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  Determining if a Hamiltonian has anti-unitary symmetries

+ 1 like - 0 dislike
92 views

I'm trying to understand the topological properties of different systems and where they fall on the Periodic Table of Topological Phases.  Such systems might include the Quantum Anomalous Hall Effect, Su-Schrieffer-Heeger model of trans-polyacetylene, Kitaev chain, or B-phase of He-3.

Once a resource gives me the right matrices, I can verify that indeed, there does indeed exist a U such that

\(U H U^{-1} = -H \qquad U U^{\dagger} = \mathbb{1}\)

(or respective for the T and C symmetries) .

But how do I go about showing that a system doesn't have any U such that this holds? Or go about finding such a U if I didn't read it in a paper?

My attempt using eigenvectors seems to be telling me that things I know aren't symmetric are.

asked Apr 1 in Theoretical Physics by anonymous [ revision history ]

1 Answer

+ 1 like - 0 dislike

You're right in the sense that only the eigenvectors of $H$ are relevant.

If you can arrange the set of eigenvectors in pairs such that the corresponding eigenvalues are opposite then you can build a unitary or anti-unitary $U$.

It is an equivalence,

What do you mean then by "things I know aren't symmetric" ?

answered Apr 1 by anonymous [ revision history ]
edited Apr 1

It is not quite an equivalence. Necessary and sufficient is that the nonzero eigenvalues come in opposite pairs, and a similar requirement (but more complex to formulate) for the continuous part of the spectrum if there is one.

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