Field theories like QED/QCD are a priori non-perturbative theories. Perturbatively you can describe them by Feynman diagrams which essentially sum over all topologies of virtual particle creation and annihilation processes.

In string theory descriptions through Conformal Field Theory (CFT) the Feynman diagrams are replaced by sums over topologies of the string world-sheet. Because of this, CFT descriptions are said to be inherently perturbative.

Given this clear analogy between CFT and QED, my question is how it can be understood that CFT descriptions are inherently perturbative while QED is not.

Secondly I would like to know how in principle we can make non-perturbative calculations in string theory. I am satisfied with "in principle" as I know that in the strong-coupling versions of string theory, namely F-theory and M-theory, as of today no amplitudes can be computed in practice.

(Note: I am not asking about a true non-perturbative calculation and not about dualities between weakly and strongly coupled theories.)


QED is non-perturbative because it is not defined by the sum over Feynman diagrams. Standard QFT does not rely on the perturbative expansion to define scattering amplitudes, it only uses it to compute them.

String theory through CFT, on the other hand, defines the string scattering amplitude through the sum over worldsheets. This is not a perturbative expansion of some non-perturbative expression for the amplitude, it is its definition. Therefore, this approach is inherently perturbative. The "perturbativeness" does not lie in the usage of CFT, but in the fact that we use a "perturbative sum over CFTs" as our definition of the string scattering amplitudes.

  • $\begingroup$ Thanks. So how can we get around this? String field theory? Describing 7-branes with 6D SCFT's? $\endgroup$ – quantifant Apr 11 '16 at 14:34
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    $\begingroup$ @quantifant: I think "how can we get around this" is a question of current research - M-theory and F-theory are such non-perturbative programmes, but I don't think a definite answer to "What is the correct non-perturbative formulation of string theory?" can be currently given. $\endgroup$ – ACuriousMind Apr 11 '16 at 14:51
  • $\begingroup$ Amplitudes is defined as a sum over scatterings, for any QFT. If that is "non-perturbative", the same applies to string theory, where different scatterings are identified with their topologies $\endgroup$ – CGH May 6 '16 at 19:21

There is a confusion and misunderstanding about what "perturbative" means and its relation with the power series expansion around certain small parameter.

If you state that QED/QCD is non-perturbative, because amplitudes are defined exactly, and not as a power expansion, then the same applies to string theory.

Amplitudes, in any QFT, is defined a sum over all possible scatterings. That is not a perturbative definition. The same happens in string theory, where amplitudes are also defined as the sum of all posible scatterings, with the difference now being that different scattering are identified with different topologies.

When you say that QED/QCD "Perturbatively you can describe them by Feynman diagrams", you are confusing the concept of perturbative. Feynman diagrams are not perturbation theory, perturbation theory is to consider a small parameter (the coupling constant) and take the scattering as a power expansion of that small parameter. But that does not mean that you are doing perturbations over "Feynman diagram."

The same applies to string theory: to sum over topologies is not the same as to perform a perturbation calculation. To sum over topologies is equivalent as a sum over all possible scatterings. What happens in string theory is that there is a parameter associated with the topologies, its genus, which can be used as a perturbation parameter, just as the coupling constant in QFT.

Finally, QED is "inherently perturbative": it is based on a path integral formalism, and the path integral only makes sense when there is a classical limit ($\hbar\to0$) which is equivalent to a week-coupling limit. This makes any theory based on the path integral "inherently perturbative", even string theory. Also, QED is not even well defined, it is IR divergent, because it is a effective theory of another theory, known as the electro-weak model.

Your second question is too broad. How in principle can you compute instantons in QFT?, what about E-M duality?, what about dualities in the Ising model? In "principle" is just too broad. If you ask about, T-duality, S-duality, instanton computation, brane computations, you could get an answer. But all of them are different topics by itself.

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    $\begingroup$ "Amplitudes, in any QFT, is defined a sum over all possible scatterings." I don't think that's correct. Amplitudes are defined by time ordered correlation functions, which have non-perturbative definitions via the path integral. In particular, amplitudes can have genuinely non-perturbative contributions, that are not just the sum over all scatterings. $\endgroup$ – Thomas May 6 '16 at 21:10
  • $\begingroup$ Thomas, the "time ordered correlation function" is just a sum over all possible scatterings. $\endgroup$ – CGH May 6 '16 at 22:41
  • $\begingroup$ 1) QED isn't an effective field theory for QCD. They describe different physical phenomena. $\endgroup$ – user1504 May 6 '16 at 23:41
  • $\begingroup$ 2) Your answer to the main question obfuscates a critical difference between string theory and QFT. In QFT, we have an essentially complete description of scattering amplitudes and we have perturbative approximations to them. In string theory, we're missing much more. In most situations, the perturbation theory is all we have. (There's an essentially complete definition in matrix theory, but the form of it is in general unknown.) $\endgroup$ – user1504 May 6 '16 at 23:46
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    $\begingroup$ @CGH : Correlation functions have non-perturbative contributions, such as instantons. In QCD we have a definition of the full non-perturbative correlation function, thanks to Wilson's lattice formulation of the path integral. In string theory, no analogous definition exists. $\endgroup$ – Thomas May 7 '16 at 1:24

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