# 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. Oct 27 '19 at 5:12
• "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. 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.
• sorry, I am confused by what you said here: $$\forall v,w\in H:~~\langle C(v) | w\rangle~=~\overline{\langle v | C(w)\rangle}\tag{A}$$ This is indeed true for complex conjugation $C$. Roughly we have $$(v^*)^*w = (v^* w^*)^*\tag{B}$$ on the left and right sides equal to $vw$ But should a Hermitian operator say $O$ instead have a different condition? $$~\langle O(v) | w\rangle~=~ {\langle v |O(w)\rangle}?\tag{C}$$ Roughly we have $$(O v )^\dagger w = (v^\dagger O^\dagger w )= (v^\dagger O w )\tag{D}$$ so we have Hermitian condition $$O^\dagger=O ?\tag{E}$$ What am I missing? >< Sep 21 '20 at 22:03
• Hi @annie marie heart. Thanks for your feedback. Is your operator $O$ linear? The above is correct if $O$ is linear, but $C$ is anti-linear. Sep 21 '20 at 22:30
• You may spell out the reason for \overline in $\overline{\langle v | C(w)\rangle}$? or maybe give an example? Sep 21 '20 at 22:43