Let $\mathscr{H}$ be a complex Hilbert space and denote $\mathscr{H}_{n}$ the tensor product $\overbrace{\mathscr{H}\otimes\cdots\otimes\mathscr{H}}^{\text{n}}$. Denote by $\Pi_{\pm}$ the projection operators onto symmetric and anti-symmetric projections on $\mathscr{H}_{n}$: $$\Pi_{+}(\psi_{1}\otimes \cdots \otimes \psi_{n}) = \frac{1}{n!}\sum_{\pi}\psi_{\pi(1)}\otimes\cdots\otimes \psi_{\pi(n)} \quad \mbox{and} \quad \Pi_{-}(\psi_{1}\otimes\cdots\otimes \psi_{n}) = \frac{1}{n!}\sum_{\pi}\operatorname{sign}(\pi)\psi_{\pi(1)}\otimes\cdots\psi_{\pi(n)}$$ where $\sum_{\pi}$ is a sum over all permutations $\pi$ of $\{1,...,n\}$. These operator can be extended to all $\mathscr{H}_{n}$ and I'll denote their images $\Pi_{\pm}(\mathscr{H}_{n})$ by $\mathscr{H}_{n}^{\pm}$.
Now, if $f \in \mathscr{H}$, let's take the usual creation and annihilation operators $b^{\dagger}(f)$ and $b(f)$, defined by means of the following conditions: $$b^{\dagger}(f)(\psi_{1}\otimes \cdots \otimes \psi_{n}) = \sqrt{n+1}(f \otimes \psi_{1}\otimes \cdots \otimes \psi_{n})$$ $$b(f)(\psi_{1}\otimes \cdots \otimes \psi_{n}) =\sqrt{n}\langle f,\psi_{1}\rangle (\psi_{2}\otimes \cdots \otimes \psi_{n})$$ Again, these extend to bounded linear operators on $\mathscr{H}_{n}$. Depending on what subspace $\Pi_{\pm}(\mathscr{H}_{n})$ one wants to consider, one can define $a^{\dagger}(f) = \Pi_{\pm}b^{\dagger}(f)$ and $a(f) = \Pi_{\pm}b(f)$. One can prove: $$a^{\dagger}(f)\Pi_{n}^{\pm}(\psi_{1}\otimes \cdots \otimes \psi_{n}) = \sqrt{n+1}\Pi_{n+1}^{\pm}(f\otimes \psi_{1}\otimes \cdots \otimes \psi_{n}). \tag{1}\label{1}$$
We can extend our analysis to symmetric and anti-symmetric Fock spaces $\mathcal{F}^{\pm}(\mathscr{H}) = \bigoplus_{n=0}^{\infty}(\mathscr{H}_{n}^{\pm})$. In this case, we extend both $a^{\dagger}(f)$ and $a(f)$ to a dense subspace of $\mathcal{F}^{\pm}(\mathscr{H})$, which is the space of finitely many particles $\mathcal{D}_{0} := \{\psi \in \mathcal{F}^{\pm}(\mathscr{H}): \mbox{there exists $N \in \mathbb{N}$ such that $\psi_{n} = 0$ for every $n \ge N$}\}$, by letting each of these operators to act componentwise. By using (\ref{1}), we see that this dense set can be built from sucessive creation of particles, i.e: $$\Pi^{\pm}(f_{1}\otimes \cdots \otimes f_{n}) = \frac{1}{\sqrt{n!}}a^{\dagger}(f_{1})\cdots a^{\dagger}(f_{1})\Omega_{0} \tag{2}\label{2}$$ where $\Omega_{0} = (1,0,0,...) \in \mathcal{F}^{\pm}(\mathscr{H})$ is the vacuum state.
Question 1: It is usually used, specially in the physics literature, that every linear operator $A$ on $\mathcal{F}^{\pm}(\mathscr{H})$ can be written in terms of creation and annihilation operators. However, I have never seen any proof of this result, and it does not seem trivial to me. Is this a provable fact?
Question 2: How can we change the above scenario to include time evolution? More especifically, let $H$ be a Hamiltonian on $\mathscr{H}$. At least if the particles of the system are noninteracting, the time evolution operator on $\mathscr{H}_{n}$ is $e^{-itH_{n}} = e^{-itH}\otimes \cdots \otimes e^{-itH}$, where there are $n$ such tensor products. In Heisenberg's picture, one defined time-dependent operators, i.e: $$a^{\dagger}(f,t) := e^{itH_{n}}a^{\dagger}(f)e^{-itH_{n}} \tag{3}\label{3}$$ On the other hand, we can evolve elements: $$e^{-itH_{n}}(\Pi^{\pm}(f_{1}\otimes \cdots \otimes f_{n})) \tag{4}\label{4}$$ Does it follow from (\ref{1}) and (\ref{2}) that every such time-evolved element can be built from sucessive applications of $a^{\dagger}(f,t)$?
Add: Question 2 is better rephrased as follows. If time is considered, does (\ref{2}) still holds when the creation operators $a^{\dagger}(f_{i})$ on the right hand side of this equation is replaced by the time-dependent (\ref{3}) and the left hand side of (\ref{2}) is replaced by (\ref{4})?
