Is the Operator C Hermitian and What are the Eigenfunctions and Eigenvalues?

[Domnu]

Problem
Consider the operator \hat{C} which satisfies the property that \hat{C} \phi (x) = \phi ^ * (x). Is \hat{C} Hermitian? What are the eigenfunctions and eigenvalues of \hat{C}?

Solution
We have

\hat{C} \phi = \phi ^ *
\iff \phi^* \hat{C}^\dagger = \phi

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

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

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

Could someone verify that the above is true? It seems too simple =/

 

[Dick]

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.

 

[Domnu]

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.

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

\langle a | C b \rangle = \langle a | b^* \rangle = a^* b^*
\langle Ca | b \rangle = \langle a^* | b \rangle = ab

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

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}

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