Perturbation of the simple harmonic oscillator

  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. nrqed

    nrqed 3,048
    Science Advisor
    Homework Helper

    What makes you think this is not right?
    Have you tried to compute the integral? It's a standard one.
  4. [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. CompuChip

    CompuChip 4,299
    Science Advisor
    Homework Helper

    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. Ok.

    Thank you both!
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