How does one rigorously define two-point functions?

Question

Let $\mathscr{H}$ be a complex Hilbert space, and $\mathcal{F}^{\pm}(\mathscr{H})$ be its associated bosonic (+) and fermionic (-) Fock spaces. Given $f \in \mathscr{H}$, we can define rigorously the creation and annihilation operators $a^{*}(f)$ and $a(f)$ (see, e.g. Bratteli & Robinson). These operators are densely defined on $D(\sqrt{\mathcal{N}})$, where $\mathcal{N}$ denotes the number operator.

Given a second quantized Hamiltonian $H$ on $\mathcal{F}(\mathscr{H})$, the time evolution of such operators in the Heisenberg picture are given by: $$e^{iHt}a^{*}(f)e^{-iHt} \quad \mbox{and} \quad e^{iHt}a(f)e^{-iHt}$$

To make the connection between these definitions and the usual definitions in the physics literature, consider the set $\mathcal{D}_{0}$ formed by those vectors $\psi = (\psi_{n})_{n\in \mathbb{N}} \in \mathcal{F}^{\pm}(\mathscr{H})$ with only finitely many nonzero entries $\psi_{n}$ and: $$D_{\mathscr{S}} := \{\psi \in \mathcal{D}_{0}: \mbox{$\psi_{n} \in \mathscr{S}(\mathbb{R}^{3n})$ for every $n$}\}$$ Then, we formally write: $$a^{*}(f) = \int dx f(x)a_{x}^{*} \quad \mbox{and} \quad a(f) = \int dx \overline{f(x)}a_{x}$$ and the formal integral operators are well-defined as quadratic forms on $D_{\mathscr{S}}$.

Analogously, one can consider formal operators: $$a^{*}_{x}(t) := e^{itH}a_{x}^{*}e^{-itH} \quad \mbox{and} \quad a_{x}(t) := e^{itH}a_{x}e^{-itH}$$

I have two related questions regarding these objects.

1. The quadratic forms $\langle \psi, a_{x}^{*}(t)\varphi\rangle$ and $\langle \psi, a_{x}(t)\varphi\rangle$, with $\psi, \varphi \in D_{\mathscr{S}}$, are not well-defined like this, because these need to be integrated out in $x$. But then, how to properly define/understand two point (or Green) functions: $$G(x,t,y,t') = -i\theta(t-t')\langle \Omega_{0}, a_{x}(t)a_{x}^{*}(t')\Omega_{0}\rangle$$ where $\Omega_{0}$ is the vacuum state $\Omega_{0}$ of $H$ and $\theta$ the Heaviside theta function. Is this well-defined as a distribution? To me this is not clear.
2. How are these Green functions $G(x,t,y,t')$ related to the (two-point) Wightman functions of axiomatic QFT? Are these the same?

Answer 1

Firstly as quadratic forms, $a^\ast(x)$ and $a(x)$ make perfect sense on the form domain generated by finite linear combinations of n-particle wavefunctions lying in Schwartz space. In fact any normal ordered product of such operators defines a quadratic form. Informally speaking, the creation operator takes an element of the domain into a distributional wavefunction, which you can then pair with an element of the domain to get a number. It is only when you want operators that smearing is required.

Now, the answer to the second part of this question is yes, although this is a bit more nontrivial. Informally this should look like a tensor product of distributions. Forget the time dependence for a moment and consider $G(f, g) = \langle \Omega_0, a(f)a^\ast(g)\Omega_0\rangle$. By the Schwartz nuclear theorem this gives a distribution in double the number of variables, so it's sufficient to show that it's a distribution in each component separately. By the CCRs: $= \langle f, g\rangle$. The fact that it's a distribution is then follows from the inner product on the single-particle space. For nonrelativistic quantum mechanics it's just an $L^2$ inner product which of course gives a distribution. In the Fock space of a relativistic QFT, this is a sobolev inner product, which is of course also continuous.

As for the conection to the Wightman functions, these are basically a special case. It's less natural to use Fock space terms in interacting rel-QFTs because of Haag's theorem, but for a free field $\phi = a+a^\ast$, you do in fact have that $\langle\phi(f)\phi(g)\rangle =\langle\Omega_0, a(f)a^\ast(g)\Omega_0\rangle= \langle f, (\Box+m^2)^{-1}g\rangle$, which is the aforementioned sobolev inner product.