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'Symmetry argument' for eigenstate superposition

  1. Dec 27, 2015 #1
    The problem statement, all variables and given/known data
    For an infinite potential well of length [0 ; L], I am asked to write the following function ##\Psi## (at t=0) as a superposition of eigenstates (##\psi_n##):
    $$\Psi (x, t=0)=Ax(L-x) $$
    for ## 0<x<L##, and ##0## everywhere else.

    The attempt at a solution
    I have first normalized this function, which gave me a certain value for A.
    Next, I used the following reasoning:
    $$\Psi> = \sum_n \psi_n><\psi_n . \Psi> = \sum_n <\psi_n . \Psi> . \psi_n> $$
    Where:
    $$<\psi_n . \Psi> = \int_0^L \psi_n(x)^*.\Psi(x) dx$$
    I have calculated these integrals to be:
    $$C_n.(1-cos(n.\pi ))$$ where ##C_n## is a constant combination of L's, n's and numbers, not really relevant to my question I think.


    Here I see that only odd (uneven) eigenstates will contribute to the superposition that makes up the function ##\Psi##. Now I am asked to support this with a symmetry argument, but I don't know how this should be done.
    I suspected I had to prove ##\Psi## is an uneven function, but it isn't as far as I can see. In class my teacher said something vague about the "completeness" of the Hilbertspace that prohibits even eigenstates to contribute to uneven wavefunctions, but I'm not too sure what that means. I'm not completely sure my integral calculation is correct, but any general explanation about this sort of symmetry argument would be much appreciated.

    Thanks in advance, Lennart :)
     
  2. jcsd
  3. Dec 27, 2015 #2

    Orodruin

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    You are probably looking at odd and even wrt x = 0, this is not what the potential is symmetric around ...
     
  4. Dec 28, 2015 #3
    Oh I see! ##\Psi## is indeed symmetric/even around ##x=L/2##. So is it then a valid argument to say that only even eigenstates contribute to even functions?
     
  5. Dec 28, 2015 #4

    Orodruin

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    Yes, by the symmetry the inner product with the odd states is zero.
     
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