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In the CFT book by Di Francesco et al. they use conventions such that part of the conformal algebra (see eq. 4.19) is $$ [D,P_\mu]=iP_\mu, \\ [D,K_\mu]=-iK_\mu, \tag{1} $$ where $P_\mu$, $D$ and $K_\mu$ are the generators of translations, dilatations and SCTs respectively. $D$ and $K_\mu$ act on a scalar field $\phi(x)$ as (see eq. 4.31) $$ D\phi(x) =-i(x^\nu \partial_\nu +\Delta)\phi(x), \\ K_\mu \phi(x) =-i(2x_\mu \Delta +2x_\mu x^\nu \partial_\nu -x^2 \partial_\mu)\phi(x), \tag{2} \label{2} $$ where $\Delta$ is the scaling dimension of $\phi$.

Then they define (scalar) primary fields as those transforming under conformal transformations as $$ \phi'(x') =\bigg[ \mathrm{det}\bigg( \frac{\partial x'}{\partial x} \bigg) \bigg]^{-\Delta/d} \phi(x). \tag{3} $$ In many texts (e.g. Simmons-Duffin's notes https://arxiv.org/abs/1602.07982) primaries satisfy - or rather are defined as - $K_\mu \phi(0) =0$, and this is consistent with \eqref{2} when $x=0$. One thing I find inconsistent is that using Di Francesco's conventions $$\tag{4} DP_\mu \phi(0) =[D,P_\mu]\phi(0) +P_\mu D\phi(0) =iP_\mu \phi(0) -i\Delta P_\mu \phi(0) =-i(\Delta-1)P_\mu \phi(0), \\ DK_\mu \phi(0) =[D,K_\mu]\phi(0) +K_\mu D\phi(0) =-iK_\mu \phi(0) -i\Delta K_\mu \phi(0) =-i(\Delta+1)K_\mu \phi(0), $$ i.e. $P_\mu =-i\partial_\mu$ is the lowering operator for dimension, and not $K_\mu$ as usual.

Intuitively, this doesn't even make sense looking at the form of these generators on the fields. $P_\mu$ is proportional to $\partial_\mu$, which is the inverse of a length: $\partial_\mu \phi$ should have dimension $\Delta+1$. Similarly, $K_\mu$ has the dimension of a length and should thus lower $\Delta$ by one.

Am I missing something?

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  • $\begingroup$ Have you tried also taking into account the different definitions of the operators $D$, $K$, and $P$ are between your two references? That is, comparing Eq. 4.18 in Di Francesco with Eqs. (21)-(24) in Simmons-Duffin? $\endgroup$ Aug 8, 2022 at 17:33
  • $\begingroup$ Yes, the conventions are completely different. But I thought that the statement "$K_\mu$ is the lowering operator for the dimension" was true in any choiche of convention, sign, i factors, etc. $\endgroup$
    – Rubilax96
    Aug 9, 2022 at 7:45
  • $\begingroup$ I am very confused by this too! A way out would of course be to assume $D\phi(0)=i\Delta\phi$. This would fix the result. However, I think that the problem is deeper. Computing as shown in arxiv.org/abs/1602.07982 the action of the operators on $\phi(x)$ is wrong. I asked this in physics.stackexchange.com/questions/659957/… and physics.stackexchange.com/questions/659607/…. At the time I thought the questions were answer but I am not sure anymore. $\endgroup$ Oct 27, 2022 at 1:47

1 Answer 1

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Ok, I finally understood what is going on. For the sake of brevity, let me focus on dilations and translations. Dilations and translations are generated as vector fields in spacetime by $$id=x^\mu\partial_\mu\quad\text{and}\quad ip_\mu=\partial_\mu.$$ So, for example, if we wish to translate $x$ by an infinitesimal amount $\epsilon^\mu$, we can do it by $$x^\mu\mapsto\tilde{x}^\mu=e^{i\epsilon^\nu p_\nu}x^\mu=x^\mu+\epsilon^\nu\partial_\nu x^\mu=x^\mu+\epsilon^\mu.$$ In particular, it is these vectors who satisfy the commutation relations $$[id,ip_\mu]=-ip_\mu.$$

On the other hand, the action of dilations and translation on fields, is generated by $iD$ and $iP_\mu$ defined so that, for example, if we translate the field by an infinitesimal $\epsilon^\mu$ we have $$\phi(x)\mapsto \phi(x)-i\epsilon^\mu P_\mu\phi(x)=e^{-i\epsilon^\mu P_\mu}\phi(x)=\tilde{\phi}(x)=\phi(x-\epsilon)=\phi(x)-\epsilon^\mu\partial_\mu\phi(x).$$ We then conclude that $$iP_\mu\phi(x)=\partial_\mu\phi(x)=ip_\mu\phi(x).$$ It would be wrong to conclude form this that $iP_\mu=\partial_\mu=ip_\mu$. Indeed, the first is a vector on the space of fields, while the later is a vector in spacetime. The correct conclusion is $$iP_\mu=\int\text{d}^Dx\,\partial_\mu\phi(x)\frac{\delta}{\delta\phi(x)}.$$

In particular, while the action of $iD$ and $iP_\mu$ are very similar to those of $id$ and $ip_\mu$, they actually satisfy the opposite commutation relations $$[iD,iP_\mu]=iP_\mu,$$ so that $iP_\mu$ is a raising operator for $iD$. Indeed, this stems from the fact that both $x\mapsto\tilde{x}$ and $\phi\mapsto\tilde{\phi}$ form representations of the conformal group. However, the first acts by $e^{i\epsilon^\mu p_\mu}$ while the second by $e^{-i\epsilon^\mu P_\mu}$. This minus sign of difference is the culprit.

To see this explicitly, consider the case of a dimension $0$ field. Then $$iD\phi(x)=x^\mu\partial_\mu\phi(x).$$ Thus $$iDiP_\mu\phi(x)=iD(\partial_\mu\phi(x))=\int\text{d}^Dyy^\nu\partial_\nu\phi(y)\frac{\delta}{delta\phi(y)}\partial_{x^\mu}\phi(x)=\int\text{d}^Dyy^\nu\partial_\nu\phi(y)\partial_{x^\mu}\delta(x-y)=\partial_{x^\mu}\left(\int\text{d}^Dyy^\nu\partial_{y^\nu}\phi(y)\delta(x-y)\right)=\partial_\mu(x^\nu\partial_\nu\phi(x))=ip_\mu id\phi(x).$$ In other words $iD(\partial_\mu\phi(x))=\partial_\mu(iD\phi(x))$ because $iD$ is only a functional change, it doesn't change the point in spacetime. We see that the action of the functional changes in terms of the differential operators is reversed. This is the comment made in section 3.2 of https://arxiv.org/abs/1602.07982

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