# Eigenvalues of antiunitary operators

I have sometimes come across the statement that antiunitary operators have no eigenvalues. For example, on page 34 in the book "Topological Insulators and Topological Superconductors" by Bernevig and Hughes, it is stated that

The preceding ( $$T i T^{-1} = -i$$ ) makes it clear that the time-reversal operator $$T$$ must be proportional to the operator of complex conjugation. Such operators are called antiunitary and, unlike unitary (sic.) operators, do not have eigenvalues.

I do not understand this statement. For example, consider the antiunitary operator $$\sigma_x K$$ where $$K$$ corresponds to complex conjugation and $$\sigma_x$$ is a Pauli matrix, then

$$\begin{equation} \sigma_x K \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix} = \pm \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix} \end{equation}$$

Naively, I would therefore conclude that $$\left( 1, \pm 1 \right)^T$$ is an "eigenstate" of $$\sigma_x K$$ with "eigenvalue" $$\pm 1$$. If we multiply this eigenstate by a phase $$e^{i\phi}$$, it remains an eigenstate but its "eigenvalue" changes by $$e^{-2i\phi}$$. Hence, it seems that one can have eigenstates of an antiunitary operator but their eigenvalue is not a single scalar. Instead the eigenvalue corresponds to a circle.

Edit

I have found this paper which deals with the subject, but seems to contradict the original statement: https://arxiv.org/abs/1507.06545

If we consider the time-reversal operator again, since for spinless particles $$T^2=1$$, there exist eigenstates of $$T$$ without unique eigenvalues. However, for spin 1/2 particles, $$T^2 = -1$$ and there exist no eigenstates (see the answer of CosmasZachos).

• Ellipticity is not a virtue on this cite. I read your question several times, but it lacked the background and context to allow the reader to guess where you were coming from, and would certainly profit from specifics referred to your belated reference. Apologies if you read it as idle snarkiness, but… Jul 20, 2021 at 11:55
• @CosmasZachos Thank you for your comment. I will try to add more context to my question. Keep in mind that I am not a mathematical physicist and what might be obvious to you is not at all obvious to me. For example, I have no idea what you mean with ellipticity in this context. I am guessing the answer to my question is most likely completely trivial to you. Jul 20, 2021 at 12:34
• I meant ellipticity as the heavy-handed application of ellipsis. The average reader, like me, has no access to the book whose language puzzles you. It is self-evident that IK has eigenvalues on a circle for complex numbers, but I wager that's not what the authors have in mind; they may have folded it into their framework. Jul 20, 2021 at 14:06

But the counterexample you chose is of the $$\vartheta ^2={\mathbb I}$$ variety, and so $$\vartheta$$ does have the obvious eigenvectors: that's the point of Proposition 2.3 , corollary 2.4 !
Check your $$\sigma_x K \sigma_x K ={\mathbb I},$$ in sharp contrast to $$i\sigma_y K i\sigma_y K =-{\mathbb I}.$$ The first has eigenvectors with $$\vartheta^2$$ having a positive semidefinite spectrum, but the second doesn't.
• I did read the arXiv version of the linked paper (see edited answer) and the section you refer to. However, I could not reconcile this with the original statement "antiunitary operators have no eigenvalues". I guess it is simply very imprecise and only truly holds for the case $(UK)^2=-1$ (e.g. the time-reversal operator for spin 1/2 particles). Indeed, one finds a contradiction $|\lambda|^2 = -1$ where $\lambda$ is the supposed eigenvalue. Jul 20, 2021 at 18:13