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Is there any possibility in any field theory model that at coupling >= 1.0, perturbation theory can be used to calculate, say field propagators beyond tree level? What is the room left for perturbation theory in this situation? In other words, is perturbation relevant at all at the above mentioned couplings?

Nihar Karve
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Iosuf
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    Related: https://physics.stackexchange.com/q/622974/ – Andrew Apr 30 '22 at 01:57
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    Our perturbative expansions almost always have zero radius of convergence, so it's even "worse" than you're suggesting. On the other hand, this is part of what makes QFT at the first few orders so accurate. I'd say it remains a mystery how to construct the whole theory from the perturbative expansion. Some modern researchers are hopeful it can be done, using things like Borel summation, since many examples of these asymptotic series are resurgent. – Ryan Thorngren Apr 30 '22 at 02:04
  • Thank you so much for your answer, i sincerely appreciate it. May i conclude that in scientific community there is no recognized way to apply perturbation theory when coupling is as high as i mentioned. In general, one should better avoid it. – Iosuf Apr 30 '22 at 02:13
  • see also https://physics.stackexchange.com/q/141733/84967 – AccidentalFourierTransform Apr 30 '22 at 08:10

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Yes it is still relevant in some sense. An analogy is the expansion of $(1-g)^{-1}$ as a formal power series $\sum_{k=0} g^k$ for $g>1$. Truncating the power series is not itself useful as an approximation since each next order term gets larger and larger. But we can use methods such as Borel summation to recover the exact result $(1-g)^{-1}$ from the formal power series, so information about the exact result is still hidden in the power series even though the power series is not practically useful as an approximation method.

From a theoretical point of view the real problem with the perturbative expansion is that it misses corrections going schematically like $\sim e^{-1/g}$ which are often important in the exact result. This problem has nothing to do with the magnitude of $g$ per se.

octonion
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