Scaling dimension in statistical field theory

I got stuck in understanding the scaling dimension in statistical field theory. Currently I am reading the statistical field theory written by Prof. David Tong. In his note(p.63), it states that the naive dimensional analysis is not applicable to describe the critical exponent $$\eta$$ appear in the correlation $$\langle \phi(x) \phi(y) \rangle = \frac{1}{|x-y|^{d-2 + \eta}}$$. Suppose we consider the energy functional like: $$$$S[\phi] = \int d^{d}x \frac{1}{2} \nabla \phi \cdot \nabla \phi + \ldots$$$$ From naive dimensional analysis ( $$[\partial_{x}] = 1$$ ; [S] = 0 ; $$[d^{d}x] = -d$$), we know that the dimension of $$[\phi] = \frac{d -2}{2}$$. However, this differs from the dimensional analysis of the correlation function $$[\phi] = \frac{d -2 + \eta}{2}$$. Prof. Tong says that we need to think about the dimensional analysis in terms of scaling. Under scaling $$x \rightarrow x' = x/\zeta$$, our field changes as $$\phi(x) \rightarrow \phi'(x') = \zeta^{\Delta_{\phi}} \phi(x)$$, where $$$$\Delta_{\phi} = \frac{d -2 + \eta}{2}$$$$ My problem is that I do not understand why there is an extra $$\eta/2$$ in the $$\Delta_{\phi}$$. As Prof. Tong says in his note, during scaling the formula is still invariant. It implies that $$E[\phi] = E[\phi']$$. $$$$E[\phi'] = \int d^{d}x' \frac{1}{2}[\partial' \phi'(x')]^{2} = \int d^{d}x (\zeta)^{-d} (\zeta)^{2} \frac{1}{2}[\partial \phi'(x')]^{2} = \zeta^{-d + 2} \int d^{d}x \frac{1}{2} [ \partial \phi'(x')]^{2} \rightarrow \phi'(x') = \zeta^{ \frac{d+2}{2}} \phi$$$$ If $$\phi'(x') = \zeta^{ \frac{d+2}{2}} \phi$$, it does cancel the extra scaling $$\zeta^{-d+2}$$ from $$d^{d}x$$ and $$\partial_{x}$$. It leaves the energy functional invariant under scaling. However, in my above derivation, the $$\eta$$ term does not appears in the exponent of the scaling factor $$\zeta$$. Therefore, what is the mistake of my derivation such that my result ($$\Delta_{\phi} = \frac{d-2}{2}$$) differs from Prof. Tong result $$( \Delta_{\phi} = \frac{d - 2 + \eta}{2} )$$? I appreciate any comment.

• RG steps rescale distances as you've done but they also renormalize the field by a power of $a$. So naive dimensional analysis can only constrain the sum of the $\zeta$ exponent and the $a$ exponent. Jul 9 at 18:04
• Thank you for your comment @ConnorBehan. If I understand correctly, RG literally means the "picture" share self-similarity in all scales. After each rescaling $x \rightarrow x'=x/\zeta$, we need to renormalize the filed strength $\phi$ by a factor in order to tune the contrast. However, I do not understand why the factor is $a$ and why there is an exponent $\eta$ on $a$. Jul 10 at 4:18