How can perturbativity survive renormalization?

The most usual way to renormalize quantum field theories is by re-writing the Lagrangian in terms of physical (finite) parameters plus counter-terms. Take $\lambda \phi^4$ theory for instance:

$${\cal L} = {1\over2}(\partial_{\mu}\phi)^2-{m^2 \over 2} \phi^2 - {\lambda\over 4!}\phi^4 + {\cal L}_{CT}$$

$${\cal L}_{CT} = {\delta Z\over2}(\partial_{\mu}\phi)^2-{\delta m \over 2} \phi^2 - {\delta\lambda\over 4!}\phi^4$$

All parameters with $\delta$ in ${\cal L}_{CT}$ are divergent quantities. Then what we do is to treat everything in ${\cal L}_{CT}$ as interactions and calculate it perturbatively

My question is: how can we do that? Considering that the "couplings" in this case ($\delta Z$, $\delta m$ and $\delta \lambda$) are huge numbers?

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This is the subject of almost any QFT textbook, and is way outside the scope of a few paragraphs. For a renormalizable theory, each infinity is pushed off into the next order, leaving only finite pieces –  Columbia Oct 7 '12 at 9:03
The question is more specific than "explain me renormalization". Also, if the infinity is just "pushed to the next order", how can you explain truncating the series? –  Forever_a_Newcomer Oct 7 '12 at 12:28
Your question is very much at the heart of renormalization, and with all due respect to the answers below, really deserves a textbook treatment. The typical regularization and renormaliztion scheme and book treatment seems a bit magical at first, but it makes perfect sense once you get to the renormalization group and Wilsonian treatment where the approximation makes more sense –  Columbia Oct 9 '12 at 2:21

This also bothered me briefly when learning renormalization. You first assume that $\delta\lambda$ is small throughout, which means that you make a cutoff first, and take $\lambda$ small enough so that $\lambda\log(\Lambda)$ is not large. This is not particularly difficult--- logs are never large unless the cutoff is astronomical.

Then you rewrite the series in terms of the physical couplings, and you notice that the resulting series is independent of the bare couplings and masses, and then you make the justified hypothesis that this expansion is valid only under the assumption that the physical coupling is small, even if you go ahead and make the cutoff so big that the bare coupling is large.

The justification for this is the local scale-running of the coupling. Suppose for definiteness that you make an Ising model at some scale (which is the infinite coupling limit of the Euclidean $\phi^4$ theory in the near-critical symmetry broken phase), if you go to the critical point of the Ising model, you are in the limit that the long-wavelength theory is described by perturbation theory.

If you start with the Ising model and do a few iterations of a real-space renormalization group block spinning, a few Migdal-Kadanoff-like real-space renormalization steps using average fields, you get a physical coupling for the average field which is no longer infinite-- the average field is not stuck at -1 or 1, but is averaged into blobs around these values. If you go to larger scales, eventually the effective average field coupling always gets small gradually as your momentum cutoff gets smaller. Since the long-distance theory is universal, meaning you only need to adjust one parameter, and then everything else is determined by the continuum theory, you have to get the same answer for long-range correlations whether the microscopic theory is cut off at the Ising scale (Landau-pole scale), where the coupling is infinite, or at some much larger scale where the coupling stays small throughout the entire range. Since the physics is independent of cutoff, for the regime where the cutoff produces an everywhere small coupling, the perturbation theory should be reliable.

One should imagine the cutoff to be at a physical scale, at a real length, not zero, and the bare coupling to be only at most a factor of 2 different from the physical coupling. This is usually the case for physical log-running theories, like the Higgs sector in the standard model.

The argument above requires log-running to keep $\delta\lambda$ small everywhere in the range from zero momentum to the cutoff, and this doesn't naively work in 3 dimensions or 2 dimensions, where you have strong power-law running, because the coupling is dimensional. In these lower dimensions, the renormalization process is not so compatible with perturbativity, because the coupling in $\phi^4$ theory goes as a power, it is dimensional. In this case, you have three tools, resummation, real-space renormalization, and $\epsilon$ expansion.

In the $\epsilon$ expansion you make an ansatz that the coupling goes as a power of the scale, and that the coefficient has a fixed point which is detemined by the structure of the 4 dimensional log-running theory. This works to predict critical exponents to a certain precision, but in epsilon expansion, the perturbation theory is not as useful as a calculation tool for correlation functions, because the coupling is power-law scale dependent. But once you have the critical exponents, you can try to construct the conformal field theory of the critical point in 3d by doing Kadanoff-Polyakov operator tricks. This is an industry, and in 2d, it is essentially understood, with many conformal points resolved and studied using advanced mathematical tools.

The resummation methods are harder--- these involve Borel analysis, and this gives very little insight per hour invested. But this was the only tool until the 1970s, so all the old literature devotes most of the time to this.

The real space methods allow you to sidestep the whole issue by defining the renormalization on lattice theories Lagrangians directly, without any k-space expansion. You replace the field variables with block variables, and you shift the couplings for the block variables and look for a fixed point. This idea is due to Kadanoff, but the result is often called Wilsonian because the Nobel committee was monumentally stupid as usual.

But if you are in 4d, just do what people do in books, keeping the cutoff large but the couping small throughout, and you will never get into trouble.

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In the usual textbook explanations, you begin with infinities and push them under a renormalization carpet, and it nowhere becomes clear why perturbation theory is justified with such infinite perturbations.

However, this is just a problem of the current generation of textbooks which usually try to recreate some of the historical process of discovering renormalization.

There are alternative ways of doing renormalization that never encounter an infinity, so that perturbation theory is meaningful. On the level of ordinary quantum mechanics, this is fairly simple to understand; see my paper http://www.mat.univie.ac.at/~neum/ms/ren.pdf

A sound treatment on the quantum field level is given in Salmhofer's book http://books.google.at/books?hl=en&lr=&id=nAXncL7_KrQC&oi=fnd&pg=PA1&dq=salmhofer+renormalization&ots=w9TM3hYqi0&sig=1zJAQirvfNmmoe4qAxN7mtBK9Hw. But this is already far more technical.

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The perturbation is not just ${\cal L}_{CT}$, but $-{\lambda\over 4!}\phi^4 + {\cal L}_{CT}$. The first term gives also infinite contributions and the counter-terms are added precisely with the purpose to make the difference finite. The first term $\propto \phi^4$ gives infinite corrections not because the physical $\lambda$ is infinite, but because such an "interaction" gives infinite corrections to the fundamental constants and without subtracting them the calculations results are useless.

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But, being more specific: when you do perturbation you have to expand $Exp(- i S_{I}- i S_{CT})$ (S are the action of the interaction and of the counterterms), and pick the first terms. I understand doing that for $S_{I}$ but not $S_{CT}$. It seems that this step in the calculation is invalid. Of course the infinities will cancel but that seems to me like "holding" a limit - it seems like the order in which you take limits ($\lambda$ small vs. huge regulator) make a difference in the result –  Forever_a_Newcomer Oct 7 '12 at 12:34
You may consider the counter-terms as proportional to $\lambda^2$ and join their contributions with the (divergent) second-order perturbative contributions of the "main" interaction $\propto \phi^4$. –  Vladimir Kalitvianski Oct 7 '12 at 13:14
Anyone who has downvoted should at least leave a comment so that others may know what is wrong with this answer. I personally think it is correct. +1 –  user10001 Oct 8 '12 at 15:20