The Wikipedia page for Feynman Diagrams claims that

Thinking of Feynman diagrams as a perturbation series, nonperturbative effects like tunnelling do not show up, because any effect that goes to zero faster than any polynomial does not affect the Taylor series. Even bound states are absent, since at any finite order particles are only exchanged a finite number of times, and to make a bound state, the binding force must last forever.

But this point of view is misleading, because the diagrams not only describe scattering, but they also are a representation of the short-distance field theory correlations. They encode not only asymptotic processes like particle scattering, they also describe the multiplication rules for fields, the operator product expansion. Nonperturbative tunnelling processes involve field configurations that on average get big when the coupling constant gets small, but each configuration is a coherent superposition of particles whose local interactions are described by Feynman diagrams. When the coupling is small, these become collective processes that involve large numbers of particles, but where the interactions between each of the particles is simple.

This means that nonperturbative effects show up asymptotically in resummations of infinite classes of diagrams, and these diagrams can be locally simple. The graphs determine the local equations of motion, while the allowed large-scale configurations describe non-perturbative physics. But because Feynman propagators are nonlocal in time, translating a field process to a coherent particle language is not completely intuitive, and has only been explicitly worked out in certain special cases. In the case of nonrelativistic bound states, the Bethe–Salpeter equation describes the class of diagrams to include to describe a relativistic atom. For quantum chromodynamics, the Shifman Vainshtein Zakharov sum rules describe non-perturbatively excited long-wavelength field modes in particle language, but only in a phenomenological way.

This passage confuses me. Does it mean that non perturbative effects can be calculated using Feynman Diagrams? I thought that Feynman diagrams were by definition perturbation series.

  • $\begingroup$ For 1D stongly correlated electrons (assuming that backscattering is supressed) you have an 1-1 maping to a representation in terms of free bosons which pertubation series is trivial and yields all the information necessary to get a all non-perturbative results for your original problem $\endgroup$ – tired Sep 28 '17 at 20:24

If the theory is Borel summable, you can recover non-perturbative information from the perturbative series. This can be shown explicitly for example by calculating the exact effective action in the presence of a constant electromagnetic field, à la Schwinger. You can find a very clear exposition in A.R.Bogojevi's lecture notes, chapters 24&25. For other examples of Borel summable results, see Heisenberg-Euler Effective Lagrangians: Basics and Extensions, by G. V. Dunne. See also Why is the Borel summation relevant for asymptotic series of physical observables?, from Physics Overflow, for other relevant references.

Another well-known example where you can extract non-perturbative information from the perturbative series is zero-dimensional QFT. See e.g. A Non-perturbative Solution of the Zero-Dimensional $\lambda\phi^4$ Field Theory, by Malbouisson, Portugal, Svaiter. The calculation in $d=0$ is particularly illuminating because we can obtain explicit expressions, but in principle the analysis holds for higher $d$ (cf. A. Neumaier's answer in the post on Overflow).

Moreover, it is usually the case that objects that are protected by topological considerations, perturbative calculations are in fact exact. Such is the case of the axial anomaly, which can be analised non-perturbatively (by means of, say, a change of variables in the path integral as Fujikawa taught us to), and the result agrees with a one-loop calculation. You can find a nice discussion in the PSE post Instantons, anomalies, and 1-loop effects.

Finally, let me mention that for objects protected by symmetries, such as BPS states, you can usually extract reliable information valid in the strongly coupled regime by studying the same object in the weakly coupled regime. The philosophy behind this is that the symmetries are sometimes so constraining that the properties of the object evolve rigidly from one regime to the other. In the same vein, in theories related by dualities you can obtain results valid in the strongly coupled theory by studying the weakly coupled one. Whether this qualifies as extracting non-perturbative information from a perturbative result is a matter of opinion, so this may be taken as a more contrived examples than those from before.

  • $\begingroup$ See also 1511.05977 and 9704351. $\endgroup$ – AccidentalFourierTransform Sep 27 '17 at 18:35
  • 1
    $\begingroup$ But in 4D relativistic QFT, the perturbative series is not Borel summable, due to the presence of renormalons. Getting nonperturbative information from perturbative calculations is therefore more a black art than science. $\endgroup$ – Arnold Neumaier Sep 28 '17 at 12:04
  • $\begingroup$ @ArnoldNeumaier hmm I was under the impression that some 4D perturbative series are non-Borel-summable, but some others are. For example, 0406216 claims that in the purely magnetic case, the Heisenberg-Euler effective action is Borel summable and agrees with the exact result. Of course, when the series is non-Borel-summable, you are right, but it may be to pessimistic to say that 4D is never summable, right? $\endgroup$ – AccidentalFourierTransform Sep 28 '17 at 12:18
  • 1
    $\begingroup$ I had qualified my statement by the word relativistic. The nonrelativistic case generally seems to be better behaved, though I don't know of any theorem in this direction. $\endgroup$ – Arnold Neumaier Oct 2 '17 at 14:01
  • $\begingroup$ In reading about soliton solutions in field theory, it's pretty frequently referenced that such solutions are entirely "missed" by perturbative techniques. This also goes for other things like color confinement (hence the difficulty in proving it) ui.adsabs.harvard.edu/abs/1979PhR....49...91B/abstract $\endgroup$ – R. Rankin Dec 9 '20 at 6:29

No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.