Path integral zero dimensional QFT

Question

We consider the following partition function$$ \mathcal{Z}[\lambda] = \int{dx \; \exp\left(-\frac{1}{2}x^2-\frac{\lambda}{4!}x^4\right)} $$

Which is basically $\phi^4$ theory in 0+0 dimensions. The Feynman rules for such a theory are just 1 for a propagator and $-\lambda$ for a vertex. Now, when we consider $$\mathcal{W}[\lambda] = \log\left(\frac{\mathcal{Z}[\lambda]}{\mathcal{Z}[0]}\right)$$ in a perturbative expansion up to order $\lambda^2$, what are the Feynman diagrams we need to include and why?($\mathcal{Z}[0]$ is the partition function for the free theory)

As far as I understand we are interested in connected diagrams only. Thus for 1st order we require one vertex and 4 propagators.

FIRST ORDER

What's the physical difference between the following diagrams? Should I include all of them when calculating $\mathcal{W}[\lambda]$ in first order?

SECOND ORDER

I understand the difference between the first and the second diagram. Then the 3rd is an un-amputated 1st order diagram if I'm correct. Should we include that in the $\mathcal{W}[\lambda]$ calculation? I would say not?

Answer 1

Your diagrams that you have in the first order are all equivalent. However you shouldn't lose sight of the fact that Feynman diagrams are about pairing fields in Wick's theorem. There are 3 ways to pair the four $x$ fields in $x^4$ so this Feynman diagram will carry a factor of 3. We don't actually draw 3 diagrams but this combinatorial factor is still important.

In your second order diagrams, the third one is a disconnected diagram. If you expand ${Z[\lambda]}/{Z[0]}$ to 2nd order you indeed do get this disconnected diagram since it is a valid way of pairing the fields. However, you are interested in calculating $W$, $$\frac{Z[\lambda]}{Z[0]}=\exp[W]=\exp\left[\lambda W_1+\lambda^2 W_2+\dots\right]=1+\lambda W_1 +\lambda^2(\frac{1}{2!}W_1^2 + W_2)+\dots$$ The disconnected diagram is already accounted for by the $\lambda^2W_1^2/2$ term and so does not contribute to the new $W_2$ term.

This will be true in general for $W$. You can just ignore any disconnected diagram.