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According to the Wightman axioms, for Wightman fields $\phi_1,\dots,\phi_n$, the vacuum expectation value $$\langle\Omega, \phi_1(f_1)\dots\phi_n(f_n)\Omega\rangle$$ is a multilinear continuous map from an n-tuple product of Schwartz spaces to the complex numbers, i.e.: $$\langle\Omega, \phi_1(f_1)\dots\phi_n(f_n)\Omega\rangle : \underbrace{\mathcal{S}(\mathbb{R}^4) \times \dots \times \mathcal{S}(\mathbb{R}^4)}_{n \text{ times}} \rightarrow \mathbb{C}. $$

Then the nuclear theorem states that there is a unique tempered distribution $\mathcal{W}: \mathcal{S}(\mathbb{R}^{4n}) \rightarrow \mathbb{C}$ such that: $$\langle\Omega, \phi_1(f_1)\dots\phi_n(f_n)\Omega\rangle = \mathcal{W}(f), $$

where $f(x_1,\dots,x_n) = f_1(x_1)\cdot f_2(x_2) \cdot \dots \cdot f_n(x_n)$.

This distribution is then suddenly written as $\mathcal{W}(x_1,\dots,x_n) = \langle \Omega, \phi_1(x_1)\dots\phi_n(x_n) \Omega \rangle$.

This does not make sense as $\mathcal{W}$ is not a function of $n$ four-vectors $x_1,\dots,x_n \in \mathbb{R}^4$. What does this notation mean?

user353840
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1 Answers1

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It just means that $$\int \mathcal{W}(x_1,\dots,x_n) f(x_1)\cdots f(x_n) dx_1\cdots dx_n= \langle \Omega, \phi_1(f_1)\dots\phi_n(f_n) \Omega \rangle\:.$$ in distributional sense: there is a Schwartz distribution of ${\cal S}'(\mathbb R^4 \times \cdots \times \mathbb R^4)$ denoted by $\mathcal{W}$ whose action on functions of ${\cal S}(\mathbb R^4 \times \cdots \times \mathbb R^4)$ with the special form $f_1\otimes \cdots \otimes f_n$ gives $\langle \Omega, \phi_1(f_1)\dots\phi_n(f_n) \Omega \rangle$. Another (perhaps less ambiguous) form for this statement is the following one.

There exists $\mathcal{W} \in {\cal S}'(\mathbb R^4 \times \cdots \times \mathbb R^4)$ such that $$\mathcal{W}(f_1\otimes \cdots \otimes f_n)= \langle \Omega, \phi_1(f_1)\dots\phi_n(f_n) \Omega \rangle\:,$$ for every choice of $f_k \in {\cal S}(\mathbb R^4)$, $k=1,2,\ldots, n$.

Valter Moretti
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