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Time Independant Pertubation Theory - QM

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data

    An electron is confined to a 1 dimensional infinite well [itex]0 \leq x \leq L[/itex]
    Use lowest order pertubation theory to determine the shift in the second level due to a pertubation [itex]V(x) = -V_0 \frac{x}{L}[/itex] where Vo is small (0.1eV).


    2. Relevant equations

    [1]
    [itex]E_n \approx E_n^{(0)} + V_{nn}[/itex]

    [2]
    [itex]V_{nn} = \int_{-\infty}^{\infty} \psi_{n}^{(0) *} (x) V(x) \psi_{n}^{(0)} (x) dx[/itex]

    the following integral may be useful:

    [3]
    [itex]\int_{0}^{2\pi}\phi sin^2 \phi d\phi = \pi^2 [/itex]



    3. The attempt at a solution

    From [1] and the known result for E2 of an infinite well
    [itex]E_2 = \frac{4\hbar^2 \pi^2}{2mL^2} - \frac{2V_0}{L^2}\int_{0}^{L} x sin^2\left(\frac{2\pi x}{L}\right) dx[/itex]

    I cant see a substitution that will get it into the form in [3], anyone have any ideas?
    Also, is equation [1] a general result for the time independant case for first order pertubations?

    Thanks
     
  2. jcsd
  3. Dec 16, 2011 #2
    If i were to make the substitution:

    [itex]\phi = \frac{2\pi x}{L}[/itex]

    [itex]\frac{\phi}{x} = \frac{2\pi}{L}[/itex]

    does this imply that the limits of integration change from L to 2π ?
     
    Last edited: Dec 16, 2011
  4. Dec 16, 2011 #3

    vela

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    Yes. Simply plug in the limits for x to find the limits for ɸ.
     
  5. Dec 17, 2011 #4

    vela

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    No, it isn't. It applies when the energy eigenstates of the unperturbed Hamiltonian states are non-degenerate, like in this problem. You'll have to use a different approach for the degenerate case.
     
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