I would like to clarify the following statement:

Perturbation theory (PT) in QFT is derived with several assumptions such as: adiabatic interaction, spectrum is bounded downward...

This statement I know from my teacher and therefore it is not a complete statement. But I do not understand where I can find this conceptual details. In Peskin-Shroeder detailed discussion is omitted and there are only implicit mentions of the PT assumptions. Can anyone help me to find these conceptual details? I mean list of assumptions for PT in QFT.

As I understand, some of these assumption are

  1. spectrum is bounded downwards (it can be captured during the derivation of PT in P&S)
  2. adiabatic switching on/off of interaction (it guarentees that all the evolution of the system is nothing more than phase factor $e^{iL}$)

Then, I can also derive PT for QFT from the path integral, expanding the exp into series by coupling constants. And using this approach, I do not see where the assumptions appear. Indeed, I can obtain all the correlation functions (connected, disconnected, amputated connected) from the specfic generating functional and there is no any assumptions in this derivation. How can I see the assumptions during the derivation of PT with path integral?

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    $\begingroup$ Well, perhaps somebody else can give you a more formal answer, but I believe it all boils down to why the free theory should be a reasonable starting point for the perturbation series at all. If it is fundamentally unstable (spectrum is not bounded from below) or if the interactions are so strong the "free" states look nothing like the real states (non-adiabatic interaction), then you cannot do perturbation theory. An example of the latter would be trying to describe QCD at low energies by a perturbative series from free quarks. $\endgroup$
    – Void
    May 11, 2019 at 10:41
  • $\begingroup$ Dear @Void , thank You for an answer! I understand why the free theory should be the starting point but, as I understand, these assumptions come not only from this fact, don't they? You example gives me nothing more than in QCD the expansion should be constructed as the series by $1/\text{coupling}$ $\endgroup$ May 11, 2019 at 10:52
  • $\begingroup$ @Void That looks like an answer, not a comment ;) $\endgroup$
    – ACuriousMind
    May 11, 2019 at 11:20
  • $\begingroup$ @Void I kind of get your point about the bottomless pit of the spectrum but I just calculated the three-point function of the phi-cubed theory for my homework and they worked just fine? :P What do those finite answers mean if perturbation theory is not valid for a phi-cubed theory? $\endgroup$
    – user87745
    May 13, 2019 at 16:18
  • $\begingroup$ @DvijMankad , importantd comment. Indeed, phi-3 theory is one of the basic examples in QFT textbooks for diagrammatic technique $\endgroup$ May 13, 2019 at 16:20

2 Answers 2


There is no (general) rigorous non-perturbative definition of a QFT, so there is no rigorous proof of perturbation theory either. Therefore, it makes no sense to claim that the perturbative expansion rests on some analytic assumptions. It rests on no assumptions, because it cannot be derived from anything. There is nothing "more fundamental" that, when expanded in power series, yields a perturbative QFT. That being said, you can proceed as follows:

  1. You define a QFT through its perturbative series (say, in the causal approach if you want to be mathematically rigorous). Here, and when regarded as a formal power series, the perturbative expansion is well-defined regardless of any analytic properties of the Hamiltonian, so there are essentially no conditions on the operators.

  2. You analyse the problem in standard QM (one-dimensional QFT, if you will: the only spacetime coordinate is time), and assume that the same formalism should hold in QFT, provided we eventually find a good formulation. A canonical reference for rigorous perturbation theory in QM is Kato's Perturbation Theoryfor Linear Operators. It is a tough route, so have fun if you want to go there; no guarantee you will find what you're looking for, but it is hard to imagine you will find anything more explicit than this.

  3. Some very specific (lower dimensional) QFTs have been constructed rigorously, from where the perturbative expansion can be derived. The canonical example is Glimm & Jaffe's Quantum physics: A functional Integral point of view. Here the authors deal with two-dimensional (Euclidean) $\phi^4$ theory, which has the key property that normal-ordering is all you need to render it finite. Therefore, you cannot really hope to draw general conclusions from this example but, sadly, we don't have many more rigorous (interacting) QFTs that can be analysed explicitly.

Finally, let me mention that a heuristic reason the conditions in the OP are usually assumed is the so-called Gell-Mann and Low theorem, which is sometimes used to justify perturbation theory. This theorem does require the spectrum to be bounded from below, and that interactions are switched on and off adiabatically.

  • $\begingroup$ If You can add something about this topic it will be beatiful. I think that Yours answer is comprehensive, thank you! $\endgroup$ May 13, 2019 at 20:56
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    $\begingroup$ @ArtemAlexandrov Thank you! 1) I'd be glad to add some more details. Is there anything in particular you'd like me to elaborate on? Any of the topics I mentioned? 2) Thank you for accepting my answer, but we usually recommend people to wait one or two days before accepting any answer. That way, you encourage other users to post answers, and maybe you'll like one of those new answers better! Maybe you could unaccept my answer for now, and come back in a couple of days and accept it again, if no new answers have been posted. Cheers! $\endgroup$ May 14, 2019 at 0:19
  • $\begingroup$ I have heard this many times now that you define a QFT through its perturbative series. But I don't understand it. Sure, we can calculate $n-$point functions only perturbatively but the $n-$point function is defined exactly, right? If an alien knows how to solve the crazy functional integral then it can write down all $n-$point functions in explicit closed form, right? This is puzzling me because we can't exactly solve equations of motion even in classical mechanics for all sorts of potentials, but we don't really say that a certain classical theory is "defined" only perturbatively. $\endgroup$
    – user87745
    May 14, 2019 at 3:10
  • $\begingroup$ @DvijMankad ODEs (& PDEs) are perfectly well-defined mathematically, and existence and uniqueness theorems guarantee that a solution exists, even if we cannot express it in terms of elementary functions. Path integrals are a whole nother story: there is no rigorous mathematical definition of this object, and so it is not true that the $n$-point function is defined exactly. In order for an alien to solve the PI, they must define it first. For now, we don't have a formal definition, we only have a vague idea of what it is meant to represent. For us physicists, that is more than enough though $\endgroup$ May 14, 2019 at 13:16
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    $\begingroup$ @AccidentalFourierTransform Ah, I see. Thanks! $\endgroup$
    – user87745
    May 14, 2019 at 18:49

The assumption for a valid perturbation theory in quantum mechanics is that the Hamiltonian of the interacting theory differs from the free theory around which the perturbation is done by a relatively compact term (see the treatise by Reed and Simon), and that the interaction strength (the multiplier of this term) is small enough. [This requires as a necessary condition that the spectrum is bounded from below (since that of the free theory is), and one way of proceeding technically is by switching the interactions on and off adiabatically, thus explaining the comment of your teacher.]

In relativistic quantum field theory, the deviation form a free field is not given by a relatively compact term, making perturbation theory strictly speaking inapplicable. This shows through the fact that all terms in the perturbation series except for those corresponding to the tree diagrams diverge, hence give meaningless results.

What is done instead is to truncate the field theory using some hard or soft cutoff at some energy scale $\Lambda$, do perturbation theory on the truncated level (for which the above assumption can be proved in certain cases), and then adjust the interaction constants as a function of $\Lambda$ in such a way that order by order, the limit $\Lambda\to\infty$ can be taken. This gives a renormalized perturbation series which is then used to extract physcial information, using heuristic resummations and other tricks. Nothing is said about the convergence of the series, which in 4D is unlikely even after Borel summation. The formal renormalized perturbation series exists even when the Hamiltonian is unbounded below, though such a theory (e.g., $\phi^3$ in 4 dimensions) has no sensible physical interpretation and is used only for toy calculations to get practice.

In causal perturbation theory one avoids the noncovariant truncations by working with axioms for the S-matrix that are then shown to be satisfiable by formal power series in terms of a mathematically well-defined construction using careful distribution splitting. This gives the same perturbation series, but without the need to take limits (and hence avoiding having to move across Landau poles when resumming the series). Again, nothing is said about the convergence of the series.

In the path integral approach, one starts with a mathematically ill-defined path integral (which simply ignores all possibly needed assumptions) and works with formal rules valid for discretized variants where space-time is replaced by a finite set of points, pretending that they remain valid for a putative functional integral on Euclidean or Minkowski space-time. This explains why one cannot see the assumptions during the derivation of perturbation theory with path integral.

  • $\begingroup$ Thank Your for the answer! I am only student but from at least one book "Introduction to the Functional Renormalization Group" it seems that path integral (PI) gives something more than formal rules, doesn't it? $\endgroup$ May 16, 2019 at 9:19
  • $\begingroup$ @ArtemAlexandrov: There is no rigorous path integral for non-free fields; thus all derivations, powerful as they are, are ill-defined. $\endgroup$ May 17, 2019 at 8:58

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