# Is complex conjugation operator Hermitian?

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?

• Are you sure the eigenvalues are not necessarily real? – Alex Oct 27 '19 at 4:46
• The weird thing is that any c-number a+ib is an "eigenstate" of the operator with an eigenvalue (a-ib)/(a+ib), which is not necessarily real. – fieryslug Oct 27 '19 at 5:12
• Looks to me like this belongs on Math SE – Time4Tea Oct 27 '19 at 11:58
• "more generally, how do we define Hermiticity for antilinear operators?" $-$ who says that we do? I wouldn't consider using such notation at all, and I would class its usage in any text in the literature as misleading unless the text was extremely clear about what they meant by the notation and how the notion interacts with the usual notion of hermiticity. – Emilio Pisanty Oct 27 '19 at 15:44

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.