1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted