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I heard about the argument that the perturbative expansion in QFT must be asymptotic, such as http://ncatlab.org/nlab/show/perturbation+theory#DivergenceConvergence

Roughly this can be understood as follows: since the pertrubation is in the coupling constant about vanishing coupling, a non-zero radius of convergence would imply that the theory is finite also for negative coupling (where “things fly apart”), which will not happen in realistic theories.

My question is, why the negative coupling causing "things fly apart" will lead to asymptotic series? Suppose we have an electron and positron, if they fly apart, it corresponds to electron and electron, and vice versa.

  • $\begingroup$ Maybe because the instanton part diverges for $g \to 0^-$, because of the exponential. $\endgroup$
    – Trimok
    Commented Aug 9, 2014 at 12:04
  • $\begingroup$ An asymptotic series happens when you expand a function around a singular point (that's why it has zero radius of convergence). The point of zero coupling is usually a branch point, which is singular. The reason for the appearance of the branch point is that for negative coupling the vacuum is unstable because the vacuum will create pairs of particles and antiparticles and "they will fly apart" forever. This leads to an imaginary part (i.e. a branch cut) in correlation functions. That's why (in your language) "things flying apart will lead to asymptotic series." $\endgroup$
    – QuantumDot
    Commented Aug 9, 2014 at 19:17

1 Answer 1


I think both the link and the question refer to Dyson's heuristical argument on why the perturbative series in QED could not be convergent. It goes somewhat like this:

Suppose the series in $\alpha$ converges in some radius. The it converges also for negative values of the coupling constant inside that radius. Consider now what kind of theory is QED with a negative $\alpha$. In that theory like charges attract and opposite sign charges repel each other. Now take the vacuum of the non-interacting theory. This state is unstable against formation of electron-positron pairs, because said pairs would repel indefinitely leading t a lower energy state. You can make an even lower energy state by adding pairs that would separate in two clusters of electrons on one side and positrons on the other. Therefore this theory does not possess a ground state, since the spectrum is unbounded from below. Hence there is no consistent QED for negative coupling constant. And so the perturbative series cannot converge.

As far as I know this argument is strictly heuristical, but shortly after it appeared (in the 1950s) Walter Thirring proved the divergence for a particular scattering process (I'm not in my office so I don't have the correct reference, but I'm positive the paper is in Thirring's selected works as well as explained in his autobiography).

Note that this question of convergence was proeminent in a period where people tried to define QFT in terms of the perturbative expansion. The advent of non-perturbative effects (instantons, confinements, pick your favorite...) coupled with renormalization group showed that this was the wrong approach for QFT.

But note also that the argument of vacuum instability depends on the interaction. It does not preclude the possibility of designing a QFT with convergent perturbative expansion, it just shows that it it not to be expected in a general theory.

  • $\begingroup$ If there is no stable ground state is a problem of perturbative expansion, the $\phi^3$ theory was extensively studied in Srednicki's book, e.g. sections 16-19, does it mean the perturbative computations on $\phi^3$ making no sense? $\endgroup$
    – user26143
    Commented Aug 10, 2014 at 17:13
  • 1
    $\begingroup$ The absence of a ground state is not a problem of perturbation theory in this case, but rather the converse, the independent argument for no ground state implies a problem in the perturbative expansion. Since we know that the theory does not exist we get that the series must not be convergent. The $\phi^3$ in Srednicki's is different. In page 71, section 9, he explicitly mentions that although there is no stable vacuum this is invisible by perturbation theory. In any case he also says that he is only interested in giving an example, and not concerned with overall consistency of the theory. $\endgroup$ Commented Aug 11, 2014 at 19:41
  • $\begingroup$ My point is that in Srednicki's the argument is that suppose we have a physical system roughly described by $\phi^3$ theory. We can still use perturbation theory to describe some scattering for example, but since non-perturbatively there is no stable ground state what we can infer is that $\phi^3$ does not describe completely the system, but it may be adequate in a limited sense when restricted only to perturbation. The negative coupling QED is different, Dyson's argument regards perturbative processes and therefore already at the level of perturbative expansion the theory is nonsense $\endgroup$ Commented Aug 11, 2014 at 19:45
  • $\begingroup$ Excuse me, I still have a question. Does "formation of electron-positron pairs" mean virtual, off-shell particle? Since on-shell electrons have rest mass, the repulsion lead to lower energy may not be larger than the rest mass. $\endgroup$
    – user26143
    Commented Sep 26, 2014 at 5:06
  • 1
    $\begingroup$ In the heuristical picture yes, I was implying that by starting with vacuum and considering virtual pairs it would be better for them to separate than to annihilate from energetic arguments $\endgroup$ Commented Sep 27, 2014 at 3:31

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