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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: \begin{equation} S[\phi] = \int d^{d}x \frac{1}{2} \nabla \phi \cdot \nabla \phi + \ldots \end{equation} 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 \begin{equation} \Delta_{\phi} = \frac{d -2 + \eta}{2} \end{equation} 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 $E[\phi] = E[\phi']$. \begin{equation} 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 \end{equation} 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.

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    $\begingroup$ 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. $\endgroup$ Jul 9 at 18:04
  • $\begingroup$ 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 $a$. $\endgroup$
    – Ricky Pang
    Jul 10 at 4:18

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