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Let $\lambda$ be the coupling constant of a quantum field theory. It is said that

  1. Perturbation theory is only valid when the theory is weakly coupled ($\lambda \ll 1$).
  2. In most cases, the series of Feynman diagrams is divergent.

I would like to know the reasoning behind the above two statements.

Since the exponential function has an infinite radius of convergence one would think that by truncating sufficiently far in the series we can obtain a good approximation to the theory. Is the statement (1) better phrased as "the first few terms in the sequence are a good approximation only when the theory is weakly coupled"?

Similarly, why should the series of Feynman diagrams diverge if they are obtained from a convergent power series (of the exponential functional)?

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    $\begingroup$ The series are often not convergent, but rather asymptotic $\endgroup$
    – DanDan面
    Commented May 2 at 16:43

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Yes, the problem with strong coupling is typically that you will need many terms to get a decent approximation. However, it is also true that the perturbative series often does not converge. Hence, if you need way too many terms, perturbation theory will just fail, because the series will eventually start to diverge.

The most famous example is probably the $\lambda \phi^4$ theory. For $\lambda < 0$, the theory is unstable (the potential does not have a global minimum), while this is false for $\lambda > 0$. Hence, the theory is not analytic at $\lambda = 0$ due to this stability problem. Hence, perturbation theory will eventually diverge.

Another way of understanding this is noticing that adding loops to Feynman diagrams increases the number of diagrams very fast. You get way too many diagrams for many loops and the series end up growing too fast.

While the series for the exponential function is convergent, calculating physical quantities from it is non-trivial and involves other mathematical manipulations. These manipulations end up spoiling the convergence properties.

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  • $\begingroup$ Thank you! So in principle a strongly coupled theory can still be described perturbatively if one can compute sufficiently far in the power series? I am also curious about the $\lambda \phi^4$ extrema you mentioned. Why is it necessary for the potential to have a global minimum for the theory to be stable or for perturbation theory to be valid? $\endgroup$
    – CBBAM
    Commented May 2 at 16:51
  • $\begingroup$ @CBBAM consider a potential that has more than one minimum value that are separate from one another, and understand that tunneling can happen between minima, especially in the direction of lower potential. $\endgroup$
    – Triatticus
    Commented May 2 at 18:29
  • $\begingroup$ @CBBAM Yes, if you can compute sufficiently far perturbation will work: but the series might stop representing the correct function at some point. I recall someone telling me that you can trust the series up to $1/\lambda$ terms or something, so for QED you can trust up to $1/\alpha \approx 137$ terms, but I never saw this made rigorous (or better defined). About the $\lambda \phi^4$ example, you need a minimum because otherwise you cannot define the vacuum state. Without the vacuum state, nothing else makes sense. $\endgroup$ Commented May 3 at 0:03
  • $\begingroup$ @NíckolasAlves What is the relationship between the minimum and the vacuum state? How does one imply the existence of the other? $\endgroup$
    – CBBAM
    Commented May 3 at 3:38
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    $\begingroup$ @CBBAM The vacuum state is, by definition, the state that minimizes the Hamiltonian. If the potential is unbounded from below, then you can always find a state with less energy, and hence there is no state with minimum energy, i.e., there's no vacuum $\endgroup$ Commented May 3 at 13:42

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