# Hermitian pickle

1. Sep 14, 2008

### mhazelm

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*> = $$\int$$phi* psi* dx

which would have to also equal

< C^t phi| psi> = < C phi| psi> = < phi* | psi> = $$\int$$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?

2. Sep 14, 2008

### Dick

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
3. Sep 14, 2008

### mhazelm

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.

4. Sep 14, 2008

### Dick

Right! And if lambda=1 what does that mean about the eigenfunction? Ditto for -1. Note all of this only works for lambda real.

5. Sep 14, 2008

### mhazelm

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:

6. Sep 15, 2008

### Dick

Yes, if psi is real the eigenvalue is 1. Is psi is purely imaginary then the eigenvalue is -1.

7. Sep 15, 2008

### mhazelm

that's fabulous! this problem has made me happy. There is hope for the operators that aren't Hermitian!

Thanks for the help!!

8. Sep 15, 2008

### Dick

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.