I apologize if this question ends up being too basic, but I couldn't find the answer by myself and I'm sure you can help me.
In quantum mechanics, one takes a complex Hilbert space $\mathscr{H}$ as the state space, i.e. elements $\psi \in \mathscr{H}$ of norm one describe quantum systems. Most of the time, it is enough to take $\mathscr{H} = L^{2}(\mathbb{R}^{3};\mathbb{C})$, so let's focus on this case. A state is then a complex-valued and square-integrable function $\psi = \psi({\bf{x}})$. Now, my problem arises when time-dependence is considered. The time-evolution of a state $\psi$ is a one-parameter unitary group $U(t)$ which, by Stone's Theorem, can always be written as $U(t) = e^{-itH}$, where $H$ is the Hamiltonian of the system. The time-evolved state is then: $$\psi(t,{\bf{x}}) = U(t)\psi({\bf{x}}) \tag{1}\label{1}$$ Now, for each $t \in \mathbb{R}^{+} := [0,+\infty)$, equation (\ref{1}) makes sense so the time-evolved state is a function of both $t$ and ${\bf{x}}$ so it looks like an element of $L^{2}(\mathbb{R}^{+}\times \mathbb{R}^{3};\mathbb{C})$, not of $L^{2}(\mathbb{R}^{3};\mathbb{C})$. But the state space was assumed to be $L^{2}(\mathbb{R}^{3};\mathbb{C})$. So, where is the gap? Shouldn't one take $\mathscr{H} = L^{2}(\mathbb{R}^{+}\times \mathbb{R}^{3};\mathbb{C})$ right from the start?
In addition, what is the difference between this picture and the Heisenberg's picture? In Schrödinger's picture the states are time-dependent. In Heisenberg's picture, the operators should be time dependent: an operator $A$ on $\mathscr{H}$ is evolved in time by $A(t) = U(t)AU(t)^{-1}$. But when applied to a state $\psi \in \mathscr{H}$, the state $\psi$ becomes time-dependent $\psi(t,{\bf{x}}) = A(t)\psi({\bf{x}})$ and it looks like we're back to the first scenario, where the states depend on time and the state space should be $L^{2}(\mathbb{R}^{+}\times \mathbb{R}^{3};\mathbb{C})$. But isn't it Schödinger's picture, instead of Heisenberg's?
