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Perturbation Theory

  1. Dec 16, 2006 #1
    Please help me try to understand this problem. It deals with the quantum-confined Stark effect in nanoparticles.

    For odd n, n = 1, 3, 5, ...
    [tex]\psi_{n}(x) = \sqrt{\frac{2}{a}} \cos (\frac{n \pi x}{a})[/tex]

    and for even n = 2, 4, 6, ...
    [tex]\psi_{n}(x) = \sqrt{\frac{2}{a}} \sin (\frac{n \pi x}{a})[/tex]

    and the zeroth order energy levels are

    [tex]E_{n} = \frac{h^2 \pi^2 n^2}{2ma^2}[/tex]

    The external field pertubation, H' = -qFx , where q is the charge and F is the applied electric field strength.

    Now here's my work for the first order correction to the energy levels.

    For odd n:
    [tex]E_{n} = < \sqrt{\frac{2}{a}} \cos (\frac{n \pi x}{a})| H' | \sqrt{\frac{2}{a}} \cos (\frac{n \pi x}{a})> = 0[/tex]

    For even n, I still get 0 for the first order correction. I just know that isn't right, and I think I know why:

    Am I treating H' = -qFx correctly by assuming q and F are constants and x as the operator?

    Thanks for the help. :shy:
    Last edited: Dec 16, 2006
  2. jcsd
  3. Dec 17, 2006 #2


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    I'm actually not surprised you get 0 in first order. There is a mathematical reason, and a physical reason. The mathematical reason is that all the stationary solutions of the unperturbed system (which are the ones you are considering) are symmetrical, or anti-symmetrical, which means that after squaring, they are symmetrical. Now, your perturbation term is anti-symmetrical, so the overall product is anti-symmetrical, which gives you 0 after integration.

    The physical (somewhat handwaving) reason is this: for stationary states, there is a symmetry around the origin. If you calculate the probability density for the particle to be somewhere, then you have just as much chance to be at -x than you have to be at +x. The "center of gravity" of the probability density of your particle, in a stationary solution, is at x=0. Now, for x=0, your perturbation is 0. So in first order, your perturbation does not alter "the center of gravity". It will be due to higher-order distributions, which will slightly deform (and not displace) the wavefunctions, that you will get effects - but that's something that is not seen in first-order perturbation.
    You can get a better grasp for this as follows: imagine your perturbation was not Fx but rather F(x-x_0). Then there would be a net displacement of the center of gravity, and you would get a first-order effect, of the order of F x_0.
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