3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The exact same thing $\delta(x)$ denotes: it is a symbolic notation, whose actual meaning is $\delta(f):=f(0)$. $\endgroup$ – AccidentalFourierTransform May 31 '18 at 21:58
  • $\begingroup$ But for the Dirac delta, there's a clear intuitive meaning that's lacking here. $\endgroup$ – user353840 Jun 1 '18 at 6:41
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Jun 1 '18 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. $\endgroup$ – Valter Moretti Jun 1 '18 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 Jun 1 '18 at 7:36
  • $\begingroup$ It is nothing but the way used by physicists.... $\endgroup$ – Valter Moretti Jun 1 '18 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 Jun 1 '18 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.