Let $\mathcal L$ be a orthomodular sub-$\sigma$-lattice of the lattice ${\rm L}(H)$ consisting of all closed subspaces of the separable Hilbert space $H$,
(precisely
- $\mathcal L$ is a set of closed linear subspaces of $H$, including $0:=\{0\}$ and $1:=H$,
- for members $L_0,L_1,...$ of $\mathcal L$, the intersection $\bigcap_{n\in\mathbb N}L_n$ is a member of $\mathcal L$,
- and for members $K,L$ of $\mathcal L$ such that $K\subset L$, the relative orthocomplement $K^{\perp}\cap L$ is a member of $\mathcal L$,)
and $t\mapsto U(t)=e^{itA}$ be a strongly continuous semigroup of unitary operators on $H$, then I expect the following result:
$U(t)(L)\in\mathcal L$ for any $L\in\mathcal L$ and for any $t$, if and only if the spectral measure of the self-adjoint operator $A$ maps any Borel set of $\mathbb R$ to the orthogonal projection whose image is a member of $\mathcal L$.
The above statement can be interpreted as the correspondence between continuous symmetries and observables of the quantum system $\mathcal L$. When the system is irreducible, it becomes the whole ${\rm L}(H)$ and every self-adjoint operator on $H$ is a observable. The correspondence is the well-known physics interpretation of Stone's theorem in this case.
It seems also easy when the superselection rule is discrete, i.e. $H$ is the Hilbert orthogonal sum of closed subspaces $H_0,H_1,...$ and $\mathcal L=\{\overline{\sum_nL_n}|\ L_n\in{\rm L}(H_n)\}$. What about the general case? Or maybe extra assumptions about $\mathcal L$ are needed?
I find the original question stupid for very simple counterexamples, like $\mathcal L=\{0,L,L^{\perp},1\}$ with $L$ a proper closed subspace and $U$ is a nontrivial unitary semigroup mapping $L$ to $L$ bijectively. So more constraints are necessary.
I would like to provide a proof for the case that $H=\hat{\bigoplus}_nH_n$ is the Hilbert sum of closed subspaces and $\mathcal L$ is the $\sigma$-lattice generated by $\bigcup_n{\rm L}(H_n)$, although it is not hard.
Let $t\mapsto U_t=e^{itA}$ be a strongly continuous semigroup of unitary operators, such that $U_t(L)\in\mathcal L$, $\forall L\in\mathcal L$. Then it is clear that $U_t(H_n)\subset H_n$, $\forall t$, by considering 1-dimensional subspaces of $H_n$, $\forall n$. Using Stone's theorem for every unitary semigroup $U|_{H_n}:[0,+\infty)\times H_n\to H_n$, it holds $A=\sum_nA_nP_n$, where $iA_n$ is the infinitesimal generator of $U|_{H_n}$ and $P_n$ is the orthogonal projection onto $H_n$, $\forall n$. So the spectral projection ${\rm P}^A=\sum_n{\rm P}^{A_n}P_n$ and its image ${\rm P}^A(E)H=\overline{\sum_n{\rm P}^{A_n}(E)H_n}\in\mathcal L$ for every Borel set $E\subset\mathbb R$.
Conversely, if ${\rm P}^A(E)$ is the orthogonal projection onto $\overline{\sum_nL_n}\in\mathcal L$ with $L_n\in{\rm L}(H_n)$, then every $v\in H_k$ is projected into $L_k$ by ${\rm P}^A(E)$ in fact, especially ${\rm P}^A(E)v\in H_k$. So the signed measure $\langle{\rm P}^Av,w\rangle$ vanishes for $w\in H_k^{\perp}$, then $\langle e^{itA}v,w\rangle=\int_{\mathbb R}e^{itx}d\langle{\rm P}^Av,w\rangle(x)=0$. Hence $e^{itA}H_k\subset H_k$, $\forall k$, which implies that $e^{itA}L\in\mathcal L$, $\forall L\in\mathcal L$, $\forall t$.