# When does QFT perturbation theory stop being valid?

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?

• Perturbation theory in ordinary QM (even the simplest anharmonic oscillator) leads to asymptotic series too, for what it's worth. Mar 22, 2021 at 14:52
• You are mixing two different types of expansions. One is called semi-classical or loop expansion (the one concerning hbar), the other one is a perturbation expansion in the couplings. As such they have different ranges of validity. Mar 22, 2021 at 15:12

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.]
• This is one the reasons I am confused. I get that there isn't a sharp change from valid to not valid but my prof's comments suggest that the change in behaviour of convergence occurs roughly around $\Lambda_{QCD}$ when the coupling diverges not when it gets to around 1. This is what made me think that the situation may be different for an asymptotic series. Mar 22, 2021 at 15:27
• @Dan It is but a matter of definitions. When $\alpha=\infty$ the series is definitely wrong. (It is not self-consistent). When $\alpha=1$ it is most likely bad, but it may still capture some generic features of the dynamics. So there are two natural scales: $\mu=\Lambda$, where $\alpha$ blows up, and $\mu=\Lambda e^{2\pi/b_0}$, where $\alpha$ equals $1$. They both parametrize how bad the expansion is, but with different criteria. And for real-life systems they are of the same order of magnitude anyway. In any case, recommended reading: Landau pole Mar 22, 2021 at 15:33