I am having this silly confusion! Suppose I have a system (a Hamiltonian or an action say) and also suppose I have a perturbation parameter present (say only one in sight) in there, using which I can find out some kind of perturbative solution of that problem. Suppose now we also found out a non perturbative solution to that problem (by non-perturbative, I have in mind that the solution is not an expansion using the perturbation parameter) (well, origin of this question is that in string theory I have $\alpha^\prime$ and also $g_s$ as perturbation parameter and then we also see $D$-branes as its non perturbative d.o.f. But I am asking here as a general scenario).

But is it guranteed (I thought its not) that you have ``solved'' the problem once you know that particular perturbative solution and non-perturbative solution?

Let me rephrase: Unless I know the exact solution of a problem and the only way to solve that problem to us is such perturbative, non perturbative methods, how can we ever be sure that we've solved the problem and there are no more solutions? (probably it will vary from problem to problem, but can there be a problem with such ambiguity?)


1 Answer 1


I think that there may be some confusion about the term "non-perturbative solution" in the question. Perturbative methods are methods based on Taylor expansions in a small parameter, typically a coupling constant. These methods fail to capture some phenomena - e.g. instantons - which are non-perturbative phenomena.

Non-perturbative methods are any methods that don't rely on the perturbative expansions. They can usually produce insights or terms that cannot be extracted by perturbative methods themselves. The lattice QCD calculations or interactions induced by instantons are examples.

But if you say that you have a "non-perturbative solution", the only thing that it can mean in my understanding of the words is that you simply have the exact solution. The adjective "non-perturbative" means that the solution is not just a "perturbative approximation". That really means that it's exact, doesn't it? Assuming that it deserves to be called a solution at all.

So if you have a non-perturbative solution, then you have the exact solution, and you don't even need any additional perturbative solution; a perturbative solution may be obtained by Taylor-expanding the exact one, right?

Also, I think that the question about the uniqueness is a completely different question. Some equations or conditions have a single solution; others may have many solutions. That's true perturbatively; and that's true exactly. Sometimes people know all the solutions; sometimes they may be missing some of them. Two different perturbative solutions - unless they're equivalent by some transformation i.e. a change of the renormalization scheme - are approximations of two different exact solutions. The opposite is not quite true; two different exact solutions may accidentally reduce to the same perturbative approximation when Taylor-expanded.

But I may have misunderstood your question. Maybe by "non-perturbative solutions", you meant some strong coupling expansions from the opposite side? This is not the same thing as "non-perturbative". If one has two expansions from the opposite sides and they are consistent with each other, as in the S-dualities in field theories and string theory, it is a strong hint that the theory is OK everywhere in between but I would agree with you - if you wanted to say it - that it is not quite a proof. In supersymmetric theories, many important quantities may be calculated exactly anywhere in between and the function agrees with expansions from both sides. It's an even stronger circumstantial evidence that the theory is OK and solutions exist in the middle - however, it's not a full proof and it doesn't tell us what the solutions to any questions are "in between".

Cheers, LM

  • $\begingroup$ You're right about what I meant and thanks for a formal explanation of the term. So, yes, let me rephrase it in terms of coupling. Say, the coupling parameter was small so we expanded perturbatively. Now we found some S type duality to get a strong coupling result. But at this time do we have any way to understand that those two solution exhausts all possibility of solution (actually if there's a matching in the middle, that can be a signal of full solution..right?)? Sorry, I can't much probably elaborate.I have few characters left in typing. $\endgroup$
    – user1349
    Jan 17, 2011 at 0:06

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