# What's the relation between perturbative and nonperturbative QFT?

In case of any miscommunication let me describe my understanding of the meaning of "perturbative" and "non-perturbative", and correct me if something is wrong: In a perturbatively defined QFT the fields are quantized as free fields, and the interaction is constructed by multiplying free fields operators, hereafter particles can scatter with each other; non-perturbatively defined QFT is any QFT in which the interaction is not constructed by the above-described method, e.g. a lattice QFT.

Now I'm curious about the connection between two constructions. The possibilities I can imagine are the following:

1. Perturbative and non-perturbative QFT are complementary to each other, they are both approximations of some underlying theory.There are things perturbative QFT can cover which non-perturbative QFT can't, and vice versa.

2. Perturbative QFT is contained in non-perturbative QFT, thus can be derived from it as some kind of limiting situation, however some calculations are more efficient in perturbative QFT.

Which one is correct? My sanity favors (1), but "non-perturbative" really sounds like a more powerful word to me so I can't help thinking (2) is possible. I'd really appreciate a comprehensive explanation on the issue.

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Related: physics.stackexchange.com/q/31930/2451 and links therein. – Qmechanic Jul 31 '12 at 8:42

Regarding your specific question: It's (2). Pertubative QFT arises from non-perturbative QFT by Taylor series approximation in the coupling coefficients.

Really, non-perturbative QFT should just be called "QFT". But people often try to define a QFT by writing down a perturbative approximation, so we are stuck with this weird terminology.

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Yes, for example an axiomatic, mathematically rigorous field theory is certainly non-perturbative and we should be able to derive perturbative QFT from it. But this is not what I'm asking for, since none of the realistic QFT has such rigorous foundation to date, yet we still have lattice QFT which is also called "non-pertrubative". So let's just replace every "non-perturbative" by "lattice" in my original post, what would be the answer now? – Jia Yiyang Jul 31 '12 at 12:40
You can recover perturbative QFT from lattice QFT by Taylor expanding and taking the continuum limit. This is standard in lattice field theory; its the basis of "perturbation-improved lattice gauge theory", where one uses the continuum perturbation theory to work out the right coefficients on the lattice. (Also, the question of mathematical rigor is mainly orthogonal to the one you've asked. When rigor can be had, the mathematical story is the one I've outlined.) – user1504 Jul 31 '12 at 21:20
I see, but can one get lattice QFT by discretizing a perturbatively defined QFT? – Jia Yiyang Aug 2 '12 at 1:04
In general, no. In perturbative QFT, all you get is a formal power series. In most cases, this series has radius of convergence equal to zero. (Dyson's argument: if changing the sign of the coupling constant makes the energy unbounded below, the power series can't converge, because the path integral itself doesn't converge.) You can sometimes find a clever resummation of this power series, but doing so amounts to introducing additional non-perturbative data. – user1504 Aug 2 '12 at 1:16
Ok, all the above together seem to answer my whole question, thanks. – Jia Yiyang Aug 2 '12 at 1:38