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Wilsonian versus old-fashioned renormalization points of view

+ 5 like - 0 dislike

Some participants point out that Wilsonian point of view is much better than all old-fashioned renormalization ideologies. So I would like to learn what makes you feel more comfortable with Wilsonian POV? What makes you accept it without hesitation? What is its advantage?

asked Aug 16, 2014 in Theoretical Physics by Vladimir Kalitvianski (22 points) [ no revision ]

Here's a good explanation of the Wilsonian point of view by @LubosMotl on TRF. Kenneth Wilson, RIP.

Thank you, dimension10, I read it when Lubosh wrote it. Still, there might be "personal" differences in different people. I would like to learn as much as possible about the "convincing part".

I accept it simply because it is logically more coherent and natural than the "subtracting infinity" picture, and it reproduces perturbative renormalizability criteria in a natural way. It doesn't tell us what the underlying short distance theory really looks like, but it tells us it's ok not knowing what the underlying theory is.

Thanks, Jia. By the way, I am not going to argue at all; I just want to listen to argumentations/insights that make people convinced in advantages of the Wilsonian POV on QFT.

A wise decision!

I am sorry, I cannot (may not) vote up, guys, so I write a "thanks" comment when I want to vote up your answers/comments.

1 Answer

+ 4 like - 0 dislike

The ''old-fashioned renormalization" presumably refers to renormalization by subtracting infinities. This is just a loose way of talking, and not really done on the formal level. Indeed, there is no subtraction of infinities once one introduces a cutoff $\Lambda$ - one only subtracts huge finite numbers, and takes the limit after the subtracted numbers are of physical size. This is just like calculating a limit $\lim_{\Lambda\to\infty} \sqrt{\Lambda^2+\Lambda}-\sqrt{\Lambda^2-\Lambda}$ by expanding the expression into powers of $\Lambda^{-1}$ (including negative powers, i.e., a term linear in $\Lambda$, which cancels) before taking the limit, thereby getting in place of $\infty-\infty$ the finite result $1$. The procedure is completely unambiguous.

The magic and uneasiness enters the QFT version of this simple and undisputed toy problem in calculus because in QFT the parameters of the Lagrangian are an (often diverging) function of the cutoff $\Lambda$, and the quest is for understanding how this comes about. 

Wilson's picture of integrating out high energy degrees of freedom (cf. http://motls.blogspot.in/2013/06/kenneth-wilson-rip.html) gives an intuitive understanding of what happens - it naturally explains why at a different cutoff scale one should expect to get different values of the parameters of the resulting effective theory, and hence allows an interpretation of the bare parameters in the Lagrangian without cutoff as parameters of an effective theory. It also suggest a systematic way of obtaining a differential equation for the cutoff-dependence of the bare parameters - the Wilson renormalization group. In fact, the latter is only a semigroup, as one loses information in the process of integrating out high energy degrees of freedom, and in general cannot go backwards. cf. http://www.physicsoverflow.org/6902/under-conditions-renormalization-group-equations-reversible.

Wilson's picture is physical only in the quantum field theory of thermodynamics and solid state physics, where the cutoff is an observational scale and hence has a physical meaning. In relativistic QFT, the cutoff is a computational device as it destroys Poincare invariance and locality, which are both valid rigorously in a QFT. But the analogy is fairly compelling when one interprets the cutoff as resulting from an unknown but neglected submicroscopic structure (e.g., a grand unified theory, effects of quantum gravity, or a string theory). Integrating out these submicroscopic degrees of freedom would indeed result in a nonlocal cutoff in the resulting effective field theory.

On the other hand, one expects that at least asymptotically free theories such as QCD are well-determined even without an existing substructure. (The Yang-Mills millennium problem asks to settle a related problem, cf. the discussion in http://www.physicsoverflow.org/21784/yang-mills-millenium-question-and-dynin-formalism). In this case, Wilson's picture is just that - a picture to motivate the cutoff-dependence of the coupling that leads to a finite limit when the cutoff is removed.

However, asymptotically free theories show an additional phenomenon called dimensional transmutation. For simplicity, I discuss massless QCD, which depends on just a single bare parameter $g$. Renormalization requires [notation as in Chapter 17 of Peskin/Schroeder]  that the bare coupling constant in QCD is a function $g(\Lambda,\alpha_s,M)$ of three arguments, the cutoff $\Lambda$, the dimensionless strong fine structure constant $\alpha_s$ (the QCD analogue of the electromagnetic fine structure constant), and an arbitrary renormalization scale $M$. Wilson's renormalization semigroup shows how  $g(\Lambda,\alpha_s,M)$ changes as $\Lambda$ is reduced, but because we know - or at present rather assume - that the exact, local theory at $\Lambda=\infty$ exists, we can look for the limit $\Lambda\to\infty$ of the family of Wilson theories when $\Lambda$ increases beyond limits.. This is believed to give a well-defined QFT, the theory we call QCD. It still has a dependence on two parameters $\alpha_s$ and $M$, although there is only a 1-parameter family of  QFTs (since we have a limit of a 1-parameter family of bare theories with cutoff). Therefore there must be a definite functional relation between $\alpha_s$ and $M$, usually expressed by writing $\alpha_s$ as a function of the mass scale $M$. [Since massless QCD is scale invariant, one may scale the fields such that $\alpha_s=1$ and gets ''the mass scale $M$ of QCD'' - to be fitted to experiments. However, then quark masses must also be taken into account, which complicates the story.) The dependence  $\alpha_s(M)$ (and of other observables) is called the running coupling constant, and there is a second Stueckelberg renormalization group (this time a true group) that expresses how quantities depend on $M$. In causal perturbation theory, where there is no cutoff and hence no Wilson renormalization semigroup, the Stueckelberg renormalization group is still present.

Note, however, that the Wilson renormalization semigroup and the Stueckelberg renormalization group have nothing in common except for some formal analogy. In particular, in relativistic QFT, Wilson's point of view is an interpretation aid, no essential ingredient (since the true theory has no cutoff), while the Stueckelberg renormalization group is a property of the renormalized equations, and choosing $M$ of the order of the experimental energy is important to keep the approximation errors in truncated computations small. (In exact computations to all orders, the value of $M$ would be immaterial.) 

answered Aug 17, 2014 by Arnold Neumaier (12,365 points) [ revision history ]
edited Aug 17, 2014 by Arnold Neumaier

Thank you, Arnold, too!

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