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Expansion of a wave fuction in energy eigenfunctions

  1. Sep 9, 2011 #1
    Hi, suppose we have an unidimensional finite square well potential and we want to expand an arbitrary wave function in terms of energy eigenfunctions but considering the possibility of bounded (discrete) AND unbounded (continue) states. How do you express the expansion?. The problem is that each set of eigenfuntions (correponding to bounded or unbounded states) is complete so i think i just can't add one set of eigenfunctions (with their coefficients) to the other because there would be no way of finding the coefficients.
    To motivate this suposse we have an uncertainty in the energy (dE) such that <E>+dE > Vo and <E>-dE < Vo, where V0 is the depth of the well.

    Last edited: Sep 9, 2011
  2. jcsd
  3. Sep 9, 2011 #2


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    Why should both sets of eigenfunctions be complete?
  4. Sep 9, 2011 #3
    Every bound state is orthogonal to every unbound state (they must be, since they have different eigenvalues of the Hermitian operator H). So neither set of eigenfunctions is complete of itself. A general wave function can be expressed as

    [tex]\psi(x) = \sum_n c_n \psi_n(x) + \int_0^\infty c(E)\psi_E(x) dE[/tex]

    for some set of coefficients c_n and some function c(E). Here the sum on n is over the finite number of bound states psi_n and the integral is over the unbound states, where psi_E is the eigenfunction for energy E > 0.
  5. Sep 9, 2011 #4
    Ok, I get it now but I have another question: when you write the unbounded eigenfunctions how should I obtain them? I mean, how should i consider incident particles?.

  6. Sep 9, 2011 #5

    Unbound states is superposition of both coming and going waves. Because it is unbound, it spreads out to infinity.

  7. Sep 10, 2011 #6
    Thanks, but I don't know if I understood that. I think you say that when writing a specific eigenfunction you must solve the well with coming waves from infinity and -infinity at the same time, is that ok?.
  8. Sep 10, 2011 #7
    Hi, zalook

    For example a typical eigenfunction of unbound state out side the well is cos kx = (e^ikx + e^-ikx) /2 which shows superposition of wave from left to right e^ikx k>0 and wave from right to left e^-ikx k>0. Time does not appear in formula because it is stationary state or energy eigenstate.

  9. Sep 10, 2011 #8
    Just solve the time independent Schrodinger equation with an energy E > 0. When you do this (it may be necessary to use some sort of approximation) you will find that as x goes to plus or minus infinity, the solution becomes a superposition of incoming and outgoing free particle wave functions.

    I think most intro QM books should have an introductory treatment of scattering in one dimension that shows how to use this approach to determine what happens to an unbound particle coming from infinity incident on your potential.
  10. Sep 10, 2011 #9

    Ken G

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    Perhaps another way to sum that up is, you need some kind of boundary condition to get an actual solution, if you include particles propagating from infinity.
  11. Sep 10, 2011 #10
    Ok, thanks. I understand what you are saying but in all the cases I have seen scattering treated they took only ONE incident particle, so they obtain superposition only in its side of the barrier (for example). My question is how should I take incident particle/s to get the eigenfunctions?. I think I must take incident on both sides (barrier in (0,a): (A e^ikx + B e^-ikx) for x<=0, and (C e^ikx + D e^-ikx) for x>=a).
    By the way, i know that in order to normalize the unbounded wavefunctions I would need to take a special care, but I'm not interested in that now (thanks anyway).
  12. Sep 10, 2011 #11
    Hi, zalook.

    Let me ask you question how should you put particle inside the well to get eigenfunction of bounded states.
    Preparing a particle inside the well for bounded state and preparing a particle overriding the well for unbinding states do not differ so much, do they?

  13. Sep 10, 2011 #12
    I know that, but I'm not sure if what I have said (barrier or well, it doesn't matter) is ok. I only want someone to confirm that. Thanks.
  14. Sep 10, 2011 #13
    Hi, zalook.

    As for a unbinding state, ONE particle exists in all the region, i.e. RHS, LHS and inside of the well with both coming and leaving motion.

    I get your question right?

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