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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?

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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 $\phi_{\rm bg}$).

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  • $\begingroup$ 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? $\endgroup$
    – saad
    Aug 18, 2020 at 16:39
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Aug 18, 2020 at 16:47

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