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In Coleman's "Aspects of Symmetry", chapter 7, section 3.2, he makes a claim that configurations of finite action form a set of zero measure and are therefore unimportant. Further, he goes on to prove the claim in Appendix 3 of the same chapter and says that the finite action contribution to the path integral must be zero. This really confuses me.

When we perform a saddle point approximation about a finite action field configuration, we do it based on the fact that the maximum contribution to the path integral comes from configurations around this saddle point. If what Coleman claims is true, then the saddle point approximation makes no sense.

What am I missing here? It would be great if someone could clarify this.

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Consider the path integral for a lattice QFT. On a finite lattice, all configurations have finite action. What happens in the continuum limit?

If we apply the saddle-point approximation first and then take the continuum limit, no problems arise. Coleman's comments apply to the continuum limit of the original path integral and so do not contradict the legitimacy of the saddle point approximation. Actually, Coleman's comments also apply to the result of the saddle-point approximation: after applying the saddle-point approximation to the lattice QFT, the only path integral remaining is a gaussian path integral, and Coleman's comments apply to the continuum limit of that gaussian path integral.

A similar comment applies to renormalization. The traditional approach to renormalization tries to take the continuum limit with the coefficients in the action held fixed, then notices that the results are undefined, and then tries to repair them using ad-hoc shenanigans. The mathematically legitimate way to do renormalization is to acknowledge that the coefficients in the action are not constants: they depend on the lattice scale and so do not remain constant during the process of taking the continuum limit. In fact they may diverge, and that's fine, just like $\lim_{x\to 0} x^{-1} x$ is perfectly well-defined despite the presence of the divergent $x^{-1}$ factor. We run into trouble only when we try to take the limit before we calculate the product.

The guiding principle is: taking the continuum limit should be the last thing we do, if we ever do it at all.

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I've always thought Coleman's phrasing here is a bit misleading. The point is that there are two kinds of integrals you're interested in when doing perturbation theory. Consider the finite-dimensional case as an analogy. The stationary phase approximation says that $$ \int_{\mathbb{R}^{n}} dx\, g(x)\, e^{iS(x)/\hbar} \simeq \left(2\pi\hbar\right)^{n/2}\int_{\text{Crit}(S)} \frac{e^{i\frac{\pi}{4}\text{sgn}(\text{Hess}(S))}}{\sqrt{\lvert\det(\text{Hess}(S))\rvert}}\, g(x_{0})\, e^{iS(x_{0})/\hbar} $$ to lowest order in $\hbar$ in the small-$\hbar$ limit. That is, the integral on the left over $\mathbb{R}^{n}$ can be approximated by the integral on the right over the set $\text{Crit}(S)$ of critical points of $S$. This is the precise form of the claim that the region around the critical points of $S$ gives the main contributions to the total integral.

Now $\text{Crit}(S)$ is a subset of $\mathbb{R}^{n}$, and indeed a finite subset. This means that $\text{Crit}(S)$ has measure zero in $\mathbb{R}^{n}$, and so the analogy of Coleman's claim is that the set of critical points is "unimportant" to the integral on the left. If we removed $\text{Crit}(S)$ from the domain of integration on the left hand side then it would make no difference to the value of the integral on the left. But this doesn't mean that the stationary points of $S$ are "unimportant" full stop. $\text{Crit}(S)$ is the entire domain of integration on the right hand side, so it's not unimportant there.

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  • $\begingroup$ So is it essentially saying that once you perform the saddle point approximation, then in the remaining gaussian path integral, there are no finite action contributions as @Chiral_Anomaly mentioned in their answer? $\endgroup$ – adithya Apr 16 '20 at 22:15
  • $\begingroup$ I would interpret "perform the saddle point approximation" to mean "replace the total integral on the left with the integral over the critical points on the right". When you perform the saddle point approximation you throw out everything except for the critical points, which have finite action. So once you perform the saddle point approximation there are only finite action contributions to the new integral. $\endgroup$ – John Dougherty Apr 16 '20 at 22:24

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