# Hermitian Operator or not?

1. Jun 24, 2008

### 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 =/

2. Jun 24, 2008

### 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.

3. Jun 24, 2008

### Domnu

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!