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Consider the quantum effective action of a fixed QFT. If we compute it perturbatively to finite loop order $\ell$, we get a sum over an infinite number of Feynman diagrams. For example, the 1-loop quantum effective action of QED contains contributions from all diagrams in which a single electron loop is connected to k external photon legs.

What is known about the convergence of this sum? Does it converge "on the nose"? Does it at least converge after some formal manipulations a la Borel summation?

Also, are there examples where the quantum effective action can be written down "explicitely", in some sense? I.e. as an analytic expression of a (non-linear) functional?

EDIT: Evidently I haven't expressed myself clearly. Let's take $\phi^3$ theory for example. It's inconsistent beyond perturbation theory because the vacuum is unstable but it doesn't matter. The effective action is a functional $I(\phi)$ given in perturbation theory by an infinite sum over 1-particle irreducible Feynman diagrams. For example, consider a diagram with a loop to which 4 external legs are attached. The diagram evaluates to a function $f(p_1, p_2, p_3, p_4)$ of the external 4-momenta. If the function was polynomial the resulting term in the effective action would be the integral of a quartic differential operator. Otherwise something more complicated results. To describe it we need to consider the Fourier transform phi^ of phi. The diagram's contribution is roughly

$$\int f(p_1, p_2, p_3, -p_1-p_2-p_3) \phi^{p_1} \phi^{p_2} \phi^{p_3} \phi^{-p_1-p_2-p_3} dp_1 dp_2 dp_3$$

If we compute the effective action to some fixed finite order in $\hbar$, it corresponds to restricting the sum to diagrams of limited loop order. However, the sum is still infinite. For example, to 1-loop order we have all of the diagrams with a loop and $k$ external legs attached. The question is whether this sum converges to a well-defined functional $I(\phi)$. In other words, I want to actually evaluate the effective action on field configurations rather than considering it as a formal expression.

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I don't understand the question. Do you have an IR cutoff, i.e. are you asking about summing diagrams with soft photons? –  Pavel Safronov Dec 10 '11 at 0:23
    
@Pavel, it has no direct relation to IR issues. I tried to articulate it better in the edit –  Squark Dec 10 '11 at 11:06
    
Sorry, @Squark, -1 for the question because you if you mean neither the sum of the perturbation series, nor the sum over soft photons related to IR divergences, there can't possibly exist anything else that makes sense and that you could meant. You may only add a large number of vertices to a diagram if 1) you also add a high number of loops (and you have uniform external legs) or 2) you are computing an inclusive cross section (you sum over different final states). The former leads to my answer; the second leads to the discussion of IR divergences. No other potentially divergent sum exists. –  Luboš Motl Dec 11 '11 at 18:55
    
Dear @Lubos, the quantity I am computing is I[phi] where I is the quantum effective action and phi is a fixed field configuration. It's not a Green function, it's not a scattering amplitude and it's not a term in the effective action. Like the word "term" suggests I[phi] is a sum over all terms and their number is infinite already at finite loop order (since they are of growing order in phi) –  Squark Dec 11 '11 at 21:18
    
Dear @Squark, I didn't mean just one term, like one diagram. I meant all terms with a particular choice of the external legs. That's what you seem to be computing. Then this is equivalent to calculating a Green's functions (up to some universal factors) and be assured that a very high power of the coupling constant requires a very high number of loops, just like in the calculation of the scattering amplitude. If you want to instantly include arbitrary powers in $\phi$, well, then you are computing the full effective action and you must include all loop corrections, too. –  Luboš Motl Dec 12 '11 at 7:54

1 Answer 1

The amplitudes in generic QFTs behave like $$ {\mathcal A} \sim \sum_{L=0}^\infty L! \cdot A_L \cdot g^{2L} $$ where $A_L$ has a slower dependence on $L$ than the factor $L!$. This fact may be obtained by counting Feynman diagrams (permutations of vertices and loops... many types of Feynman diagrams) or by solving analytically solvable examples.

Because of the $L!$ increase, it doesn't converge. If you try to find the smallest term in the series – which ultimately diverges for large $L$ (which is why the smallest term, either included or not, measures the minimum error of the resummation) – it will be the term with $L$ scaling like $1/g$ and this term is of order $\exp(-C/g^2)$, as you can see from a minimization problem and Stirling's formula, comparable to the instanton (leading nonperturbative) corrections to the amplitude.

Pretty much the only exceptions in which the divergence above is avoided are finite QFTs where the amplitudes typically terminate after a finite number of terms or have other special properties.

The divergence also exists, to the same extent, in other field theories and in string theory with the open/closed couplings with the maps $$g_{\rm closed} \sim \lambda_{\phi^4}\sim g^2 \sim g_{\rm open}^2 $$ Again, even in string theory, it's true that the leading nonperturbative corrections, now of order $\exp(-C/g_{\rm closed})$, like D-brane instantons, are of the same order as the minimal error of the resummation of the divergent series.

The series are mathematically known as asymptotic series

http://en.wikipedia.org/wiki/Asymptotic_series

There is no unique well-defined sum. Indeed, that's a good thing because the ambiguity of the perturbative sum – which is smaller than any finite term of the perturbative expansion, much like $\exp(-C/g)$ – is linked to the ambiguities of how you exactly include the non-perturbative corrections.

The divergence may also be justified by a heuristic argument. The radius of convergence in the expansion in $g$ e.g. in QED or any QFT has to be zero because the theory is strictly inconsistent for an infinitesimal negative fine-structure constant. If the electrostatic force for like charges were attractive, big chunks of positive matter and negative matter could form in the Universe. The interaction energy would be negative so this could be created out of vacuum and the vacuum would be unstable, which should really mean that all the amplitudes between the seemingly well-defined excitations of the vacuum should be calculated as inconsistent. And indeed, they are: $\exp(-C/g^2)$ is very small for small and positive $g^2$ but it diverges for a small and negative $g^2$.

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Lubos, if I understand you correctly you are talking about the convergence of the loop order expansion. On the other hand, I am talking about the quantum effective action for a fixed finite loop order –  Squark Dec 9 '11 at 21:52
    
OK, then sorry, @Squark, but in that case, I have no idea what you could be possibly asking. You are talking about a convergence of a "sum". The only sum that is at risk of being divergent (and it, indeed, is divergent) is the loop order expansion. There is no "infinite sum" at a finite order. At a finite order, there is always a finite number of diagrams/terms and the sum of a finite number of finite terms is always convergent. It seems that you think that some methods to "complicate" the diagrams (ext. legs?) don't add any $e$; but all of them do. –  Luboš Motl Dec 10 '11 at 18:39
    
the expansion I'm considering has a fixed number of loops but a growing number of vertices. That is, it is an expansion in the coupling constant but not an expansion in hbar. I'm considering the quantum effective action evaluated on a given field configuration rather than an n-point function or a scattering amplitude. –  Squark Dec 10 '11 at 19:49
    
Dear @Squark, as I said in the previous sentence, it's not possible. If you compute any particular amplitude (Green's function, scattering amplitude, or a term in the effective action), a large number of vertices may only be achieved by adding a high number of loops. You can't add vertices without adding loops, unless you are adding applies and oranges which you shouldn't. As some other people mentioned, you could also compute inclusive cross sections with may include a very high number of soft photons (without loops), but you said it's not what you meant, either. –  Luboš Motl Dec 11 '11 at 18:54

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