Propagator in $\phi^3$ Theory in $d=0$ Spacetime Dimensions I am working through some problems in Srednicki's book in QFT (page 70).  In one of the problems we work with the following integral:
$$Z[g,J]=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}dx \; \exp \bigg[-\frac{1}{2}x^2 + \frac{1}{6}gx^3 + Jx \bigg].$$
The solution to the problem uses the fact that this is really the path integral of $\phi^3$ theory in $d=0$ spacetime dimensions, hence in its Feynman-diagram expansion each propagator is of the form
$$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}dx \; e^{-x^2/2}x^2.$$
Could someone explain why this is the propagator in this case?  (Apologies if this is trivial as I have not been studying QFT for long).
 A: The partition function $Z[J]$ is the generating function of correlation functions:
$$
\frac{\delta^{n} Z[J]}{\delta J (x_1) ... \delta J (x_n)}\biggl\vert_{J=0}
=
\langle \phi(x_1) \ldots \phi(x_n) \rangle,
$$
which also means
$$
Z[J] = 1 + \int dx \langle \phi(x) \rangle J(x) 
+\int dx dy \, \langle \phi(x) \phi(y) \rangle J(x) J(y) + \ldots.
$$
In particular the propagator is defined as the two-point correlation
of the free theory:
$$
\Delta (x_1 - x_2) := \langle \phi(x_1) \phi(x_2) \rangle 
= \frac{\delta^{2} Z_{0}[J]}{\delta J(x_1) \delta J(x_2)}\biggl\vert_{J=0}.
$$
In your case, considering 
$$
Z[J,g=0] = Z_{0} [J] = 
\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}dx \; \exp \bigg[\frac{1}{2}x^2 + Jx \bigg],
$$
we get
$$
\Delta (x_1 - x_2) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}dx \; x^2 \exp \bigg(\frac{1}{2}x^2 \bigg).
$$
Let me finish with a small comment: A free theory is given by the quadratic part of the action. The definition of propagator that I gave fits nicely into Feynman's diagrammatic approach because $Z_0$ corresponds to the zeroth order expansion in the coupling constant $g$.
