Complex extension of the Stone - von Neumann Theorem?

Question

I have two questions related to an extension of the Stone - von Neumann Theorem:

(1) Are there unitary groups with uncountably many elements indexed over the complex plane?

(2) Can the Stone - von Neumann Theorem be formulated over $\mathbb {C}$ instead of over $\mathbb {R}$, to establish a one-to-one correspondence between self-adjoint operators on a Hilbert space $\mathcal {H}$ and one-parameter families of continuous unitary operators $\{U_{z}\}_{z\in \mathbb {C}}$:

(1)$\forall z_{o}\in \mathbb {C} ,\ \psi \in {\mathcal {H}}:\ \lim _{z\to z_o}U_{z}(\psi )=U_{z_o}(\psi), $

(2)$\forall z_1,z_2\in \mathbb {C} :\ U_{z_1+z_2}=U_{z_1}U_{z_2}.$

I'm interested whether its possible to derive a one-to-one correspondence between a (complex) parameter strongly continuous unitary groups of operators $U :\mathcal{H}\to\mathcal{H}$ and linear operators (not necessarily self-adjoint/Hermitian!) $A_U :\mathcal{H}\to\mathcal{H}$ by

$$A_U\ \psi = \lim_{|z|\to 0}{\frac{U(z)\psi - \psi}{iz}}\\ U_A(z)=e^{izA}\ ,$$

where the domain of the operator $A$ is $D(A) = \left \{\psi\in\mathcal{H}: \lim_{|z|\to 0}{\frac{U(z)\psi-\psi}{iz}}\ {\text{exists}}\right \}$.

And conversely, for any a given operator $A_o$, there exists a (complex-indexed) strongly continuous unitary group $\{U_{A_o}(z)\}_{z\in\mathbb{C}}$, such that this relation between the operator $A_o$ and the unitary group $U_{A_o}$ is satisfied on the domain of $A_o$.

Answer 1

(I do not understand well where the Stone-Von Neumann Theorem enters the issue, maybe the right theorem is the Stone Theorem, not the SvN one?)

Suppose that a strongly continuous one-parameter group of unitaries $\mathbb{C} \ni z \mapsto U_z$ exists in the complex Hilbert space $\cal H$ with the properties you described. In particular $A_U : D(A_U) \to {\cal H}$ exists with domain $D(A_U)\subset {\cal H}$ defined as you indicated.

Take $\phi \in \mathbb{R}$. Then $\mathbb{R} \ni t \mapsto U_{e^{i\phi}t}$ is a strongly continuous one-parameter group of unitaries in standard sense and every $\psi\in D(A_U)$, per definition, belongs to the domain of its selfadjoint generator $A_\phi$ according to the Stone theorem. As a consequence, computing the $t$-derivative of both sides of $$U_{e^{i\phi}t} \psi = e^{itA_\phi}\psi$$ we have $$\tag{1} e^{i\phi}A_U\psi = A_\phi \psi\:, \quad \forall \psi \in D(A_U)\:.$$
In particular $$e^{-i\phi'} \langle \psi, A_{\phi'}\psi\rangle = \langle \psi, A_U\psi \rangle = e^{-i\phi}\langle \psi,A_\phi\psi\rangle \:.$$ In other words $$\langle \psi, A_{\phi'}\psi\rangle = e^{i(\phi'-\phi)} \langle \psi, A_{\phi}\psi\rangle\:.$$ The two operators are selfadjoint and thus $\langle \psi, A_{\phi'}\psi\rangle, \langle \psi, A_{\phi}\psi\rangle \in \mathbb{R}$. The found identity is therefore true only if $$\langle \psi, A_{\phi}\psi\rangle =0\:, \quad \forall \psi \in D(A_U)\:, \quad \forall \phi \in \mathbb{R}\:.$$ By polarization we also have $$\langle \psi', A_{\phi}\psi\rangle =0\:, \quad \forall \psi, \psi' \in D(A_U)\:, \quad \forall \phi\in \mathbb{R}\:.$$ At this point (1) yields $$\langle \psi', A_U\psi\rangle =0\:, \quad \forall \psi, \psi' \in D(A_U)\:, \quad \forall \phi \in \mathbb{R}\:.$$ If $D(A_U)$ is dense, this identity immediately proves that $A_U=0$.

I should check but I am very confident that the density property of $D(A_U)$ can be proved exactly as in the case of the proof of the standard Stone theorem (using the Garding vectors on the additive group $\mathbb{C}$).

With this caveat (but it is easy to check), the result is disappointing: if a representation as the one you defined exists, the generator is always trivial $A_U=0$. Standard properties of $C_0$ semigroups imply that the representation $U$ itself is trivial as well: $U_z=I$ for all $z\in \mathbb{C}$.

I stress that the overall obstruction is the request that all $U_z$ of the group representation are unitaries if $z\in \mathbb{C}$.