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I'm currently studying the Wilsonian effective theory and I'm a bit confused about relevant/irrelevant/marginal operators. I understood why they are called in this way, in particular that the coefficients of relevant operators increase as we reduce the scale at which we observe the system and vice versa for the coefficients of irrelevant operators, while the coefficients of marginal operators don't rescale at all.

My question has to do with the following statement: "Relevant operators drive the system away from a UV stable fixed point."

Assuming that our system is in a UV stable fixed point in the first place, this seems understandable to me, but I'm trying to apply all of this to a different situation.

Suppose we start from the free real scalar boson theory in 4 dimensions, which is at an IR stable fixed point. Now, suppose we introduce the interacting term $\lambda \Phi^4$.

My reasoning is the following: classically the dimension of this operator is $4$ (it coincides with the number of dimensions), so the operator is marginal. However, on a quantum level the dimension of the field gets the anomalous correction, so it becomes $1+\gamma(\lambda)$.

Suppose for a second that $\gamma(\lambda)>0$:

  1. When we rescale the system going to lower energies, an operator is relevant if its dimension is $<4$, so in this case $\lambda \Phi^4$ is irrelevant.
  2. When we rescale the system going to higher energies, an operator is relevant if its dimension is $>4$, so in this case $\lambda \Phi^4$ is relevant.

However, at the fixed point $\lambda_F=0$ from which we are starting, $\gamma(\lambda_F)=0$, therefore at the end of the day $\lambda \Phi^4$ remains marginal.

I think this reasoning must be flawed, since when we study the renormalization group equations for $\lambda \Phi^4$ theory we find out that the coupling constant runs, therefore the system doesn't always remain at its fixed point.

Could you help me understand where the mistake is?

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    $\begingroup$ I think I don't get your point but just in case: physically, you always start not exactly at the fixed point but arbitrary close. In which case, $|\gamma|$ is arbtrarly small, but non zero. Implying that in the IR or UV limit, if the operator is relevant, you'll see it's effect. The coupling constant runs, as long as you have a small perturbation around the gaussian fixed point $\endgroup$ Commented Jul 19 at 13:46
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    $\begingroup$ I'm also not sure I understand the question, but I don't think $\lambda \Phi^4$ has a fixed point, except for $\lambda=0$, which is a free field theory. So in the case of a free field theory, the coupling constants don't run, but that's a trivial case. For $\lambda \neq 0$, there is no fixed point, so it doesn't make sense to say that the theory "doesn't stay at the fixed point." If you look at a theory at a fixed point, then the theory will stay there. If you perturb the theory away from a fixed point, you can ask whether the RG flow will take the theory toward or away from that point. $\endgroup$ Commented Jul 19 at 13:53
  • $\begingroup$ I was assuming you actually start at the fixed point and then add the coupling term, so that's why I was thinking that the operator remains just marginal. I didn't understand you already have to start not exactly at the fixed point but just close. So, once we assume this, can we conclude that $\Phi^4$ is relevant in the UV and irrelevant in the IR? $\endgroup$ Commented Jul 19 at 14:02
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    $\begingroup$ Saying relevant in the UV might be meaningless in this context depending on whether the Landau pole is real. But $\Phi^4$ is definitely irrelevant in the IR. Since this requires a check of the anomalous dimension, some books will further specify it as marginally irrelevant. $\endgroup$ Commented Jul 19 at 15:23

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When we say an operator is relevant or irrelevant, we always mean relative to a specific fixed point. The same operator can be relevant at one fixed point and irrelevant at a different one. Relevant/irrelevant is only defined at the fixed point, it becomes meaningless in between. Equivalently, scaling dimensions are only defined at a fixed point: $\gamma$ is only meaningful when you have conformal symmetry.

Consider $L=\frac12(\partial\phi)^2$. This is a CFT, so it makes sense to ask whether various operators such as $\phi^2,\phi^4$, etc. are relevant or not. The mass term is obviously relevant: if you turn it on, you are no longer at the fixed point, the RG flow begins, and at long distances the theory becomes trivial. The quartic is a little more subtle: classically it is marginal, but a one-loop computation shows that it is in fact irrelevant. So if you turn it on, the theory is no longer conformal, you generate an RG flow, which at long distances gives you back the original free CFT.

So the theory $L=\frac12(\partial\phi)^2+\phi^4$ has a single fixed point, which occurs at $L=\frac12(\partial\phi)^2$. This is an infrared fixed point, it appears at long distances. The operator $\phi^4$ is irrelevant at this fixed point. There is no "UV fixed point", so it is meaningless to ask whether $\phi^4$ is "UV relevant" or not. The theory $L=\frac12(\partial\phi)^2+\phi^4$ is not UV complete, there is no conformal symmetry at short distances, you cannot flow up the RG flow, only down.

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