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Help solving Eigenvalue problem

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data

    solve the eigenvalue problem

    (-∞)x dx' (ψ(x' ) x' )=λψ(x)

    what values of the eigenvalue λ lead to square-integrable eigenfunctions?

    3. The attempt at a solution

    (-∞)xdx' (ψ(x' ) x' )=λψ(x)

    differentiate both sides to get

    ψ(x)x=λ d/dx ψ(x)

    ψ(x)x/λ= d/dx ψ(x)

    2xe x^2 =d/dx e x^2

    so ψ(x) = e x^2 and λ = 2

    but this is not square integrable so either this is incorrect or there are other solutions I am not seeing

    Can anyone help me find what I am missing?

    Thank you
  2. jcsd
  3. Sep 23, 2008 #2


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    You want to solve the differential equation, ψ(x)x/λ= d/dx ψ(x) Separate the variables. You should get a solution with a lambda in it. Figure out what values of lambda make it square integrable.
  4. Sep 23, 2008 #3
    I think there are other solutions.
    Take your equation,
    [tex] x\psi\left(x\right) = \lambda\frac{d}{dx}\psi\left(x\right)[/tex]
    and substitute [tex] \psi\left(x\right) = \exp\left(f\left(x\right)\right)[/tex] and see if you don't get an equation for [tex]f\left(x\right)[/tex] which has a solution that depends on [tex]\lambda[/tex].

    Also, don't forget your solution has to be finite at the lower endpoint of the integral ([tex]-\infty[/tex]) in the original problem statement.
  5. Sep 23, 2008 #4
    ok i separated variables, integrated and got

    ψ(x)=e1/(2λ) x2

    however I don't see how this can be square integrable since for any value of λ I can think of the integral of the square will equal infinity. I feel like I am missing something obvious here but I don't know what it is
  6. Sep 23, 2008 #5


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    Will it equal infinity even if lambda is negative?
  7. Sep 23, 2008 #6
    of course! im so embarrassed, thank you
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