# Non-Perturbative Effects in Quantum Field Theory

In elementary QFT we only really know how to solve free theories. These require quantizing infinitely many harmonic oscillators, which is an easy problem. But the real world is described by interacting theories with Lagrangians like $\mathcal{L}_{\textrm{QED}} = \overline{\psi}(i\not D-m)\psi - \frac{1}{4}F^2$

in which the fields are mixed up together. The standard way of dealing with these is to write them as $\mathcal{L} = \mathcal{L}_{\textrm{free}}+\mathcal{L}_{\textrm{int}}$

and then split the path integral as $\exp(i\int d^4x \mathcal{L}) = \exp(i\int d^4 x \mathcal{L}_{\textrm{free}})(1+i\int d^4x\mathcal{L}_{\textrm{int}}+\dots)$

The first factor explicitly yields free theory propagators. The second factor may now be expanded as a Taylor series in some coupling constant $g$. Often this is trivial since $g$ appears in $\mathcal{L}_{\textrm{int}}$ precisely to first order. For example in QED we have $\mathcal{L}_{\textrm{int}} = g\overline{\psi}\gamma^{\mu}\psi$

with $g$ being the electric charge. This means that the perturbation expansion is exactly the usual exponential series.

At present it looks like perturbation theory must capture all information in a QFT, provided you sum up all the terms. However, we’ve missed a crucial mathematical point. We’ll view the scattering amplitude $\Gamma$ for a given process as a complex-valued function of a real coupling $g$; that is to say $\Gamma \equiv \Gamma(e)$. We construct this function in (somewhat) trivial way from smooth functions, so it should be smooth.

However, smoothness does not guarantee analyticity! Put another way, we cannot be sure that the Taylor series for $g$ will converge for $g\neq 0$. Even if it does, it’s not guaranteed to converge to the function $\Gamma$ itself. This means that there are effects that lie outside the realm of perturbation theory, even if you could sum every term.

Let’s take an example. Suppose for instance that $\mathcal{L}_{\textrm{int}}$ has a $1/g^2$ dependence. If you’re balking at this “unphysical” choice, then rest assured that such Lagrangians do crop up. Now try to compute the Taylor expansion around $g = 0$ for $exp(-1/x^2)$. A little thought shows that this is identically zero.

Such non-perturbative effects require new approaches. One fruitful method involves solving the full theory classically then then considering quantum perturbations around such solutions. For Yang-Mills theories in particular this yields the notion of an instanton.

Finally let’s go back to QED and work out whether we should see non-perturbative effects there. Although not immediately as obvious as the simple $1/g^2$ example, one can argue that the perturbation series has $0$ radius of convergence. We follow Dyson and observe that if the radius of convergence were not zero, we’d be able to “reverse” the electric field by flipping the sign of $g$. This would render the vacuum unstable against decays into electrons and positrons, since these would repel rather than annihilate within a “Heisenberg allowed” time period. To rule out this option, we must take $g = 0$ formally.

This argument is not mathematically rigorous, and I don’t know whether one can make it so. Please comment if you know! Thankfully there’s a more modern perspective that sheds new light. We can view the sickness in QED in the context of Wilsonian renormalization. In particular, the theory has no well-defined high energy limit, as the coupling constants flow to infinite values at finite energy. This Landau pole behaviour is perhaps evidence that non-perturbative effects rule high energy QED. Again, I don’t whether this argument can be made watertight, so please take a rain-cheque on that one!