The idea of a perturbation series in powers of a coupling $\alpha\ll1$ (for example, the fine structure constant in QED) make sense if the contribution of $(n+1)^{th}$ term in the series is smaller than the contribution of $n^{th}$ term. Now, the tree-level amplitude for a QED process is finite. However, at $\mathcal{O}(\alpha^2)$, the contribution from the S-matrix to the amplitude is divergent (all diagrams with loops are divergent because loop momentum can be arbitrarily high). How does perturbation theory make sense in QED (or in any quantum field theory)?
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4$\begingroup$ In any case, this seems to a be a duplicate of Why Does Renormalized Perturbation Theory Work? and How can perturbativity survive renormalization?. $\endgroup$– AccidentalFourierTransformCommented May 30, 2017 at 13:54
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$\begingroup$ What you are asking for is a rigorous mathematical justIfiction for the renormalization perscription. I don't believe one exists, however, QED demonstrates that the result agrees with experiment to more than ten significant digits. That is enough justification for most physicists. $\endgroup$– Lewis MillerCommented May 30, 2017 at 14:10
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$\begingroup$ BTW there are renormalization perscription in some QFT models that do not rely on perturbation theory. The infinitis can be eliminated in loop expansions of infinite order. $\endgroup$– Lewis MillerCommented May 30, 2017 at 14:15
3 Answers
A sketch of the philosophy:
Step one: Introduce a regulator. Everything is finite.
Step two: Set up the perturbative expansion. This is a purely algebraic step, where there is no notion of "this is small, this is big". The series is a formal power series, in the strict mathematical sense.
Step three: Renormalise observables. Adjust the coefficients of the Lagrangian, as a function of the regulating parameter, so that every observable is finite in the physical limit. Residual (finite) renormalisations are in principle arbitrary, and can be determined by means of some (physical or unphysical) prescription. After renormalisation is done, every term in the series is finite and, if you wish, you can remove the regulator. The end result is a perfectly healthy formal power series, where all the terms are finite and well-defined.
Step four: Is the series good? In a wide class of theories, we observe that the first few orders in perturbation theory are indeed decreasing in magnitude as you increase the order. We deem the series asymptotic, and call it a day. If the different terms do not decrease, we declare that the theory is not weakly interacting and resort to non-perturbative methods.
Every step is well defined and consistent. If you want an ab initio formal power series that omits the regularisation + renormalisation steps, you have to work with some formulation where there are no divergences, such as the causal formulation of QFT. Here the different terms of the formal power series are calculated unambiguously from the vertices of the Lagrangian, but in such a way that all integrals are convergent from the very beginning.
Whether you set up your theory in a manifestly finite framework, or use one where the regularisation + renormalisation steps are required, the output of perturbation theory is always a formal power series, where no notion of convergence is required. The perturbation parameter has no numerical value, so it makes no sense to state that the $(n+1)$th term is smaller than the $n$th one. It is a purely algebraic procedure. The $(n+1)$th term comes after the $n$th term because it has one more power of the perturbation parameter, not because it is smaller in magnitude.
The formal power series can be recast as an asymptotic series by giving the perturbation parameter a numerical value, but there is no guarantee that the series is going to be in any way useful. You don't know a priori whether the series is going to be numerically well-behaved. In fact you typically expect it to start growing at some order in perturbation theory, so the best-case-scenario is that this turning-point order is large, so that your predictions are meaningful.
Therefore, to sum up: The perturbation series makes sense because it is an algebraic process -- the series is a formal power series, with no notion of convergence. Sometimes the series is useful, sometimes it is not. But the framework is consistent either way.
The divergence is due to a poor starting point for setting up the theory; see my paper ''Renormalization without infinities - a tutorial''.
No problems arise with the expansion into the fine structure constant when one does everything with the proper mathematical care, as in causal perturbation theory. See my review on causal perturbation theory.
Perhaps a simpler way of making sense of perturbation theory (in the context of usual QFT, not causal formulations) is to understand that the procedure of renormalization described already by AccidentalFourierTransform ends up making the coupling (the coefficient of the interaction term in your Lagrangian(s)) a function of some energy scale, turning it into a running coupling. This energy scale at the same time controls where your theory is applicable, at least with a perturbative treatment. So at the end of the day this is saying that modes that are well beyond such energy scale should be controlled (regulated) or incorporated into the theory through some prescription. This is a very rough qualitative description of the Wilsonian approach to renormalization, the full details are worth writing a full chapter which you may find in Peskin and Schröder's classic "An Introduction to Quantum Field Theory".
The moral of the story as has been mentioned is that: first, those divergences are an artifact of some of the mathematical "violations" committed (specifically multiplication of distributions, also known as generalized functions), second those violations can be fixed through the renormalization procedure (sometimes one has to do this order by order even), third one may attempt to give a physical justification to such procedure as in the Wilsonian view.