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In Theorem 1 of this paper Segal stablish a relation between states and generating functionals. He assert that in order to $\mu$ be a generating functional must satisfy $$ \sum_{j,k\in F} \mu (z_j-z_k)e^{iB(z_j\cdot z_k)}\bar{\alpha}_k\alpha_j\ge 0 $$ Then, as an example he show the functional $$ \mu(z)= e^{-\frac{1}{4}|z|^2} $$ is the zero-interaction vacuum generating functional.

The question is: Why this functional satisfies the desidered condition?

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Just repeating my answer of the crosspost: https://math.stackexchange.com/questions/4198282/a-condition-for-the-vacuum-generating-functional

According to the article (first paragraph of Sec. 4) the logical order of things is as follows: you define $C(z)$ (and thus $R(z)$), as well as $E$, then you let by definition $\mu(z):=E[e^{iR(z)}]$ (Eq. I). Then you prove that an explicit formula for $\mu(z)$ is $\mu(z)=e^{-\frac{1}{4}|z|^2}$ (Eq. II). The reason $\mu$ satisfies the positivity of Theorem 1 is because of (Eq. I) and not because of (Eq. II). The explanation for positivity via (Eq. I) is the "obvious part" top of page 9 of the article. It is possible to prove positivity from (Eq. II), but that uses the Schur Product Theorem. This is, for instance, done in the book on quantum physics by Glimm and Jaffe (Theorem 6.2.2 in 1987 Edition). If the positivity condition seems hard for you to fathom, then look up Bochner's Theorem and the notion of functions of positive type, for more background. Finally: not sure this article by Segal is what you want to read.

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  • $\begingroup$ Then look at the result I cited from the book of Glimm and Jaffe. They do it for Osterwalder-Schrader positivity instead of the state positivity (Nelson-Symanzik) but all you have to do is remove theta from all their formulas. The proof works just the same. $\endgroup$ Jul 14, 2021 at 22:47

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