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Regarding the propagator $\mathcal{G}(k,i\omega,r)$ of a Euclidean scalar real Gaussian quantum field theory $$\mathcal{Z_0}=\int\mathcal{D}[\phi]e^{-\mathcal{S}[\phi]}$$ $$\mathcal{S[\phi]}=\int d{\vec{q}}\int d{\omega} \, \mathcal{G}^{-1}(\vec{q}, i\omega,r) \tilde{\Phi}(\vec{q},i\omega)\tilde{\Phi}(-\vec{q},-i\omega) $$ $$\mathcal{G}(\vec q,i\omega,r)=\frac{1}{q^2+\omega^2+r} \ .$$ There is a RG fixed point at $r=0$ describing a spontaneous $\mathbb{Z}_2$ symmetry breaking. For $r>0$ the expectation value of the field $\langle\phi\rangle=0$ while for $r<0$ there is a non-zero expectation value due to the spontaneous symmetry breaking.

I have several questions regarding this framework:

  1. Is the propagator $\mathcal{G}(\vec q, i\omega,r)$ meaningful for $r<0$? The problem I see is that the poles of the propagator are on the real axis and that for a Gaussian FT the field is unbounded in the symmetry broken case.
  2. In case the propagator is meaningful in the case $r<0$. Is the propagator associated with the connected or unconnected correlation function? By connected correlation function we mean the correlation function from which the mean field value is subtracted $\langle \phi(x)\phi(y)\rangle-\langle\phi\rangle^2$ while with unconnected correlation function we refer only to the $\langle \phi(x) \phi(y) \rangle$ term.
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  • $\begingroup$ The Gaussian integral is not convergent for $r<0$. $\endgroup$
    – Qmechanic
    Commented Feb 22, 2022 at 14:09
  • $\begingroup$ That is kind of the point. What is the interpretation of $\mathcal{G}(\vec q, i\omega,r)$ in if $r<0$? Is it at all sensible to use the above Gaussian theory in the case $r<0$? How would one proceed to formulate a theory for the vicinity of the Gaussian fixed point for $r<0$? $\endgroup$
    – jan0155
    Commented Feb 22, 2022 at 15:34
  • $\begingroup$ It's not even correct to say that $\langle \phi \rangle \neq 0$ for $r<0$ - the expectation value is undefined. I'm not sure why you expect this path integral to have any physical interpretation at all for $r<0$? $\endgroup$ Commented Feb 24, 2022 at 15:39

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