Is complex conjugation operator Hermitian?

Question

I wonder whether the complex conjugation operator, defined on a wavefunction as

$$ C \psi(x) = \psi^*(x), $$

is Hermitian?

On one hand, its eigenvalues are not necessarily real. On the other hand, since $C$ is an antilinear operator, its adjoint must satisfy

$$ ⟨u|Cv⟩ = ⟨C^\dagger u|v⟩^*. $$ In $x$-space this would be

$$ \int dx u^*(x)Cv(x) = \bigg[\int dx [C^\dagger u(x)]^* v(x)\bigg]^*. $$ But since the LHS is just $\int dx u^*(x)v^*(x) = [\int dx u(x)v(x)]^*$, $C^\dagger$ must be $C$ itself.

So the question is whether $C$ is Hermitian, or, more generally, how do we define Hermiticity for antilinear operators?

Answer 1

1. There is no unique canonical notion of complex conjugation $C:H\to H$ of vectors in an abstract complex Hilbert space $H$. However, given a notion of complex conjugation $C:H\to H$, it is naturally to demand that it is an antiunitary map $$\forall v,w\in H:~~\langle C(v) | C(w)\rangle~=~\overline{\langle v | w\rangle}.\tag{1}$$ (This is e.g. the case for the Hilbert space $L^2(\mathbb{R}^3)$ equipped with the standard sesquilinear form $\langle \cdot | \cdot\rangle$ and complex conjugate.)

2. Since $C$ is an involution, eq. (1) is equivalent to the definition that the antilinear map $C$ is "Hermitian" $$\forall v,w\in H:~~\langle C(v) | w\rangle~=~\overline{\langle v | C(w)\rangle},\tag{2}$$ in the sense that $C^{\dagger}=C$, cf. e.g. this Phys.SE post.