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Setup:
Let $(\mathcal{H},\langle \cdot |\cdot\rangle)$ be a separable complex Hilbert space, and $A:D(A)\to \mathcal{H}$ be a densely defined symmetric unbounded operator in $\mathcal{H}$ (i.e. $D(A)\subset \mathcal{H}$ is dense and $\forall \,u,\,v\in D(A):\,\langle u|Av\rangle = \langle Au|v\rangle$). The operator $(A,D(A))$ is not self-adjoint, but possesses self-adjoint extensions (its deficiency indices are equal).
Next, let $T:[0,\infty)\to B(\mathcal{H})$ be a strongly continuous semigroup such that for any $u\in D(A)$ the mapping $[0,\infty)\ni s\mapsto T(s)u\in \mathcal{H} $ is an element of $C([0,\infty),\mathcal{H})\cap C^1((0,\infty),\mathcal{H})$, and satisfies the differential equation $$ \forall \,v\in D(A):\quad \left \langle \frac{d}{ds}T(s)u\,\middle|\, v \right\rangle+i\big\langle T(s)u\big |A v \big\rangle=0\,,\quad \forall s\in (0,\infty)\,. $$
Question:
Is the map $T$ the restriction to $[0,\infty)$ of a unitary group $U$ generated by some self-adjoint extension of $(A,D(A))$?