For many operators, their adjoint can be expressed as a function of other known operators, for example $$\hat{T}_a^\dagger = \hat{T}_{-a} \\ \hat{p}_x^\dagger = \hat{p}_x$$ where $\hat{T}_a \psi (x) = \psi (x+a)$ and $\hat{p}_x \psi(x)= -ih \frac{\partial \psi(x)}{\partial x}$.

But if we consider the operator $\hat{K}\psi (x) = \psi^* (x)$ (here $^*$ denotes complex conjugate) then, from definition,

$$ \langle \phi |\hat{K}^\dagger |\psi \rangle = \langle \psi |\hat{K} |\phi \rangle^* = \langle \psi |\phi^* \rangle^* = \langle \psi^* |\phi \rangle $$

Can it be expressed as $\langle \phi |\hat{A}|\psi \rangle$ for some operator $\hat{A}$? If not, why?

Also, what does this result imply? For example, is $\hat{K}$ Hermitian?

$$ \langle \phi |\hat{K}^\dagger |\psi \rangle = \langle \phi |\hat{K} |\psi \rangle \\ \therefore \langle \psi^* |\phi \rangle = \langle \phi |\psi^* \rangle$$

So it is Hermitian only for certain wavefunctions?


2 Answers 2


The operator $K$ on the Hilbert space $H$ is antilinear and satisfies $$(\eta,K^*\psi)=\overline{(\eta,K\psi)},\qquad\forall \eta,\psi\in H,$$ which is not what you would expect for self-adjointness.

You can consider the conjugate Hilbert space $\overline H$ and turn $T$ into a linear map by regarding it as $T:H\to\overline H$. If you then take $$(\eta,\psi)_{\overline H}:=\overline{(\eta,\psi)}_H$$ as the inner product on $\overline H$, the above identity becomes $$(\eta,K^*\psi)_{\overline H}=(\eta,K\psi)_H,\qquad\forall \eta,\psi\in H,$$ but since the operator is now between two "different" Hilbert spaces, this doesn't imply that $K$ is self-adjoint.

  • 1
    $\begingroup$ Just a very pedantic remark: it is customary for the adjoint of antilinear operators to take a different definition, so the adjoint is again antilinear. The adjusted definition is that for any $x,y\in H$, $\langle x, Ay\rangle=\overline{\langle A^{a-*} x,y\rangle}$ (where $a-*$ stands explicitly for this antilinear adjoint). In that case your first equation should become $\langle \eta, K^{a-*}\psi\rangle=\langle \eta, K\psi\rangle$. Therefore the conjugation is antilinearly self-adjoint, i.e. self-adjoint w.r.t. the proper antilinear definition of adjoint ;-) $\endgroup$
    – yuggib
    Commented Mar 30, 2015 at 15:12

$\hat{K} \Psi = \Psi^*$ is not a well defined operator at all. On the one hand it is clear that $\hat{K^2} = 1$ and $\hat{K}$ is therefore real. On the other hand by your definition $$ <\Psi| K |\Psi> = \int \Psi^{*2} d^3r \not\in \mathbb{R} $$ It is ill defined because it does not map from the space of the Kets into the space of the kets but instead from Ket to Bra space.

  • $\begingroup$ Sure it is a well-defined operator, it just happens that it is not linear but rather anti-linear. $\endgroup$ Commented Jan 4, 2022 at 1:50

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