I am trying to grasp the basics of rigorous quantum field theory. Let me summise how the setup of non-interacting quantum field theories look like to me.

Let $\mathcal{H}$ be a Hilbert space in which the state of a single particle of some type resides. For a particle moving in $\mathbb{R}^d$ with some spin $k$ this will be $\mathcal{H} = \operatorname{L}^2(\mathbb{R}^d) \oplus \mathbb{C}^{2k+1}$. Then to build a theory of arbitrarily many indistinguishable particles of that type we pass over to either the fermionic (antisymmetric) or bosonic (symmetric) Fock space. To simplify notation, let us assume that we the particles in question are bosons so we get the symmetric Fock space $\operatorname{Sym}(\mathcal{H})$ which is a quotient of the tensor algebra $\bigoplus_{n \in \mathbb{N}} \mathcal{H}^{\otimes n}$.

Now in single particle quantum mechanics a state is a unit vector $\psi \in \mathcal{H}$, an observable is a self-adjoint unbounded operator $T$ on $\mathcal{H}$ and the expecation value of $T$ in the state $\psi$ is given by the number $\langle T \psi, \psi \rangle \in \mathbb{R}$.

In the many particle case this should be the same: a state is an element of $\operatorname{Sym}(\mathcal{H})$ and observables are self-adjoint unbounded operators on the Hilbert space $\operatorname{Sym}(\mathcal{H})$. For example if $\psi$ is some single particle state, the element $\frac{1}{\sqrt{2}}\psi \otimes \psi \in \operatorname{Sym}(\mathcal{H})$ will be the two-particle state corresponding to two particles being in the state $\psi$. So far so good.

But now in quantum field theory one does not focus on unit vectors and self-adjoint operators, but rather on operator-valued distributions on $\operatorname{Sym}(\mathcal{H})$. A quantum field $\Phi$ then returns for every suitable test function $f$ on $\mathbb{R}^d$ an operator $\Phi(f)$ on $\operatorname{Sym}(\mathcal{H})$. In general this operator need not be self-adjoint if one has a “charged field”.

Now to my question: Why are quantum fields now suddenly these complicated operator-valued distributions and not just states on $\operatorname{Sym}(\mathcal{H})$? Since states on $\operatorname{Sym}(\mathcal{H})$ still make sense and correspond to so-and-so many particles being in specific states, how do these connect to the quantum field $\Phi$? In particular given a state $\psi \in \operatorname{Sym}(\mathcal{H})$ one can form the quantity $\phi(f)\psi$. How should one interpret this new state?

Since most of the times $\Phi$ seems to be built out of creation and annihilation operators on the Fock space, $\phi(f)$ creates and destroys particles inside of $\operatorname{supp}(f) \subseteq \mathbb{R}^d$. What is the meaning in passing from the particles $\psi$ to the particles $\phi(f)\psi$? I guess the question lurking behind this vague uncomfortableness of the deviation of quantum field theory from the standard quantum mechanics axioms is the question why it does not suffice to just build $\operatorname{Sym}(\mathcal{H})$, its operator algebra and the states on that operator algebra. Why operator valued distributions?

I am trying to grasp the basics of rigorous quantum field theory. Let me summise how the setup of non-interacting quantum field theories look like to me.

Let $\mathcal{H}$ be a Hilbert space in which the state of a single particle of some type resides. For a particle moving in $\mathbb{R}^d$ with some spin $k$ this will be $\mathcal{H} = \operatorname{L}^2(\mathbb{R}^d) \otimes \mathbb{C}^{2k+1}$. Then to build a theory of arbitrarily many indistinguishable particles of that type we pass over to either the fermionic (antisymmetric) or bosonic (symmetric) Fock space. To simplify notation, let us assume that we the particles in question are bosons so we get the symmetric Fock space $\operatorname{Sym}(\mathcal{H})$ which is a quotient of the tensor algebra $\bigoplus_{n \in \mathbb{N}} \mathcal{H}^{\otimes n}$.

Now in single particle quantum mechanics a state is a unit vector $\psi \in \mathcal{H}$, an observable is a self-adjoint unbounded operator $T$ on $\mathcal{H}$ and the expecation value of $T$ in the state $\psi$ is given by the number $\langle T \psi, \psi \rangle \in \mathbb{R}$.

In the many particle case this should be the same: a state is an element of $\operatorname{Sym}(\mathcal{H})$ and observables are self-adjoint unbounded operators on the Hilbert space $\operatorname{Sym}(\mathcal{H})$. For example if $\psi$ is some single particle state, the element $\frac{1}{\sqrt{2}}\psi \otimes \psi \in \operatorname{Sym}(\mathcal{H})$ will be the two-particle state corresponding to two particles being in the state $\psi$. So far so good.

But now in quantum field theory one does not focus on unit vectors and self-adjoint operators, but rather on operator-valued distributions on $\operatorname{Sym}(\mathcal{H})$. A quantum field $\Phi$ then returns for every suitable test function $f$ on $\mathbb{R}^d$ an operator $\Phi(f)$ on $\operatorname{Sym}(\mathcal{H})$. In general this operator need not be self-adjoint if one has a “charged field”.

Now to my question: Why are quantum fields now suddenly these complicated operator-valued distributions and not just states on $\operatorname{Sym}(\mathcal{H})$? Since states on $\operatorname{Sym}(\mathcal{H})$ still make sense and correspond to so-and-so many particles being in specific states, how do these connect to the quantum field $\Phi$? In particular given a state $\psi \in \operatorname{Sym}(\mathcal{H})$ one can form the quantity $\phi(f)\psi$. How should one interpret this new state?

Since most of the times $\Phi$ seems to be built out of creation and annihilation operators on the Fock space, $\phi(f)$ creates and destroys particles inside of $\operatorname{supp}(f) \subseteq \mathbb{R}^d$. What is the meaning in passing from the particles $\psi$ to the particles $\phi(f)\psi$? I guess the question lurking behind this vague uncomfortableness of the deviation of quantum field theory from the standard quantum mechanics axioms is the question why it does not suffice to just build $\operatorname{Sym}(\mathcal{H})$, its operator algebra and the states on that operator algebra. Why operator valued distributions?