Convergence of quantum effective action to finite loop order

Question

Consider the quantum effective action ofa fixed QFT. If we compute it perturbatively to finite loop order l, 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(p1, p2, p3, p4) 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

integral f(p1, p2, p3, -p1-p2-p3) phi^(p1) phi^(p2) phi^(p3) phi^(-p1-p2-p3) dp1 dp2 dp3

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.

This post has been migrated from (A51.SE)

Comments

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.

This post has been migrated from (A51.SE)

---

In other words, your question assumes that you isolate some infinite subset of diagrams and pretend that they're "of the same order in the loop expansion". But they're not. You're using an inconsistent rule for counting the loops. If you have a generic nonlinear effective action of the fields, its curved/nonlinear character does arise from summing multiloop diagrams (if you can derive it from something else at all). Any truncation of the full action must be "with respect to a well-defined limit", e.g. $g\to 0$ (my answer) or $E_\gamma\to 0$ (IR debates). You don't seem to have any new limit.

This post has been migrated from (A51.SE)

---

@Lubos, the limit I'm using is hbar -> 0. As far as I understand (of course I might be confused about something) the order in hbar is the number of loops whereas the order in the coupling constant is the number of vertices. Hence if I compute I[phi] to finite order in hbar I'm getting diagrams with any number of external legs (since phi is finite) and any number of vertices (since g is finite) but only a limited number of loops (since hbar is infinitesimal)

This post has been migrated from (A51.SE)

---

Dear @Squark, I don't think it's possible to treat these two limits differently in any real situation. The real expansion parameter measuring "how much quantum" the theory is - how important the loops are - is always the same. For the electromagnetic interaction, it's the fine-structure constant $e^2/(4\pi \epsilon_0)\hbar c$. This is small, $1/137$, and that's why the loops - or additional external legs - bring suppression. Your unusual asymmetric limit seems to depend on an arbitrary renormalization of the fields by a power of $\hbar$ which is unphysical.

This post has been migrated from (A51.SE)

---

@Lubos, I think this limit is natural from a purely mathematical point of view and thus bound to appear somewhere. But I don't really know. In some special cases there is another limit which leads to the same thing. For example in QED we can take the mass of the electron -> infinity. Expanding to finite order in 1/m yields finite loop order.

This post has been migrated from (A51.SE)

---

@Pavel, it has no direct relation to IR issues. I tried to articulate it better in the edit

This post has been migrated from (A51.SE)

---

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.

This post has been migrated from (A51.SE)

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 $g^2$.

This post has been migrated from (A51.SE)

Comments

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

This post has been migrated from (A51.SE)

---

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.

This post has been migrated from (A51.SE)

---

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.

This post has been migrated from (A51.SE)

---

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.

This post has been migrated from (A51.SE)