In other words, does a finite set correlation functions sufficient to determine a theory? Is there a chance correlation functions are more fundamental then the lagrangian?

  • $\begingroup$ Related: physics.stackexchange.com/q/13488/2451 $\endgroup$
    – Qmechanic
    Dec 19, 2013 at 21:29
  • $\begingroup$ Comment to the question (v2): Are you assuming that there exists a Lagrangian to begin with? $\endgroup$
    – Qmechanic
    Dec 19, 2013 at 21:31
  • $\begingroup$ I remember I heard that not all theories have a lagrangian so it would be better not to assume its existence. $\endgroup$
    – Yair
    Dec 19, 2013 at 21:36

1 Answer 1


The observables of the theory are, first of all, those in the algebra $\cal A$ (technically a $^*$-algebra with unit) of objects generated by the smeared fields $\phi(f)$. I mean linear combinations of $I$ and products of smeared fields $\phi(f)$, where $f$ is a complex valued compactly supported smooth function. This algebra can be enlarged by including renormalised objects like $\phi^n(f)$ of $T_{\mu\nu}(f)$ and so on. When you are equipped with the correlation functions you are able to compute all expectation values of the elements of $\cal A$. E.g., $$\langle \phi(f)\phi(g) \rangle = \int_{M\times M} G(x_1,x_2) f(x_1)g(x_2) d^nx_1 d^nx_2\:.$$ So the correlation functions determine a state $\langle \cdot \rangle$ on the algebra $\cal A$. There is a corresponding theorem that assures it provided the linear map $\langle \cdot \rangle : {\cal A} \to C$ is positive, i.e. $\langle A^*A \rangle \geq 0$ for $A\in \cal A$, the so-called GNS theorem. That theorem also guarantees that there is a (uniquely defined up to unitary equivalences) Hilbert space where everything can be represented in the standard way (the elements of $\cal A$ are operators, $\langle \cdot \rangle$ corresponds to an expectation value od the form $\langle \Psi| \cdot |\Psi\rangle$). The found state can be extended to the extended algebra including renormalized objects, but here the procedure becomes complicated and I do not enter into details. Summarizing: Yes, under certain mild hypotheses, the class of correlation functions uniquely determines the quantum theory. There is no guarantee, however, for the existence of a Lagrangian describing the found theory. In this sense, the correlation functions are more fundamental than the Lagrangian.


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