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Hermitian Operator or not?

  1. Jun 24, 2008 #1
    Consider the operator [tex]\hat{C}[/tex] which satisfies the property that [tex]\hat{C} \phi (x) = \phi ^ * (x)[/tex]. Is [tex]\hat{C}[/tex] Hermitian? What are the eigenfunctions and eigenvalues of [tex]\hat{C}[/tex]?

    We have

    [tex]\hat{C} \phi = \phi ^ *[/tex]
    [tex]\iff \phi^* \hat{C}^\dagger = \phi [/tex]

    Substituting back into the first equation,
    [tex]\hat{C} (\phi^* \hat{C}^\dagger) = \phi ^ * = \hat{C} \phi[/tex]
    [tex]\iff (\hat{C} \phi)(\hat{C} - I) = 0[/tex]

    Now, we know that [tex]\hat{C} \phi \neq 0[/tex], since (we assume that) [tex]\phi[/tex] isn't zero... if [tex]\hat{C} = 0,[/tex] then the original property that [tex]\hat{C}[/tex] satisfied couldn't possibly be true. Thus, we have that [tex]\hat{C} = I[/tex], which is clearly Hermitian.

    Since [tex]\hat{C}[/tex] is just the identity operator, we know that all functions are the eigenfunctions of [tex]\hat{C}[/tex] and the only eigenvalue of [tex]\hat{C}[/tex] is 1.

    Could someone verify that the above is true? It seems too simple =/
  2. jcsd
  3. Jun 24, 2008 #2


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    C is OBVIOUSLY not the identity. It's the complex conjugation operator. It takes a function into it's complex conjugate. The identity takes a function into itself. Are you sure you can't think of any eigenstates of C?? They are pretty easy. Then write down the definition of Hermitian without trying to think of C as a matrix.
  4. Jun 24, 2008 #3
    Heh, whoops... I see that all real functions are eigenstates for C. And C is not Hermitian... my calculation messed up in the factorization, which is incorrect. It's not Hermitian, because

    [tex]\langle a | C b \rangle = \langle a | b^* \rangle = a^* b^*[/tex]
    [tex]\langle Ca | b \rangle = \langle a^* | b \rangle = ab[/tex]

    We can see that all forms of [tex]e^{i\alpha}[/tex] are eigenvalues, where [tex]\alpha[/tex] is some real number. This is because

    [tex]C \phi = p \phi \iff \phi^* = p \phi \iff \phi = p^* \phi^* \iff \phi^* = pp^* \phi^* \iff p^*p = 1 \iff p = e^{i \alpha}[/tex]

    for some [tex]\alpha \in \mathbb{R}[/tex]. Thanks a bunch for the help and motivation to solve the problem!

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