A very basic question on renormalization group

I am trying to understand the meaning of the names 'relevant', 'irrelevant' and 'marginal' in RG. So far, it seemed to me that these are defined with respect to the behaviour of the coupling constant in IR. For example: $$\lambda(\mu)\xrightarrow[\mu\to0]{}0\qquad \text{irrelevant (???)}$$

Now, assume I add a term to the action of the form $$\delta S = \int \operatorname{d}^D x \,\lambda \,\hat{\mathcal{O}}$$

The dimensions of the operator and the coupling constant are: $$[\hat{\mathcal{O}}] = \Delta \quad,\qquad [\lambda] = D - \Delta$$

One can introduce the dimensionless coupling $$\lambda_0$$ via $$\lambda = \mu^{D-\Delta} \lambda_0$$

And, if one wishes, to calculate the corresponding $$\beta$$-function: $$\beta_\lambda = \mu \dfrac{\operatorname{d}\lambda}{\operatorname{d}\mu} = (D-\Delta)\lambda$$

Just by looking at $$\lambda = \mu^{D-\Delta} \lambda_0$$, one can deduce that if, say, $$D>\Delta$$, we have the exponential growth, which means: $$D>\Delta \Longrightarrow \begin{cases}\lambda(\mu)\xrightarrow[\mu\to0]{}0 \\ \lambda(\mu)\xrightarrow[\mu\to\infty]{}\infty\end{cases}$$

This operator is clearly irrelevant in IR. But why do people call it relevant?

In the Wilson's path integral formalism, we consider a theory regularized by the (Euclidean) 4-momenta cutoff $\Lambda$:

$$Z[J] = \intop_{\Lambda} D\varphi e^{i S[\varphi]} e^{i \intop J(x) \varphi(x)},$$

where the path integral is over the modes with 4-momenta constrained by $\Lambda$.

The theory is defined by a number of coupling constants, as well as by the cut-off scale $\Lambda$.

Suppose that you have a coupling of dimension $$[\lambda] = D - \Delta = \delta.$$

Let's choose it to be relevant: $$\delta > 0.$$

You want to understand why it is relevant in the IR. For this, consider a one-parametric group of rescalings of space-time (imagine zooming out space-time).

The value of $\lambda$ obviously decreases since it has positive mass dimension. The value of $\Lambda$ also decreases for the same reason.

Now comes the important part: the cut-off mustn't change! Changing the cut-off implies modifying the original theory, since the theory is dependent on the value of this cut-off just as well as it is dependent on the value of coupling constants.

But the cut-off changes. What can we make of this?

To answer this, first remember how thinking of our theory as dependent on $n + 1$ parameters ($n$ coupling constants + a single cut-off) is naive, because of the renormalization group action. It is always possible to change the value of the cut-off, compensating it with a change of couplings such that the quantum theory with the cut-off remains the same, that is, gives the same correlations. The renormalization group flow is thus a symmetry of the perturbative quantum theory, at least when the external momenta are far from reaching the value of $\Lambda$.

Back to our situation. We have to compensate the change in $\Lambda$ introduced by the space-time rescaling with a transformation from the renormalization group flow. We want to keep the original value of $\Lambda$.

The real question is - how does the value of the coupling change under a combination of these two transformations? Or, equivalently, what are the changes of the effective couplings in the effective action?

An even better question is - what are the changes of the dimensionless couplings (divided by an appropriate power of the relevant energy scale $\mu$)? In the classical theory those don't change under space-time rescalings at all, but in quantum theory the ones which come from couplings with positive dimensions of energy become more and more relevant in the IR. It is this quantum behaviour of dimensionless versions of couplings that determines whether it is relevant or irrelevant.

This question is answered in Peskin & Schroeder, chapter 12. I am not going to copy-paste the derivation here, because I feel like there is no need for this: I've aimed to address the confusion that you had and it should be gone by now :)

UPDATE: The source of confusion, I think, lies in the fact that you expect the cut-off to change upon space-time rescalings among other parameters. But this can't be right, because the theory is not well-defined this way. The cut-off is used when we define the finite theory and it must stay invariant under all external symmetry transformations as a part of the definition of the theory.

If it weren't for this, there would be no RG flow at all. A theory with no fixed dimension-full parameters would behave under space-time rescalings trivially (like the classical theory, the behaviour of couplings of which is determined only by their scaling dimensions). In another words: you have a single fixed parameter with dimensions of mass, the relevant scale $\mu$. Thus all unfixed dimension-full couplings must grow with $\mu$ according to the values of their classical scaling dimensions.

When we define the regularized quantum theory, we introduce this fixed dimension-full cut-off $\Lambda$, and we can no longer claim that the behaviour of the theory upon rescalings is trivial. In fact, it can depend arbitrarily on $\mu / \Lambda$! The exact behaviour is to be determined by the theory (you calculate its predictions, specifically, the effective action, and look for how they change upon rescalings).

• What would you say about the following logic? Assume the interaction is happening at the energy scale $E$, while the dimensionless coupling constant $\lambda_0$ is defined with respect to the cutoff $\Lambda$. Then ${\underbrace{\lambda}_{\delta} \underbrace{\int \operatorname{d}^D \hat{\mathcal{O}}}_{-\delta} \approx\lambda_0 \,\Lambda^\delta \,E^{-\delta} = \lambda_0 \left( \dfrac{\Lambda}{E} \right)^\delta}$. From where we see that $\delta>0$ indeed corresponds to large $\lambda$ in the $E\to0$ limit, and so $\delta>0$ corresponds to relevant coupling. – mavzolej Nov 23 '16 at 20:16
• I'd say that it probably lacks concreteness, but will do as a useful analogy. – Prof. Legolasov Nov 23 '16 at 20:43