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Perturbation Theory energy shift

  1. Jan 6, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm trying to calculate the energy shift given an electron in a 1D harmonic potential has a wavefunction

    [tex]\Psi_{0}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}exp\left(\frac{-m\omega x^{2}}{2\hbar}\right)[/tex]

    The shift in [tex]E_{0} = \frac{\hbar\omega}{2} = 2eV[/tex]

    due to the time independent perturbation
    [tex]V(x) = V_{0}cos\frac{\pi x}{L}[/tex]
    where [tex]V_{0}=1eV, L = 5x10^{-10}m[/tex].

    I'm told that [tex]\int^{\infty}_{-\infty}e^{-a^{2}x^{2}}cos(bx)dx = \frac{\sqrt{\pi}}{a}e^{-b^{2}/(4a^{2})}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    OK, here's what I have:

    [tex]\Delta E_{0} = V_{00} = \int^{\infty}_{-\infty}\Psi_{0}^{*}V\Psi_{0} dV[/tex]

    [tex]= V_{0}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2} \int^{\infty}_{-\infty}exp\left(\frac{-m\omega x^{2}}{\hbar}\right)cos\left(\frac{\pi x}{L}\right)dx[/tex]

    Which, using the integration provided leads to


    Now the exponential I have = 1 using the values provided, leading me to [tex]\Delta E_{0} = V_{0} = 1eV[/tex]

    HOWEVER, the answers I've been provided with show that:

    [tex]\Delta E_{0} = \sqrt{2} exp\left(\frac{\pi^{2}\hbar}{2L^{2}m\omega}\right)[/tex]

    The exponential still goes to one, leaving delta E at 1.41 eV, but I can't see for the life of me how it's been arrived at... Help!!!
  2. jcsd
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