# What is the purpose of the renormalization group?

1. Apr 7, 2012

### AuraCrystal

Hello,

I've been reading a book on QCD on I have a question: what is the purpose of the renormalization group? Is it to remove the large logs so that we can use pertubation theory (at least for large $-q^2$)? And what is the physical significance of the renormalization scale $\mu^2$?

2. Apr 8, 2012

### vanhees71

QCD is a very good example to explain the meaning or renormalization and how an energy-momentum scale comes into a theory that has no dimensionful parameters to begin with (considering the limit of massless quarks, or the chiral limit).

In perturbation theory you split the Lagrangian or Hamiltonian in a free part, which are bilinear (or sesquilinear) functionals of the fields, leading to the description of free particles with definite masses and a standard normalization of the field operators. The rest of the Hamiltonian, i.e., the interaction part, is treated in perturbation theory, and the ultimate goal in vacuum qft is to calculate the S-matrix, i.e., the transition-probability rates for processes, where asymptotically free particles are prepared, undergo a reaction (scattering process) and asymptotically free particle states are measured. Of course, in QCD you have the problem that in reality you never have asymptotically free quarks and gluons, but this confinement property cannot be described by perturbation theory, so we just discuss perturbative QCD in a (very) naive way as if you could use standard asymptotic theory.

Now, taking into account the interactions in the perturbative sense, a single quark is surrounded by a cloud of quantum fluctuations, particularly its own color-Coulomb field. This changes its mass. In perturbation theory this is even diverging, and you have to make sense of this mass change. The idea behind renormalization is that in fact there is no such thing as a non-interacting quark, and its finite measurable mass (to measure quark masses is another difficult topic in itself!) is the one including the self-interaction of the quark with its own color field. At the same time due to interactions the field normalization changes compared to free particles. The asymptotic field has a renormalization factor, usually called Z, compared to a non-interacting field, since a Heisenberg field operator has overlap not only with the single-particle but also with multi-particle asymptotic states. It can be shown that if Z=1, the theory is non-interacting. So you have to renormalize also the asymptotic wave functions.

In practice of perturbation theory this amounts to the evaluation of the quark self-energy diagrams. It is linearly divergent and leads to the wave-function and the mass renormalization counter terms. Now gluons are massless, and you cannot use an on-shell renormalization scheme. If you calculate the one-loop diagram as the most simple example, the loop consists of a quark and a gluon propagator. At least the gluon is necessarily massless since it's the gauge boson of an un-Higgsed local gauge symmetry (color SU(3)). Thus the branch cut of the quark's self-energy function starts at s=m^2 (where m denotes the quark mass). In order to get a real counter term in the Lagrangian you thus have to introduce an arbitrary spacelike scale, where to subtract the divergences of your self-energy. In this way a momentum scale enters the game, even if you consider the limit of massless quarks.

The same holds true for the gluon. The only difference is that there is a Slavnov-Taylor identity that tells you that the gluon self-energy is transverse and that it never builds a mass term, i.e., there's no mass-counter term necessary as it should be not to break gauge invariance. For gluons you must subtract the wave-function renormalization counter term at a space-like momentum, again introducing an energy-momentum scale.

The same procedure is necessary also when you calculate corrections to the interaction vertices. The Slavnov-Taylor identities again ensure that the additional coupling-constant counterterms of the qqg-, the ggg, and the gggg vertices are the only ones you need and that they fulfill the constraints in order to keep the couplings universal at the appropriate powers (g for the three-point and g^2 for the gluon four-point function) as again dictated by the non-abelian color-gauge symmetry. Also the coupling has to be specified at a certain momentum scale.

Of course, if you could calculate the transition matrix elements exactly, i.e., to all orders of the coupling, there shouldn't be any dependence of the result on where you chose this renormalization scale. After all the S-matrix elements are measurable quantities and should have a well-defined value for any physical process considered. This is precisely what's described by the renormalization group equations: It describes the change of wave-function normalization factors, masses, and couplings as functions of the renormalization scale such that the S-matrix elements stay invariant under this change. At any order of perturbation theory that's of course only approximately the case.Of course perturbation theory only applies if the coupling is small, and for QCD, as the renormalization group equation for the coupling constant tells you, that's the case at large renormalization scales (which can be taken as the typical range of exchanged momentum in a reaction), and this is the celebrated property of asymptotic freedom. That means you should do a calculation of a process at large renormalization scales. All loops lead then to terms with logarithms of the type $\ln(q^2/\Lambda^2)$, where $q$ is a combination of the external four-momenta of the diagram and $\Lambda$ is the renormalization scale, where you define the constants of your theory. There you see that you encounter large contributions, if $q^2$ differs a lot from the renormalization scale $\Lambda$. The renormalization-group equation (RGE) then takes care to resum these "leading logs". Thus it can be a better approximation to first calculate the loop diagrams at large $\Lambda$ and then "scale" the result with help of the RGE to another typical scale of your problem, because it leads to a resummation of the leading logs.

This very physical picture of renormalization (in the sense of relativistic vacuum quantum field theory) is often hidden a bit by the modern treatment of the perturbative treatment of gauge theories, which immediately uses dimensional regularization as an intermediate step to first define converging Feynman rules in terms of the regularization parameter $\epsilon=(4-d)/2$, where $d$ is the space-time dimension considered in the evaluation of loop diagrams. Here the scale comes in a somewhat more hidden way: In order to keep the coupling constant and the wave-function normalization factors dimensionless quantities and at the same time also the action dimensionless (in natural units, where $\hbar=c=1$), you must introduce an energy-momentum scale, usually called $\mu$. Then you only subtract the divergent terms (going in powers of $1/\epsilon$), applying the socalled minimal-subtraction scheme MS (or you include some additional finite terms, leading to the modified minimal-subtraction scheme $\overline{\text{MS}}$. This scale takes the role of renormalization scale in this special renormalization schemes. The good thing of this procedure is that it is a very convenient scheme for perturbative calculations, leading to a mass-independent renormalization scheme and thus simple homogeneous RGEs.

One should also stress that renormalization has also another much more physical meaning than this rather involved and formal mathematical renormalization techniques. This comes from the consideration of QFT as a many-body theory and goes back to L. Kadanoff and K. Wilson. It's explained, e.g., in Peskin-Schroeder's textbook (which is however not a good source for the above described technical aspects of renormalization theory since it's too sloppy and even gives equations with dimensionful arguments in the logarithms, which must not occur in any meaningful expression, and that's precisely where the necessity of a renormalization scale becomes obvious from the above given formal arguments about renormalization).

3. Apr 8, 2012

### AuraCrystal

So is it just to help with the large logs? (So that the coupling constant becomes small, justifying the use of perturbation theory?)

4. Apr 8, 2012

### geoduck

Is the momentum scale that you're referring to commonly called μ?

By standard normalization of the field operators, do you mean Z=1, and by definite mass, do you mean m0?

Does this mean the definite mass you referred to earlier (with Lagrangians involving only bilinear fields) is in fact infinite, to cancel the infinite change δmo in the mass due to quantum fluctuations arising from (non-bilinear) interactions, to get a finite mass m=mo+δmo? Don't all interactions, not just color interactions, contribute to δmo? Also, what really is mo, and why does it have a value of infinity? It can't be due to interactions, because interactions tell you what m=mo+δmo is, but mo comes from just the free bilinear theory with no interactions? Is it the Higgs that gives mo, which is subsequently changed by interactions?

I thought a rescaling of the fields ∅ → C*∅ where C is a constant and ∅ is the field, makes no alteration of the S-matrix, so why do we even have to consider ∅ → sqrt[Z]*∅ ?

But this space-like scale you introduced came out of thin air, and can be any value you want. I thought this scale was introduced so that you can use experimental measurements of the 1P1 diagrams to replace bare couplings. So isn't the mass scale something that is introduced so that experimental measurement can inform the theory, and not a property of the theory itself? The theory still has no intrinsic momentum scale.

What was the reason for space-like momentum? Do all scales have to be space-like? Are space-like momentums even physically achievable?

So does this mean that it is true that there is no intrinsic momentum scale with massless particles, but one is introduced because we can't add to all orders in perturbation theory? So the theory intrinsically has no scale (what about the cutoff?), but when calculating with it, for practical purposes, you have to introduce a scale?

So the renormalization group is basically a way to get results approximately, because we can't calculate infinite numbers of diagrams, and we don't know how to solve the system exactly without perturbative methods? So really that's all the renormalization group method is, an approximation technique?

I'm not sure I understand. Are you saying that you can calculate a 1-loop result at high energies (which is only good at high energies in order for higher order terms to be smaller than lower order terms), and take the derivative of it with respect to scale to get the renormalization group equation, then solve this differential equation for the coupling at a low-energy scale, and if you use this low-energy coupling then magically it incorporates more than 1-loop accuracy at this low-energy scale?