In the Heisenberg picture, observables, rather than states, undergo unitary evolution. Can an observable turn into an $un$observable as a result of unitary time evolution? That is, If $U$ is a unitary transformation and $H$ a Hermitian operator representing an observable, can $UHU^{\dagger}$ end up being a non-Hermitian?

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    $\begingroup$ Set $Y=UHU^\dagger$ and try calculating $Y^\dagger$. You might need to use $(AB)^\dagger=B^\dagger A^\dagger$, $(U^\dagger)^\dagger=U$ and $H=H^\dagger$. What do you get? Is $Y^\dagger$ related to $Y$? $\endgroup$
    – Andrew
    Dec 21, 2022 at 1:27
  • $\begingroup$ I do not understand the downvotes really. $\endgroup$ Dec 21, 2022 at 2:21
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    $\begingroup$ @SongofPhysics : I downvoted because a) this is homework-like (in the sense that I wouldn't hesitate for a second to assign this question for homework) and the OP gave no indication of any attempt to solve it; b) it is therefore off topic; and c) off topic questions give a misleading impression of what this site is about, so I think it is a good idea to keep them off the front page, so I often downvote them. $\endgroup$
    – WillO
    Dec 21, 2022 at 3:07
  • $\begingroup$ @WillO, well I am relatively new here, but this doesn't seem like homework given some rather obvious misconceptions in the question itself. However, I do get your point, so thanks for clarifying! $\endgroup$ Dec 21, 2022 at 3:13
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    $\begingroup$ @SongofPhysics I wouldn't characterize it as homework as much as I would characterize it as almost completely trivial, given the definitions of the words OP is using, such as "Hermitian" and "Unitary." OP is using the words so I assume OP knows what they mean, which makes this question (almost completely) trivial. $\endgroup$
    – hft
    Dec 21, 2022 at 3:15

1 Answer 1


For starters, let me just say that the condition on an observable being Hermitian is sufficient, but not necessary. You may well have non-Hermitian observables, and these have been studied quite extensively.

Now, for any Hermitian operator $H: H = H^\dagger$, under a unitary evolution $U$, we must have $(UHU^\dagger)^\dagger = UH^\dagger U^\dagger = UHU^\dagger$, which is clearly Hermitian. In fact, the Schrödinger and Heisenberg pictures are unitarily equivalent, i.e, the pictures are equivalent upto the action of a unitary transformation (time evolution). Thus, whatever holds true about an observable in one picture, holds true in the other, besides of course artefacts resulting from the choice of picture (such as which among states/operators undergo time evolution).

For a deeper understanding about why (anti-)unitary transformations do what they do in QM, you could check out a more general result, i.e, the so-called Wigner's Theorem.

Hope this helps!


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