Hermitian Operators and Eigenvalues

AI Thread Summary
The discussion centers on the operator C, which transforms a function into its complex conjugate. Participants analyze whether C is Hermitian, concluding that it is not, as the integrals derived from the Hermiticity condition do not hold for all functions. They explore the implications of C being non-Hermitian, noting that eigenvalues can still be determined, with the modulus of the eigenvalue constrained to 1. The expected value of a squared operator is discussed, affirming that it remains positive due to the squaring of the constant. The conversation highlights confusion over the properties of Hermitian operators and their implications for eigenvalues and orthogonality of eigenfunctions.
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Homework Statement


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?


Homework Equations


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

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?
 
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can you run through the proof of why C is hermitian please?
 
I'm sorry, it's not hermitian. Got confused using bra-kets to prove it.

C(psi)=(psi*)

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.
 
yes. i also found it to be non-hermitian. but I am pretty sure its meant to be because otherwise the rest of the question doesn't make any sense as its all based around the properties of hermitian operators
 
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?
 
/Cpsi>=Ipsi*>
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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