The transformation law of quasi-primary operators in CFT

Question

In the Big Yellow Book(BYB) of CFT by P.D. Francesco et al, the authors claim in Sec.4.2 that under a conformal transformation (CT) $\psi: x\to x'$, a spinless field $\phi(x)$ transforms as (Eq.(4.32) in BYB) $$ \phi(x)\to\phi'(x')=\Big|\frac{\partial x'}{\partial x}\Big|^{-\Delta/d}\phi(x),\tag{1} $$ where $\Delta$ is the scale dimension of $\phi$, $d$ is the spacetime dimension, and $\Big|\frac{\partial x'}{\partial x}\Big|$ the Jacobian of the conformal transformation of the coordinates.

I'm trying to reproduce the above transformation law (1) in the theory of massless free boson (MFB) in $d$-dimensions, and here is what I did:

Consider the action of the MFB $$ S=\int_{\mathbb{R}^d} d^d x(\partial\phi)^2, $$ and suppose that the metric component after the coordinate transformation $\psi$ reads $$ g'_{\mu\nu}(x')=\psi_*\Big(\delta_{\mu\nu}(x)\Big)=e^{f(x')}\delta_{\mu\nu}. $$ Then we have (the principle of general covariance) $$ \begin{align} S=\int_{\mathbb{R}^d} d^d x\partial_{\mu}\phi\partial_{\nu}\phi \delta^{\mu\nu}=&\int_{\mathbb{R}^d}\sqrt{|g'|}d^d x'\partial'_{\mu}\phi_s'\partial'_{\nu}\phi_s'g'^{\mu\nu}\nonumber\\ =&\int_{\mathbb{R}^d}e^{(d-2)f/2} d^d x'\partial'_{\mu}\phi_s'\partial'_{\nu}\phi_s'\delta^{\mu\nu}.\tag{2} \end{align} $$ where $\phi_s':=\psi_*\phi$ denotes the standard push forward field of $\phi$ generated by $\psi$. Since $\phi$ is a scalar, we simply have $\phi'_s(x')=\phi(x)$.

On the other hand, since we want the $d$-dimensional MFB to be a conformal field, we impose that $$ S=\int_{\mathbb{R}^d} d^d x\partial_{\mu}\phi\partial_{\nu}\phi \delta^{\mu\nu}=S'=\int_{\mathbb{R}^d} d^d x'\partial'_{\mu}\phi'\partial'_{\nu}\phi' \delta^{\mu\nu}\tag{3}, $$ where $\phi'$ is the field after the conformal transformation. Comparing (2) and (3), we get a constraint equation of $\phi'$ $$ \sum_{\mu}(\partial'_{\mu}\phi')^2=e^{(d-2)f/2}\sum_{\mu}(\partial'_{\mu}\phi_s')^2, \tag{4} $$ which may give that $$ \begin{align} \partial'_{\mu}\phi'=&e^{(d-2)f/4}\partial'_{\mu}\phi_s',\quad(\mu=0,1,...d-1).\tag{5}\\ &\Downarrow\\ \phi'=&\int dx'^{\mu}e^{(d-2)f/4}\partial'_{\mu}\phi_s' \quad(\text{no summation for $\mu$}).\tag{6} \end{align} $$ However, in term of (1) (Eq.(4.32) in BYB), we have $$ \phi'(x')=\Big|\frac{\partial x'}{\partial x}\Big|^{-\Delta/d}\phi(x)=e^{(d-2)f/4}\phi'_s(x')\tag{7}. $$ Apparently, (6) equals (7) if and only if $f=const.$, but for general conformal transformations, we should allow $f$ to be a function of $x'$.

What's wrong with my derivation? Any help is appreciated!

Note added: One might think that my question is similar to this one. But I believe that my question is different from that one. The OP of that question confused conformal transformations in field theory with diffeomorphism in gravity. I do understand that a CFT is an FT that is invariant under (conformal) diffeomorphism + Weyl transformation, and I hope my derivation shows this well.

Note added 2: Why does someone always think that his abstract answer to that question answers my specific question here?

Answer 1

Okay, I know what's wrong with my derivation.

Let me first show the conclusion: following the logic in my problem description, we can indeed derive the correct transformation law for the spinless operator under global conformal transformation.

The problem first arises in eq.(4). Actually, we can't go from (2) and (3) to (4). This is because we allow the integrands to differ by a divergence term. Let me first define $a:=(d-2)/4$, and restore Einstein's summation convention (i.e., $(\partial_{\mu}f)^2\equiv\sum_{\mu} (\partial_{\mu}f)(\partial_{\mu}f),~~\partial^2_{\mu}f\equiv\sum_{\mu} (\partial_{\mu}\partial_{\mu}f)$). $$ \begin{align} (\partial'_\mu\phi')^2=&\big(e^{af}\partial'_{\mu}\phi_s'\big)^2+\text{possible div. term}\nonumber\\ =&\Big(\partial'_{\mu}\big(e^{af}\phi_s')-\phi'_s\partial_{\mu}'e^{af}\Big)^2 +\text{possible div. term}\nonumber\\ =&\big(\partial'_{\mu}\big(e^{af}\phi_s')\big)^2+\Big(\phi_s'^2(\partial_{\mu}'e^{af})^2-2\phi'_s(\partial_{\mu}'e^{af})\partial'_{\mu}\big(e^{af}\phi_s')\Big)+\text{possible div. term}.\tag{i} \end{align} $$ Hence, if we can show that the second term in (i) is just a divergence term, then we can deduce that

$$ \begin{align} (\partial'_\mu\phi')^2=\big(\partial'_{\mu}\big(e^{af}\phi_s')\big)^2+\text{possible div. term}\Longrightarrow\phi'(x')=e^{af}\phi_s'(x')=e^{(d-2)f/4}\phi(x),~(\text{if there actually is no div. term}),\tag{ii} \end{align} $$ which is exactly the conformal transformation of a spinless field (Eq.(4.32) in BYB).

So let us show that the second term in (i) is indeed a divergence term (In the following derivation, $\cong$ means that two expressions differ by at most a divergence term)

$$ \begin{align} &\phi_s'^2(\partial_{\mu}'e^{af})^2-2\phi'_s(\partial_{\mu}'e^{af})\partial'_{\mu}\big(e^{af}\phi_s')\nonumber\\ =&\phi_s'^2(\partial_{\mu}'e^{af})^2-2\phi'^2_s(\partial_{\mu}'e^{af})^2-2e^{af}\phi'_s\partial'_{\mu}\big(e^{af})\partial'_{\mu}\phi_s'\nonumber\\ =&-\phi_s'^2(\partial_{\mu}'e^{af})^2-\frac{1}{2}\partial'_{\mu}(e^{af})^2\partial'_{\mu}(\phi'_s)^2\nonumber\\ \cong&\frac{\phi'^2_s}{2}\partial'^2_{\mu}(e^{af})^2-\phi_s'^2(\partial_{\mu}'e^{af})^2\nonumber\\ =&\phi'^2_se ^{af}\partial'^2_{\mu} e^{af}.\tag{iii} \end{align} $$

The exciting thing is that we can show that $\partial_{\mu}'^2 e^{af}=0$. This is because our push forward metric $g'_{\mu\nu}=\psi_*\delta_{\mu\nu}=e^{f}\delta_{\mu\nu}$ is indeed flat, and it places many limitations on the choice of function $f$. One limitation comes from the fact that the Ricci scalar of $g'_{\mu\nu}$ should vanish.

In the case of $d\geq3$, if a metric $g_{\mu\nu}$ is conformally flat, that is $g_{\mu\nu}=e^{f}\delta_{\mu\nu}$, then the Ricci scalar of $g_{\mu\nu}$ reads (which can be found in Wiki)

$$ R(g)=-4\frac{(d-1)}{(d-2)}e^{-\frac{d+2}{4}f}\partial_{\mu}^2 e^{\frac{d-2}{4}f}. $$

Therefore we have $$ R(g)=0\Longrightarrow \partial_{\mu}^2 e^{\frac{d-2}{4}f}=0, $$ which indeed tells us that (iii) is zero. We thus show that the second term in (i) is indeed a divergence term, we then prove (ii).