Why does the divergence of a QFT's coupling constant under RG flow trivialize the theory if it occurs in the UV but not in the IR?

Question

When you first learn quantum field theory, at some point you calculate the beta function (to leading order) for a renormalizable coupling constant of some theory like $\varphi^4$ theory, Yukawa theory, or QED. You find that it takes the form

$$\mu \frac{dg}{d\mu} := \frac{dg}{d\log(\mu)} = \beta(g) = c\, g^p + o\left(g^{p+1} \right),$$

where $c$ is a positive numerical constant that takes a lot of work to calculate, and $p$ is either 2 or 3 (depending on your conventions and the particular theory). You solve this ODE (neglecting the subleading corrections) with the initial condition $g(\mu_0) = g_0$, to get $$g(\mu) = \left[ \frac{1}{g_0^{p-1}}-c(p-1) \ln\left( \frac{\mu}{\mu_0} \right) \right]^{-\frac{1}{p-1}}.$$ This function increases with $\mu$ and diverges at a finite value $\mu = \mu_0 \exp\left[ \frac{1}{c(p-1)g_0^{p-1}} \right] > \mu_0$ that is called the "Landau pole" (at least in the context of QED). Your teacher gravely explains that this Landau pole is a Very Serious Problem: if $g_0 \neq 0$ then the theory becomes internally inconsistent at sufficiently high energies (unless new physics kicks in at those high energies to modify and possibly salvage the theory). Therefore, the only logically consistent choice is $g_0 = 0$, which implies that $g(\mu) \equiv 0$ and the theory is "quantum trivial". If your teacher is particularly careful, he or she may mention in passing that we can't necessarily trust this formula for $g(\mu)$ at high energies, because as $g$ becomes large, the neglected terms in the beta function become important.

Later on in the course, you calculate the beta function (to leading order) for nonabelian gauge theory. You get the same result as above, except that $c$ is negative (unless the gauge field is coupled to a sufficiently high number of flavors of matter field). Therefore $g(\mu)$ decreases with $\mu$, and the function diverges at some value of $\mu$ less than $\mu_0$. Your teacher cheerfully explains that this is a neat feature of nonabelian gauge theory called "confinement", and it indicates that at sufficiently low energies (or equivalently far separations), particles become so strongly interacting that perturbation theory breaks down, and the theory changes qualitatively (as the natural choice of constituent matter "particles" changes from color-charged quarks to color-neutral hadrons).

Why the asymmetry between these two cases? In both cases, perturbation theory breaks down as you flow in some direction and the coupling constant becomes strong and formally diverges. Why is the former case described as a fundamental pathology that "trivializes" the entire theory at all energy scales, while the latter case is described as simply indicating a qualitative change in the physics described by the theory?

Answer 1

Why is the former case described as a fundamental pathology that "trivializes" the entire theory at all energy scales, while the latter case is described as simply indicating a qualitative change in the physics described by the theory?

Because of history. The Landau pole was discovered in the 1950s, when renormalization was not well-understood, and even divergences in general were sometimes viewed as a reason to throw out QFT. Landau was personally strongly against QFT and used this as a bludgeon against it. Asymptotic freedom was discovered in the 1970s, when both the divergences of QFT and RG flow were understood, and was a crucial part of understanding the strong interactions, so it was regarded as a triumph of QFT. This history was fossilized and preserved in the standard QFT course, which has been largely handed down from teacher to student unchanged for decades.

In both cases, the theory gets strongly coupled at some scale, and we can't perform calculations reliably. In both cases, we still want to be able to do physics, and usually do this by finding a new set of variables which are weakly coupled. If there is an asymmetry, it is that the new degrees of freedom in the UV case could be ones that weren't accounted for in the original theory at all, while in principle everything in the IR case is contained in the original theory.

Answer 2

The description of "trivialization" is outdated and not very sharp. In a more modern language, it just means that the low-energy description you have, say a $\phi^4$ theory in $4d$, is not by itself enough to define a weakly coupled continuum QFT unless the coupling is zero. However, a key point is that the RG flow (integrating out degrees of freedom as you go from high to low energies) erases information. That means that as a matter of principle, you can't flow upwards in energy. So it's very well possible that there exists a continuum QFT with a lot of interesting dynamics (many new fields) at high energies, but at long distances this putative theory looks just like $\phi^4$ theory. The interesting and sharp question is therefore not whether "$\phi^4$ theory is trivial", it's whether $\phi^4$ theory admits a UV completion, and if so, which one(s). That question is however very complicated and can't be answered using known methods.

Going in the other direction (flowing down) is a well-defined procedure: if you know a theory at an energy scale $E_0$, at least in principle you can compute an effective action at any energy scale $E < E_0$. Perhaps your theory will become more and more strongly coupled as you decrease $E$ so perturbation theory becomes useless, but in that case you can at least in theory use a numerical method to compute whatever you want (in practice, it's not always so easy).