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Two different notions of (ir)relevant and marginal operators ?

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4710 views

On p10 of these EFT lecture notes, the "relevance" of operators in a Lagrangian is determined by comparing their mass dimension to the spacetime "d" one considers such that an operator is

  • Relevant if its dimension is $ < d$
  • Marginal if its dimension is $ = d$
  • And irrelevant if its dimension is $ > d$

This means for example for the action of e scalar field in $d=4$\(S[\phi] = \int d^d x\left( \frac{1}{2} \partial_{\mu}\phi \partial^{\mu} \phi - \frac{1}{2} m^2\phi^2- \frac{\lambda}{4!} \phi^4 - \frac{\tau}{6!} \phi^6 + ...\right)\)

that the mass term is relevant, the $\phi^4$ coupling is marginal, and the $\phi^6$ coupling is irrelevant, etc

However, when analyzing the RG flow around a fixed point $S^{*}[(\phi)]$, the (ir)rrelevance of an operator is determined by linearizing the RG flow equation around this fixed point ($M$ is the linearized right hand side of the for example Wilson RG flow equation, $t$ is the RG time),

\(\frac{\partial S}{\partial t} = M S^{*}[\phi]\)

solving the corresponding Eigenvalue problem

\(M O_i(\phi) = \lambda_i O_i(\phi)\)

and looking at the sign of each Eigenvalue $\lambda_i$.

The action around the fixed point can then be approximated as

\(S[\phi] = S^{*}[\phi] + \sum\limits_i \alpha_i e^{\lambda_i t} O_i(t)\)

and the operator $O_i$ (or direction in coupling space) is said to be

  • Relevant, if $\lambda_i > 0$ (leads away from the fixed point)
  • Marginal, if $\lambda_i = 0$
  • Irrelevant, if $\lambda_i < 0$ (these operators describe the fixed point theory)

So my question is: What is the relationship between these two notions / definitions of (ir)relevant and marginal operators in an effective field theory? Are they they equivalent, and if so how can I see this (mathematically) ?

asked Apr 15, 2014 in Theoretical Physics by Dilaton (4,295 points) [ revision history ]
edited Apr 15, 2014 by Dilaton

1 Answer

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A scale transformation changes $dx$ to $dx(s)=e^{s}dx$ and $\phi$ to $\phi(s)=e^{-s}\phi$, hence $S=\int dx^d \sum_k g_k \phi^k$ to $S(s)=\int dx^d \sum_k g_k e^{(d-k)s}\phi^k$. Thus your $\lambda_k$ is $d-k$, which is positive if $k<d$ zero if $k=d$, and negative if $k>d$.

answered Apr 15, 2014 by Arnold Neumaier (12,425 points) [ no revision ]

Ha thanks Arnold, feeling that my question was a bit too stupid now ...

If the answer to a question provides new insight, it is never stupid.

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