I know for example that theories like QED and QCD when expended perturbatively in terms of Feynman diagrams produce asymptotic series that don't converge after a large number of terms and in fact begin to diverge. (Although if you stop after about 100 terms you should get something reasonably accurate).

If you instead use a non-perturbative approach like lattice gauge theory, do these theories converge to finite results?

For example using QCD to calculate the Proton mass, does this calculation converge or is it asymptotic?

Then QCD lattice theory has the fermion-doubling problem.

Does this mean that QED and QCD cannot by themselves produce mathematically precise results? Or must we think of them as low energy approximations to some bigger theory?

Are there any quantum field theories that give convergent series to questions within the scope of the theory with potentially infinite numbers of significant figures?

  • $\begingroup$ I think OP may also be asking about the existence of "closed" form solutions to the equations of motion in QFTs $\endgroup$ Commented Dec 25, 2018 at 3:42
  • $\begingroup$ @Dan Yard. well-defined series that are not convergent are not well defined. As in it can only give an approximation to a certain number of decimal places before the series diverges again. I'm not sure what you mean by "well-defined but non-convergent". Do you mean there is another way to get the values that does converge to a single value? Do you mean the lattice theory gives precise answers but the perturbation method doesn't? $\endgroup$
    – user84158
    Commented Dec 25, 2018 at 15:22
  • $\begingroup$ @InitialObserver nope, not closed form, just convergent and non-asymptotic methods of getting precise values. $\endgroup$
    – user84158
    Commented Dec 25, 2018 at 15:25
  • $\begingroup$ @Dan Just if QED are well defined on the lattice really. $\endgroup$
    – user84158
    Commented Dec 27, 2018 at 17:27
  • $\begingroup$ @Dan In my view not having a continuum limit means the theory is not well defined because only in the limit are all symmetries manifest. We may differ on terminology. "Probably" is not really mathematically rigorous! I'm not concerned with physical reality, just if the mathematics is well defined. $\endgroup$
    – user84158
    Commented Dec 28, 2018 at 1:37

1 Answer 1


Short answers to the various questions, which I can expand if there is interest and I have an opportunity. Many of these have probably been discussed here before, but I'm not in a good position to track down links at the moment.

The number of terms you might want to retain in an asymptotic expansion depends on the coupling $\alpha$. If I recall correctly, $n$-loop perturbation theory starts to break down for couplings roughly $\alpha \gtrsim 1 / n$. So keeping about 100 terms would only be reasonable for $\alpha \lesssim 0.01$.

Lattice calculations are not series expansions, so saying they "converge" may not be the best terminology. They produce finite numbers, which do need to be extrapolated to the continuum limit where the spacing between lattice sites, $a$, is taken to zero, corresponding to the removal of the effective UV cutoff $1/a \to \infty$. The older lattice QCD literature may talk about a "scaling regime" in which this approach to the continuum limit is well under control (with standard observables depending linearly on either $a$ or $a^2$); most literature over the past twenty years or so generally takes it for granted that the lattice calculations are in this regime.

For vector-like theories like QCD, the fermion-doubling problem has also been solved for the past twenty years or so.

QCD is a UV-complete theory that can by itself produce mathematically precise results. QED is not: it becomes part of the electroweak theory at high energies, and if we were to ignore that then it would hit a Landau pole at very high energies.

Obtaining an infinite number of digits from lattice calculations would require an infinite amount of computing, but one can generally estimate how much computing is needed to obtain a given observable with a given precision. Since experiments obtain finite-precision results, this is generally all we're interested in on the theory side.

You might enjoy searching for information about "integrable" quantum field theories, for which the path integral can be evaluated analytically. Most of these are lower-dimensional, but 4d maximally supersymmetric Yang--Mills theory in the $N = \infty$ planar limit is known to be integrable at zero and infinite 't Hooft coupling, and is conjectured to remain integrable at intermediate values of the coupling.

  • 1
    $\begingroup$ Great answer. Regarding the Landau pole in QED, there's an interesting discussion in this post: Does QED really break down at the Landau pole?. $\endgroup$ Commented Dec 26, 2018 at 1:26
  • $\begingroup$ Good answer. So QED is not well defined mathematically but QCD is using lattice procedures. But one question is if it has been proved that lattice QCD produces finite answers in the continuum limit? Or does such a thing make sense to ask? I guess if the answer depends linearly on $a$ then the linear term goes to zero right? (We hope!) $\endgroup$
    – user84158
    Commented Dec 27, 2018 at 17:31

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