# Intuitive interpretation of the scaling dimension of an operator?

I am reading Field Theories of Condensed Matter Physics by Fradkin and in equation (4.10) it shows that an operator transforms irreducibly under scalings as

$$\phi_n(xb^{-1}) = b^{\Delta_n}\phi_x(x)$$

where $$\Delta_n$$ in the so called scaling dimension of the operator. I am confused as to how exactly I should be thinking about this quantity. I understand homogeneous functions and scale invariance, but I am just wondering how the scaling dimension fits into it. Is it telling you the degree to which the range of the operator expands or contracts based on a scaling of the domain? How is this quantity thought of in practice?

For example, I could define in one dimension $$\phi(x) = x^2$$, and scale it as $$\phi(xb^{-1}) = (xb^{-1})^2 = b^{-2} \phi(x) = b^{\Delta_n}\phi(x)$$ and so the scaling dimension would be $$\Delta_n = -2$$. I can see its value and that its related to the exponent in $$x^2$$, but should I be thinking of this as any deeper?

Thanks!

The equation in Fradkin's book does not make much sense as written. He is trying to write that under a renormalization group transformation, the field transforms according to a particular rescaling. The field on the left of the equation is therefore not the same as the field on the right; if the equation were taken too literally, then one could differentiate both sides with respect to $$b$$ and obtain a differential equation giving an explicit form of the field. But clearly this does not make sense, these are supposed to be integration variables in the path integral!
Here's how I would fix the equation. In one iteration of the momentum-shell RG, one first writes $$\phi(x) = \phi^<(x) + \phi^>(x)$$, integrates out the fast-momenta contained in $$\phi^>(x)$$, and then rescales $$x \to x' = x/b$$ to restore the original cutoff. Upon rescaling, we obtain a new field $$\phi'(x')$$ defined by $$\phi'(x') = b^{\Delta} \phi^<(x)$$ This new field contains Fourier modes up to the original cutoff. To zeroth order in the perturbative RG, a coupling of such a field will transform under a single RG step as $$\int d^d x \ \lambda \phi(x) \to \int d^d x\ \lambda \phi^<(x) = \int d^d x' \ b^{d-\Delta} \lambda \phi'(x') = \int d^d x' \ \lambda' \phi'(x')$$ where I've introduced $$\lambda' = b^{d-\Delta} \lambda$$.