If an operator only commutes with Hamiltonian can we call that operator a symmetry, If the operator would not be a unitary or anti-unitary operator?


1 Answer 1


Yes. These go by the name of non-invertible symmetries. In particular, you can find operators that commute with the Hamiltonian, but have no inverse at all, e.g. if you apply them successively you might get something like $$U\times U = \mathbb{I} + V,$$ where $V$ also commutes with the Hamiltonian (and might or might not be unitary or anti-unitary). This immediately shows that there does not exist such a thing as $U^{-1}$ and hence it cannot be identified with $U^\dagger$ or $-U^\dagger$.

  • $\begingroup$ Can you explain why those operators deserve to be called symmetries ? $\endgroup$ Aug 25, 2022 at 14:45
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    $\begingroup$ @SolubleFish In general, lately, people tend to call symmetry any operator that is topological (which is equivalent to saying it commutes with the Hamiltonian). When there exists a Noether current, a charge operator is topological, and thus you call that a symmetry. When there isn't you associate topological operators to symmetries, extending the continuous case. These operators have the correct behaviour under RG flows that symmetries should have, e.g. if they commute with the operator that triggers the RG flow, they are preserved all the way in the RG flow. $\endgroup$ Aug 26, 2022 at 12:57
  • $\begingroup$ Dear ɪdɪət strəʊlə, thanks for your comment, is there any refrence to confirm your word? $\endgroup$ Aug 26, 2022 at 14:49
  • $\begingroup$ @MiladJangjan Sure, although it is in the context of quantum field theory, rather that quantum mechanics. A nice example is arxiv.org/abs/2205.05086, see e.g. the discussion towards the middle of page 2 up to page 3. $\endgroup$ Aug 26, 2022 at 16:00
  • $\begingroup$ Dear ɪdɪət strəʊlə, thanks for your explanation and your example. As a result, for a operator which is neither unitary nor anti-unitary one can call that operator as symmetry if operator commute with Hamiltonian. Also I should say the mentioned operator is not operator of chiral, particle-hole, time reversal and inversion symmetry. $\endgroup$ Aug 26, 2022 at 16:23

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