When does QFT perturbation theory stop being valid?

Question

When introduced to the concept of perturbation theory in Quantum Mechanics we split the hamiltonian $H= H_0 + \delta H$ where $\delta H$ is small in some manner, ie if say $\epsilon$ is the relevant dimensionless coupling associated with $\delta H$ then we expect $\epsilon \ll 1$ .

In QFT the idea seems slightly different. In the path integral approach we expand the partition functional as a formal power series in $\hbar$. By expressing the interacting generating functional as an asymptotic series of the free generating functional we then yield a power series in $\hbar$ that has zero radius of convergence (by the definition of the asymptotic series). We can show that the generating functional can be well approximated via resummation techniques or just by truncating the asymptotic series at an appropriate point despite being divergent as a power series in $hbar$.

With this in mind consider a lagrangian written in terms of dimensionless couplings $\{g_i(\mu)\}$ that will a function of an arbitrary regulatory mass scale $\mu$.

For perturbation theory to be valid does $\{g_i(\mu)\}$ have to be less than 1, like for any relevant dimensionless scales in QM perturbation theory? The fact that the asymptotic series has a zero radius of convergence regardless of the values of $\{g_i(\mu)\}$ makes me think that we don't actually need the couplings to be small (as the couplings can never be small enough for the partition function to 'actually converge').

This thought was motivated by my university lecture notes. In QCD we calculate the beta function to leading order. We integrate the beta function and obtain the result

$$\alpha_s(\mu) = \frac{2 \pi}{\beta_0 \log(\mu \space / \Lambda_\text{QCD})}$$

with $\alpha_s = g^2 / 4 \pi$ and $\beta_0 >0$

This is all fine, but then my professor says we think of the scale $\Lambda_\text{QCD}$ (the scale where the coupling constant diverges) as a border between perturbative and non-perturbative QCD. I found it odd that we would talk about this result for $\alpha_s(\mu) > 1$, let alone at a scale where the expression completely diverges. It has made me question when perturbation theory is actually valid. So the main question is does $\{g_i(\mu)\}$ have to be <1 for the perturbative method to be valid? If not then I guess we just don't want the couplings to diverge?

Answer 1

"Perturbation theory works" is not a binary statement. Sometimes the expansion is excellent, sometimes it is good, sometimes it is bad, and sometimes it is terrible. There is no point where the expansion changes from "it is valid" to "it is not valid". It is a spectrum.

If the effective coupling constant $x$ (which could be $g(\mu)$, but also $\alpha(\mu)=g(\mu)^2/4\pi$, or more complicated expressions, depending on what exactly you are computing) is very small, $x\ll 1$, then the series is most likely very good. If it is very large, $x\gg 1$, it is most likely very bad. Anything in between, it is hard to say. The scale at which the behaviour changes is, non-surprisingly, around $x\sim 1$. But don't take this as a rigorous prescription. There is no sharp change at $x=1$.

[Note: I say above "most likely" because there are always exceptions. For example, some objects are zero to all orders in perturbation theory (e.g., beta functions in supersymmetric theories), in which case the perturbative expansion is bad regardless of how small $x$ is. Conversely, some objects are one-loop exact (e.g., anomalies), in which case the perturbative expansion is correct regardless of how large $x$ is.]