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In perturbation theory, we introduce a small perturbation to the system. This perturbation is usually a small parameter that slightly modifies the Hamiltonian or the equations of motion.

We assume that the solution to the perturbed system can be expressed as a series expansion in terms of the small parameter. The first term in this series is the known solution of the unperturbed system, and the subsequent terms represent corrections due to the perturbation.

we solve for each term in the series iteratively, starting with the zeroth-order term (the unperturbed solution) and then finding the first-order correction, the second-order correction, and so on.

The crucial point is the assumption that the solution to the perturbed system can be expressed as a series expansion in terms of the small parameter.

We could have assumed so many things else. Why is it that this assumptions works so well and actually approximates well the perturbed solution ?

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    $\begingroup$ What else are you expecting us to assume? We tried this because it was useful in celestial mechanics and other branches of physics, and just plain hoped that it works in quantum theory, only to realise that it works even better in quantum theory than in classical. The mathematical justification is just not there; all of these things are asymptotic. There is some theory there, but it is not exactly proven. $\endgroup$
    – naturallyInconsistent
    Commented Aug 2 at 1:34
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    $\begingroup$ As a theorist by training, it is really silly hearing people who do not understand the actual state of affairs arguing about this. It is not just perturbation in QED that is divergent. It is that all perturbation as used in physics are so. The perturbation theory in QM is also divergent, and the perturbation theory in classical mechanics is chaotic. It just so happens that all the basic functions we want to play with in classical mathematics are trivially extend-able to complex, and thus inherit the analyticity. Those functions then have the nice Taylor behaviour. $\endgroup$
    – naturallyInconsistent
    Commented Aug 2 at 3:06
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    $\begingroup$ @FlatterMann Physics is a science, which means it follows the scientific method. In other words, the question is whether the predictions made with perturbation theory agree with experimental observations. In the case of QED (and other theories based on QFT), the answer is in general a resounding yes, which justifies the OP statement that "it works so well." $\endgroup$
    – flippiefanus
    Commented Aug 2 at 3:58
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    $\begingroup$ @FlatterMann Phase transitions are non-perturbative effects. Nobody is suggesting that one tries to study them with perturbation theory. $\endgroup$
    – flippiefanus
    Commented Aug 3 at 3:34
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    $\begingroup$ @FlatterMann: "It's a tool with limited application." Not true. Take a look at the particle data book to see how many experiments have been done to compare with results obtained from perturbative calculations, with excellent agreement! $\endgroup$
    – flippiefanus
    Commented Aug 4 at 4:26

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Perturbation theory is based on the notion of an asymptotic series. It differs to some extent from the idea of a Taylor series, which is an infinite sum of terms converging to the function it represents within a certain region. The expansion parameter of a Taylor series therefore need not be small.

An asymptotic series on the other hand is truncated to have a finite number of terms. Provided that the expansion parameter is small enough, one can truncate a Taylor series to get the asymptotic series. Terms that are excluded are then sufficiently suppressed to give negligible contributions.

This property also explains why perturbation theory works so well. If the subsequent terms are small enough then the terms that are retained in the series would be accurate enough.

Those issues that are raised in the comments about all the weird functions that exists are issues that may interest mathematicians, but they are not generally relevant in physics, because they do not describe things in the physical world. For example, it is often argued that such weird functions must also be elements in the sets over which path integrals are integrated. That is not true. The path integral, as used in physics, is purely based on an assumed solution for an integrant in Gaussian form. The question is then what would be the set of functions for which the functional integral would produce such a solution for a Gaussian integrant? It turns out that one can safely discard all such weird functions and still get the required solution.

In response to some comments, let me add some discussion about the divergence/convergence of asymptotic series. It is true that there are asymptotic series that have divergent tails. In other words, at some point beyond where one would normally truncate the series the terms can start to grow larger with the result that the series as a whole does not converge.

First, obvious not all asymptotic series are divergent in this way. A simple example is a Taylor series with a small expansion parameter within its convergence region. It can be used as an asymptotic series and it does not have a divergent tail.

The important question is, what about the asymptotic series in perturbation theory. In this case, we can use the scientific method. If the asymptotic series that we are using to compute a measurable quantity is one of those that diverge, then the experimental result won't agree with the computed result. So the fact that there is agreement between the computation and the measured result indicates that the series did not diverge.

It actually runs deeper than that. All divergences must cancel when one calculates a measurable quantity. If that does not happen, then the theory is useless. The fact that we can compute results that agree with the quantities we measure indicates that the asymptotic series is convergent. Even if there are terms at large orders that become very large, they must cancel among themselves to give negligible contributions. How this happens is very difficult to determine without actually calculating such higher order terms, which is simply not practically feasible.

In the end, the whole issue about the apparent divergences in asymptotic series in perturbation theory far beyond the truncation point is not a problem in physics. The calculation procedure works. Therefore, it is a problem in pure mathematics. Mathematician are trying to understand why it works.

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  • $\begingroup$ The well know mathematical counterexamples for functions that don't have convergent Taylor series are not all that "weird" and they aren't rare, either. Most (more generally almost all) functions don't have a convergent Taylor series. These functions are actually highly relevant as long as we are doing quantum field theory with Feynman's path integral, which is one of the ugliest constructions in all of science. The problem is that nobody seems to have come up with a better way to do some calculations in quantum field theory. That's probably just a matter of time, though. $\endgroup$
    – FlatterMann
    Commented Aug 2 at 7:58
  • $\begingroup$ "If the subsequent terms are small enough then the terms that are retained in the series would be accurate enough" But if you even know about asymptotic series, you would also know that the subsequent terms are exploding to infinity. The convergence radius is zero and the mathematical meaning of all these series are tenuous. We all agree that physics are incredibly well-described by those truncated asymptotic series, and we even have prescriptions for dealing with the divergent tails, but we do not really know how to prove that things work. They just do. $\endgroup$
    – naturallyInconsistent
    Commented Aug 2 at 8:24
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    $\begingroup$ @FlatterMann Functions that are relevant in physics have a finite region within which their Taylor series around a point in that region converges. Those pathological cases of functions that do not have convergent Taylor series anywhere are not relevant for physics. If you disagree, then please provide an example of such a function. What is so ugly about Feynman path integrals? $\endgroup$
    – flippiefanus
    Commented Aug 3 at 3:41
  • $\begingroup$ @flippiefanus If we are performing an infinite number of integrations over positions or momenta from minus infinity to plus infinity, then we are necessarily picking up some mighty unpleasant functions. I do agree that that's not a good idea, but that is what the naive path integral does. That is why I am saying that we just don't have the right idea, yet, on how to define quantum field theories correctly. Nature does, by the way, not calculate path integrals. They are ensemble propagators and nature does not have ensembles. $\endgroup$
    – FlatterMann
    Commented Aug 3 at 5:30
  • $\begingroup$ @FlatterMann The scientific methods shows that the path integral approach provides a successful method to perform calculations with the aid of perturbation theory. The issue about how to define the path integral mathematically is a pure mathematics issue and not an issue in physics. Path integrals compute superpositions, not ensembles. Nature works with superpositions. $\endgroup$
    – flippiefanus
    Commented Aug 4 at 4:50

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