# Perturbation sum converge?

1. Feb 25, 2010

### fermi

I heard that there are semi-rigorous arguments that the sum of all (renormalized) terms in QED or QCD perturbation expansions do not converge, even though each term in the sum is systematically renormalized and rendered finite. This is suspected to be the case even if the coupling constant is weak. In other words even though we have fantastic experimental agreement for the electron (g-2) calculated through two loops, the agreement will get worse if could calculate a thousand loops.

So, I am looking for references where this is proven, or at least discussed in detail. I searched the forums and could not find a previous discussion on this topic.

2. Feb 25, 2010

### blechman

that is correct - perturbation theory is an asymptotic expansion. This was proved for QED by Freeman Dyson in Phys. Rev. 85, 631 (1952).

The summary of the proof is as follows:

The perturbation expansion is a Maclaurin expansion in $\alpha$ (the fine-structure constant). IF a Maclaurin expansion converges, then it must converge in some interval $\alpha\in(-R,R)$ for some R > 0. However, if $\alpha<0$, then the effective potential is not bound from below and the theory does not make sense! So it must be that R=0 and the series has no radius of convergence. Thus the series is (at best) only an asymptotic series. QED. (no pun intended!)

It can also be shown that we expect the series to start to break down at order $\alpha^{-1}$; in other words, it would show up near the $\alpha^{137}$ terms of the expansion. Since even the very best calculations are 5th order I believe (that is, $\alpha^5$), this divergence is completely irrelevant in QED.

HOWEVER, it is a very serious problem in QCD!!

You can also prove that ordinary nonrelativistic perturbation theory you learn in Quantum Mechanics suffers similar disasters. In fact, it was a HW problem when I taught the class to show that x^4 perturbation to the SHO does not converge (that is, for large enough n, second-order correction > first-order correction).

Last edited: Feb 25, 2010
3. Feb 25, 2010

### arivero

4. Feb 25, 2010

### blechman

The same dyson paper i already referenced.

also, the textbook by Negele and Orland does it for phi^4 theory.

5. Feb 25, 2010

### Count Iblis

The problem is not really that the series does not converge, rather that resummation methods are tricky to apply. If you obtain a diverging series for a mathematically well defined quantity, you must have made some sort of mistake in the process. This is usually caused by illegally exchanging series expansions with integrations. But in this process, no information is really lost. So, the terms in the divergent series should still contain the exact answer. A strategy to extracting the real answer from a divergent series is called a resummation method.

Borel resummation is one such method that is widely used. But this doesn't work in many cases in QFT. I think http://arxiv.org/abs/hep-ph/0510142" [Broken] has a dissenting view on this matter.

Last edited by a moderator: May 4, 2017
6. Feb 25, 2010

### Haelfix

The rough sketch of the proof is like so

A generic QFT has an asymptotic expansion that looks something like the sum ((A^n *n^alpha*n!^s) = Z(n). Where alpha is the coupling constant and A is an order (1) number and im ignoring various numerical factors like color and other things.

If the theory is renormalizable there will be only one factorial contribution (the factorial coming from the obvious numerical perponderance of feynman diagrams), so s=1. If it is nonrenormalizable you might have extra renormalon factorial powers.

We are interested in comparing the nth term to the (n+1)th term and note that an asymptotic series is roughly good so long as the next term is smaller than the preceding one. We expect a turning point when the next term is roughly the same magnitude as the preceding point (and subsequently the next terms will be bigger, and the approximation will break down)

so z(n+1)/z(n) ~ A*n*alpha ~ 1 (where by the above rule of thumb should yield a turning point when its roughly equal to 1). So n ~ 1/(A alpha) ~ 1/alpha for a renormalizable theory.

So it says that for instance in QED, you can trust roughly 137 orders in perturbation theory, before the expansion starts to diverge.

7. Feb 25, 2010

### fermi

Thank you Blechman for the reference and for the short but clear explanation. However I have two objections, or more precisely I have two followup remarks and questions.

The first one is on general mathematical grounds. If a Taylor (or MacLaurin) series converges in a given real interval, it simply does. However, this statement says nothing about whether a particular solution makes physical sense. Conversely, one should not be able to say that the series does not converge only because the solution makes no physical sense on the entire interval. I would say that mathematics does not care. Physics is full of examples where mathematics gives us too many solutions and we weed out some on physical grounds. Therefore, just because the field theory is inconsistent for negative $\alpha$ should not mean that the series does not converge. It simply means that the solutions with negative $\alpha$ are not physically interesting solutions.

The second issue I would like raise relates to what happens when we extend the amplitudes to the complex $\alpha$ domain. Now the Taylor series has to be replaced by a Laurent series, which now may include a few (but finite) number of negative powers of $\alpha$. There could be poles of various ordes, in other words. These poles give infinite contributions as $\alpha$ approaches to zero, but still define an analytic complex function in the punctured complex disk. In fact, the statement you made about about what happens in order $\alpha^{137}$ makes me suspect that that the perturbation expansion has some (at least one) pole terms which should be part of the calculation but ignored. Finding such terms might restore the finality of the perturbation sum. Do you have further insights along these lines?

8. Feb 25, 2010

### blechman

I don't remember the exact argument, but negative $\alpha$ is bad, even from the naive, mathematics point of view. I don't have access to the original paper right now, so I'll let someone else answer this question...

This isn't right, at least not in the sense you mean. The series DOES converge (rather trivially!) AT THE POINT $\alpha=0$, so extending to the complex plane does not suddenly introduce poles in $\alpha$. That is not what happens to asymptotic series.

There is an analysis like this that you can do, called "Borel Transformations". Then you can get poles in the variable that is "Borel-conjugate to $\alpha$" (like momentum is the Fourier-conjugate variable to position). These poles are called "renormalons" and are basically the poles you are talking about, but as you can see, they are a little more subtle. This is a very powerful technique used in QCD to deal with these issues.

I leave it to you to look up "renormalons" and "Borel Transformations" to see how this happens. VERY BRIEFLY: These "renormalons" are poles on an integration contour, and the answer results in an ambiguity coming from how you choose to deform the contour. These ambiguities are basically the manifestation of the non-convergence of the series.

9. Feb 25, 2010

### Ben Niehoff

Two separate points:

1. A nice toy problem to understand why the perturbation series doesn't converge is to attempt to solve the integral

$$\int_{-\infty}^{\infty} e^{-ax^2 - \lambda x^4} \; dx$$

The quadratic part we know the answer to, and the quartic part can be expanded in a Taylor series. However, you'll find you need to exchange the order of integration and summation, at which point the series no longer converges. And as you can see from the formula above, if you put $\lambda < 0$, then obviously the integral diverges. This integral is analogous to the path integral, and this process of solving it is analogous to the perturbation expansion.

2. If you think about it, the perturbation expansion shouldn't even be expected to converge, because the parameter we are expanding in is a pure, dimensionless number! There is no sense in which a dimensionless number can be "small", because it is simply an absolute number; there is no scale to compare it to. 1/137 might seem "small", but only when you compare it to 1; the real line does not have a natural scale to allow you to do this.

Compare to the case of, say, electrodynamics, where we expand the potential in powers of r. R is a dimensionful parameter, so we can set a scale for "smallness" by comparing it to another dimensionful parameter. If we are expanding inside a sphere, we might choose A, the radius of the sphere. Then, the "limit as R becomes small" is well-defined. We may take either R << A, or we can take A to infinity.

But we cannot consider the "limit as $\alpha$ becomes small", because $\alpha$ is a fixed number.

10. Feb 26, 2010

### tupos

Not correct. It happens because you take incorrect zero order wave functions, that is why your series doesn't converge for every alpha. However if you take correct zero order wave functions you will find complete spectrum.

This is a similar to polaron problem. It was proved that you can't use perturbation theory as well. Except you can use Operator Method, invented by Feranchuk and Komarov in 1982 and find the whole spectrum for HO and for polaron.

For Harmonic oscillator I.D.Feranchuk, L.I.Komarov, I.V.Nichipor and A.P.Ulyanenkov. Operator method in a problem of quantum anharmonic oscillator. Annals of Physics, V.238 (1995) P.370-440.

Polaron solution http://arxiv.org/abs/cond-mat/0510510

11. Feb 26, 2010

### blechman

this is what Negele and Orland do. Thanks for writing it out.

This confuses me a little bit. I have no problem with the last sentence you wrote, as you wrote it. But I would disagree with your earlier statements: in fact, the ONLY quantities we can define as "small" or "large" are dimensionless numbers, and that is: small or large compared to unity (1). When a quantity has dimensions, you can only talk about "small or large compared to something else" (as you say), and that means, taking the ratio of the two dimensionful quantities and doing an expansion in that DIMENSIONLESS quantity.

So the fact that $\alpha$ is dimensionless means that you CAN do an expansion. However, you are right when you say that it doesn't really make sense to talk about a "Taylor expansion in a constant" - that is a mathematically silly thing to do.

To emphasize my point, look at your first example: there $\lambda$ does have dimensions, and yet your argument about non-convergence still holds!

12. Feb 26, 2010

### blechman

Neat! I'll take a look...

13. Feb 26, 2010

### blechman

So I'm looking at the first paper you cite, and I don't understand why you say what I wrote is incorrect. Their method looks like a way to extend the power of perturbation theory. But in the weak-coupling limit, where Raileigh-Schrodinger PT (that is, the usual PT you see in textbooks) should be valid, they prove that their results concur.

In other words: RSPT fails to converge for large n, and other methods must be used. See, for example, the paragraph above Fig 2 on p 383.

This is all I was saying!

14. Feb 26, 2010

### arivero

Temporary here: http://lxbifi11.bifi.unizar.es/~arivero/p631_1.pdf I will delete it next monday. Consider it restricted to the participants in this thread.

While my memory about this article was that it was very short, I feel to remember that it was not exactly organized in this way, so perhaps there is a companion article somewhere.

Last edited by a moderator: Apr 24, 2017
15. Feb 26, 2010

### blechman

that may be. it's been a long time since i've thought about the details.

EDIT: After scanning the article quickly, it looks pretty self-contained. It's a straightforward Reductio-Ad-Absurdum proof as I sketched above. It also points out the $\alpha^{137}$ breakdown quickly, but that was very nicely explained above in Haelfix's post. This is just an exercise in the analysis of asymptotic expansions, there is no physics involved.

Thanx for posting the article!

Last edited: Feb 26, 2010
16. Feb 26, 2010

### Count Iblis

The book by Itzykson & Zuber also discusses the resummation of the series for the harmonic oscillator.

17. Feb 26, 2010

### blechman

Yes, another wonderful book!

18. Feb 26, 2010

### arivero

Last edited by a moderator: Apr 24, 2017
19. Feb 26, 2010

### tupos

In this sense of course. But if you use operator method you can find the whole spectrum for anharmonic oscillator. But standard PT can't be used for any coupling constant.

So it is incorrect to say that PT is not converge, because it can't be used. Anyway it is my own opinion.

Regards.

20. Feb 26, 2010

### blechman

Why do you say it can't be used?! It can be used as long as it's used appropriately!

As long as you stick to low enough n, then RSPT is perfectly self-consistent. The problem is only when you go to larger values of n. Furthermore, the smaller $\lambda$ is, the larger n you can go. But it DOES break down at some point, and therefore RSPT cannot be used whenever you are studying a process that might be sensitive to the range of n where it breaks down (roughly $n_c\sim\lambda^{-1}$.

And how do you KNOW that it breaks down? Well, because the expansion fails to converge!

So if $\lambda=10^{-3}$ and you are only interested in the behavior of, say, the first 10 energy levels, then RSPT is perfectly acceptable! However, if your observable is sensitive to very high-energy states, then you're in trouble. You cannot used RSPT and you must use some other method (your OM, or lattice, or variational techniques, or .....)

21. Feb 26, 2010

### tupos

It can't be used for arbitrary coupling constant.

But I can say the same for simple harmonic oscillator, as its energy becomes large as n is large, that is why there is no sense of trying find a correction for almost infinite value, moreover with that big n one can use quasi classical approach.

Of course for each lambda one must investigate the convergence of PT.

22. Feb 26, 2010

### blechman

OK, I am implicitly assuming that $\lambda$ is sufficiently small, otherwise it never makes any sense to talk about PT at all! The point is that even for "sufficiently small" coupling, there is always a "sufficiently large" energy level at which PT breaks down. This is manifest in the observation that the series fails to converge. And the key point is that it ALWAYS fails to converge, even for arbitrarily small coupling.

Conclusion: perturbation theory (or if you wish, Raileigh-Schrodinger PT) is at best an asymptotic expansion.

I think we agree on this.

This was the OP's question (as I read it).

23. Feb 26, 2010

### arivero

The last article I quoted can help to agree: it explains that then perturbation theory can not be used to answer "global" (my wording here) questions, because some relationship which is true order by order is not implied to be true in the "total" (again my wording).