# What does a non-perturbative theory mean?

I'm a science writer and I'm having difficulty understanding what a non-perturbative approach means. I thought I understood what perturbative meant, but in looking for explanations of non-perturbative, I'm just confused about both concepts. Please explain as simply as possible.

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Hi John, and welcome to Physics Stack Exchange! It's great to see you doing the research to make sure you understand what you're writing about. To help us give you a more useful answer, could you edit the question to describe what your current understanding of a perturbative approach is? –  David Z Jul 13 '12 at 3:22
Part of your confusion could come from the fact that "non-perturbative" has more than one meaning. It can, as anna v explains in her answer, simply mean "exact solution to a diff. eq." i.e. "not an expansion in a small parameter". But, your wikipedia link discusses a different meaning of the phrase -- namely "that which cannot be described using a perturbative expansion". This latter usage typically refers to features of an exact solution that are not present in the perturbative solution (even if it is carried out to all orders). –  kbeta Jul 13 '12 at 6:43
I'm humbled that you would all take so much trouble to answer my question. My question came in the context in learning about QCD, which is described as as non-perturbative theory. But the answer from Ron Maimon below has helped me understand. Or at least it will after I read it several more times. Now I've found this site, I will be back to pester you a lot, I'm sure. Thanks very much. –  John Boyd Jul 13 '12 at 8:53
@JohnBoyd: Don't be so humbled, although it saves you a heck of a lot of time, it only takes about 5 minutes to give an answer. –  Ron Maimon Jul 13 '12 at 9:03

A nonperturbative theory means a theory where all the results can be calculated in principle to arbitrary accuracy on a computer. This is really just the same as a well-defined theory, a theory which is mathematically ok, and which makes sense, which isn't incomplete in some way at some high energy or when you do measurements at some high accuracy.

An example of a well-defined nonperturbative theory is QCD, where you can put the theory on a space-time grid, simulate it on a computer, and take the limit of small grid size to extract all the predictions of the theory (in principle--- this is a difficult computational task). The limit isn't simple, you have to make the lattice smaller and adjust the couplings so that the physical masses stay fixed, but the procedure is known (by physicist standards of rigor) to converge in the limit of small grids to a something sensible.

A perturbative theory means a theory where you start with an approximation that particles do not interact, and add the particle interactions by allowing them to scatter a little bit, then correcting for the scattering of the scattered particles, then correcting for the scattering of the doubly-scattered particles, and so on. This is the most convenient calculation method, so most theories are formulated this way. But there is a problem that the scattering of the scattered particles which are themselves products of scattering, and so on, requires an infinite summation, and the summation is divergent, it only produces results that get better for a while, as you include more scatterings, then at high enough order (enough rescattering processes), the results you calculate start to run away from the correct answer, and diverge.

This looks like a purely mathematical annoyance at first, something which can be fixed by a better method of summing infinite series. But it isn't purely mathematical, it has a physical interpretation. Consider a theory with one kind of scalar particle which can scattering off itself. This theory can be formulated perturbatively with a nonzero scattering rate, but if you put it on a grid, as you take the grid size to zero, the scalar particles created from the vacuum at small distances sheild each other from feeling the interaction, and this makes the scattering rate at long distances go down to zero. This is also only known at physicist standard of rigor. So the perturbative theory with a nonzero scattering rate really doesn't make sense, it breaks down if the grid-size is small enough.

There is a way to understand the Landau pole in scalar field theory qualitatively which is mathematically accurate in predicting the way in which the interaction vanishes. If you make a grid, and draw a random walk on the grid (mark a point, move to one of the neighbors at random and mark that point, and go on marking a random path), then two such walks will intersect with a probability that always goes to zero as you make the grid finer in 4 dimensional grids. In less than 4 dimensions, the particles can find each other. The probability that the particles find each other goes to zero as the lattice spacing to the power 4-d, so it goes to 0 for 5,6,7 dimensions, it goes to some finite value in 1,2,3 dimensions, and in 4 dimensions, you need a better analysis, and there it goes to zero as the logarithm of the lattice spacing. This means that if you thing of particles finding each other to interact, in 4 dimensions, random-walking particles will only be able to find each other to the extent that the grid size isn't zero. If you see point particles interacting by self intersection (like the Higgs boson in the standard model), you can predict that the lattice must be bigger than a certain amount just from the observed interaction. This size is exponentially tiny in the coupling, and for the standard model Higgs interactions (with the mass and couplings of the Higgs now known) it is much smaller than the Planck length.

The same effect, vanishing of the long-distance scattering rate, or a "Landau pole", is conjectured to happen in quantum electrodynamics and in the standard model (due to the Higgs self-interaction and also to the U(1) gauge group). Nobody worries about it, because the scattering only goes to zero as the logarithm of the grid-size, so the grid where you get problems is smaller than the Planck scale. The perturbative approximation is then fine, since it is not approximating a theory defined on the continuum, but it is approximating something else that takes over at high energies.

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Outstanding! Thanks very much indeed. –  John Boyd Jul 13 '12 at 8:47

Ideally, when we have a differential equation to solve, we will try to solve it analytically. Find explicit functions that encode the variables. The solutions for a harmonic oscillator, for example. The solutions of a potential in Schroedinger's equation. Those are examples of non perturbative solutions. They satisfy the differential ( or integral) equations.

Some solutions though can only be found as an expansion in a sum of series of terms; if each higher term is smaller than the previous and the series can be proven to converge this is as valid a solution, adding higher terms if more accuracy is needed. In every day problems these are useful for programming them numerically and getting the answer.

In particle physics this is codified into a "perturbative expansion" coming from "perturbation theory". Initially the potentials added to a theory were assumed to perturb the free system solutions, thus the nomenclature. It led to the Feynman diagram formulation of cross sections for interacting particles.

So non perturbative should mean a clean analytic solution. It will depend then on the context where the term is used.

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