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First-order perturbation for a simple harmonic potential well

  1. Dec 17, 2014 #1
    1. The problem statement, all variables and given/known data
    The ground state of the wavefunction for an electron in a simple one-dimensional harmonic potential well is

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

    By employing first-order perturbation theory calculate the energy shift [tex]E_{0}= \frac{\hbar \omega}{2} = 2eV[/tex] caused by the time-independant perturbation.

    [tex]V(x)= V_{0} x^{3}[/tex]

    with V0= 2eV and L= 5.10^-10 m

    2. Relevant equations
    E = E0 + V00

    3. The attempt at a solution
    The problem I'm having is that no integrals are given for this question (it's a past exam question). If I follow the process of finding V00

    [tex]V_{00}= \left \langle \Psi _{0}| V_{0} x^{3} | \Psi _{0} \right \rangle [/tex]

    [tex]
    \int_{0}^{L} (\frac{m\omega }{\pi \hbar} \right ))^{1/2} x^{3} exp (-\frac{m\omega x^{2}}{\hbar})[/tex]

    if I integrate by parts between 0 and L I get an integral of exp (-y^2) which ordinarily between minus infinity and infinity would be root pi, but between 0 and L gives ERF. Any Ideas of what to do in this situation?
     
  2. jcsd
  3. Dec 17, 2014 #2

    DrClaude

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    Staff: Mentor

    You should be able to get that by integrating by parts, provided you choose the correct parts! What did you use?
     
  4. Dec 17, 2014 #3
    Using [tex]\int u dv = uv - \int vdu[/tex]

    I set u = x^3 in order to reduce it to a lower power of x (and continue to so with the next by parts integrals) and dv = exp (-ax^2) where a is a constant.

    The problem is when I integrate with 0 to L limits using substitution of y^2, it comes out as ERF.

    http://www.integral-calculator.com/#expr=e^(-y^2)&intvar=y&ubound=l&lbound=0

    Thanks for the reply.
     
  5. Dec 17, 2014 #4

    DrClaude

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    Staff: Mentor

    Try ##u = x^2##.
     
  6. Dec 17, 2014 #5
    thanks, yes thats what I did but this gives an exponential of a square for dv, which when I integrate with 0 to L limits using substitution, namely exp (-y^2), it comes out as ERF.
     
  7. Dec 17, 2014 #6

    TeethWhitener

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    Gold Member

  8. Dec 17, 2014 #7

    DrClaude

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    If you set ##u = x^2##, then ##dv = x e^{-a x^2} dx##, which can be easily integrated.
     
  9. Dec 17, 2014 #8
    ahhhh, thats a clever little trick! i see now, many thanks for your help!!

    it's okay, I've got it now, but thanks!
     
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