Effective action as a generating functional and its derivative expansion

Question

On page 381 of Peskin and Schroeder, equation (11.90) reads

$$ \frac{\delta^2 \Gamma}{\delta \phi_{cl}(x)\delta \phi_{cl}(y)} = iD^{-1}(x,y).\tag{11.90}$$

I am having a bit of trouble interpreting this formula. On the left-hand side, we have the second derivative of the quantum effective action. However, on page 130 of Srednicki, the quantum action is written in terms of a derivative expansion, $$\Gamma[\phi_{cl}] = \int d^4x \,\left[-{\cal U}(\phi_{cl}) -\frac{1}{2}{\cal Z}(\phi_{cl})\partial^{\mu}\phi_{cl}\partial_{\mu}\phi_{cl}+\ldots\right],\tag{21.19}$$ suggesting that the quantum action is local, and hence $$\frac{\delta^2 \Gamma}{\delta \phi_{cl}(x)\delta \phi_{cl}(y)} \propto \delta(x-y).$$

However, the propagator on the right-hand side is not proportional to the delta function. So what is going on?

Answer 1

The classical action is local, while the quantum effective action is very much not so.

The quantum effective action is the sum over one-particle irreducible diagrams. You can easily see that these diagrams are not polynomial in the momenta. In fact, they can contain poles and logs.

Answer 2

The effective action $\Gamma[\phi_{\rm cl}]$ is the generator of 1PI-diagrams with bi-local propagators, and hence a non-local functional by construction.

1. Refs. 1-2 are less ambitious and do not discuss a derivative expansion per se, but instead consider $x$-independent field configurations $\phi_{\rm cl}$. From spacetime translation invariance it then follows immediately that the functional $$\Gamma[\phi_{\rm cl}] ~=~-\text{Vol}_4~{\cal V}(\phi_{\rm cl})\tag{11.50/16.2.3}$$ reduces to a function, known as the effective potential. This is of course the leading term in a derivative expansion.

2. The conceptionally simplest way to understand the derivative expansion (21.19/9-116) in Refs. 3-4 is to introduce a coupling constant $\lambda$ in front of each spacetime derivative $\partial_{\mu}\to \lambda\partial_{\mu}$ in the action $$S[\phi]~=~S_{\rm free}[\phi]+ S_{\rm int}[\phi].$$ Then the kinetic derivative terms get demoted to the interaction part $S_{\rm int}[\phi].$ Since the free part should remain non-degenerate, perturbation theory would then only work for massive fields. In this case the free propagators $$G_0(x\!-\!y)~\propto~ \delta^4(x\!-\!y),$$ and hence all connected Feynman diagrams become localized in position-space, so that the generator $\Gamma[\phi_{\rm cl}]$ of 1PI-diagrams can be written as a spacetime integral over (infinitely many) mono-local terms, i.e. the aforementioned derivative expansion.

   However, this approach is in many way too$^1$ radical, and one would have to re-sum diagrams just to get the leading term, i.e. the effective potential.

3. A less radical approach would be to consider the tree-level formula $$\Gamma_0[\phi_{\rm cl}]~=~S[\phi_{\rm cl}]\tag{0}$$ and 1-loop formula $$\Gamma_1[\phi_{\rm cl}] ~=~\frac{i\hbar}{2}\ln {\rm Det}\left(\frac{1}{i}\frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k \delta \phi_{\rm cl}^{\ell}}\right),\tag{1}$$ which are manifestly local functionals, cf. e.g. my Phys.SE answer here.

   To write the higher-loops $\Gamma_{\geq 2}[\phi_{\rm cl}]$ as a spacetime integral over (infinitely many) mono-local terms, it seems one would have to resort to methods mention under point 2. For a recursive formula, see e.g. my Phys.SE answer here.

4. By the way, it is interesting to compare the non-locality of the 1PI effective action with the non-locality of the Wilsonian effective action, see e.g. my Phys.SE answer here.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; eq. (11.50).

2. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; eq. (16.2.3).

3. M. Srednicki, QFT, 2007; eq. (21.19). A prepublication draft PDF file is available here.

4. Itzykson & J.-B. Zuber, QFT, 1985; eq. (9-116).

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$^1$ Parts of the 1-loop formula (1) would become $\lambda$-dependent. The functional determinant (1) can in many case be performed without expanding in $\lambda$,