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Is there a non-perturbative remormalization? If so, how does it work?

+ 4 like - 0 dislike

Is there a method to renormalize a theory without using perturbative expansions for the divergences? For example, is there a method to get masses and other renormalized quantities without using expansions and counterterms?

I have heard about Lattice Gauge theory,

but what other solutions of examples of non perturbative renormalization (numerical or analytical ) are there?

This post imported from StackExchange Physics at 2014-06-06 02:42 (UCT), posted by SE-user Jose Javier Garcia

asked Aug 14, 2012 in Theoretical Physics by Jose Javier Garcia (70 points) [ revision history ]
retagged Jun 6, 2014

3 Answers

+ 5 like - 0 dislike

The standard nonperturbative way (that provided rigorous constructions in 1+1D and 1+2D QFTs) is constructing the Euclidean (imaginary time) field theory as a limit of lattice theories, and then using analytic continuation to real time via the Osterwalder--Schrader theorem.

In 1+3D, there is so far no rigorous construction of an interacting QFT, but neither is there a corresponding no-go theorem.

In 1+1D, there are also lots of exactly solvable QFTs, where the nonperturbative solution is obtained by the quantum inverse scattering method.

This post imported from StackExchange Physics at 2014-06-06 02:42 (UCT), posted by SE-user Arnold Neumaier
answered Aug 14, 2012 by Arnold Neumaier (12,385 points) [ no revision ]
Is there a rigorous interacting QFT in euclidean 4 dimensions?

This post imported from StackExchange Physics at 2014-06-06 02:42 (UCT), posted by SE-user drake

No known one with $O(4)$ invariance and satisfying reflection positivity, which are necessary and sufficient conditions for a Euclidean theory corresponding to a physical QFT.

This post imported from StackExchange Physics at 2014-06-06 02:42 (UCT), posted by SE-user Arnold Neumaier

+ 4 like - 0 dislike

Actually, Wilson (who received the Nobel for his work on renormalisation -- though that's not to say he was the first one to think of it) worked on numerical renormalisation. The starting point is to formulate (in an abstract and somewhat formal (i.e. not very computable) way) the Exact Renormalisation Group Equation.

There are various ways to formulate this, one choice is to either fix bare action and integrating out the higher frequencies to get an effective action, or to fix the desired action and take the cut-off to infinity; the other choice is in the form of the cut-off --- one might want nice properties such as the regulator obeying the symmetries of the action (Poincaré, gauge, etc.).

Up to this point, everything is exact, and in the words of a lecturer of mine, "therefore useless". The other problem is then to solve said equation, which is actually a set of uncountably infinitely many coupled non-linear differential equations (or even differentio-integral equations). The solution, numerically or otherwise, is then a separate problem of approximating suitably. On this front there is not a clear answer, and is the main focus of research. One may easily implement the kinds of analytic approximations that one finds in textbooks (e.g. $n$-th loop, $n$-th order in perturbation, etc.) but there is a lack of real understanding of how the various choices affect the outcome, and when they are appropriate for a given problem.

This post imported from StackExchange Physics at 2014-06-06 02:42 (UCT), posted by SE-user genneth
answered Jan 15, 2012 by genneth (565 points) [ no revision ]
+ 4 like - 0 dislike

Non-perturbative RG methods exist and are useful. They have been used to understand several phenomena in (nonequilibrium) statistical mechanics.

  1. NPRG of the KPZ equation
  2. Reaction-Diffusion processes
  3. Critical dynamics
  4. Nonequilibrium steady-states
  5. Voter models
  6. Branching and annihilating random walks
  7. 3D Ising model

There are also nice reviews and books on the subject.

EDIT (11 Apr 2012)

Another recent review.

This post imported from StackExchange Physics at 2014-06-06 02:42 (UCT), posted by SE-user Vijay Murthy
answered Mar 6, 2012 by Vijay Murthy (90 points) [ no revision ]

The problem with renormalizations in QFT is that, even though the conter-terms are "known", they cannot be joined with the bare Lagrangian terms in order to express everything in physical terms, but they are supposed to be treated perturbatively. A quote of Lagrangian from A. Zee (Edition of 2003, page 175):$$L= \left [ \frac{1}{2}[(\partial \phi)^2 -m_{Ph} ^2\phi^2]-\frac{\lambda_{Ph}}{4!}\phi^4 \right ] + A(\partial \phi)^2 +B\phi^2+C\phi^4\quad (1)$$ Lagrangian in the square brackets only contains physical constants (subscript "Ph"), but it gives itself wrong equations with wrong solutions. The wrongness can be represented as harmful "corrections" to the mass, charge and the fields strength due to (obviously wrong) interaction term $\propto\phi^4$. The rest is OK (in a renormalizable theory). In order to subtract these unnecessary and harmful corrections in a systematic way, one introduces the counter-terms - everything that is out of big square brackets in (1) - and joins them to the perturbation $\frac{\lambda_P}{4!}\phi ^4$ in the perturbation theory. Note, the fields $\phi$ and alike are unknown in the Lagrangian (they are formal variables before varying the action), but the coefficients$A$, $B$, and $C$ are already solution-dependent. Then, with appropriate choice of $A$, $B$, and $C$, one can cancel or subtract the harmful corrections in each perturbative order appearing due to the "initial interaction" $\frac{\lambda_P}{4!}\phi ^4$. These heavy techniques ("know how") are offered nowadays because of rejecting recognition of the wrongness of the original Lagrangian which is based on the wrong understanding of physics of coupled things and on mathematical errors.

An example of exact renormalization is presented in my toy models here and here. The subtraction of counter-terms is done exactly in the wrong (model) Lagrangians, so the remainder is a physical Lagrangian with no problem. It also shows how different physically and mathematically are the initially wrong formulations from the correct formulations.

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