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Consider a quantum system with Hilbert space $\mathcal{H}$ and Hamiltonian $H$. Let $G$ be a Lie group and $U$ a unitary representation of $G$ on $H$. What are the most general conditions that $H$, $G$ and $U$ must fulfil so that $G$ can be considered a symmetry of the system?

I know, for example, that $G$ is a symmetry if $$ [U(g), H] = 0\quad\forall g\in G\quad, $$ but this is not the most general condition. In a relativistic QFT the representations of Lorentz boosts do not commute with the Hamiltonian. In the case of Lorentz symmetry the criterium seems to be that $H$ must transform as the time component of a four-vector, as explained, for instance, in https://physics.stackexchange.com/a/568141/197448. But how does this generalise to arbitrary groups? Is there a broader notion of symmetry in QM for which both Lorentz symmetry in QFT and $[U(g), H]=0$ are just special examples?

For example, would it be valid to say that the system is symmetric under $G$ if $H$ is a linear combination of the generators of the representation $U$ (so that it transforms as a vector in the adjoint representation of $G$)? This would cover the case of Lorentz boosts in QFT if we take $G$ to be the Poincare group. It would also cover the cases where $H$ commutes with $U(g)$ if we let $G$ be the product of the time translations and some other group of transformations which do not involve time.

Qmechanic
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Martin Wiebusch
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    Related: (1) Symmetry transformations on a quantum system; Definitions ... (2) References for prerequisite material for understanding papers on Generalized Global Symmetries – Chiral Anomaly Jul 03 '21 at 16:19
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    While perhaps not useful in practice, I would say you can call $G$ a symmetry if there exists some $U: G \rightarrow \mathrm{End}(\mathcal{H})$ which is both unitary and faithful. – Connor Behan Jul 03 '21 at 17:46
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    @ConnorBehan That's what I would call a faithful representation of $G$. But to decide if $G$ is a symmetry the Hamiltonian $H$ should come into play somehow, right? By your definition a system of a particle in a potential would be rotationally symmetric even if the potential is not rotationally symmetric. – Martin Wiebusch Jul 03 '21 at 18:36
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    It's what everyone would call a faithful representation of $G$ :). But the idea of symmetry is general enough that it shouldn't depend on locality or a Hamiltonian. The connection with conserved quantities is where I think $H$ needs to come into play. – Connor Behan Jul 03 '21 at 19:12
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    And I'm not seeing what you mean about rotational symmetry. We can consider a particle in a 2D box where a basis for $\mathcal{H}$ is labelled by integers $n_1$ and $n_2$ with $n_i$ and $-n_i$ identified. In that case, independent sign flips and $n_1 \leftrightarrow n_2$ represent the dihedral group faithfully as expected. But what is a permutation of these states that respects the group law of $SO(2)$? – Connor Behan Jul 03 '21 at 19:15
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    @ConnorBehan So, I have a 3D particle in a potential $V(\vec r)$ which is not rotationally symmetric. The Hilbert space is $L^2(\mathbb{R}^3)$ and the rotation group is faithfully represented by $U(R)\psi(\vec r)=\psi(R^{-1}\vec r)$. By your definition of symmetry this system is symmetric under rotations. Are you claiming that this is the case? – Martin Wiebusch Jul 03 '21 at 21:26
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    You're right. I was trying to take a time agnostic viewpoint where there is no Heisenberg picture. And that would make the Hilbert space consist of $L^2(\mathbb{R}^3)$ valued functions of $t$. This space would be invariant under rotations and time translations and whether or not they commute would only be known after we specify how the latter acts (i.e. after we specify $V(\vec{r})$). – Connor Behan Jul 04 '21 at 00:54
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    So I agree with you now. The definition I gave was overly broad and trying to "improve" it by restricting to transformations that commute with $H$ would just take us in a circle. I'll watch this question to see if someone gives a precise answer which involves choosing a certain set of "worldsheet related" transformations to be privileged, but that's just a guess. – Connor Behan Jul 04 '21 at 01:00
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    Did you see the paper that Chiral Anomaly linked? – Nihar Karve Jul 11 '21 at 10:00
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    Yes, I've been reading it for the last few days. (Thanks @ChiralAnomaly) The paper gives a very clean definition of space-time and internal symmetries in the Hamiltonian formalism. I haven't (yet?) found the answer to my question in it since I'm looking for a definition that does not depend on structures other than the Hilbert space and the Hamiltonian. It has helped to clear things up a lot though. – Martin Wiebusch Jul 12 '21 at 11:56

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