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  Do fixed points of the RG flow field depend on the renormalization scheme applied?

+ 3 like - 0 dislike

In my textbook from p 438 on it is explained, that for example for the Ginzburg-Landau model defined by the effective Lagrangian

\[\mathcal{L}_{eff} = \mathcal{L}_{kin} -\frac{1}{2}(T-T_C)\phi^2 -\frac{1}{4!}\lambda\phi^4 +\cdots\]

the position of the  non-trivial (Wilson-Fisher) fixed point which is found in this case by an $\epsilon$ expansion and dimensional regularization
\[\lambda_* =\frac{16\pi^2\epsilon}{3}, \quad m^2_* = 0\]

depends on the RG scheme applied and is therefore not physical. Only the critical exponents (that describe the behavior of the RG flow near the fixed point) are universal.

Is it always the case that the position of fixed points (or even more generally the structure) of the RG flow field depends on the renormalization scheme applied?

Concerning the fixed points, I dont understand why their position (and maybe even their presence or absence?) should depend on the scheme, as I always thought that fixed points of the RG flow corresponds to a scale (or even conformal) invariant theory, their basins of attraction define universality classes etc, so they should be physical and not depend on the exact renormalization method (scheme) applied?

asked Jul 10, 2015 in Theoretical Physics by Dilaton (6,240 points) [ no revision ]

At the end of the calculation, shouldn't you set $\epsilon$ to a specific value (say 0 or -1)?

@RyanThorngren in this calculation of the fixed point $d = 4 - \epsilon$ is assumed, with $0 < \epsilon \ll 1$

The dimreg values are only physical after fixing epsilon, usually to zero, at the end of the computation.

1 Answer

+ 3 like - 0 dislike

The beta function (and others running functions) does indeed depend on the renormalization scheme (RS), as the renormalized parameters (couplings, masses) are different for distinct schemes. Thus, fixed points have different expressions in distinct schemes. 

There are however theorems that guarantee that the beta function is the same for some set of RSs up to some given order in perturbations theory. For example, the beta function is the same for all mass-independent RSs at first and second order in perturbation theory. The proof only involves two facts: in a mass-independent RS the beta function is just a power series in the coupling constant and all coupling constants are the same at tree level.

answered Jul 14, 2015 by drake (885 points) [ no revision ]

Thanks Drake!

Can you also explain what using different RSs means in the RG flow point of view, when for example looking at a two-dimensional parameter space spanned by the mass and a coupling constant? Does applying different RSs in this picture mean, that you approach a fixed point when going from higher to lower energy along different trajectories of the RG flow, such that when you apply a mass independent RS you follow a trajectory that is orthogonal to the mass axis?

@Dilaton, I don't have access to your textbook so I don't claim to understand the context of your question fully. However, a likely explanation is that, in different schemes the definitions of a coupling constant may differ. For example, in a phi^4 theory one may define $\lambda$ as the scattering amplitude of $\phi\phi \to \phi\phi$ at any external momenta(even off-shell momenta are acceptable! For details see Weinberg Vol1 page 515), hence you may say $\lambda$ can be defined differently, and all this does not matter as long as you have the same S-matrix in the end. Thus it simply doesn't make much sense to draw the RG trajectories from different schemes(if they define couplings differently) on the same graph.

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