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First order perturbation theory, quantum physics

  1. Sep 19, 2008 #1
    I have an infinite potential well with length L. The first task was to calculate the eigenvalues and -functions for the energy of a particle in the well. The requirements were
    [tex]\psi(0, L) = 0[/tex] and there is no time-dependence.

    I've calculated:

    [tex]\hat{H}\psi(x) = E\psi(x)[/tex]

    [tex]E = \frac{(\pi\hbar n)^{2}}{2 m L^{2}}[/tex]

    [tex]\psi(x) = sin(\frac{\pi\n x}{L})[/tex]

    Now the question. We add a small potential "rectangle" V(x) at the center of the potential well. It has length a and height q, a << L. What are the new eigenvalues and -functions for the perturbed case? Im supposed to use first order perturbation theory.

    I've done like this:

    There's a formula saying:

    [tex]E_{n_p} = \int \overline{\psi_{n}} H' \psi_{n} dV =[/tex]
    [tex]\int \overline{\psi_{n}} H' \psi_{n} x^{2} dx[/tex]

    Which gives the new eigenvalues. I tried just using H' = V(x) = q at first. But the dimension of E didn't match. If you substitute V(x) (which is an energy) into the above equation you see that the dimension will be Energy*distance^4 after integration. So what to do with H' ?
    Last edited: Sep 19, 2008
  2. jcsd
  3. Sep 19, 2008 #2
    So, two things:
    First make sure that your eigenfunctions are properly normalized, e.g. in one dimension:
    [tex]\int \overline{\psi_n}\psi_n\,dx = 1[/tex].
    Also, what does this equation imply about the units of [tex]\psi[/tex]? (hint: the dx infinitessimal has units of length)

    Second, what is the volume element dV? If we are in one dimension then dV = dx. If we are in cartesian coordinates in 3 dimensions then dV = dxdydz and if we are in spherical coordinates then [tex]dV = r^2drd\theta d\phi[/tex] which appears to be the volume element you are using without the angular terms.

    Hope this helps
  4. Sep 21, 2008 #3
    Thanks! It helped. Normalizing psi and replacing x^2 dx with just dx did the trick :)
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