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Hermitian Operators and Eigenvalues

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    C is an operator that changes a function to its complex conjugate
    a) Determine whether C is hermitian or not
    b) Find the eigenvalues of C
    c) Determine if eigenfunctions form a complete set and have orthogonality.
    d) Why is the expected value of a squared hermitian operator always positive?

    2. Relevant equations
    If C is hermitian, then <C(psi1)\(psi2)>=<(psi1)\C(psi2)>
    For eigenvalues: C(psi)=a(psi), where a is a constant

    3. The attempt at a solution
    I don't know even if I'm doing wrong but using the condition for hermiticity described above I get the integrals for the products (psi*)(psi*) and (psi)(psi) are equal. (Being the terms with "*" the complex conjugate)
    For d), if the operator is squared, then the constant is squared too, but how do I know "a" is not a complex constant?

    Ok, I guess I was a little desperate ad didn't check my results as I had to, from the beggining.
    Simply substituting C(psi) for (psi*) and (psi*) for C(psi) makes evident C is hermitian.
    Then, there's a theorem stating eigenvalues of hermitian operators are real, because average values are always real numbers. Squaring the constant makes for a positive number. Yet I'm not sure how to get the eigenvalues of b). How "far" can I go with this information?
    Last edited: Feb 4, 2009
  2. jcsd
  3. Feb 4, 2009 #2
    can you run through the proof of why C is hermitian please?
  4. Feb 4, 2009 #3
    I'm sorry, it's not hermitian. Got confused using bra-kets to prove it.


    If it were hermitian then the integral (psi*)C(psi) would be equal to that of (psi)(C(psi*)), but substituting C(psi) in the integrals yields (psi*)(psi*) is equal to (psi)(psi). I guess it would be an hermitian operator in case (psi) is a real function, but that would be a specific case. So, in general it's not hermitian.

    Sorry again for spreading my own confusion.
  5. Feb 5, 2009 #4
    yes. i also found it to be non-hermitian. but im pretty sure its meant to be because otherwise the rest of the question doesnt make any sense as its all based around the properties of hermitian operators
  6. Feb 5, 2009 #5
    Well, I think you can still get eigenvalues for non-hermitian operators. If you calculate the modulus of the eigenvalue and the eigenfunction (multiplying by its complex conjugate) you'll find you need the eigenvalue's modulus to be 1. So you can propose it to be exp(-i*alpha). It should make sense, because the eigenvalues of non-hermitian operators are not necessarily real numbers, and its magnitude is 1. So this alpha constant depends on the form of psi, and it should be chosen so that the operator turns psi into psi*.

    I can't figure out the answer about the complete set and their orthogonality. If C were hermitian there's no problem, the functions of different states are orthogonal with each other. But what can you tell for this case?
  7. Feb 15, 2009 #6
    multiply by <psi/

    <psi/Cpsi>=<psi/psi*> ------ 1
    <psi/Cpsi>=<psi/psi>* ------2

    can we say from 1 & 2 that it is Hermitain
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