# Why are many physicists trying to develop non-perturbative quantum theories? [closed]

I would like to briefly know where (and why) does perturbation theory fail and why are physicists so desperate looking for non-perturbative theories. (No Mc A, it's not obvious to me, is it not obvious to you that if it were obvious to me I wouldn't ask the question?)

## closed as too broad by John Rennie, Daniel Griscom, user36790, Kyle Kanos, GertJan 12 '16 at 3:23

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• 1. I have never met a truly "desperate" physicist. 2. Have you tried a Taylor-expansion of say $f(x)=\exp{-1/x^2}$? – Bort Jan 11 '16 at 15:58
• Whenever the coupling constant is large, the most prominent example being the low energy limit of QCD and confinement of quarks to hadrons. – Photon Jan 11 '16 at 15:58
• @Bort in what problems do those kind of functions appear? – SaudiBombsYemen Jan 11 '16 at 16:03
• @Pablo I remember foolishly trying to expand that function on a statistical mechanics final exam in grad school. I don't remember the exact context but I think it had something to do with Bose-Einstein condensation. – DanielSank Jan 11 '16 at 16:44
• Is it not obvious to you that a non-perturbative exact theory would be better than a perturbative theory with limited radius of convergence (or even no convergence)? – ACuriousMind Jan 11 '16 at 16:54

There are many phenomenon in quantum field theory that falls outside the understanding of perturbative analysis. To understand the nonperturbative physics is to understand the full dynamics of the theory whereas perturbation theory is not reliable.

The basic examples are 1) dynamical symmetry breaking (supersymmetry or chiral symmetry) which is usually accompanied by some fermionic condensate $\langle \lambda \lambda \rangle \neq 0$. The condensation cannot be seen in perturbative physics. 2) Structure of the vacuum of the theory. In the famous ${\cal N}=1$ SQCD introduced by Seiberg, there are cases where vacuum is there classically but completely eliminated by non-perturbative quantum effects. 3) There are certain theories that admit no perturbative description. The assumption of perturbation theory is that various couplings are small, but this cannot be satisfied always.

There are also mathematical motivations. The nonperturbative physics in four dimensions are related what physicists call instantons", and is mathematically labelled by second Chern class $c_2(M)$. Studying instantons is closely related to Donaldson theory that gives invariant of a large class of four manifold $M_4$, which cannot be obtained by usual methods.

Perturbative theory is understood as an application of the Taylor expansion around a linearity $g=0$ (unperturbed theory) in an underlying theory: $$f(g)=\sum_{n}A_n g^n$$ The unperturbed theory could be, for example, an Hydrogen Atom for the underlying QED. Then, the QED perturbative theory is a prescription of how inserting $g\neq0$ corrections of the type: $$(A_0+A_1g)+A_2g^2+A_3g^3+...$$ in the Hydrogen atom when $g$ is small.

You see that the perturbative approach are applied in a state-dependent manner. You initiate in some state and then calculates the corrections. This unperturbed states almost always are not connected themselves by perturbative corrections: You can't build the Hydrogen atom by perturbative calculations of the free theory (proton+electron).

Actually, the perturbative calculations always misses some part of the underlying theory when exist functions that are insensitive under taylor expansions: $$f(g)=\sum_{n}A_ng^n+e^{-\frac{1}{g}}\sum_nB_ng^n+...$$ where a Taylor expansion of $e^{-\frac{1}{g}}$ is $$0+0\times g+0\times g^2+...$$ and you know that for $g\neq0$, $e^{-\frac{1}{g}}\neq0$. So, bound states, tunneling effect and other effects may comes from this small (if $g$ small) exponential $e^{-\frac{1}{g}}$. And one of the the more clear limitations is when $g$ is not small, then the exponential $e^{-\frac{1}{g}}$ is important.

Note that if $g$ is small, the same is for the exponential, the non-perturbative term can be important under some marginal process through the scale like a formation of a bound state or a tunneling effect.

• Nice! I just noticed that your example for non-expandable functions solved a completely unrelated problem (in economics) for me. – CuriousOne Jan 11 '16 at 17:53
• This function is very important indeed. Maybe is a lack of knowledge of my part but a think that new things are behind of this function $e^{-\frac{1}{g}}$. I think that is something related to problems of the type $\int DXe^{-S[X]}$, where $S[X]=H_0-gH_1$, where $H_1$ contain some non-linearities. One way to get this function is by a marginal contribution of the type: $1/g(\mu)=1/g_0\pm\beta log\,\Lambda/\mu$, when $g(\mu)\sim 1$ – Nogueira Jan 11 '16 at 18:02
• I believe to have seen this function discussed in single variable calculus books (was it in Courant's classic?), so the example has been around a lot longer than since the discovery of field theories, but your example puts it into a solid physical context. I will certainly pay more attention the next time I do a naive expansion. – CuriousOne Jan 11 '16 at 18:12
• Assuming $g>0$, $e^{-\frac1g}$ is bounded above by $1$. Is $1$ what you mean by big? – Ruslan Jan 11 '16 at 18:46
• @Ruslan When $g\sim1$, $e^{-\frac{1}{g}}\sim0.5$, so you have an error of $\sim\pm0.5$. The taylor serie start to diverge as you compute correction of the order $e^{-\frac{1}{g}}$, so you are clearly in a non-perturbative scenario. – Nogueira Jan 11 '16 at 18:50