# Ladder Structure of $d>$2 CFT

The action of the dilation generator $D$ on a primary operator of dimension $\Delta$, $\mathcal{O}_{\Delta}(0)$, is given by $$\left[D,\mathcal{O}_{\Delta}(0)\right] = -i\Delta\mathcal{O}(0). \,\,\,\, (1)$$ But the ladder operator subalgebra $$[D,P_\mu] = iP_\mu\,\,\,\, (2a)$$ $$[D,K_\mu]=-iK_\mu\,\,\,\, (2b)$$ $$[K_\mu,P_\nu] = 2i(g_{\mu\nu}-M_{\mu\nu}),\,\,\,\, (2c)$$ seems to imply that $P_\mu$ lowers the scaling dimension, instead of raising it: $$DP_\mu|\Delta\rangle = ([D,P_\mu]+P_\mu D)|\Delta\rangle = (iP_\mu-i\Delta P_\mu)|\Delta\rangle=-i(\Delta-1)P_\mu|\Delta\rangle. \,\,\,\, (3)$$

Things work out as they should if we consider that $$D|\Delta\rangle=i\Delta|\Delta\rangle,\,\,\,\, (4)$$ as is done in, e.g. Qualls' lecture notes on page 39: https://arxiv.org/abs/1511.04074. He also uses $(1)$ on page 30.

I thought for a time that it was just a minor convention change, since the scaling dimension must be positive (otherwise correlations would grow with distance), then would seem natural to make definition $(4)$, although during the discussion of the representation of $D$ it would be defined with a minus, see $(1)$. If this was the case, then this global sign could be absorved as a global phase, but because the ladder algebra is unnafected, the relative minus in $\Delta-1$ on eq. $(3)$ would persist.

So what is going on? Any guidance would be appreciated!

You are confusing them with the Hilbert space operators, for which the commutation relations have the opposite sign. Suppose you have operators $A$ and $B$, represented by $$[A,O(x)]=\mathcal{A}O(x),\\ [B,O(x)]=\mathcal{B}O(x),$$ where $\mathcal{A}$ and $\mathcal{B}$ are differential operators, then $$[A,[B,O(x)]]=[A,\mathcal{B}O(x)]=\mathcal{B}[A,O(x)]=\mathcal{B}\mathcal{A}O(x),$$ so the differential operators are multiplied in the opposite order.