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?

  • $\begingroup$ The exact same thing $\delta(x)$ denotes: it is a symbolic notation, whose actual meaning is $\delta(f):=f(0)$. $\endgroup$ Commented May 31, 2018 at 21:58
  • $\begingroup$ But for the Dirac delta, there's a clear intuitive meaning that's lacking here. $\endgroup$
    – user353840
    Commented Jun 1, 2018 at 6:41

1 Answer 1


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 $f_k \in {\cal S}(\mathbb R^4)$, $k=1,2,\ldots, n$.

  • $\begingroup$ But what does that integral mean? I would just write $\mathcal{W}(f)$, where $f$ is defined as in the question. Writing $\mathcal{W}(x_1,\cdots,x_n)$ implies that $\mathcal{W}$ is dependent on the vectors $x_i$, even if you write in in an integral, so I don't understand what it's supposed to mean. $\endgroup$
    – user353840
    Commented Jun 1, 2018 at 6:40
  • $\begingroup$ ${W}(x_1,\dots,x_n) $ is the formal kernel of a distribution. That is not an integral in proper sense. TSchwart's nuclear theorem implies that the map $f_1\otimes \cdots \otimes f_n \mapsto \langle \Omega, \phi_1(f_1)\dots\phi_n(f_n) \Omega \rangle$ is the restriction of a distribution whose action is formally indicated by an integration. $\endgroup$ Commented Jun 1, 2018 at 6:52
  • $\begingroup$ So it's just an informal way of writing things? I think I get it, but I still think it's a weird way of writing things. $\endgroup$
    – user353840
    Commented Jun 1, 2018 at 7:36
  • $\begingroup$ It is nothing but the way used by physicists.... $\endgroup$ Commented Jun 1, 2018 at 11:21
  • $\begingroup$ It's not even that weird. Schwarz functions are dense in the space of distributions. Take an approximating sequence, and the natural pairing is approximated by integration. $\endgroup$
    – user1504
    Commented Jun 1, 2018 at 11:34

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