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Hermitian pickle

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Ok, here is another little pickle. I am trying to determine what the eigenfunctions and eigenvalues are for the operator C that is defined such that C phi(x) = phi*(x).
    Part a wants to know if this is a Hermitian operator. Parts b,c want eigenfunctions and eigenvalues.

    2. Relevant equations

    If an operator is Hermitian, then C=C^t, where ^t is the adjoint symbol.

    3. The attempt at a solution

    From parts b and c and the fact that the section preceding this problem is called "Properties of Hermitian Operators", we might expect C to be Hermitian. BUT - here's what I found:

    If we assume C is Hermitian, that would mean that

    < phi| C psi> = <phi | psi*> = [tex]\int[/tex]phi* psi* dx

    which would have to also equal

    < C^t phi| psi> = < C phi| psi> = < phi* | psi> = [tex]\int[/tex]phi psi dx

    and I would argue that in general these two integrals are strictly not the same!

    This would imply that C is not Hermitian, but if C isn't, then I haven't the foggiest idea how to speculate what the eigenfunctions and eigenvalues are. It seems like there's a mistake and it should be Hermitian so that I can just assert that it has real eigenvalues and orthogonal (normalizable) eigenfunctions.

    What's the deal? :rolleyes:
  2. jcsd
  3. Sep 14, 2008 #2


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    You are certainly correct that C is not hermitian. That doesn't mean it doesn't have eigenvalues. The eigenvalue equation is C(psi(x))=psi^*(x)=lambda*psi(x). Write that as (a(x)-b(x)*i)=lambda*(a(x)+b(x)*i) where a(x) and b(x) are real. Now I hope they are only asking for real eigenvalues. If that's the case what does that tell you about a and b and lambda? If they weren't careful to specify real eigenvalues then there are lots of them for psi(x)=constant. But they aren't very interesting.
    Last edited: Sep 14, 2008
  4. Sep 14, 2008 #3
    Hm, well, if we have two complex numbers set equal to each other I think that means that their real and imaginary parts must be equal. This would imply that lambda = 1 or lambda =-1.
  5. Sep 14, 2008 #4


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    Right! And if lambda=1 what does that mean about the eigenfunction? Ditto for -1. Note all of this only works for lambda real.
  6. Sep 14, 2008 #5
    not as sure this time... can we substitute in each eigenvalue and see what happens?

    If I use lambda =1, we get b(x) = 0, so the eigenfunction is related to a(x) (or the real part of phi).

    If we try lambda = -1, then a(x) = 0 and the eigenfunction is related to b(x) (imaginary part of phi).

    is that right? this is all very exciting.. I love it when things begin to work! :tongue:
  7. Sep 15, 2008 #6


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    Yes, if psi is real the eigenvalue is 1. Is psi is purely imaginary then the eigenvalue is -1.
  8. Sep 15, 2008 #7
    that's fabulous! this problem has made me happy. There is hope for the operators that aren't Hermitian!

    Thanks for the help!! :smile:
  9. Sep 15, 2008 #8


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    If C had only real eigenvalues it would be hermitian and we know it's not. Can you also characterize the complex eigenvalues and eigenfunctions? Try writing the function in polar form psi(x)=r(x)*exp(i*theta(x)). I think we actually should have gone that way to begin with, come to think of it.
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