# What is the objective of renormalization group theory?

1. Jul 14, 2011

### ndung200790

It seem to me that the objective of renormalization were the exclusion the infinities.But in renormalization group theory,they consider the dependence of physics parameters(e.g the interaction constant lamda,the mass parameter) on momentum p.Then I do not understand what is the objective(meaning) of renormalization group theory?
Thank you very much in advance.

2. Jul 14, 2011

### tom.stoer

Renormalization inorer to remove infinities is just a dirty trick. The renormalization group itself says that you can look at a theory at different energy scales E and E' (or at different length scales which is the same) and renromalize masses and coupling constants g(E) and g'(E') such that the predictions of both theories are the same. Essentially they are THE SAME theory.

This can e.g. be interpreted as "integrating out" degrees of freedom and hiding their contribution in the renormalization of parameters. Have a look at http://en.wikipedia.org/wiki/Renormalization_group#Block_spin_renormalization_group

Last edited: Jul 14, 2011
3. Jul 15, 2011

### ndung200790

Then,it seem to me there is an unclear in ''finite'' characteristic(''already renormalization'') in renormalization group theory.They pose a renormalization ''condition'' p^2=-M^2,then leading to Callan-Symazik...ect.So,the ''finite characteristic'' of the theory depends on the renormalizable nature of concrete QTF theory that we consider.

4. Jul 15, 2011

### Demystifier

I would put it this way.

To make renormalization, you first need to do a regularization. Any regularization involves some arbitrary energy-scale parameter, such as the cut off. The physical result should not depend on the choice of the value of that parameter. But this requirement is not trivial at all. It turns out that this requirement has a consequence that some physical quantities depend on energy in a physical process in a specific way. This dependence is what the renormalization group equation describes.

5. Jul 15, 2011

### tom.stoer

... and this dependence does exist even if regularization is NOT required.

6. Jul 15, 2011

### vanhees71

Renormalization is not (only) a "dirty trick" to get rid of infinities in S-matrix elements or other observable quantities but necessary even for perfectly finite quantities. The reason is that in perturbation theory one uses parameters of fictitious objects called "bare particles", i.e., of free particles, and one has to adapt the wave-function normalization when considering interactions. This you can see even in usual quantum mechanics when you do, e.g., the "old fashioned" perturbation theory of stationary states (Schrödinger perturbation theory).

In quantum field theory, in addition one can absorb the infinities into the "bare parameters", that are unobservable, but as Demystifier pointed out, that requires the introduction of a energy/momentum scale. It does not come from the regularization, because the renormalized theory should be independent of the regularization used. A clearer picture is BPHZ renormalization, where you don't go over an intermediate step of a regularization, but interpret the Feynman rules as expressions for the integrands of loop integrals and subtract the divergences and subdivergences systematically before you do the integrations by using Zimmermann's Forest Formula.

These subtractions can usually be done only at a non-zero spacelike four momentum due to the analytic properties of the vertex functions and the demand that the counterterms, contributing to the wave-function-normalization factors, masses, and coupling constants must be real in order to have a self-adjoint Hamilton operator which guarantees a unitary S matrix. In this way you always introduce an energy-momentum scale, and this choice is in principle arbitrary, but the values of the "dressed parameters" depend on this choice of the scale. Changing the scale thus leads to finite renormalizations of these parameters, such that the prediction for observable quantities (S-matrix elements) don't change. The renormalization-group equations describe this change of the renormalized parameters of the theory such that the observable quantities do not change.

In the context of perturbation theory that independence in principle holds true only up to the order of the expansion parameter (e.g., some coupling or [itex]\hbar[/tex], etc.) taken into account in the calculation. The solution of the renormalization group equation with the "anomalous dimensions" given by perturbation theory are however usually to any order or the expansion parameter. One can show that this corresponds to a certain resummation of perturbation theory, summing up leading logarithms (see, e.g., Weinberg, Quantum Theory of Fields, Vol. II).

Another interpretation, originating from the work of K. Wilson, is that "irrelevant degrees of freedom" are integrated out in the path integral of the action. This can, e.g., be "hard modes" and thus the meaning is to coarse grain over small-range correlations lower than a certain energy scale. This is particularly intuitive in the context of many-body applications, where one looks at collective modes of a system rather than at the many microscopic degrees of freedom making up this collective behavior since if one restricts oneself only to observations below a certain energy or momentum scale (equivalent to the resolution of time and space dependences of observables) the complicated microscopic degrees of freedom become irrelevant concerning the collective behavior at the resolution of the observation.

7. Jul 15, 2011

### Demystifier

True. But when regularization is not needed (as in solid state physics) then even renormalization is not necessary. In this case, renormalization is only a convenient trick that simplifies certain calculations.

8. Jul 15, 2011

### DrDu

Let me ask you a question, too: Which other methods besides renormalization group theory do you know to deal with singular perturbation problems?