# Why does the background field effective action generate only vacuum graphs?

I refer to LF Abbott's "Introduction to the background field method". The background field generating functional is

$$\tilde{Z}[J,\phi] = \int \mathcal{D}Q \exp i[S[Q+\phi] + J.Q], \text{ where } J.Q := \int d^{d}x J(x) Q(x).$$

The generator of connected diagrams is:

$$\tilde{W}[J, \phi] = -i \log \tilde{Z}[J,\phi]$$

and

$$\tilde{\Gamma}[J,\phi] = \tilde{W}[J,\phi] - J.\tilde{Q} \text{ where }\tilde{Q} := \frac{\delta{\tilde{W}}}{\delta J}$$ by analogy with $$W[J]$$ and $$\Gamma[\bar{Q}]$$. To get the effective action, as is shown in the paper,we use

$$\tilde{\Gamma}[0,\phi] = \Gamma[\phi], \text{ and evaluate } \tilde{\Gamma}[0,\phi] .$$

The fact that $$\tilde{\Gamma}[0,\phi]$$ generates 1PI graphs with no legs (vacuum graphs) makes calculations much easier. My question is: How does does the fact that $$\tilde{\Gamma}$$ is independent of $$\tilde{Q}$$ lead to only vacuum graphs?

The effective action $$\Gamma[\phi_{\rm cl}; \phi_{\rm bg}]$$ (in a background $$\phi_{\rm bg}$$) is the generating functional of 1PI correlation functions, cf. e.g. this Phys.SE post. In particular, $$\Gamma[\phi_{\rm cl}\!=\!0; \phi_{\rm bg}]$$ is the 1PI 0-pt correlator, i.e. consists of 1PI vacuum diagrams (in a background $$\phi_{\rm bg}$$).
• So $\because$ there is now no dependence of $\Gamma$ on $\phi_{cl}$, the correlation function on the RHS in the linked question is 0 and we have a vacuum graph?
• $\uparrow$ Yes. Aug 18, 2020 at 16:47