpha x^2/2}V_0e^{-ax^2}(\frac{\alpha}{\pi})^{1/4}e^{-\alpha x^2/2}dx=(\frac{\alpha}{\pi})^{1/2}V_0\int ^{\infty}_{-\infty} e^{(-\alpha -a)x^2}dx[/tex]
    So I suppose this is not what is wanted.
  2. jcsd
  3. Dec 3, 2007 #2


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    What makes you think this is not right?
    Have you tried to compute the integral? It's a standard one.
  4. Dec 3, 2007 #3
    [tex]\int e^{-\xi x^2}=\frac{\sqrt{\pi}\ erf(\sqrt{\xi}x)}{2\sqrt{\xi}}\ \ \ ?[/tex]
    I've never encountered an error function before in any homework problem, so I automatically assumed that I had done something wrong.
  5. Dec 3, 2007 #4


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    That's right, the indefinite integral contains an erf.
    But you have more information: you know the boundary conditions and (you should have) [itex]\xi > 0[/itex]. Using
    [tex]\lim_{x \to \pm \infty} \operatorname{erf}(x) = \pm 1[/tex]
    you can calculate it, and in fact it is just a Gaussian integral,
    [tex]\int_{-\infty}^\infty e^{-\xi x^2} dx = \sqrt{\frac{\pi}{\xi}}[/tex]
    Remember this -- it's ubiquitous in physics (at least, every sort of physics that has to do with any sort of statistics, amongst which QM, thermal physics, QFT, SFT).
    Last edited: Dec 3, 2007
  6. Dec 3, 2007 #5

    Thank you both!
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