1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Hermitian pickle
  1. Hermitian conjugate (Replies: 6)

  2. Hermitian operator (Replies: 10)

  3. Hermitian Operators (Replies: 5)

  4. Hermitian Operators (Replies: 15)