Solutions to the Reconstruction Problem

Abstract

In this chapter we return to a foundational issue raised in Sect. 1.3 called the reconstruction problem. In the discussions that follow we phrase all arguments in terms of a single-component scalar field so that none of the extra structure that comes along when dealing with metric degrees of freedom plays a role.

Appendix 1: Why a UV Regulated Effective Average Action Must Depend on the UV Regulator

It is clear that at least for a general form of UV cutoff, the effective average action $\hat{\Gamma }_{k}^\Lambda [\varphi ]$ must depend on the UV regulator $\Lambda$ as indicated. Indeed if we embed the UV cutoff in the free propagator as done in (2.2.9) then the Feynman diagrams that follow from its perturbative expansion will evidently have all free propagators $1/p^2$ replaced by $\Delta ^\Lambda _k(p)$. The fact that $\hat{\Gamma }_{k}^\Lambda [\varphi ]$ thus depends on two scales, means that a bare action cannot be reconstructed which would directly give the continuum version $\hat{\Gamma }_k$ in the usual way. This is the first “severe issue” outlined above (2.1.2).

Following Ref. [2], a sharp UV cutoff and infrared optimised cutoff would appear to provide an exception however. With a sharp UV cutoff in place, (2.1.1) can alternatively be written

$$\begin{aligned} \frac{\partial }{\partial k}\hat{\Gamma }_{k}^\Lambda [\varphi ]=\frac{1}{2}\text {Tr}\bigg [\bigg (R_{k}+\frac{\delta ^{2}\hat{\Gamma }_{k}^\Lambda }{\delta \varphi \delta \varphi }\bigg )^{\!-1}\frac{\partial R_{k}}{\partial k}\bigg ]-\frac{1}{2}\text {Tr}\bigg [\theta (|p|-\Lambda )\bigg (R_{k}+\frac{\delta ^{2}\hat{\Gamma }_{k}^\Lambda }{\delta \varphi \delta \varphi }\bigg )^{\!-1}\frac{\partial R_{k}}{\partial k}\bigg ],\nonumber \\ \end{aligned}$$

(2.8.1)

where the first space-time trace leads to an unrestricted momentum integral

$$\begin{aligned} \int \!\!\frac{d^{d}p}{(2\pi )^{d}}\, \bigg (R_{k}+\frac{\delta ^{2}\hat{\Gamma }_{k}^\Lambda }{\delta \varphi \delta \varphi }\bigg )^{\!-1}\!\!\!\!\!\!(p,-p)\,\,\frac{\partial R_{k}(p)}{\partial k}, \end{aligned}$$

(2.8.2)

and we mean that the second term, the “remainder term”, has the momentum integral defining the trace restricted to $|p|>\Lambda$ as indicated. With the optimised IR cutoff profile we have $\partial R_k(p)/\partial k = 2k\theta (k^2-p^2)$ and thus, since $k\le \Lambda$, the remainder term vanishes in this case. At first sight this would appear then to allow us to consistently set $\hat{\Gamma }_{k}^{\Lambda }[\varphi ] = \hat{\Gamma }_{k}^{}[\varphi ]$ in (2.8.1) (providing only that we restrict flows to $k\le \Lambda$), meaning that for these choice of cutoffs, the dependence of the effective average action on $\Lambda$ disappears. This is not correct however as can be seen by expanding the inverse kernel. Define the full inverse propagator as

$$\begin{aligned} \hat{\Delta }^{\!-1}(p) := R_k(p) +\frac{\delta ^{2}\hat{\Gamma }_{k}^\Lambda }{\delta \varphi (p)\delta \varphi (-p)}\bigg |_{\varphi =0}, \end{aligned}$$

(2.8.3)

(temporarily suppressing the k and $\Lambda$ dependence) and similarly define $\Gamma ^{\prime }[\varphi ]$ to be the remainder after the term quadratic in the fields is removed (which thus starts at $\mathcal{O}(\varphi ^3)$ in a field expansion). Then

$$\begin{aligned} \Bigg (R_{k}\,\, +&\,\, \frac{\delta ^{2}\hat{\Gamma }_{k}^\Lambda }{\delta \varphi \delta \varphi }\Bigg )^{\!-1}(p,-p)\\ =&\, \left( \hat{\Delta }^{\!-1}+\frac{\delta ^{2}\Gamma '}{\delta \varphi \delta \varphi }\right) ^{\!-1}(p,-p)\nonumber \\ =&\,\,\, \hat{\Delta }(p)- \hat{\Delta }(p)\frac{\delta ^{2}\Gamma ' }{\delta \varphi (p)\delta \varphi (-p)}\hat{\Delta }(p)\nonumber \\&+\int ^\Lambda \!\!\!\!\frac{d^{d}q}{(2\pi )^{d}}\, \hat{\Delta }(p)\frac{\delta ^{2}\Gamma '}{\delta \varphi (p)\delta \varphi (-p-q)}\hat{\Delta }(p+q)\frac{\delta ^{2}\Gamma '}{\delta \varphi (p+q)\delta \varphi (-p)}\hat{\Delta }(p)-\cdots .\nonumber \end{aligned}$$

(2.8.4)

The momentum q is the external momentum injected by the fields remaining in $\Gamma '$:

$$\begin{aligned} \frac{\delta ^{2}\Gamma '}{\delta \varphi (p)\delta \varphi (-p-q)} = \Gamma ^{(3)}(p,-p-q,q;k,\Lambda )\varphi (-q)+\mathcal{O}(\varphi ^2), \end{aligned}$$

(2.8.5)

where we have displayed as a simple example the 1PI three-point vertex defined as in (2.5.2). (The higher point vertices will have an integral over the field momenta with a delta-function restricting the sum to $-q$.) With a sharp UV cutoff in place, not only are the external momenta $|q|\le \Lambda$ restricted, but the momentum running through any internal line is also restricted, thus here we also have $|p+q|\le \Lambda$. This is because ultimately all the free propagators come (via Wick’s theorem) from a Gaussian integral over the fields $\phi (r)$ in the path integral whose momenta $|r|\le \Lambda$ have been restricted by the sharp UV cutoff. Although the momentum p already has a sharp UV cutoff k provided by $\partial R_k(p)/\partial k$ which means the overall UV cutoff $\Lambda$ is invisible for it, this invisibility does not work for the other internal momenta, such as $p+q$, hidden in the construction of the inverse kernel. In other words even if the argument p above is freed from its UV cutoff at $\Lambda$, this cutoff remains inside the construction in all the internal propagators, such as displayed in (2.8.4), and thus despite appearances the first term on the right hand side of (2.8.1) actually still does depend non-trivially on $\Lambda$, implying also that $\hat{\Gamma }_{k}^\Lambda [\varphi ]$ is a non-trivial function of $\Lambda$.